On the finite irreducible subgroups of GL(11, ℂ) Geoffrey R. Robinson Institute of Mathematics, University of Aberdeen, Aberdeen AB24 3UE

Foreword: The document that follows is largely based on my 1981 Oxford D.Phil. Thesis, in which the finite irreducible subgroups of GL(11, ℂ) were classified. The main result of the thesis has remained unpublished for the intervening 30 years, although some of the general theory presented made its way into some of my later publications. The reason that the main result remained unpublished was that it was effectively superseded by the Classification of Finite Simple Groups (which was announced at around that time) in that, by invoking the Classification, it became a much more routine matter to determine, for any given n, whether a finite simple group had a faithful irreducible character of degree n, and when n is prime, it is the case that primitive finite subgroups of GL(n, ℂ) either have an irreducible normal n-subgroup, or else are close to simple. Given the journal space required to produce a proof which did not rely on the Classification, the matter rested there. However, if one is interested in particular representations, it may be a long journey from knowing that an irreducible representation of a given degree exists, which can be determined from the character table, to explicitly constructing a representation.

The number of possibilities for finite irreducible subgroups of GL(n, ℂ) depends to some extent on the prime factorization of n. So, prior to the classification of finite simple groups, the (irreducible) primitive subgroups of GL(n, ℂ) for n = 8, 9 or 10 had not been explicitly completed. When n is an odd prime p > 3, there are three possible structure for irreducible subgroups G of GL(n, ℂ). The group G may (up to conjugacy) be a monomial group. Or the group G may have an extra-special normal subgroup of order p^{3}, in which case G/F (G) is isomorphic to a subgroup of SL(2, p), where F (G) is the largest nilpotent normal subgroup of G (in this case, this is Z(G)O_{p}(G)). Or G′ may be a finite simple group which is also primitive as a linear group. It is sometimes convenient to note that when considering a finite subgroup G of GL(n, ℂ) (for general n), we may replace the generators of G by scalar multiples and suppose that G is a subgroup of SL(n, ℂ), without disturbing the derived group G′ or the isomorphism type of G/Z(G).

The “expected” finite simple primitive subgroups of GL(11, ℂ) are as follows:

i) The alternating group A_{12}, the projective special linear group PSL(2, 11) and the Mathieu group M_{12}. All of these groups have doubly transitive permutation representations on 12 points, so in each case the non-trivial irreducible constituent of the associated permutation character is irreducible of degree 11. None of these groups is imprimitive, since none of them has a subgroup of index 11 with a non-trivial linear character, although PSL(2, 11) does have subgroups of index 11. Hence these groups even occur as subgroups of GL(11, ℤ). Note that there is a chain of inclusions PSL(2, 11) ⊆ M_{12} ⊆ A_{12} as permutation groups on 12 points (suitably labelled), which leads to the same inclusion between the primitive linear groups.

We remark, incidentally, that M_{11} is a finite irreducible subgroup of GL(11, ℂ) which is imprimitive. The 11-dimensional irreducible character of M_{11} is induced from a non-trivial linear character of a subgroup H of order 720 with H′ ≅ PSL(2, 9). Hence M_{11} also occurs as an irreducible (in the lattice sense) subgroup of GL(11, ℤ), and even occurs as an integral monomial group.

ii) The projective special linear group PSL(2, 23). There are no inclusions in either direction between this group and any of the groups from case i).

iii) The projective special unitary group PSU(5, 2). This group has order 2^{10}.3^{5}.5.11, and has two algebraically conjugate irreducible characters of degree 11, each afforded by a representation which may be realised over the Eisenstein integers, so this group may be realised as a subgroup of GL(11, ℤ[ω]), where ω = e^{\frac{2πi}{3}} . The representation is primitive. The alternating group A_{12} has a Sylow 2-subgroup of order 2^{9}, so that PSU(5, 2) certainly has no subgroup of index 12 or less, so must be primitive as an irreducible subgroup of GL(11, ℂ), and, furthermore, it is not conjugate to a subgroup of A_{12}, when viewed as an irreducible subgroup of GL(11, ℂ).

This means that, up to conjugacy, the maximal finite primitive subgroups of GL(11, ℂ) have as their derived subgroups one of A_{12}, PSL(2, 23), PSU(5, 2) or E.SL(2, 11), where E is extraspecial of order 1331 and the extension is split. The case of finite primitive subgroups of GL(11, ℂ) with a non-Abelian Sylow 11-subgroup had previously been dealt with by D. Sibley, in [27].

In the work to follow, the explicit identification of the groups above is suppressed. In 1981, for the sake of convenience, for the most part, we only considered groups of order divisible by 2^{11}, since we could consider a simple group whose Sylow 2-subgroup had order at most 2^{10} as “known” by virtue of the work of Beisiegel [3] and Stingl [29]. Similarly, we could regard such linear groups as “known” if they contained a non-identity element with an eigenspace of codimension at most 2 by the accumulated work of various authors, as referenced. This excluded all the groups above.

We were also able to assume that a Sylow 2-subgroup of G had a normal elementary Abelian subgroup of order at least 8, by virtue of the work of Gorenstein and Harada, [16] (and, implicitly, A. McWilliams).

Feit had shown in previous work that a finite irreducible subgroup of GL(11, ℚ) has a subgroup of index 11 or 12 ( so, if simple, has a Sylow 2-subgroup of order at most 2^{10}).

Hence what was really proved in the work to follow was that if G were a finite simple irreducible subgroup of GL(11, ℂ) not listed above, then one of the following would occur:

i) A Sylow 2-subgroup of G would have order at most 2^{10}. ii) A Sylow 2-subgroup of G would have no elementary Abelian subgroup of order 8. iii) Some non-identity element of G would have an eigenspace of codimension at most 2.

The motivation to revisit the work came from enquiries from algebraic geometers. The heavily ad hoc methods used below to determine the finite irreducible subgroups of GL(11, ℂ) may have limited interest, and the logical structure of the proof is fairly delicately balanced and intricate, so is not easy to tinker with. However, one theme which consistently emerges from the work below is that modular methods allow us to replace strong restrictions on the eigenvalues of elements of finite complex linear groups which would not be so apparent with “ordinary” representation-theoretic information alone. The theory used (and sometimes developed) here builds on the foundational work of Brauer, and later work of Feit, and of students of Brauer and Feit, in referenced work. While this will come as no surprise to mathematicians working in representation theory of finite groups, the underlying body of work may not be so familiar to number theorists, mathematical physicists, or algebraic geometers. While in recent years, several mathematicians have used the classification to show that in quasi-simple finite complex linear groups, the spectrum of an element of prime order p (and sometimes of prime power order) is (in a sense which can be made precise) usually maximal or has at most one potential eigenvalue omitted, it is not so clear what the optimal statement is for elements is for elements whose order is divisible by more than one prime. In the work to follow, the interplay between different primes is often exploited in ways which do not occur so often in the mathematical literature. We have previously published the results of Section 1.4 and some of those of Section 1.5 in some form (those of section 1.4 concern a single prime p, and may be regarded as theorems of Hall-Higman type for groups which are not p-solvable). We do believe it is important that the body of work underlying the results presented here, which dates back through Feit, Brauer (and even Blichfeldt, who had already shown by methods which might be called modular, that if a finite subgroup G of GL(n, ℂ) has order divisible by a prime p > n(2n + 1). then G has an Abelian normal Sylow p-subgroup) should not be forgotten by future generations of mathematicians, and we also believe that, even if they are, they will need to be rediscovered in some form.

Another theme which we believe merits further examination by future workes is the use of class algebra constant calculations. These were extensively used by D. Sibley in his work in isometries and coherence. In this work, we also exploit these in unusual ways, though specific to the problem at hand. We did publish some later work on this topic, but we believe that this a rich, as yet virtually untapped, vein.

Finally, in reviewing work of this length and complexity, an urge to amend or update it is occasionally irresistible. Hence the work presented here has deviated from the original thesis, although the structure of the original proof remains intact in essence. However, the original text is held by the University of Oxford, and differs in places from what follows here.

Some Problems in the Character Theory of Finite Groups Geoffrey Raymond Robinson, Worcester College A Thesis sumitted for the degree of Doctor of Philosophy, Hilary Term 1981.

Abstract : The main aim of this thesis is to determine the structure of the finite irreducible complex linear groups of degree 11. Some results of David Sibley allow us to concentrate on groups which have an Abelian Sylow 11-subgroup.

We prove the following:

THEOREM A: Let G be a finite irreducible complex linear group of degree 11 with Abelian Sylow 11-subgroups. Then either:

a) G has an Abelian normal subgroup A such that G/A is isomorphic to a transitive permutation group of degree 11, or:

b) G′ ∩ Z(G) = 1, [G : G′ × Z(G)] ≤ 2 and G′ is isomorphic to one of the following groups: A_{12}, M_{12}, PSL(2, 11), PSL(2, 23) or PSU(5, 2).

Remarks: Note that the 11-dimensional complex irreducible representation of the Mathieu group M_{11} can be induced from a non-trivial linear character of a subgroup of index 11, so that this group is covered by case a). When combined with Sibley’s results for groups with a non-Abelian Sylow 11-subgroup, this Theorem completes the determination of the structure of finite irreducible complex linear groups of degree 11.

Acknowledgements

I acknowledge the patient support of my supervisor, Dr. Michael J. Collins, and the support and encouragement of Professor George Glauberman, who supervised me during a two-term absence of Dr. Collins. I also gratefully acknowedge the financial support of the Science Research Council, via Research Studentship B/76/307744.

Notation

Our notation will largely follow that of Gorenstein ([15]), except that we will occasionally omit subscripts from centralisers if it is clear which group we are working in. We will denote the generalized Fitting subgroup of a finite group G by F^{∗}(G). The generalized Fitting subgroup does not appear in [15]. We have F^{∗}(G) = F (G)E(G), where F (G) is the Fitting subgroup of G and E(G) is the central product of the quasisimple subnormal subgroups of G. Furthermore, [F (G), E(G)] = 1.

Unless stated explicitly to the contrary, all groups considered in this thesis will be finite. When n is a positive integer, we denote the number of primes less than or equal to n by π(n).

INTRODUCTION

Ever since the representation theory of finite groups began, there has been some interest in determining the structure of the finite subgroups of GL(n, ℂ) for small values of n. C. Jordan demonstrated the existence of a function f : ℕ → ℕ such that whenever G is a finite subgroup of GL(n, ℂ), then there is an Abelian normal subgroup A of G with [G : A] ≤ f(n). Frobenius and Blichfelt obtained explicit bounds, but (at the time of writing) there has been little improvement on the bound of Blichfeldt. The finite subgroups of GL(n, ℂ) for n ≤ 4 were determined in the late 19-th and early 20-th century ( the case n = 4 being handled by Blichfeldt). From that time, until 1965, little progress was made in this area. In 1965, Brauer determined the structure of the primitive subgroups of GL(5, ℂ). Brauer’s theory of modular characters was very useful for this result.

Since that time, further progress has been made. John H. Lindsey II determined the structure of quasiprimitive irreducible finite subgroups of GL(6, ℂ). Among the groups he found was a central extension of the Hall-Janko simple group of order 604800 which had only been discovered shortly before that time. D.B. Wales determined the finite primitive subgroups of GL(7, ℂ) and W.Feit, W.C. Huffmann and Wales between them determined the finite quasiprimitive subgroups of GL(8, ℂ) and GL(9, ℂ). Feit also made some progress towards determination of the finite quasiprimitive subgroups of GL(10, ℂ). Using completely different methods, Feit also determined the finite irreducible subgroups of GL(11, ℚ).

In this thesis, we will complete the determination of the finite irreducible subgroups of GL(11, ℂ). We also prove some results which may be of use in the study of complex linear groups in general. The proof of our main theorem uses methods from many areas of finite group theory, and occasionally we prove results which might of interest in a wider context.

Definition: An irreducible representation ρ of a finite group H is said to be primitive if ρ is not equivalent to a representation which may be induced from a proper subgroup of H. An irreducible representation τ of H is said to be quasiprimitive if Res_{N}^{H}(τ) is the sum of equivalent irreducible representations of N whenever N is a normal subgroup of H.

By Clifford’s theorem, a primitive representation is quasiprimitive. On the other hand, all irreducible representations of quasisimple groups are quasiprimitive, but not all are primitive. Also, in view of these definitions, we may speak of primitive and quasiprimitive linear groups. We remark that all Abelian normal subgroups of quasiprimitive linear groups are central.

Section Two: Some Known Results

In this section, we state some results which will frequently be required and may occasionally be quoted without reference.

Lemma 1.2.1 (Feit-Thompson,[13]): Let H be a finite complex linear group of degree d and let p be a prime. Then H has an Abelian normal Sylow p-subgroup for each prime p > 2d + 1.

Lemma 1.2.2 (Brauer,[5]): Let H be a primitive complex linear group of degree \frac{p−1}{2} for some prime p. Then either H/Z(H) ≅ PSL(2, p) or H has an Abelian normal Sylow p-subgroup.

Lemma 1.2.3 (Hayden,[17]): Let H be a primitive complex linear group of degree \frac{p+1}{2}, where p is a prime greater than 7. Then either H has an Abelian normal Sylow p-subgroup or else H/Z(H) ≅ PSL(2, p).

Lemma 1.2.4 (Feit,[9]): Let H be a finite complex linear group of degree n and let p > n + 1 be a prime. Let P be a Sylow p-subgroup of H. Then P is Abelian and [P : O_{p}(H)] ≤ p. Furthermore, there is at one most prime q > n + 1 for which H does not have a normal Sylow q-subgroup.

Lemma 1.2.5 (Feit): Let H be a finite irreducible complex linear group of degree p − 2, where p is a prime. Then either H has an Abelian normal Sylow p-subgroup, or p is a Fermat prime and H/Z(H) ≅ SL(2, p − 1).

Lemma 1.2.6 (Lindsey,[21]): Let H be a finite complex linear group of degree p−1, where p ≥ 7 is the largest prime divisor of |H|. and let p > n+1 be a prime. Let P be a Sylow p-subgroup of H. Then P is Abelian and [P : O_{p}(H)] ≤ p.

Lemma 1.2.7 (Sibley,[28]): Let H be a finite irreducible complex linear group of degree n and let A be an Abelian p-subgroup of H such that C_{H}(a) = AZ(H) for all a ∈ A^{#}. Then either A ⊲ H or n ≥ \frac{|A|−1}{2}.

Lemma 1.2.8 (Sibley,[27]): Suppose that H is a finite irreducible complex linear group of prime degree p ≥ 5. Then p^{2} does not divide [H : O_{p}(H)].

Lemma 1.2.9 (Rigby,[25]): Let H be a finite quasiprimitive complex linear group of degree n. Let p be a prime and let Q be a normal p-subgroup of H. Then Q = ZE where Z ≤ Z(H) and either E = 1 or else E is extra-special of order p^{2k+1} where p^{k} divides n. In the latter case, E has exponent p if p is odd and exponent 4 if p = 2.

Lemma 1.2.10 (Shult,[26]): Let x be an automorphism of prime order q of the finite q′-group A. Suppose that the semidirect product A⟨x⟩ acts faithfully on some finite-dimensional complex vector space V. Then the minimum polynomial of x on V is X^{q} − 1 unless q is a Fermat prime, in which case the minimum polynomial of x on V is either X^{q} − 1 or \frac{X^{q}−1}{X−1}.

Lemma 1.2.10 yields an easy proof of the next result, which is a Theorem of N.Ito.

Lemma 1.2.11: Let H be a finite p-solvable complex linear group of degree n for some prime p. Then if n < p and p is not a Fermat prime, or if n < p − 1 and p is a Fermat prime, H has a normal Sylow p-subgroup.

Definition: Let H be a finite group, χ be a complex character, and p be a prime divisor of |H|. Let β be a primitive complex |H|_{p′}-th root of unity and let 𝕂 = ℚ[β].

We say that χ requires the p^{a}-th roots of unity if there is an element h ∈ H such that χ(h) ∉ 𝕂[ω], where ω is a complex primitive p^{a−1}-root of unity.

The next result is due to Blichfeldt, but for a shorter proof due to Brauer, see [8].

Lemma 1.2.12: Let H be finite group, χ be a complex irreducible character of H, and let p_{1}, p_{2}, …, p_{n} be the prime divisors of |H|. Suppose for each i that χ requires the p_{i}^{a_{i}}-th roots of unity. Then H contains an element of order ∏_{i=1}^{n} p_{i}^{a_{i}}.

Lemma 1.2.13(Schur): Let p be a prime and let P be a finite p-group of order p^{a} which has a faithful rational-valued character of degree n. Then a < \frac{np}{(p−1)^{2}} .

Lemma 1.2.14 (Feit,[12]): Let H be a finite group and p.q be distinct prime divisors of |H|. Let P be a Sylow p-subgroup of H and x be a q-element which commutes with no non-identity p-element of H. Let B be a p-block of ordinary irreducible characters of H and B^{∗} be a q-block of ordinary irreducible characters of H. Then if B ∩ B^{∗} ≠ ∅, we have ∑_{χ∈B∩B^{∗}} χ(1)χ(x) ≡ 0 (mod |P |). In particular, there is a non-trivial irreducible character which lies in both the principal p-block and the principal q-block.

Lemma 1.2.15 (Brauer): Let H be a finite non-Abelian simple group of order divisible by the prime p. Then no non-trivial irreducible character of the principal p-block has degree a power of p.

Lemma 1.2.16 (J.S.Leon,[20]): Let H be a finite simple group of order 2^{a}3^{b}p^{2} for some prime p and let P be a Sylow p-subgroup of H. Then either H contains a Sylow intersection of order p, or P = C_{H}(x) for all x ∈ P^{#}.

Lemma 1.2.17 (Brauer): Let H be a finite simple group of order 2^{a}3^{b}5. Then H is isomorphic to one of A_{5}, A_{6} or Sp(4, 3).

For the sake of completeness we include the following well-known result of Burnside.

Lemma 1.2.18: Let H be a finite group, χ be a complex irreducible character of H and y be an element of H such that gcd([H : C_{H}(y)], χ(1)) = 1. Then either χ(y) = 0 or [H, y] ≤ kerχ.

Section Three

In this section, we collect some known results about complex linear groups of small degree and related topics. We first state some results about finite complex linear groups of degree at most 11.

Lemma 1.3.1: Let H be a finite irreducible complex linear group of degree 2. Then either H is imprimitive, in which case H/Z(H) is isomorphic to a dihedral group, or else Z(H) ≠ 1 and H/Z(H) ≅ A_{4}, S_{4} or A_{5}.

Remark: If H/Z(H) ≅ A_{6}, then Z(H) ≠ 1. In either of the other two possibilities, Z(H) is a direct factor of H.

Lemma 1.3.2: Let H be a finite irreducible complex linear group of degree 3 which is not solvable. Then H/Z(H) ≅ A_{5}, A_{6} or PSL(2, 7).

Lemma 1.3.3 (Blichfeldt,[4]): Let H be a finite irreducible quasi-simple complex linear group of degree 4. Then H/Z(H) ≅ A_{5}, A_{6}, A_{7}, PSL(2, 7) or PSU(4, 2).

Lemma 1.3.4 (Brauer,[6]): Let H be a finite primitive complex linear group of degree 5. Then H/Z(H) ≅ A_{6}, S_{6}, PSL(2, 11), PSU(4, 2), or the semidirect product of an elementary Abelian subgroup of order 25 with a subgroup of SL(2, 5).

Remark: Although A_{5} and S_{5} appear on Brauer’s list, they are imprimitive groups.

Lemma 1.3.5 (Lindsey,[22]): Let H be a finite quasiprimitive complex linear group of degree 6 such that [H : Z(H)] is divisible by 35. Then H/Z(H) is isomorphic to one of the following groups:

i) The Hall-Janko simple group of order 604800 (Z(H) has even order in this case). ii) PSL(3, 4) (6 divides |Z(H)|). iii) PSU(4, 3) (6 divides |Z(H)|). iv) A_{7}.

v) S_{7}.

Lemma 1.3.6 (Wales,[30]): Let H be a finite primitive complex linear group of degree 7 with Abelian Sylow 7-subgroups. Then H/Z(H) is isomorphic to A_{8}, S_{8}, PSL(2, 13), Sp(6, 2), PSL(2, 7), SL(2, 8) or PSU(3, 3).

Lemma 1.3.7 (Feit,[10]): Let H be a finite quasiprimitive complex linear group of degree 9. Suppose that 35 divides |H| and that Z(H) = 1. Then E(H) ≅ A_{9}.

Lemma 1.3.8 (Feit,[10]): Let H be a finite quasiprimitive complex linear group of degree 10. Then no prime greater than 11 divides |H|. Furthermore if 7 divides |H| or if 11 divides |H| and O_{5}(H) = 1, then every composition factor of H is a known finite simple group.

Lemma 1.3.9 (Feit,[11]): Let H be a finite group with a faithful rational valued irreducible character of degree 11. Then H has a subgroup of index 11 or 12.

We next state some more general results about complex linear groups of arbitrary degree. One of the first and best known results was stated by Blichfeldt for primitive linear groups, but as the proof given in [8] shows, is valid for quasiprimitive linear groups.

Lemma 1.3.10: Let H be a finite quasiprimitive complex linear group and let x be an element of H all of whose eigenvalues lie within \frac{π}{3} of one of them on the unit circle. Then x ∈ Z(H).

A well-known consequence of Lemma 1.3.10 is:

Lemma 1.3.11: Let H be a finite complex linear group of degree n. Then H has an Abelian normal subgroup of index at most n!6^{(n−1)(1+π(n))}.

A less well-known result than Lemma 1.3.10 is sometimes useful when considering the structure of Sylow p-subgroups of complex linear groups for primes p ≤ 5.

Lemma 1.3.12 (Brauer,[6]): Let H be a primitive finite complex linear group of degree n and let p be a prime not dividing n. Then [A : A ∩ Z(H)] ≤ p^{n−1} whenever A is an Abelian p-subgroup of H.

The next few results all concern the structure of finite complex linear groups which contain a non-identity element with a very large eigenspace.

Lemma 1.3.13 (Mitchell,[23]): Let H be a finite quasiprimitive complex linear group of degree n ≥ 8. Suppose that there is a non-central element x ∈ H of prime order p which contains an element of order p with an eigenspace of codimension 1. Let M denotes the subgroup of H generated by the conjugates of x. Then p = 2 and M/Z(M) ≅ S_{n+1} unless, perhaps n = 8, in which case it is also possible that Z(M) ≠ 1 and M′/Z(M) ≅ PΩ_{8}^{+}(2).

The next result is a special case of a Theorem of Huffman and Wales and relies ultimately on deep results of Aschbacher.

Lemma 1.3.14 (Huffman-Wales,[31]): Let H be a finite quasiprimitive complex linear group of degree n ≥ 11 which contains an involution t with an eigenspace of codimension 2. Then M/Z(M) ≅ A_{n+1}, where M is the subgroup of H generated by the conjugates of t.

The previous two results may be combined to restrict the structure of a finite primitive complex linear group G of degree 11.

Lemma 1.3.13*: Let G be a finite unimodular primitive complex linear group of degree 11 with Abelian Sylow 11-subgroups and Z(G) = 1. Suppose that G contains an involution t with an eigenspace of dimension 8. Then C_{G}(t) does not have a component L containing t with L/⟨t⟩ ≅ PΩ_{8}^{+}(2).

Proof: Since PΩ_{8}^{+}(2) has no proper subgroup of index 12 or less, G is not isomorphic to A_{12}. By 1.3.14, we may suppose that no involution of G has a fixedpoint space of codimension 2. By unimodularity, the 8-dimensional eigenspace of t must be the −1-eigenspace, V say. Suppose that there is such a component L. Then L must act as the derived group if a reflection group on V. Since L is simple, but has no non-trivial irreducible representation of degree 3 or less, L acts trivially on the 1-eigenspace of t. Now L contains an involution u which is a product of two reflections of V, so acts of V with a fixed-point space of codimension 2. But then u has the eigenvalue 1 with total multiplicity 9, a contradiction.

Lemma 1.3.15 (Huffman,[18]): Let H be a finite quasiprimitive complex linear group of degree n ≥ 9 containing an element x of order 3 with an eigenspace of codimension 2. Let M be the normal subgroup of H generated by the conjugates of x. Then M/Z(M) ≅ A_{n+1}.

Lemma 1.3.16 (Huffman,[18]): Let H be a finite quasiprimitive complex linear group of degree n ≥ 5. Then H contains no element of order 4 with an eigenspace of codimension 2.

Lemma 1.3.17 (Lindsey,[21]): Let H be a finite quasiprimitive complex linear group of degree n ≥ 6. Then H contains no element of order 5 with an eigenspace of codimension 2.

Lemma 1.3.18 (Huffman-Wales,[19]): Let H be a finite quasiprimitive complex linear group of degree n ≥ 4 and let p > 5 be a prime. Then H contains no element of order p with an eigenspace of codimension 2.

The following result is often useful, and is due to Blichfeldt.

Lemma 1.3.19: Let H be a finite complex linear group, and let x, y be elements of H which each have at most two eigenvalues on the underlying space V. Then V is a direct sum ⟨x, y⟩-invariant subspaces which are each 1- or 2-dimensional.

Section Four

In this section, we discuss ways in which results from modular representation theory can be applied to obtain information about finite complex linear groups.

Let H be a finite complex linear group of degree n. Then by a suitable choice of basis we may consider H as a subgroup of GL(n, 𝔽) where 𝔽 is an extension of finite degree of the rational numbers.

Let R be the ring of algebraic integers on 𝔽, p be a prime divisor of |H| and π be a prime ideal of R containing p. Then R has a subring R_{π}, which is a principal ideal properly containing R, such that R_{π} has a unique prime ideal ρ with R_{π}/ρ ≅ R/π. We set K = R_{π}/ρ, a finite field of characteristic p.

As R is not necessarily a principal ideal domain, H need not be equivalent to a subgroup of GL(n, R). However, since R_{ρ} is a principal ideal domain whose field of fractions is 𝔽, we may regard H as a subgroup of GL(n, R_{ρ}) after a suitable choice of basis. We may suppose that 𝔽 is a splitting field for H. By taking residues (mod ρ), we obtain obtain a homomorphism from H into the finite group GL(n, K). We refer to the image of H under this homomorphism as a reduction of H (mod p).

There is a p-adic valuation ν_{π} defined on 𝔽 in terms of π and it is sometimes necessary to work with the completion of R_{π} to obtain results from modular representation theory.

Let x be a p-element of H and let x̄ denote the image of x under the above homomorphism. Then as K is a field of characteristic p, we have (x̄ − 1)^{n} = 0. Now let k be the smallest positive integer such that p^{k} ≥ n. Then the order of x̄ divides p^{k}. Hence xO_{p}(H) has order at most p^{k} in H/O_{p}(H), since the kernel of reduction (mod p) is a p-group. The next few results illustrate how to exploit better information about the minimum polynomial of x̄. The first two results of this section are quoted from Feit [12].

Lemma 1.4.1: Let H be a finite irreducible complex linear group of degree n and let p be a prime divisor of |H|. Suppose that H has a cyclic Sylow p-subgroup P and set Q = Ω_{1}(P). Then if n ≤ \frac{|P |−1}{2}, N_{H}(Q) is also irreducible.

Lemma 1.4.2: Let H be a finite irreducible complex linear group of degree n and let p be a prime. Let P ∈ Syl_{p}(H) and let M be a subgroup of H containing N_{H}(P). Then if M is reducible, there is some h ∈ H\M such that [P : P ∩ P^{h}] ≤ n^{2}.

The next result utilizes methods from (Feit, [12]) and evolved from ideas of Professor G. Glauberman,and our proof owes a great deal to that idea. Part of the proof is a special case of Theorem A1.6 of Glauberman’s article in [24].

Lemma 1.4.3: Let H be a finite irreducible complex linear group of degree n, let p be a prime divisor of |H| and let x ∈ H be an element of order p. Let S ∈ Syl_{p}(H) with x ∈ H and let Q be the weak closure of x in S with respect to H and let M = N_{H}(Q).

Suppose that the image of x in some reduction (mod p) of H acts with minimum polynomial of degree k > 1. Then we have:

a) If k ≤ \frac{p−1}{2}, M is irreducible.

b) If k ≤ \frac{p+1}{4} and H is quasiprimitive, then M is also quasiprimitive. In particular, Q is extraspecial and Z(Q) ≤ Z(H), so that O_{p}(H) ≠ 1.

c) If k ≤ \frac{p+5}{8} and H is quasiprimitive, then M = H and x ∈ O_{p}(H).

Proof: As in the previous discussion, we may consider to acting on some R_{π}Hmodule, V say, where R_{π} is a principal ideal domain having only one prime ideal domain, say ρ, and K = R_{π}/ρ is a finite field of characteristic p.

In case a) we set W = V ⊗ V^{∗}, in case b) we set W = V ⊗ V^{∗} ⊗ V ⊗ V^{∗}, and in case c) we set W = V ⊗ V^{∗} ⊗ V ⊗ V^{∗} ⊗ V ⊗ V^{∗} ⊗ V ⊗ V^{∗}.

We first claim that in each case we have W^{H} = W^{M}.

Let y_{1}, …, y_{r} be a set of representatives for the right cosets of MinH. We may define a homomorphism of R_{π}-modules ϕ : W^{M} → W^{H} via vϕ = ∑_{i=1}^{n} vy_{i} for v ∈ W^{M}. Note that ϕ is independent of the choice of coset representatives. We need to prove that kerϕ = {0}. Suppose otherwise.

Let W̄ be the KH-module obtained from W by reduction (mod ρ). There is an element w ≠ 0 in W^{M} with wϕ = 0 and, furthermore, we may, and do, suppose that w̄ ≠ 0. On the other hand, we will prove that w̄ϕ = w̄, a contradiction as w̄ ≠ 0, but wϕ = 0.

We first prove that W̄ (1 − x̄)^{p−1} = 0. This follows because if T is a linear transformation such that U(1 − T)^{r} = Y (1 − T)^{s} for vector spaces U and Y, then U ⊗ Y (1 − T)^{r+s−1} = 0.

Now we may write H as a disjoint union ∪_{i=1}^{j}Mz_{i}Q of (M, Q)-double cosets, and we may suppose that z_{1} = 1_{H}. We note that for i > 1, we have Mz_{i}Q ≠ Mz_{i}, for otherwise z_{i}Qz_{i}^{−1} ≤ M, then we have mz_{i}Q(mz_{i})^{−1} ≤ S. As Q is the weak closure of x, we have mz_{i}Q(mz_{i})^{−1} = Q, so that mz_{i} ∈ N_{H}(Q) = M. Then z_{i} ∈ M, a contradiction as i > 1. We next claim that for i ≥ 2, there is a conjugate x_{i} of x in Q such that all orbits of ⟨x_{i}⟩ on Mz_{i}Q (by right translation) have length p. For let Q_{i} be a point stabilizer in the action of Q by right translation on Mz_{i}Q. Then Q_{i} < Q, since i > 1, so there is a maximal subgroup N_{i} of Q which contains Q_{i}. Then N_{i} ⊲ Q, and since Q is generated by conjugates of x, there is a conjugate x_{i} of x which lies outside N_{i}. Then x_{i} fixes no point in the action of Q by right translation on Mz_{i}Q.

Let us write Mz_{i}Q = ∪_{s=0}^{p−1} ∪_{t=1}^{q_{i}} Mz_{i}u_{t}x_{i}^{s}. Then the contribution to w̄ϕ from the double coset Mz_{i}Q is ∑_{t=1}^{q_{i}} \overline{wu_{t}}(1 − x̄_{i})^{p−1} = 0, since x̄_{i} has minimum polynomial of degree at most p − 1 on W̄ .

Since this argument applies whenever i > 1, we have w̄ϕ = w̄, the required contradiction. Hence we do have W^{M} = W^{H}. Now let χ be the complex irreducible character afforded by V. Then in case i), we have ⟨Res_{M}^{H}(χ), res_{H}^{M}(χ)⟩ = ⟨χ, χ⟩ = 1, so that M is irreducible.

In case ii), we have ⟨Res_{M}^{H}(χ\overline{χ}), Res_{M}^{H}(χ\overline{χ})⟩ = ⟨χ\overline{χ}, χ\overline{χ}⟩. To prove that M is quasiprimitive, we need to prove that whenever N ⊲ M, all irreducible constituents of Res_{N}^{H}(χ) are equal (we already know from i) that Res_{M}^{H}(χ) is irreducible). We claim that (V ⊗V^{∗})^{N} is H-invariant. We ⟨Res_{M}^{H}(χ\overline{χ}), Res_{M}^{H}(χ\overline{χ})⟩ = ⟨χ\overline{χ}, χ\overline{χ}⟩, so that all all irreducible constituents of χχ̄ restrict irreducibly to M. Thus each irreducible constituent of Res_{M}^{H}(χχ̄) which contains N in its kernel is the restriction of an irreducible constituent of χχ̄ containing N in its kernel.

Let L be the intersection of all irreducible constituents of χχ̄ with N in their kernels. Then L ⊲ H and N ≤ L. Then (V ⊗ V^{∗})^{N} ≤ (V ⊗ V^{∗})^{L} ≤ (V ⊗ V^{∗})^{N}, so that (V ⊗ V^{∗})^{N} = (V ⊗ V^{∗})^{L} is H-invariant and Res_{N}^{H}(χ) and Res_{L}^{H}(χ) have the same norm. Since L⊲H and H is quasiprimitive, we have Res_{L}^{H}(χ) = aμ for some integer a and irreducible character μ of L. Hence Res_{N}^{L}(μ) must be irreducible and also Res_{N}^{H}(χ) = aRes_{N}^{L}(μ). Since N was arbitrary, M is quasiprimitive. Since x acts on V̄ with minimum polynomial of degree k > 1, ∉ Z(H). Since M is quasiprimitive, and Q⊲M is generated by conjugates of x, Q is non-Abelian. By Lemma 1.2.9, Q is the central product of an extraspecial p-group of exponent p and a cyclic p-group. Since Q is generated by conjugates of x and x has order p, Q is extraspecial of exponent p. Also, since M is quasiprimitive and Z(Q) ⊲ M, we see that Z(Q) ≤ Z(M) ≤ Z(H) and O_{p}(H) ≠ 1.

In case iii), we know from case ii) that M is quasiprimitive and that Q is extraspecial of exponent p. Also, Res_{Q}^{H}(χ) = aψ for some (faithful) irreducible character ψ of Q and some positive integer a. We also know that χχ̄χc̄hi and Res_{M}^{H}(χχ̄χχ̄) have the same norm. Arguing as in case ii), whenever N ⊲ M, there is L ⊲ H with N ≤ L such that Res_{L}^{H}(χχ̄) and Res_{N}^{H}(χχ̄) both have the same norm.

Now ψψ̄ is the regular character of Q/Z(Q) since Q is extraspecial. In particular, all irreducible constituents of Res_{Q}^{H}(χχ̄) are linear. There is a subgroup T ⊲ H with Q ≤ T such that Res_{T}^{H}(χχ̄) and Res_{Q}^{H}(χχ̄) both have the same norm. Also, from case ii), Res_{T}^{H}(χ) and Res_{Q}^{H}(χ) have the same norm.

Now ψ = Res_{Q}^{T}(μ) for some (faithful) irreducible character μ of T. Then all irreducible constituents of μμ̄ are linear. Now μμ̄ is a faithful character of T/Z(T), while also T′ is contained in its kernel, so that T is nilpotent and T ≤ F (H). Since Q ≤ T, we have Q ≤ O_{p}(H). In fact Q ⊲ H since Q is the subgroup generated by all H-conjugates of x within the Sylow p-subgroup S.

We single out some consequences of Lemma 1.4.3:

Corollary 1.4.4: Let H be a finite quasiprimitive complex linear group of degree n.

i) If x ∈ H\Z(H) is an element of prime order p, then x has at least \frac{p+3}{4} distinct eigenvalues. ii)If O_{p}(H) = 1 and n ≤ \frac{p(p−ε)}{4}, where ε = ±1 and p ≡ ε (mod 4), then H has elementary Abelian Sylow p-subgroups.

Proof: Let S be a Sylow p-subgroup of H and let Q be the weak closure of ⟨x⟩ in S with respect to H. If x has fewer than \frac{p+3}{4} eigenvalues, then x has minimum polynomial of degree at most \frac{p+1}{4} in any reduction (mod p) of H, so that 1.4.3 applies. Then Q is extraspecial of exponent p and Z(Q) ≤ Z(H), so that x ∈ Q\Z(Q). Furthermore, every element of Q\Z(Q) has trace zero, so every element of Q\Z(Q) has p different eigenvalues, a contradiction, proving i).

To prove ii), we first note that S is Abelian. For otherwise, there is an element x ∈ S′ ∩ Z(S) of order p. Now x acts trivially on each 1-dimensional S-invariant subspace, and x has only one eigenvalue on any irreducible pdimensional S-invariant subspace, so x has at most \frac{p+1}{4} distinct eigenvalues, and x ∈ Z(H) by part i), a contradiction. Also, S must be elementary Abelian, for if y ∈ S has order p^{2} and x = y^{p}, then the minimum polynomial of x in any reduction (mod p) of H has degree at most \frac{p+1}{4}, contrary to 1.4.3 b), since O_{p}(H) = 1.

Lemma 1.4.5: Let H be a finite quasiprimitive complex linear group of degree n > 8. Let x be an element of prime order p in H\O_{p}(H) which acts as a transvection in some reduction (mod p) of H. Then p = 2 and H has a normal subgroup N with N/Z(N) ≅ A_{n+1}.

Proof: Since x ∉ O_{p}(H), by the Baer-Suzuki theorem, there is an element h ∈ H such that X = ⟨x, x^{h}⟩ is not a p-group. Then there is only one non-trivial composition factor of X on the underlying module in characteristic p, which is 2-dimensional. Then X has a subgroup acting as SL(2, p) on this composition factor. If p > 3, then X contains an element y of order 6 which (in the reduction (mod p)) has the eigenvalue 1 with multiplicity n − 2 and further eigenvalues −ω, −ω^{2} for some primitive cube root of unity ω ( with multiplicity 1 each).

But then the complex eigenvalues on y all lie within \frac{π}{3} of 1 on the unit circle, contrary to 1.3.10. Hence p ≤ 3. Suppose that p = 3. Then a similar argument shows that X contains an element of order 4 which has the (complex) eigenvalue 1 with multiplicity n − 2 and with eigenvalues i and −i with multiplicity one each. This contradicts Lemma 1.3.16. Hence p = 2.

Now X contains a strongly real element z of odd prime order q which has the (complex) eigenvalue with multiplicity n − 2 and α and ᾱ with multiplicity one each, for some primitive complex q-th root of unity α. By 1.3.10 and 1.3.17 we must have q = 3. As n ≥ 9, the result now follows from Lemma 1.3.15.

Section Five

We have seen that when H is a finite quasiprimitive complex linear group H, any element x ∈ H of prime order p which acts with minimum polynomial of small enough degree in the reduction (mod p) of H is in O_{p}(H). We now examine ways in which we can deduce that a p-element of H has minimum polynomial of low degree in any reduction (mod p) of H.

Let us define the ring R_{π} as at the beginning of this section, and we let V be an R_{π}H-module obtained from the given representation. Let M be a subgroup of H and let B_{1}, B_{2}, …, B_{n} be the p-blocks of ordinary irreducible characters of H. For each i, let e_{i} be the primitive central idempotent of R_{π}H associated to H. Then e_{i} is expressible as \frac{1}{|H|} (∑_{χ∈B_{i}} χ(1) ∑_{h∈H} χ(h^{−1})h) , and we have e_{i}e_{j} = δ_{ij}e_{i} for each i, j. Then V = V e_{1} ⊕ V e_{2} ⊕ … ⊕ V e_{n} is a direct sum decomposition of V as R_{π}H-module (though some terms may be zero).

In particular, if M is a p′-group, then each p-block of M contains only one irreducible character and V e_{i} ⊗_{R_{π}} ℂ is a homogeneous ℂM-module (possibly zero) for each i in this case.

Suppose that M contains a non-identity p-element x. Let ρ be the unique prime ideal of R_{π}, and let bars denote images (mod ρ). Suppose that x acts with minimum polynomial of degree k on V̄ . Then x must have minimum polynomial of degree k on V e_{i} for some i. Let W = V e_{i} for this i. Now let T be a p′-subgroup of C_{M}(x).

We may decompose W as W = W f_{1} ⊕ … ⊕ W f_{m} as a direct sum of R_{π}T - modules, where each f_{i} is a primitive central idempotent of R_{π}T. Since x centralizes T, this is a decomposition as a direct sum of R_{π}(T × ⟨x⟩)-modules. Hence x must act with minimum polynomial of degree k on \overline{W f_{j}} for some j. Set U = W f_{j} ⊗_{R_{π}} ℂ for this j. Then x must have at least k distinct eigenvalues on U. Now each eigenspace of x on U is T -invariant, while U is homogeneous as ℂT -module, so U as ℂT -module is a direct sum of at least k copies of an irreducible module. As a particular case in which M = C_{H}(x), we state:

Lemma 15.5.1: Let H be a finite quasiprimitive complex linear group of degree n and let χ be the afforded by the given representation of H. Let p be a prime and x be a non-identity p-element of H which acts with minimum polynomial of degree k in some reduction (mod p) of H.

Then for any p′-subgroup T of C_{H}(x) there is a p-block B of ordinary irreducible characters of C_{H}(x) and some irreducible character ϕ of T such that

∑_{μ∈B}⟨Res_{C_{H}(x)}^{H}, μ⟩⟨Res_{T}^{C_{H}(x)}(μ), ϕ⟩ ≥ k.

We single out a particular case for special mention:

Lemma 1.5.2: let H be a finite quasiprimitive complex linear group of degree n. Let x be a non-identity p-element of H for some prime p and suppose that x acts with minimum polynomial of degree k in some reduction (mod p) of H. Let y be a p-regular element of C_{H}(x).

Then y has an eigenspace on which x acts with at least k distinct eigenvalues.

Section Six

In this section we prove two results which will be useful later and which may be be of some independent interest. The first is a character-theoretic result which may be well-known.

Lemma 1.6.1: Let H be a finite group with a cyclic Sylow p-subgroup P = ⟨x⟩ of order p for some prime p. Suppose that N_{H}(P) : C_{H}(P)] = 3 and [H : O_{p′}(H)] is even. Then all non-exceptional characters in the principal p-block of H are rational-valued.

Proof: We may suppose that O_{p′}(H) = 1, and do so. By Brauer’s theory of blocks of defect one, there are exactly three non-exceptional characters in the principal p-block of H. The principal p-block is invariant under algebraic conjugation, so , all agreeing if one of its non-exceptional characters is not rational valued, the non-exceptional characters in the principal p-block consist of the trivial character and two other algebraically conjugate characters, say χ_{1} and χ_{2}. There are also \frac{p−1}{3} exceptional characters in the principal p-block all agreeing on p-regular elements, and there is a sign ε such that the sum of the exceptional characters takes value ε on p-singular elements. Furthermore, there is a sign δ such that χ_{1}(x) = χ_{2}(x) = δ. We have 1 + 2δχ_{1}(1) + εμ(1) = 0. Then we must have ε = −δ and μ(1) = 2χ_{1}(1) + δ.

Since H has even order, there is an involution t ∈ H. By hypothesis, t inverts no p-singular element. By the block orthogonality relations, we have μ(t) = 2χ_{1}(t) + δ. Since t inverts no p-singular element, we have

1 + 2δ \frac{χ_{1}(t)^{2}}{χ_{1}(1)} − δ \frac{(2χ_{1}(t) + δ)^{2}}{2χ_{1}(1) + δ} = 0.

Then χ_{1}(t) = χ_{1}(1) and t is in the kernel of each irreducible character of the principal p-block. Hence t ∈ O_{p′}(H) = 1, a contradiction.

The next result is motivated by a situation which is frequently encountered in signalizer functor analysis.

Lemma 1.6.2: Let H be a finite group with Sylow p-subgroup P for some prime p. Suppose that P contains an Abelian normal subgroup which is not cyclic and that there is a subgroup M of H such that N_{H}(D) ≤ H whenever D ≤ P and D contains an elementary Abelian subgroup of order p^{2}. Then P ∩ H′ = P ∩ M′.

Proof: Let F denote the set of subgroups Q of P such that:

i) N_{P} (Q) ∈ Syl_{p}(N_{H}(Q)). ii)N_{H}(Q) is p-constrained. iii) Q ∈ Syl_{p}(O_{p′,p}(N_{H}(Q)).

Then by a Theorem of Goldschmidt, F is a conjugation family, and by a Theorem of Alperin, we have P ∩ H′ = ⟨P ∩ N_{H}(Q)′ : Q ∈ F⟩. Thus it suffices to prove that N_{H}(Q) ≤ M for all Q ∈ F. Suppose otherwise.

Let Q be a subgroup in F with N_{H}(Q) ≰ M. Then Q contains no elementary Abelian subgroup of order p^{2}, so Q is cyclic or generalized quaternion. Set N = N_{H}(Q) and N̄ = N/O_{p′}(N).

Now Q ≠ P from the hypotheses, so N_{P} (Q) ≠ Q. We claim that p = 2 and Q is quaternion of order 8. Since Q ∈ Syl_{p}(O_{p′,p}(N) and N is p-constrained, N̄/Z(Q̄) acts faithfully on Q̄. If Q is cyclic, then N̄/Q̄ is Abelian, so has order prime to p, contrary to the fact that N_{P} (Q) > Q. Hence p = 2 and Q is generalized quaternion. If |Q| > 8, then Aut(Q) is a 2-group, so that N̄/Z(Q̄) is a 2-group, again leading to the contradiction N_{P} (Q) = Q. Hence Q is quaternion of order 8, as claimed.

Now N̄/Q̄ acts faithfully on Q̄/Φ(Q̄), so is isomorphic to GL(2, 2) since it has even order, but is not a 2-group. In particular, N_{P} (Q) has order 16, and for any a ∈ N_{P} (Q)\Q, [Q, a] ≰ Z(Q).

Now let A⊲P be Abelian, but not cyclic. Then AQ > Q, so that N_{AQ}(Q) > Q. But N_{AQ}(Q) = QN_{A}(Q), so that N_{A}(Q) ≰ Q. Choose a ∈ N_{A}(Q) ∩ Q. Then [Q, a] ≤ A ∩ Q ≤ Z(Q), a contradiction.

Chapter Two

In this chapter, we establish further properties of finite irreducible complex linear groups of prime degree.

We assume from now on that the finite group G is an irreducible complex linear group of degree 11 which does not satisfy the conclusion of Theorem A. Then G is certainly primitive. We state some further properties of G which for the most follow from the results of Chapter One. We let χ denote the character of G afforded by the given representation.

Section One: Some Preliminary Results

The first result of the section is well-known, but we include a proof for the sake of completeness.

Lemma 2.1.1: Let H be a finite primitive complex linear group of prime degree p with Z(H) = 1. Then H has a self-centralizing Sylow p-subgroup of order p.

Proof: Let P ∈ Syl_{p}(H). Then P is Abelian, for otherwise 1 ≠ Z(P) ≤ Z(H), contrary to the hypotheses. Thus P is Abelian. Let y ∈ C_{H}(P)^{#}, and μ be the character afforded by the given representation of H. Then gcd([H : C_{H}(P)], μ(1)) = 1, so that μ(y) = 0, since Z(H) = 1. Now

|C_{H}(P)|⟨Res_{C_{H}(P)}^{H}(μ), 1⟩ = p

and |C_{H}(P)| divides p. Hence P = C_{H}(P).

We next give a different proof of a result of R. Brauer.

Lemma 2.1.2: Let H be a finite primitive complex linear group of prime degree p > 2 with an Abelian Sylow p-subgroup P. Then H′ is a non-Abelian simple group.

Proof: By transfer, we have P ∩H′∩Z(H) = 1. As in the proof of Lemma 2.1.1, μ(h) = 0 for all h ∈ P \Z(H). Hence |P |⟨Res_{P}^{H}(μ), Res_{P}^{H}(μ)⟩ = p^{2}|P ∩ Z(H)|. On the other hand, ⟨Res_{P}^{H}(μ), Res_{P}^{H}(μ)⟩ ≥ p as P is Abelian and μ(1) = p. Hence [P : P ∩ Z(H)] ≤ p.

Since p = μ(1) divides [H : Z(H)], we have [P : P ∩ Z(H)] = p. Now O_{p′} is Abelian, since all irreducible constituents of Res_{O_{p′} (H)}^{H}(μ) are linear by Clifford’s theorem. Since H is primitive, we have O_{p′}(H) ≤ Z(H), so that, in particular, H has no normal p-complement. Thus p divides |H′|. Thus P = (P capH′) × (P ∩ Z(H)). Let Q = P ∩ H′. Now C_{H}(Q) = C_{H}(P), and as before we see that μ vanishes on all elements of C_{H}(Q)\Z(H). Hence

|C_{H}(Q)|⟨Res_{C_{H}(Q)}^{H}(μ), Res_{C_{H}(Q)}^{H}(μ)⟩ = p^{2}|Z(H)|.

Hence C_{H}(Q) = P × O_{p′}(Z(H)) = Q × Z(H). Also C_{H′}(Q) = Q by transfer and unimodularity of the representation of H′.

We now have F (H′) ≤ Z(H′) = 1, so that H′ has a component L which is a non-Abelian simple group of order divisible by p, and is irreducible. Then C_{H′}(L) ≤ Z(H′) = 1, so that L = F^{∗}(H′). Now H = LN_{H}(Q) and L × Z(H) = LC_{H}(Q). Thus H′ ≤ L × Z(H), and H′ = L × (H′ ∩ Z(H)) = L.

In view of this result, and since none of the simple groups occurring in case b) of Theorem A admit an outer automorphism of order 5, we may make the additional:

Hypothesis: G is a non-Abelian finite simple group.

From now on, we will let P denote a fixed Sylow 11-subgroup of G, and we now have C_{G}(P) = P.

We now aim to place further restrictions on the proper subgroups of G which contain P. This will eventual enable us to show that any proper subgroup of G which contains P and has order divisible by 25 is itself irreducible. This will in turn allow us to obtain a good bound on the power of 5 dividing |G|.

Lemma 2.1.3: Let H be a finite simple primitive complex linear group of degree p where p is an odd prime. Let Q ∈ Syl_{p}(H) and let M be a proper subgroup of H which contains Q. Then one of the following occurs:

i) M is irreducible. ii) E(M) is isomorphic to a primitive complex linear group of degree p − 1. iii) M has a normal subgroup N ≅ PSL(2, p) with [M : N] ≤ 2. iv) Q ⊲ M. Furthermore, the fixed-point space of Q, which is 1-dimensional, is M-invariant in cases ii),iii) and iv).

Proof: Let V be the underlying ℂH-module, and let μ be the irreducible character it affords. By Lemma 2.1.1, we have C_{H}(Q) = Q and |Q| = p. Let y be a generator of Q. Since μ(y) = 0, the eigenvalue 1 of y occurs with multiplicity 1 ( as do all other p-th roots of unity). Let V_{0} be the fixed-point space of y on V. Suppose that Res_{M}^{H}(V) is reducible and that N_{M}(Q) ≠ M. Notice that Z(M) = 1 in that case. We may write Res_{M}^{H}V = W_{1} ⊕ W_{2} ⊕ … ⊕ W_{k} where k > 1 and each W_{i} is an irreducible ℂM-module, and we may suppose that V_{0} ≤ W_{1}.

We first prove that M acts faithfully on W_{j} for each j > 1. For let K_{j} be the kernel of the action of M on W_{j}. Then K_{j} ≤ O_{p′}(M) as V^{Q} ≤ W_{1}. Then K_{j} = 1, for otherwise QK_{j} is a Frobenius group with complement Q, so must act irreducibly on V, a contradiction. Since Q is not normal in M, we have dim_{ℂ}(W_{j}) ≥ \frac{p−1}{2} for each j > 1, by Lemma 1.2.1. Thus k ≤ 3.

If k = 3, then M ≅ PSL(2, p) by Lemma 1.2.2, so iii) holds. Furthermore, dim_{ℂ}(W_{1}) = 1, so that W_{1} = V_{0}. Suppose then that k = 2. Now Q acts faithfully on W_{2} so that M acts faithfully on W_{2} by an argument like that above. Since Q is not normal in M, we must have dim_{ℂ}(W_{2}) ≥ \frac{p−1}{2}. If dim_{ℂ}(W_{2}) < p − 1, then Q acts faithfully on W_{1}, so we also have dim_{ℂ}(W_{1}) ≥ \frac{p+1}{2}, for V_{0} ≤ W_{1}, so that N_{M}(Q) does not act irreducibly on W_{1}, so Lemma 1.4.1 i) may be invoked.

In that case, M ≅ PSL(2, p), but this is a contradiction, as M has irreducible characters of degree \frac{p−1}{2} and \frac{p+1}{2}, whereas PSL(2, p) has an irreducible character of degree \frac{p−1}{2} if and only if p ≡ 3 ( mod 4) and an has an irreducible character of degree \frac{p+1}{2} if and only if p ≡ 1 (mod 4). Thus when k = 2 we must have dim_{ℂ}(W_{2}) = p − 1 and dim_{ℂ}(W_{1}) = 1.

In the latter case, suppose that W_{2} is an imprimitive module, say W_{2} ≅ Ind_{N}^{M}(W_{0}) for some ℂN-module W_{0} and some proper subgroup N of M. let K = ∩_{m∈M}N^{m}. Since [M : N] < p, we have Q ≤ K. Now each irreducible constituent of res_{N}^{H}(μ) has degree at most \frac{p−1}{2}. Since Q is not normal in M, at least one of these constituents has degree \frac{p−1}{2}, using Lemmas 1.2.1 and Lemma 1.2.2. Hence N = K and we have N ≅ PSL(2, p) in this case, so case iii) occurs. Suppose then that W_{2} is primitive.

Let E = E(M). Now F (M) = 1 since O_{p′}(M) = 1 in this case, we have F (M) = 1, so that E ≠ 1 and Q ≤ E. Hence E is a non-Abelian simple group, as |Q| = p. Then Res_{E}^{M}(W_{2}) is a direct sum of isomorphic irreducible ℂE-modules. If there is more than one of of these, then E ≅ PSL(2, p) and [M : E] ≤ 2. If E is irreducible but imprimitive, then we again obtain E ≅ PSL(2.p) and [M : E] ≤ 2. Otherwise, we are in case ii).

Lemma 2.1.4: No prime larger than 11 divides |G|.

Proof: i) Suppose otherwise and that the prime q > 11 divides |G|. Then by Lemmas 1.2.1, 1.2.2, 1.2.4 and 1.2.5, we have q ∈ {17, 19} (since we have excluded G ≅ PSL(2, 23) by hypothesis). Also |G| is not divisible by both 17 and 19. Let Q be a Sylow q-subgroup of G and let M = N_{G}(Q). Now P = C_{G}(P), so that no conjugate of P can normalize Q. Hence M is not irreducible. Let y be a generator of Q. Then y acts with minimum polynomial of degree at least \frac{q+1}{2} in any reduction (mod q) of G. By Lemma 1.5.2, each q-regular element of C_{G}(y) has an eigenvalue of multiplicity 9 or more. By Lemmas 1.3.14, 1.3.15, 1.3.17 and 1.4.1 and 1.4.5, C_{G}(y) is a {7, q}-group. If C_{G}(y) contains an element z of order 7, then again using 1.4.1, we see that P normalizes a non-trivial 7subgroup of G, since z acts with minimum polynomial of degree at most 3 in any reduction (mod 7) of G. But since C_{G}(P) = 1, this implies that there is a P -invariant elementary Abelian 7-subgroup, A say, of order 7^{10}. Then A may be assumed to be diagonal in which case A consists of all unimodular diagonal matrices whose diagonal entries are 7-th roots of unity. But then A contains an element u of order 7 whose eigenvalues are 1 (with multiplicity 9), e^{\frac{2πi}{7}} and e^{\frac{−2πi}{7}} , contrary to Lemma 1.3.3. Hence we have C_{G}(Q) = Q. Then χ lies in the principal q-block. By the theory of blocks of defect 1, χ must be an exceptional character. Also, since C_{G}(Q) = Q, there is a divisor d of q − 1 such that χ(1) ≡ ±d (mod q), a contradiction since χ(1) = 11 and q ∈ {17, 19}.

Lemma 2.1.5*: i) The Sylow 7-subgroups of G are elementary Abelian of order at most 7^{8}. ii) No element of order 7 of G acts with minimum polynomial of degree less than 4 in any reduction (mod 7) of G. iii) Whenever u ∈ G has order 7, each 7-regular element of C_{G}(u) has an eigenvalue of multiplicity greater than 3. iv) If T is a Sylow 7-subgroup of G, then there is some g ∈ G\N_{G}(T) such that [T : T ∩ T^{g}] ≤ 49.

Proof: i) By Corollary 1.4.4, G has elementary Abelian Sylow 7-subgroups. Let T be one of these. By the argument of the previous Lemma, |T | ≤ 7^{9} and no 7-local subgroup of G is irreducible. Part ii) follows from Lemma 1.4.3. Part iii) follows from Lemma 1.5.1 and part iv) follows from Lemma 1.4.2. In fact, we may use part ii) to get the slightly sharper result stated in part i). For let λ_{1}, …, λ_{9} be (not necessarily distinct) constituents of Res_{T}^{G}(χ) and let θ = ∑_{i=1}^{9} λ_{i}. Since no non-identity element of T has fewer than four distinct eigenvalues, kerθ = 1 and we must have |T | < 7^{9}, otherwise there is an element t ∈ T^{#} and a 7-th root of unity α with θ(t) = 9α.

Lemma 2.1.6: No nonidentity element of G has an eigenspace of codimension 1 or 2.

Proof: Most of this is covered by the arguments used during the proof of Lemma 2.1.4, Lemmas 1.3.13 and 1.3.13* show that G does not contain an involution with an eigenspace of codimension 1.

Lemma 2.1.7: Let p be a prime and y ∈ G be an element of order p. Then y does not act as a transvection in any reduction (mod p) of G.

Proof: This follows from Lemma 1.4.5 and the assumption that G is simple but does not satisfy the conclusion of Theorem A.

Lemma 2.1.8: i) 2^{18} does not divide |G|. ii) 3^{14} does not divide |G|. iii) The Sylow 2-subgroups of G have exponent dividing 16. iv) The Sylow 3-subgroups of G have exponent dividing 27.

v) The Sylow 5-subgroups of G have exponent dividing 25.

Proof: i) Let R ∈ Syl_{2}(G). Then there is a (not necessarily irreducible) character α of degree 8 of R such that Res_{R}^{G}(χ) = α + β for some other character β of R. We claim that α is faithful. Otherwise, α contains an involution t which has the eigenvalue 1 with multiplicity at least 8. But t acts with determinant 1, so that t has the eigenvalue 1 with odd multiplicity. Hence t has the eigenvalue 1 with multiplicity 9, contrary to Lemma 2.1.6. Since R has a faithful character of degree 8, there is an Abelian subgroup A ⊲ R with [R : A] ≤ 2^{7}. Now |A| ≤ 2^{10} by Lemma 1.3.12, so |R| ≤ 2^{17}. ii) Let T ∈ Syl_{3}(G). Now Res_{T}^{G}(χ) contains no irreducible constituent of degree 9, for otherwise a non-identity element of Z(T) has an eigenspace of codimension 2, contrary to Lemma 2.1.6. We may write Res_{T}^{G}(χ) = ϕ + ψ, where ϕ is a reducible character of degree 9. Now ϕ must be faithful, otherwise an element of order 3 in its kernel violates 2.1.6. Since ϕ is reducible, but faithful, it is easy to see that T has an Abelian normal subgroup D with [T : D] ≤ 27. Since |D| ≤ 3^{10} by Lemma 1.3.12, we have |T | ≤ 3^{13}.

Parts iii),iv) and v) follow from the remarks preceding Lemma 1.4.1.

Section Two

In this section, whose methodology is rather different from that to be found elsewhere in the thesis, we take the first steps towards determining the 5-local structure of G. We prove that N_{G}(P) contains an element y of order 5 and we will eventually see that C_{G}(y) has restricted structure. This in turn provides useful bounds on the size of a Sylow 5-subgroup of G.

Lemma 2.2.1: [N_{G}(P) : C_{G}(P)] ≠ 2.

Proof: Suppose otherwise. By Lemma 2.1.3, we know that 11 is the largest prime divisor of |G|. Choose a generator x of P. Now χχ̄ has a nontrivial irreducible constituent θ of degree prime to 11. Either θ is non-exceptional in the principal 11-block and θ(1) ≡ ±1 (mod 11) or else θ is exceptional in the principal 11-block and θ(1) ≡ ±2 (mod 11).

In the latter case, all exceptional characters (there are 5 of these) occur with equal multiplicity in χχ̄, since they all agree on 11-regular elements and χχ̄ vanishes on all 11-singular elements. Hence we have 4 < θ(1) ≤ 24, so θ(1) ∈ {9, 20, 24} since 13 does not divide |G|. If θ(1) = 9, then Lemma 1.2.5 is contradicted. If θ(1) = 24, then the non-trivial non-exceptional character in the principal 11-block has degree 23, contrary to 2.1.4. Hence θ(1) = 20, and then the non-trivial non-exceptional irreducible character, ψ say, in the principal 11-block has degree 21. Since the trivial character and the five exceptional characters,say {θ_{i} : 1 ≤ i ≤ 5} of degree 20 all occur (with multiplicity 1) in χχ̄ in this case, ψ does not occur in χχ̄ and, furthermore, no other irreducible character of degree prime to 11 occurs, since only the principal 11-block has positive defect. But this is a contradiction since 1 + ∑_{i=1}^{5} θ_{i}(x) = 2 and χ(x) = 0.

Hence we may suppose that no exceptional character from the principal 11block occurs in χχ̄. Furthermore, we suppose that a non-exceptional irreducible character θ in the principal 11-block for which θ(x) = −1 occurs in χχ̄, since the trivial character occurs with multiplicity 1 in that character. Then for ψ an exceptional character in the principal 11-block, we have 1 − θ(1) + εψ(1) = 0 for some sign ε. Since 0 < ψ(1) = ε(θ(1) − 1), we have ε = 1, and θ(1)(θ(1) − 1) divides |G|. But θ(1) ∈ {120, 109, 98, 87, 76, 65, 54, 43, 32, 21, 10}. The divisibility condition reduces the possibilities to θ(1) ∈ {9, 21}. Now θ(1) = 9 contradicts Lemma 1.2.5, as before. Hence θ(1) = 21 and ψ(1) = 20. Note that θ is rational valued, since it is the unique irreducible character of degree 21 in the principal 11-block. Also, by block orthogonality relations, we have ψ(y) = θ(y) − 1 for all 11-regular y ∈ G.

Let t be an involution of N_{G}(P). Computing ⟨Res_{N_{G}(P)}^{G}(θ), 1⟩ gives 11 + 11θ(t) ≥ 0 and computing ⟨Res_{N_{G}(P)}^{G}(θ), λ⟩ for the other linear character λ of P ⟨t⟩ yields 11 − 11θ(t) ≥ 0. Hence θ(t) ∈ {1, −1}, but consideration of the determinant forces θ(t) ≡ 1 (mod 4), so that θ(t) = 1.

Now 11 conjugates of t invert x, and standard class algebra constant calculations show that

\frac{|G|}{|C_{G}(t)|^{2}} \frac{(θ(1) − θ(t))^{2}}{θ(1)(θ(1) − 1)} = 11

and that all involutions of G are conjugate to t. Now we have

[G : C_{G}(t)] = 3 × 7 × 11 × \frac{|C_{G}(t)|}{20}.

Since [G : C_{G}(t)] is odd, |G| is not divisible by 8.

Let R ∈ Syl_{7}(G) and let u ∈ Ω_{1}(Z(R))^{#}. Since [G : C_{G}(u)]\frac{θ(u)}{θ(1)} is a rational algebraic integer, and since gcd([G : C_{G}(u)], θ(1)) divides 3, by a minor variant of Burnside’s argument we have: \frac{3θ(u)}{θ(1)} ∈ ℤ, so that θ(u) = 7z for some z ∈ ℤ. Since G is simple, we have θ(z) ∈ {−2, −1, 0, 1, 2}. Hence

[G : C_{G}(u)]\frac{θ(u)}{θ(1)} − [G : C_{G}(u)]

is not divisible by 7, so that θ does not lie in the principal 7-block.

Since C_{G}(P) = P, there is a non-trivial irreducible character which lies in the principal 11-block and in the principal 7-block by Lemma 1.2.14. Hence ψ (and all its algebraic conjugates) must lie in the principal 7-block. Notice then that θ is the only character in its 7-block which lies in the principal 11-block. Applying Lemma 1.2.14 with p = 7 and q = 11, (to the 7-block of θ and the principal 11-block, and with R in the role of P), we obtain θ(1)θ(x) ≡ 0 (mod |R|), so that |R| = 7. Since we already know that the principal 7-block contains at least 6 irreducible characters (and since G has no normal 7-complement), we must have [N_{G}(R) : C_{G}(R)] = 6. By the theory of blocks of defect 1, the principal 7-block contains the trivial character, the five algebraic conjugates of ψ, and one more irreducible character β, say, of degree 99.

Now let ψ = ψ_{1}, ψ_{2}, ψ_{3}, ψ_{4}, ψ_{5} be the exceptional characters in the principal 11-block, labelled so that for some primitive complex 11-th root of unity ω, we have ψ_{i}(x) = −(ω^{i} + ω̄^{i}). Notice that ∑_{i=1}^{5} ψ_{i}(x)^{2} = 9. We have already remarked that for each i and each 11-regular element y ∈ G, we have ψ_{i}(y) = θ(y) − 1.

Let y be any non-identity 11-regular element of G. Then the usual computation shows that the coefficient of y in the square of the class sum of x is

\frac{|G|}{121} (1 + \frac{θ(y)}{21} + \frac{9(θ(y) − 1)}{20}) .

This is certainly a non-negative integer, and we know that θ(y) ∈ ℤ, so that θ(y) ≥ −1.

Now let S ∈ Syl_{5}(G) an choose s ∈ Ω_{1}(Z(S))^{#}. Then the coefficient of x in the square of the class sum of s is

\frac{|G|(21 − θ(s))^{2}}{420|C_{G}(s)|^{2}} .

Since this is an integer and s ∈ Z(S), we see that 5|S| divides (21−θ(s))^{2}. Since −1 < θ(s) < 21, this yields |S| = 5.

A similar argument with an element w of order 3 central in a Sylow 3subgroup of G, using the fact that (21 − θ(w))^{2} is not divisible by 3^{5}, shows that 81 does not divide |G|.

Now we have |G| = 4 × 3^{k} × 5 × 7 × 11 for some integer k ≤ 3. Since |N_{G}(P)| = 22, Sylow’s theorem yields 3^{k−1} ≡ 1 (mod 11), so that k = 1. However, this contradicts the fact that G has an irreducible character of degree 99.

Section Three

In this section, we digress slightly to examine the structure of finite primitive complex linear groups of degree 2q for a prime q > 3. In the context of the thesis as a whole, we are particularly interested in the case q = 5, which turns out to be the most difficult to treat.

The results of this section will enable us to conclude that any proper subgroup of G of order divisible by 275 is irreducible.

Lemma 2.3.1: Let q be a prime greater than 3, and let H be a primitive complex linear group of degree 2q with H = H′ and Z(H) = 1. Then H has an elementary Abelian Sylow q-subgroup S of order at most q^{2}. Furthermore, if |S| = q^{2}, then θ lies in a q-block of defect 1 with defect group A, say. In that case, if q = 5, then H contains a non-identity element which has the eigenvalue 1 with multiplicity 8. If |S| = q, then O_{q′}(C_{H}(S)) is cyclic of order dividing 3, and if it contains a non-identity element z, then z has only two eigenvalues, each with multiplicity q, and θ(z) = −q.

Proof: Let θ be the character of H afforded by the given representation, and let S be a Sylow q-subgroup of H. Since H is primitive and Z(H) = 1, we also have F (H) = 1, and hence S is elementary Abelian if q > 5, by Lemma 1.4.3.

Suppose then that q = 5, but that S is non-Abelian. We claim that S contains an element s which acts with quadratic minimum polynomial in some reduction (mod 5) of H. If S contains an element u of order 25, then (u^{5} − 1)^{2} acts as zero in any reduction (mod 5) of H, so s = u^{5} has quadratic minimum polynomial in any reduction (mod 5) of H. If S is non-Abelian, then choose an element s of order 5 in S′ ∩ Z(S). Then all linear constituents of Res_{S}^{H}(θ) have S in their kernel, and s is represented by scalars in any irreducible 5dimensional representation of S. Hence s has only two eigenvalues in the given representation, so acts quadratically in any reduction (mod 5) of H.

By the Baer-Suzuki theorem, there is an element h ∈ H such that ⟨s, s^{h}⟩ is not a 5-group. Set T = ⟨s, s^{h}⟩. By the Hall-Higman theorem, O_{5′}(T) ≤ Z(T). Fix a reduction (mod 5) of H and let U be the underlying module. We claim that all composition factors of Res_{T}^{H}(U) have dimension at most 2 (possibly after extending scalars to an algebraically closed field of characteristic 5). For set t = s^{h}. Let W be a composition factor of Res_{T}^{H}(U). Then s−t has an eigenvector w ∈ W. Then the subspace spanned by w and ws is the same as the subspace spanned by w and wt. Since s and t act with quadratic minimum polynomial on W, this space is T -invariant, so is all of W. Since T is not a 5-group, there is a 5-regular element y ∈ T which acts non-trivially on at least one of these composition factors, say V. Let T̄ = T/C_{T} (V). Then O_{5}(T̄) = O_{5′}(T̄) = 1, (the second equality by the Hall-Higman theorem). Hence T̄ has a component, say L̄.

Let M be a terminal member of the derived series of T. Then M ≠ 1. Let M̄ = M/O_{5}(M). Then M̄ has a component, Ȳ , say. Then Ȳ has a nontrivial 2-dimensional representation over a field of characteristic 5, so that Ȳ ≅ SL(2, 5^{m}) for some m. Since all composition factors of U are 1 or 2-dimensional, each of them either admits Ȳ faithfully or with trivial action. If the action is faithul, then, in particular, the unique involution of Ȳ acts non-trivially, since PSL(2, 5^{m}) has no non-trivial 2-dimensional representation over any field.

Let b̄ be an element of order 6 in Ȳ with a preimage b of order 6 in M. Then b̄ acts non-trivially on some composition factor of Res_{M}^{G}(U), and Ȳ acts faithfully on any such composition factor. If b̄ were to act trivially on any of the composition factors, then the (complex) eigenvalues of b in the given complex representation of H would be 1, e^{\frac{2πi}{3}} , e^{− \frac{2πi}{3}} , contrary to Lemma 1.3.10. Hence Ȳ acts faithfully on each of the composition factors. But then b^{3} acts as −I in the given representation, a contradiction. Hence S is elementary Abelian in all cases.

Let us now return to the case of general q. We claim that θ does not belong to the principal q-block. This is clear when |S| = q, so suppose that |S| > q. Then for each s ∈ S^{#}, we have gcd([H : C_{H}(s)], θ(1)) ≤ 2, so that \frac{2θ(s)}{θ(1)} is an algebraic integer, and θ(z) = qα(z) for some algebraic integer α(z). Then θ(z) has the form ∑_{i=0}^{r−1} m_{i}η^{i}, where η = e^{\frac{2πi}{q}} and the m_{i} are non-negative integers with ∑_{i=0}^{q−1} m_{i} = 2q. Then α(z) = ∑_{i=0}^{q−2} \frac{(m_{i}−m_{q−1})}{q}η^{i}. Since ℤ[η] is the ring of algebraic integers of ℚ[η], and has ℤ-basis {η^{i} : 0 ≤ i ≤ r − 2}, we see that the m_{i} are all congruent to each other (mod q). Since their sum is 2q, there are only two possibilities: the m_{i} could all be equal, in which case θ(s) = 0. Or q − 1 of the m_{i} are 1, and m_{j} = q + 1 for one value of j. In the second case, θ(s) = qη^{j} for this value of j. If the second case does not occur for any s, then we have θ(s) = 0 for all s ∈ S^{#}. In that case |S| divides θ(1) = 2q, so |S| = q, contrary to the present hypotheses. Hence we may suppose that there is an s ∈ S^{#} such that θ(s) = qη^{j} for some j. In that case

2 ([H : C_{H}(s)]\frac{θ(s)}{θ(1)} − [H : C_{H}(s)]) = [H : C_{H}(s)](η^{j} − 2)

which lies outside the ideal (1 − η). Hence θ does not lie in the principal q-block when |S| > q, as claimed.

Suppose that |S| > q^{2}, and let A be a subgroup of A of order q^{3}. Suppose that there are c cyclic subgroups of A with a generator u such that θ(u) = 0, d cyclic subgroups of A with a generator v such that θ(v) = q, and e cyclic subgroups of A with a generator w such that 0 ≠ θ(w) ≠ q. Then c + d + e = 1 + q + q^{2}. Also,

|A|⟨Res_{A}^{H}(θ), 1⟩ = q(2 + d(q − 1) − e)

and

|A|⟨Res_{A}^{H}(θ), Res_{A}^{H}(θ)⟩ = q^{2}(4 + (q − 1)(d + e)).

Since |A| ≥ q^{3}, the first displayed equation yields d + e ≡ 2 (mod q) and the second displayed equation yields d + e ≡ 4 (mod q), a contradiction. Hence |S| ≤ q^{2}.

Suppose now that |S| = q^{2}. In that case, θ lies in a q-block of defect 1 by a well-known result of Brauer. In that case, H contains a Sylow intersection of order q. The equations displayed above hold with A replaced by S, except that now c + d + e = 1 + q. It follows that d + e = 2, that

⟨Res_{S}^{H}(θ), 1⟩ = d,

and that

⟨Res_{S}^{H}(θ), Res_{S}^{H}(θ)⟩ = 2q + 2.

In fact, we are now in a position to assert that the case d = e = 1 does not occur. Suppose otherwise. Then there is a cyclic subgroup A of S such that θ(a) = q for a generator a of A, and there is a cyclic subgroup B of A such that θ(b) = qη^{j} for some i with 0 < j < q. Then clearly no non-identity power of a is conjugate to a power of b. But since θ lies in a q-block of defect 1, both a and b are conjugate to non-identity elements of a defect group of that q-block, a contradiction. Hence we either have d = 2 or e = 2.

We next claim that e = 0. For suppose that e = 2. Let A and B be two cyclic subgroups of order q of S such that θ(a) = θ(b) = qη for a and b. We claim that ab ∈ Z(N_{G}(S)), which contradicts the fact that H = H′, since S is Abelian. For choose n ∈ N_{G}(S). Then n permutes the cyclic subgroups of S such that θ does not vanish on their non-identity elements. Hence n either interchanges A and B or else normalizes both A and B. If n interchanges A and B, then we must have a^{n} = b and b^{n} = a, since a is the unique element of A on which θ takes the value qη and b is the unique element of B on which θ takes the value qη. Similarly, if n normalizes both A and B, we must have a^{n} = a and b^{n} = b. In either case, n centralizes ab.

Now we can conclude that Res_{S}^{H}(θ) = 2α_{0} +∑_{i=1}^{2q−2} α_{i}, where α_{0} is the trivial character, and the α_{i} are distinct non-trivial linear characters of S. Suppose that C_{H}(S) ≠ S (still in the case |S| = q^{2}). Suppose that C_{H}(S) contains an element z of prime order r ≠ q. Then z acts as a transvection in any reduction (mod r) of H. By the Baer-Suzuki theorem, there is an element h ∈ H such that ⟨z, z^{h}⟩ is not an r-group. Then ⟨z, z^{h}⟩ contains an element w of prime order p ≠ r ( but possibly p = q). Since ⟨z, z^{h}⟩ has a fixed-point space of codimension (at most) 2 in any reduction (mod r) of H, we see that w has the (complex) eigenvalue 1 with multiplicity 2q − 2. Hence we have p ≠ q, since we have seen that no element of order q has such an eigenspace.

Now we consider C_{H}(S) in the case that |S| = q. Let V be the underlying module afforded by some reduction (mod q) of H. Then Res_{S}^{H}(V) is projective. Let z ∈ O_{q′}(C_{H}(S))^{#}. Then (possibly after extending scalars) each eigenspace of z on V affords a projective module for S. Hence each complex eigenspace of z has dimension divisible by q. Hence z has only two (complex) eigenvalues, each with multiplicity q. Since H = H′, r is odd, as θ(1) = 2q and z acts with determinant 1. Also unimodularity implies that the two eigenvalues of z are complex conjugates of each other. Now z must have order r, since z^{r} acts trivially in any reduction (mod r) of H.

By the Baer-Suzuki Theorem, there is some h ∈ H such that L = ⟨z, z^{h}⟩ is not an r-group. By Lemma 1.3.19 and Lemma 1.4.4, we have r ∈ {3, 5}, since z ∉ O_{r}(L). If r = 5, then z acts with quadratic minimum polynomial in any reduction (mod 5) of H, a situation which can be disposed of as above. Hence r = 3, and we have θ(z) = −q. Since z was an arbitrary element of order 3 in C_{H}(S) and θ(1) = 2q, we see that C_{H}(S) has no elementary Abelian subgroup of order 9, so we now know that O_{q′}(C_{H}(S)) has order dividing 3 in the present case |S| = q.

Now let us return to the case that |S| = q^{2} and that 1 ≠ A ≤ S is a defect group for the q-block of H containing θ. Since S is elementary Abelian, we have C_{H}(A) = A × C for some subgroup C by a theorem of Gaschütz. By the theory of blocks of defect 1, and since we already know that θ is rational valued on q-elements, there is an irreducible character ψ of C_{H}(A) with A in its kernel such that θ(ac) = ψ(c) for each c ∈ C, where ψ lies in the Brauer correspondent of the q-block containing θ. Also, we have Res_{C_{H}(A)}^{H}(θ) = ψ + β where β(1) = q

and β may be afforded by an R_{π}C_{H}(A) module Y such that Res_{A}^{C_{H}(A)}(Y) is projective. Then O^{q}(C_{H}(A)) = O^{q}(C) acts as scalars on Y, and acts faithfully on a complement to Y by unimodularity. Set N = O^{q}(C), which is (isomorphic to) a finite irreducible complex linear group of degree q. Also, O_{q}(N) = 1.

We suppose now that q = 5. If N′ has even order, then N′ contains an involution t. Then t acts trivially on Y. Since t acts with determinant 1, either β(t) = 1 or β(t) = −3. If β(t) = 1, then θ(t) = 6 and t has the eigenvalue 1 with multiplicity 8. If β(t) = −3, then t commutes with all its N-conjugates, (in which case the N-conjugates of t generate an Abelian subgroup of order 16, which contains an element u with ψ(u) = 1 and θ(u) = 6, so that u has the eigenvalue 1 with multiplicity 8), or else there is an N-conjugate v of t which does not commute with t, and then tv has the eigenvalue 1 with multiplicity 8. Hence we may suppose that N′ has odd order (note that N′ has order prime to 5 since N has Abelian Sylow 5-subgroups). Then F (N′) is Abelian and is not central in N′ by unimodularity. If |F (N′)| is divisible by a prime p ∈ {3, 7, 13, 17}, then F (N′) contains an elementary Abelian subgroup of order p^{4} which contains an element of order p which the eigenvalue 1 with multiplicity 8. Now |H| is not divisible by any prime greater than 19 by Lemma 1.2.1. If 19 divides |H|, then H ≅ PSL(2, 19) by Lemma 1.2.3, since Z(H) = 1 by assumption. But |PSL(2, 19)| is not divisible by 25. Hence we may suppose that F (N′) is an 11group. By Lemma 1.2.6, |H| is not divisible by 121. Let P ∈ Syl_{11}(H), and let y be a generator of P.

Now an R_{π}H-module W affording character θ restricts indecomposably to N_{H}(P) since y has only 6 eigenvalues. The source of W has rank 1, 2, 5 or 10. If the source has rank 10, then any p-regular element of C_{H}(P) has only one eigenvalue, arguing as in section 1.5, a contradiction. If the source has rank 5, then Res_{C_{H}(P)}^{H}(θ) is irreducible by Lemma 1.4.3. This is not the case, by Clifford’s Theorem, since Res_{P}^{H}(θ) involves the trivial character and a nontrivial linear character. If the source has rank 2 or 1, then y acts with minimum polynomial of degree at most 2 in any reduction (mod 11) of H, contrary to Lemma 1.4.3, since O_{11}(H) = 1.

Corollary 2.3.2:* i) Every subgroup of G of order divisible by 275 is irreducible. ii) Let s be an element of order 5 in N_{G}(P) and let A be a subgroup of order 25 of C_{G}(s) with s ∈ A. Then ⟨Res_{A}^{G}(χ), λ⟩ ≤ 2 for each linear character λ of A. iii) The Sylow 5-subgroups of C_{G}(s) are Abelian of rank at most 3.

Proof: i) We apply Lemma 2.1.3 with p = 11 and P in the role of Q, and with M a proper subgroup of G containing P. It is only necessary to consider case ii) of 2.1.3, in which M is a 10-dimensional irreducible linear group (with associated character θ, say) and H = E(M) is a primitive complex linear group of degree 10. Certainly H = H′. Any non-identity element of Z(H) has an eigenspace of codimension 1 when viewed as an element of G, contrary to Lemma 2.1.6. Thus Z(H) = 1. If 25 divides |H|, then H contains an element h with a 1eigenspace of dimension 8 (inside H as a 10-dimensional linear group). Since G = G′, h has the eigenvalue 1 with multiplicity 9 (as an element of G), again contrary to Lemma 2.1.6. Now C_{G}(P) = P, so we certainly have C_{M}(P) = P. By Lemma 1.2.10, viewing M as a 10-dimensional group, P centralizes O_{11′}(M), so O_{11′}(M) = 1. Hence P ≤ H. We have already seen that H has no normal 11-complement. If N_{H}(P) contains an involution t, then θ(t) = 0, so that t acts with determinant −1 (inside H as a 10-dimensional group), contrary to H = H′. Thus N_{H}(P) : P] = 5. But now M/H ≅ N_{M}(P)/N_{H}(P) has order prime to 5 since |N_{M}(P)| has order dividing 110.

ii) Since P ⟨s⟩ is a Frobenius group of order 55, we see that χ(s) = 1, that s has the eigenvalue 1 with multiplicity 3, each primitive 5-th root of unity with multiplicity 2, and we also see that that s acts trivially on the fixed-point space of P. Let λ be a linear character of A. If s is not in the kernel of λ, the result is clear. Suppose then that λ(s) = 1. Then ⟨Res_{A}^{G}(χ), λ⟩ ≤ 3. If equality holds, then the fixed-point space of P is A-invariant. But then ⟨P, A⟩ is reducible, contrary to part i). iii) Let Q ∈ Syl_{5}(C_{G}(s)) and let D = Ω_{1}(Q). Then Q is Abelian, since s has no eigenvalue with multiplicity 5 or more. Then the 1-eigenspace of s is Qinvariant. By part ii), no element of D\⟨s⟩ acts as a scalar multiplication on this space, otherwise D contains an elementary Abelian subgroup A of order 25 which has a linear character λ with ⟨Res_{A}^{G}(χ), λ⟩ = 3. Hence [D : ⟨s⟩] < 125.

Section Four

We know that N_{G}(P) contains an element, s say, of order 5. In this section, we place further restrictions on the structure of C_{G}(s), and examine the behaviour of χ on the 5-section of s in G. We let Q be a Sylow 5-subgroup of C_{G}(s) and S denote a Sylow 5-subgroup of G containing Q. By Lemma 2.3.2, Q is Abelian of rank at most 3.

Lemma 2.4.1: Q ⊲ C_{G}(s) and C_{G}(s) = QY where Y is isomorphic to either a cyclic group of order at most 3 or to the symmetric group of degree 3.

Proof: We know that χ(s) = 1, and that s has the eigenvalue 1 with multiplicity 3, and has each primitive 5-th root of unity as an eigenvalue with multiplicity 2. Let p ≠ 5 be a prime and y be a p-element of C_{G}(s)^{#}. In any reduction (mod p) of G, y has minimum polynomial of degree at most 3, and furthermore, its Jordan form has at most one Jordan block of size 3, the remaining Jordan blocks being of size 1 or 2. If p = 3, then y has order 3. If p = 2 and y has order 4, then y^{2} acts as a transvection in any reduction (mod 2) of G, contrary to Lemma 1.4.5. Hence C_{G}(s) has elementary Abelian Sylow 2-subgroups. By Lemma 2.1.5, p ≠ 7. Also p ≠ 11 since C_{G}(P) = P. By Lemma 2.1.4, C_{G}(s) is a {2, 3, 5}-group.

Since s has two Jordan blocks in any reduction (mod 5) of G, then whenever 1 ≠ U ≤ C_{G}(s) is an Abelian 5′-group, either Res_{U}^{G}(χ) has two distinct linear constituents of multiplicity 5 or more, or else Res_{U}^{G}(χ) has a a linear constituent of multiplicity 10. However, the latter case is excluded by Lemma 2.1.6. If U is not cyclic, and λ, μ are distinct linear characters of U each occurring with multiplicity at least 5 in Res_{U}^{G}(χ), then there is a non-identity element u ∈ U with λ(u) = μ(u), since ker(λμ̄) ≠ {1_{U}}, again contrary to 2.1.6. Hence any such U is cyclic. Thus C_{G}(s) has cyclic Sylow 3-subgroups (of order at most 3, as remarked above). Since we already know that C_{G}(s) has elementary Abelian Sylow 2-subgroups, we see that C_{G}(s) has a Sylow 2-subgroup of order at most 2, and a normal 2-complement, M say.

By Sylow’s Theorem Q ⊲ M and then also Q ⊲ C_{G}(s). Let T ∈ Syl_{3}(C_{G}(s)) and let D = [Ω_{1}(Q), T]. Since QT leaves each eigenspace of s invariant, we see that D acts trivially on each eigenspace of s apart from the 1-eigenspace, as each of these eigenspaces is 2-dimensional. If D ≠ 1, then D is elementary Abelian of order 25, and contains an element u of order 3 which has eigenvalues 1, e^{\frac{2πi}{3}} , e^{\frac{−2πi}{3}} on the 3-dimensional 1-eigenspace of s. But then u has the eigenvalue 1 with multiplicity 9 in the 11-dimensional representation of G, contrary to Lemma 2.1.6. Hence D = 1 and then [Q, T] = 1, so that T ⊲ C_{G}(s).

It remains to prove that C_{G}(s) contains no element of order 6. Let t be an involution of C_{G}(s). Then we have already observed that t must have two distinct eigenvalues with multiplicity 5 or more. By unimodularity, t must have the eigenvalue 1 with multiplicity 5 and the eigenvalue −1 with mutiplicity 6. Furthermore, since in a reduction (mod 5) of G, there is a Jordan block of size in the representation of s on each of these eigenspaces, we see that each 5-th root of unity occurs as en eigenvalue of s on each of the eigenspaces of t. The multiplicity of that eigenvalue is 1 on each eigenspace, except for the eigenvalue 1, which occurs with multiplicity 2 in the action of s on the −1-eigenspace of t. Hence we see that st has 9 eigenvalues with multiplicity 1, and has the eigenvalue −1 with multiplicity 2. Hence an element u of order 3 in C_{G}(st) would act with eigenvalue would act as a transvection in any reduction (mod 3) of G, contrary to Lemma 2.1.7.

Lemma 2.4.2: Let y be a 5-regular element of C_{G}(s)^{#}.

a) If y has order 2 then χ(sy) = −1.

b) If y has order 3 then χ(sy) ∈ \{1, ω, ω̄\}, where ω = e^{\frac{2πi}{3}} . If χ(sy) ≠ 1, then y is not conjugate to its inverse within G and |C_{G}(s)| = 3|Q|.

Furthermore, the 5-block of G which contains χ is invariant under Gal(ℚ(χ)/ℚ) unless perhaps there is an element y ∈ C_{G}(s) of order 3 with χ(sy) ≠ 1, in which case that 5-block is invariant under Gal(ℚ(χ)/ℚ[ω]).

Proof: As in the previous result, we know that y has two different eigenvalues such that s has 5 different eigenvalues in its action on each of those eigenspaces. It y has order 2, then the remaining eigenvalue of y is −1, since s has the eigenvalue 1 with total multiplicity 3, we have χ(sy) = −1. If y has order 3, then there are two possibilities (using unimodularity): either y has the eigenvalue 1 with multiplicity 1 and both the eigenvalues ω and ω̄ with multiplicity 5, or else y has the eigenvalue 1 with multiplicity 5 and one of ω, ω̄ with multiplicity 6. In the first case, χ(sy) = 1, and in the second case χ(sy) ∈ {ω, ω̄}.

Notice that in the second case, y is not conjugate to its inverse within G, so certainly not within C_{G}(s), and C_{G}(s)/Q is not isomorphic to S_{3}. Hence |C_{G}(s)| = 3|Q| in that case.

If there is an element u ∈ C_{G}(s) of order 3 with χ(su) = ω, then whenever α ∈ Gal(ℚ(χ)/ℚ[ω], χ^{α} agrees with χ on the 5-section of s in G. Otherwise, χ^{β} and χ agree on the 5-section of s for all β ∈ Gal(ℚ(χ)/ℚ. By (the section orthogonality relations following from) Brauer’s Second Main Theorem, we see that χ^{α} is in the same 5-block as χ in the former case, and χ^{β} is in the same 5-block as χ in the second case.

Section Five

In this section, we show that for each prime q, the size of a Sylow q-subgroup of G can be further restricted if the values taken by χ on q-elements satisfy certain rationality conditions.

Lemma 2.5.1: i) Let T ∈ Syl_{2}(G). Then |T | ≤ 2^{15} when Res_{T}^{G}(χ) is rationalvalued. ii) Let U ∈ Syl_{3}(G). Then |U| ≤ 3^{5} when Res_{U}^{G}(χ) is rational-valued.

Proof: In each case, we show that the Sylow subgroup in question has a faithful rational-valued character of degree 8, and then the result follows by 1.2.13. In case i), since χ is rational-valued on T, the irreducible constituents of Res_{T}^{G}(χ) are permuted by the 2-group Gal(ℚ(η)/ℚ, where η is a primitive 2^{d}-th root of unity and T has exponent 2^{d}. The sum (without multiplicities) of constituents in any one of the orbits is rational-valued of degree a power of 2. It follows that we may write Res_{T}^{G}(χ) = α + β where α and β are rational-valued characters of degrees 8 and 3 respectively. If there is an involution t ∈ kerα, then t has the eigenvalue 1 with multiplicity at least 8. Since t acts with determinant 1, the eigenvalue 1 of t occurs with odd multiplicity, t has the eigenvalue 1 with multiplicity 9, contrary to Lemma 2.1.6.

In case ii), we note that each orbit of the analogous Galois group on nontrivial irreducible constituents of Res_{U}^{G}(χ) has length of the form 2 × 3^{k}. Once more, it follows that we may write Res_{U}^{G}(χ) = γ + ϕ where γ and ϕ are rationalvalued characters of degrees 8 and 3 respectively. If there is an element u of order 3 in kerβ, then ϕ(u) = 0, so that u as the eigenvalue 1 with total multiplicity 9, again contrary to Lemma 2.1.6.

Lemma 2.5.2: Let U be as in Lemma 2.5.1. Then |U| ≤ 3^{10} if Res_{U}^{G}(χ) is real-valued and |U| ≤ 3^{11} if Res_{U}^{G}(χ + χ̄) is rational valued.

Proof: Suppose first that Res_{U}^{G}(χ) is real-valued. We claim that U has an Abelian subgroup of index at most 3. Any non-linear irreducible constituent ψ of Res_{U}^{G}(χ) has degree at most 3 since p̄si is also a constituent of Res_{U}^{G}(χ). To establish the claim, we may suppose that there is at least one such ψ. But then ψ and ψ̄ are the only non-linear constituents of Res_{U}^{G}(χ). Then U/kerψ has an Abelian normal subgroup A^{∗} of index at most 3. Let A be its full pre-image in U. Then A′ ≤ kerχ, so A is Abelian and [U : A] = 3.

We claim that A has rank at most 4. The non-trivial linear constituents of Res_{A}^{G}(χ) occur in complex conjugate pairs, and the trivial character must also be one of its constituents. There are at most 5 of the complex conjugate pairs, counting multiplicities. If A has rank 5, then there is a ∈ Ω_{1}(A)^{#}, which is in the kernel of 4 of these pairs. Then a has the eigenvalue 1 with multiplicity 9, contrary to Lemma 2.1.6. By Lemma 2.1.8, A has exponent at most 27. We claim next that A has no subgroup A^{∗} of type (27, 27). Suppose otherwise. Now if a ∈ A has order 27 then a has 11 distinct eigenvalues, for otherwise (a − 1)^{10} acts as zero in any reduction (mod 3) of G and a^{9} acts as a transvection in the reduction (mod 3), contrary to Lemma 2.1.7. For λ any non-trivial linear constituent of Res_{A^{∗}}^{G}(χ), then there is an element c ∈ A^{∗}∩kerλ of order 27. Then since the trivial character is a constituent of Res_{A}^{G}(χ), c has the eigenvalue 1 with multiplicity 2 or more, a contradiction. Hence |A| ≤ 3^{9}, as required in this case.

Suppose than that Res_{U}^{G}(χ + χ̄) is rational valued. Now as in the proof of Lemma 2.1.8, U has an Abelian normal subgroup A of index dividing 27. We need to prove that |A| ≤ 3^{8}. If A contains an element of order 27, then Res_{A}^{G}(χ) has nine algebraic conjugates of some linear character λ of A with multiplicity 1. Then kerλ = 1, otherwise A again contains a non-identity element of order 3 which has the eigenvalue 1 with multiplicity 9, contrary to 2.1.6. Thus |A| ≤ 27 and |U| ≤ 729 in this case. Hence we may suppose that A has exponent at most 9. Suppose that A is elementary Abelian of rank 9 or more. Then Res_{A}^{G}(χ) must have 9 distinct linear constituents say {λ_{j} : 1 ≤ j ≤ 9}. There must be an element a ∈ A such that λ_{j}(a) = e^{\frac{2πi}{3}} for 1 ≤ j ≤ 9, contrary to 2.1.6.

Hence we may suppose that A contains elements of order 9. If A contains a subgroup of type (9, 9, 9) then there must be three linear constituents λ_{1}, λ_{2}, λ_{3} of Res_{A}^{G}(χ) which are not algebraically conjugate but with [A : kerλ_{j}] = 9 for each j. Then three algebraic conjugates of each λ_{j} also occur as constituents of Res_{A}^{G}(χ). By Lemma 2.1.6, ∩_{j=1}^{3}kerλ_{j} = 1. Hence |A ≤ 729 and |U| ≤ 3^{9} in this case. Suppose then that A has no such subgroup.

If A contains a subgroup of type (9, 9) then there must be two linear constituents λ_{1}, λ_{2} of Res_{A}^{G}(χ) which are not algebraically conjugate but with [A : kerλ_{j}] = 9 for j=1,2 Then three algebraic conjugates of λ_{1} and of λ_{2} occur as constituents of Res_{A}^{G}(χ). Using Lemma 2.1.6, kerλ_{1} ∩ kerλ_{2} is elementary Abelian or rank at most 3, otherwise there is an element of order 3 in kerλ_{1} ∩ kerλ_{2} which is in the kernel of the six linear constituents which are algebraically conjugate to λ_{1} or λ_{2}, and a further three linear constituents of Res_{A}^{G}(χ). Hence |A| ≤ 3^{7} and |U| ≤ 3 ∗ 10 in this case. Suppose then that A contains no such subgroup.

A similar argument using 2.1.6 shows that if A has a cyclic subgroup of order 9 which has an elmentary Abelian complement, then there is a linear constituent λ with [A : kerλ] = 9 such that kerλ is elementary Abelian of rank at most 6. Hence |A| ≤ 3^{8} and |U| ≤ 3^{11} in this case.

Lemma 2.5.3: Let S ∈ Syl_{5}(G). Then if Res_{S}^{G}(χ) is rational-valued, |S| = 5. Proof: We may suppose that S contains an element s of order 5 in N_{G}(P), where P ∈ Syl_{11}(G). We have seen before that s has a fixed=point space of dimension 3 which contains the fixed-point space of P. But if Res_{S}^{G}(χ) is rationalvalued, then its non-trivial irreducible constituents fall into orbits of length divisible by 4 under algebraic conjugation. But then trivial character occurs with multiplicity congruent to 3 (mod 4) in Res_{S}^{G}(χ). But then S has the same fixed-point space as s, so that S acts triivially on the fixed-point space of P, and ⟨P, S⟩ is reducible. Then |S| = 5 by Lemma 2.3.2.

Lemma 2.5.4: Suppose that G has a non-trivial Sylow 7-subgroup T. Then Res_{T}^{G}(χ) is not rational valued.

Proof: Suppose otherwise. Let β be an irreducible constituent of Res_{T}^{G}(χ). Then a further 5 algebraic conjugates of β also occur as constituents of Res_{T}^{G}(χ). Hence β is linear, and Res_{T}^{G}(χ) contains the trivial character with multiplicity 5 and all the algebraic conjugates of β with multiplicity 1. Thus |T | = 7.

Let us consider an underlying R_{π}G-module V, where π is a prime ideal containing 7 in a suitable ring of algebraic integers. If Res_{N_{G}(T)}^{G}(V) is indecomposable, then O_{7′}(N_{G}(T)) is Abelian (it can’t be irreducible). Since Z(G) = 1, it must be the case that O_{7′}(N_{G}(T)) = 1, for otherwise Res_{N_{G}(T)}^{G}(V) is induced from a rank-1 R_{π}C_{G}(T)-module. But in that case T ⊲ G as T acts trivially in a reduction (mod 7) of G. But if C_{G}(T) = T, then χ must be non-exceptional as Res_{T}^{G}(χ) is rational-valued. But then χ(1) ≡ ±1 (mod 7) by the theory of blocks of defect 1, a contradiction.

Hence Res_{N_{G}(T)}^{G}(V) is decomposable, so must have a projective summand, necessarily of rank 7.

Hence the Green correspondent of V, say U, for N_{G}(T) has rank 4, and the source of V is trivial. We claim that C_{G}(T) has odd order. Otherwise, C_{G}(T) contains an involution u. Now u has only one eigenvalue on the projective summand of Res_{N_{G}(T)}^{G}(V). If this eigenvalue is 1, then unimodularity forces t to have the eigenvalue −1 with multiplicity 2 on U. But then u has the eigenvalue 1 with total multiplicity 9, contrary to Lemma 2.1.6. If the eigenvalue is −1 then u must have the eigenvalue −1 with multiplicity 1 on U ( since it can’t have the eigenvalue −1 with total multiplicity 10 using Lemma 2.1.6 again). Now u is not central in N_{G}(T) since U is indecomposable. But a product of two distinct N_{G}(T)-conjugates of u has the eigenvalue 1 with total multiplicity 9, contrary to Lemma 2.1.6.

Now C_{G}(T) now has odd order. But then U is induced from a module of a (normal) subgroup of N_{G}(T) containing C_{G}(T), so a subgroup of index not divisible by 4. Hence [N_{G}(T) : C_{G}(T)] = 2 and O_{7′}(C_{G}(T)) has an irreducible character of degree 2, contrary to the fact that C_{G}(T) has odd order.

Remark: Notice that A_{12} has a rational valued irreducible character of degree 11 and has order divisible both by 7 and by 25. However, in a representation affording that character, there are elements which act with a fixed-point space of codimension 2.

Section Six

Lemma 2.6.1: Let M be a 5-local subgroup of G. Then M is reducible.

Proof: Suppose otherwise. Then we may suppose that M is of maximal order subject to being irreducible with O_{5}(M) ≠ 1 and we may suppose that P ≤ M, where P ∈ Syl_{11}(G). Let W be a minimal normal subgroup of M contained in Z(O_{5}(M)). Then W P is a Frobenius group (and is irreducible, but imprimitive). Hence Res_{W}^{G} (χ) has 11 distinct constituents permuted transitively by P. By Lemma 1.5.2, C_{G}(W) is an Abelian 5-group. Hence F^{∗}(M) is an Abelian 5group. Since O_{11′}(M) is Abelian, we note also that O_{11′}(M) = O_{5}(M).

Let S ∈ Syl_{5}(M). Suppose that M is 5-stable (in particular, this applies if M is solvable). Then Z = ZJ(S) ⊲ M, so M = N_{G}(ZJ(S)) by the maximal choice of M. Hence S ∈ Syl_{5}(G). Now C_{G}(Z) is also an Abelian 5-group, so that J(S) is Abelian and (by the definition of the Thompson subgroup), J(S) is weakly closed in S with respect to G. By the Hall-Wielandt Theorem, we have S = S ∩ G′ = S ∩ M′.

If M is solvable, then M = O_{11′}(M)N_{M}(P) = O_{5}(M)N_{M}(P). Since M has no factor group of order 5, we have O_{5}(M) = S and S is Abelian. But for an element x of order 5 in N_{G}(P), we know from Lemma 2.4.1 that the Sylow 5-subgroups of C_{G}(x) are elementary Abelian of rank at most 3. However, since P acts in a Frobenius manner on S, we have |S| ≥ 3^{5}, a contradiction. Hence M is not solvable.

Now set M̄ = M/O_{5}(M). Then O_{11′}(M̄) = O_{11}(M̄) = 1 (the second equality because N_{M}(P) is solvable, the first because O_{11′}(M) = O_{5}(M)). Now W = Ω_{1}(O_{5}(M)) is elementary Abelian of rank 5, for otherwise it has rank 10 because of the action of P, and then it contains an element of order 5 with a fixed point space of codimension 2, contrary to Lemma 1.3.17. Hence M̄ is isomorphic to a subgroup of GL(5, 5). Since C_{M}(P) = P, it follows that NM̄(P̄) has order 55 and that M̄ is a non-Abelian simple group. Note also that M̄ has an orbit of length 11 on the dual module W^{∗}, since the 11-linear constituents of Res_{W}^{G} (χ) are permuted transitively by M. By a Theorem of Burnside, M̄ acts doubly transitively on the linear constituents of Res_{W}^{G} (χ), since M̄ is not solvable. Note also that M̄ contains no element of order 7, as SL(5, 5) has no such element.

Hence M̄ is isomorphic to M_{11} or PSL(2, 11). Now ⟨x̄⟩ ∈ Syl_{5}(M̄) in either case, and since ⟨x⟩P acts as a Frobenius group of order 55 in its action on W, it follows that W is projective as GF(5)M̄-module. Let L̄ be the stabilizer of a non-trivial constituent linear constituent λ of Res_{W}^{G} (χ), and suppose without loss of generality that x̄ ∈ L̄. Then M̄ : L̄] = 11. In fact, V = ResM̄ ; L̄(W) is isomorphic to the projective cover of the trivial module for L̄, since it is projective (and must be indecomposable since it has dimension 5), and L̄ fixes a non-zero element of the dual of V (when V is written additively). Then V has a unique maximal submodule, which is kerλ, and this submodule contains the one-dimensional fixed point space of L̄. Hence there is an element y ∈ kerλ which is fixed by L̄. Since L̄ transitively permutes the remaining 199 linear constituents of Res_{W}^{G} (χ), it follows that χ(y) = 1 + 10η for some primitive 5-th root of unity η, so that y has the eigenvalue η with multiplicity 10, contrary to Lemma 2.1.6.

Corollary 2.6.2: i) No element of order 5 has quadratic minimum polynomial in any reduction (mod 5) of G. ii) Let S ∈ Syl_{5}(G) and let 1 ≠ Q ≤ S be a normal subgroup of N_{G}(S). Then [S : S ∩ S^{g}] ≤ 25 for some g ∈ G\N_{G}(Q).

Proof: i) follows from 1.4.3 and ii) follows from 1.4.2.

Lemma 2.6.3: The Sylow 5-subgroups of G are Abelian.

Proof: Suppose that S ∈ Syl_{5}(G) is non-Abelian. We may suppose that S ∩ N_{G}(P) contains an element s of order 5 and we may suppose that S contains a Sylow 5-subgroup of C_{G}(s).

We note that by part i) of Corollary 2.6.2, every element of S has at least 3 different eigenvalues. Now S′ ∩Z(S) ≠ 1. Let t ∈ (S′ ∩Z(S))^{#}. Since t has three different eigenvalues, Res_{S}^{G}(χ) must have two different irreducible constituents of degree 5, say θ_{1} and θ_{2}. Note that θ_{1} and θ_{2} are both faithful. For suppose that y ∈ kerθ_{1} ∈ Z(S)^{#}. Then by unimodularity, y has the eigenvalue 1 with multiplicity 6 and has a primitive 5-th root of unity β as an eigenvalue of mutiplicity 5, a contradiction.

Since θ_{1} is faithful, Z(S) is cyclic. Each u ∈ Z(S) has at most three eigenvalues, so has order 5, since u^{5} acts trivially in any reduction (mod 5) of G. Hence |Z(S)| = 5. Also S has an Abelian normal subgroup A of index 5.

Suppose that A ≠ J(S). Then there is another Abelian subgroup X of S with [S : X] = 5. Then X ⊲S. Hence S = XA and S′ ≤ A∩X ≤ Z(S), so S has class 2. Since S has a faithful irreducible character of degree 5, and |Z(S)| = 5, we have |S| = 5^{3} in this case.

Suppose then that A = J(S). Since A is weakly closed in S with respect to G, and since A is Abelian, any two elements of A which are conjugate in G are already conjugate in N_{G}(A).

By Lemma 1.4.2 and Lemma 2.6.2, there is an element g ∈ G\N_{G}(Z(S)) with [S : S ∩ S^{g}] ≤ 25. Now S ∩ S^{g} is Abelian, for otherwise θ_{1} and θ_{2} both restrict irreducibly to S ∩ S^{g} and Z(S ∩ S^{g}) has order 5, while also Z(S ∩ S^{g}) ≤ Z(S). In that case, Z(S) = Z(S^{g}) and g ∈ N_{G}(Z(S)), contrary to assumption. Now Z(S)(S ∩ S^{g}) is Abelian, so that Z(S)(S ∩ S^{g}) = A = J(S). Now Z(S) is the unique minimal normal subgroup of S, so since Z(S) ≰ Φ(J(S)), we see that J(S) is elementary Abelian.

Suppose now that Ω_{1}(S) is Abelian. Then s ∈ J(S) and |S| ≤ 5^{4} since S ∩C_{G}(s) is Abelian of rank at most 3. Suppose then that Ω_{1}(S) is non-Abelian. Then Ω_{1}(S) > J(S), so that Ω_{1}(S) = S. If |S| > 5^{4}, we must have s ∉ J(S). Choose an element t ∈ A\C_{A}(s) and set Q = ⟨s, t⟩. Then Q is non-Abelian, so that θ_{1} and θ_{2} both restrict irreducibly to Q. Hence Z(Q) = Z(S). Hence |C_{S}(s) ∩ C_{S}(t)| = 5. But C_{S}(t) = A, so that |C_{A}(s)| = 5. Hence s has a single Jordan block when considered as a linear transformation of A. In that case, S has maximal class.

Now we have Res_{S}^{G}(χ) = λ + θ_{1} + θ_{2} where θ_{1}(z) = 5η and θ_{2}(z) = 5ω for primitive 5-th roots of unity ω and η. Notice that λ is the unique irreducible constituent of Res_{S}^{G}(χ) which contains Z(S) in its kernel. Hence λ is N_{G}(S)- stable, and extends to a linear character of N_{G}(S). Suppose that |S| < 5^{6}. Then S = S ∩ G′ = S ∩ N_{G}(S)′, the second equality by the Hall-Wielandt theorem. But if λ is not the trivial character, then N_{G}(S) has a factor group of order 5, a contradiction.

Now we claim that Z(S) is weakly closed in S with respect to G under our present assumption that |S| < 5^{6}. Let W = C_{5} ≀ C_{5} and let w be an element of order 5 in W \J(W). The derived group W′ consists of all elements of B for which the product of the components is 1_{B}, for all elements of W′ are of this form, while [W : W′] = 5^{4}. Now W = ⟨x, y⟩ whenever x ∈ W \J(W) and y ∈ W \W′.

Suppose that z^{g} ∈ S\Z(S) for some g ∈ G. Then θ_{1}(z^{g}) ≠ 0 since z^{g} has only three eigenvalues. Hence z^{g} ∈ J(S). Now we may suppose that z^{g} has exactly two eigenvalues in the representation σ_{1} affording θ_{1} (possibly after relabelling θ_{1} and θ_{2}). Then det(σ_{1}(z^{g})) ≠ 1. Hence S ≅ C_{5} ≀ C_{5} and |S| = 5^{6} in this case.

Set M = N_{G}(J(S)) if |S| = 5^{6} and M = N_{G}(Z(S)) otherwise. By the Hall Wielandt theorem, we have S ∩ G′ = S ∩ M′ in the latter case. We claim that C_{G}(J(S)) = ZJ(S) in all cases.

If J(S) is Abelian, then both θ_{1} and θ_{2} vanish off J(S), so that they must have different restrictions to J(S), as they are different characters of S. Hence Res_{J(S)}^{G}(χ) has at least 10 different linear constituents. Hence if q ≠ 5 is a prime, then any q-element of C_{G}(J(S)) acts as a transvection in any reduction (mod q) of G, contrary to Lemma 2.1.7. Thus C_{G}(J(S)) is a 5-group. But C_{S}(J(S)) = J(S). If J(S) = S, then S has class 2 as before, and we find that C_{G}(A) = A by the same argument, where A is any maximal Abelian normal subgroup of S. Thus C_{G}(S) = Z(S) in that case.

By 1.4.2, there is some g ∈ G\M such that [S : S ∩ S^{g}] ≤ 25. In the case M = N_{G}(Z(S)) we know that M controls the fusion in S since Z(S) is weakly closed in S. Also S ∩ S^{g} does not contain Z(S) in this case. However, we must have S ∩ S^{g} ∩ Z_{2}(S) ≠ 1. Thus there is some y of order 5 in Z_{2}(S) with θ_{1}(y) = θ_{2}(y) = 0 and λ(y) = 1. Then y has the eigenvalue 1 with multiplicity 3 and all primitive 5-th roots of unity with multiplicity 4. Then C_{G}(y) contains no element of order 4 (for if u were one such element, then u^{2} would contradict Lemma 2.1.7). Thus C = C_{G}(y) has elementary Abelian Sylow 2-subgroups. Now C^{(∞)} acts trivially on each 2-dimensional eigenspace of y. Any involution of C^{(∞)} has a fixed-point space of codimension 2, contrary to Lemma 2.1.6, and likewise for any element of order 3. Hence C_{G}(y) is solvable (and has an Abelian Sylow 5-subgroup, U say). Now U centralizes O_{5′}(C_{G}(y)) since by Lemma 1.2.10, [O_{5′}(C_{G}(y)), U] acts trivially on each eigenspace of y. Since C_{G}(y) is solvable, we have U ⊲ C_{G}(y). Thus U = C_{S}(y). Now Z(S) ⊲ C_{G}(y) since Z(S) is weakly closed in S. Now g ∈ C_{G}(S ∩ S^{g})M and C_{G}(S ∩ S^{g}) ≤ C_{G}(y) ≤ M, a contradiction.

We are left with the case |S| = 5^{6}, M = N_{G}(J(S)). Since λ(z) = 1 and 5 ≠ θ_{1}(z) ≠ θ_{2}(z), it follows that C = C_{G}(Z(S)) is (isomorphic to) a monomial subgroup of GL(11, ℂ) using Lemma 1.3.4, since C acts on the eigenspaces of z. Then θ_{1} and θ_{2} each extend to irreducible characters of C, say ψ_{1} and ψ_{2} respectively. By 1.3.4, these are imprimitive characters. Hence there is an Abelian normal subgroup T of C which contains J(S) and is represented by diagonal matrices in some representation affording ψ_{i} for each i. Since C_{G}(J(S)) = J(S), we have T = J(S). Hence C_{G}(Z(S)) ≤ M. Since M controls fusion in S under the current hypotheses, whenever z^{h} ∈ S, we know that z^{h} ∈ J(S) and then h = cm for some c ∈ C_{G}(z), some m ∈ M. However, C_{G}(z) ≤ M, so that h ∈ M.

Recall that we chose g ∈ G\M so that [S : S ∩ S^{g}] ≤ 25. Under current assumptions, we have S ∩ S^{g} ≤ J(S), but Z(S) ≰ S ∩ S^{g}. Hence J(S) = (S ∩ S^{g})Z(S). Since Z_{2}(S) ≤ J(S), and θ_{1} and θ_{2} both vanish on Z_{2}(S)\Z(S), we conclude that in any case, there is an element u ∈ (J(S) ∩ S^{g}) with θ_{1}(u) = θ_{2}(u) = 0.

If S has class 2 and z^{g} ∈ S\Z(S), then θ_{1}(z^{g}) = θ_{2}(z^{g}) = 0, contrary to the fact that z^{g} has only three eigenvalues. If S has class 5, then J(S) ⊲ C_{G}(z), while if z^{g} ∈ S, then z^{g} ∈ J(S), so that z^{g} = z^{m} for some m ∈ N_{G}(J(S)). But then gm^{−1} ∈ C_{G}(z) ≤ N_{G}(J(S)) and g ∈ N_{G}(J(S)). If S has class 3 or 4, then we have seen that above Z(S) is weakly closed in S with respect to G.

Hence, setting M = N_{G}(Z(S)) if |S| < 5^{6} and M = N_{G}(J(S)) if |S| = 5^{6}. Then Z(S) ≰ S ∩ S^{g} if g ∈ G\M. We may choose g ∈ G\M such that [S : S ∩ S^{g}] ≤ 25. Since Z(S) is weakly closed in S if |S| < 5^{6}, we know that M controls strong fusion in S in that case.

Now S ∩ S^{g} is Abelian, since it does not contain Z(S). Then Z(S)(S ∩ S^{g}) is Abelian, and is a maximal subgroup of S. In all cases, there is an element w of order 5 in S ∩ S^{g} with θ_{1}(w) = θ_{2}(w) = 0 and λ(w) = 1 (note that S ∩ S^{g} contains no element of order 25, for otherwise z ∈ S ∩S^{g}). Then no conjugate of w lies in Z(S) and C_{G}(w) has a unique Sylow 5-subgroup. By the weak closure, C_{G}(w) ≤ M if |S| < 5^{6}. If |S| = 5^{6}, then J(S) ⊲ C_{G}(w) and C_{G}(w) ≤ M.

If |S| < 5^{6}, then M controls fusion and g ∈ MC_{G}(S ∩ S^{g}) ≤ M, a contradiction. if |S = 5^{6}, then ⟨J(S), J(S^{g})⟩ ≤ C_{G}(S ∩ S^{g}), so that J(S) = J(S)^{g} and g ∈ M, a contradiction. Hence S is Abelian.

Lemma 2.6.4: The Sylow 5-subgroup S is elementary Abelian.

Proof: We know that S is Abelian of rank at most 3 by 2.3.2*. If x ∈ S has order 25, then x has 11 distinct eigenvalues, otherwise (x − 1)^{10} acts trivially in any reduction (mod 5) of G, and x^{5} has quadratic minimum polynomial in any reduction (mod 5) of G. In particular, C_{G}(x) is an Abelian 5-group. Also, by Lemma 1.4.3, there is an irreducible 5-local subgroup of G, contrary to Lemma 2.6.1. Hence S contains no subgroup of type (25, 25) and |S| ≤ 5^{4}.

Suppose that S is not elementary Abelian. If S is cyclic, Lemmas 1.4.1 and 2.6.1 together yield a contradiction.

We may write S = T × E, where T is cyclic of order 25 and E is elementary Abelian, and both T and E are normal in N_{G}(S). Since C_{G}(T) = S. Wrote N_{G}(S) = SV, where V is a 5′-group. Then V acts faithfully on T, so V is cyclic of order at most 4. If V has order 4, then, since T V is a Frobenius group, Res_{T V}^{G} (χ) has three linear constituents, each of which contain T = [T, V] in their kernel, a contradiction. Hence [N_{G}(S) : S] = 2. Let N_{G}(S) = S⟨t⟩. We have S = [S, t] since S ≤ G′. Thus C_{S}(t) = 1. For each u ∈ S^{#}, we see that C_{G}(u) has a normal 5-complement. By the Hall-Higman theorem, Q = Φ(S) centralizes O_{5′}(C_{G}(u)), since a generator of Q has minimum polynomial of degree at most 3 in any reduction (mod 5) of G.

Let W = ⟨O_{5′}(C_{G}(u)) : u ∈ S^{#}⟩. Then W ≤ C_{G}(Q) < G and |W | is not divisible by 11. If W = 1, then S ⊲ C_{G}(u) for all u ∈ S^{#}, so that S ∩ S^{g} = 1 for all g ∈ G\N_{G}(S), contrary to Lemmas 1.4.2 and 2.6.1. If W ≠ 1, then N_{G}(S) ≤ N_{G}(W) and S ∩ S^{g} = 1 for all g ∈ G\N_{G}(W), so that N_{G}(W) is irreducible by Lemma 1.4.2.

Since no p-local subgroup of G is irreducible for p ∈ {2, 5, 7}, we see that V = Ω_{1}(W) is an elementary Abelian 3-group of order 3^{5} (by Lemma 2.1.6, we can’t have |V | = 3^{10}). Now C_{G}(V) is an Abelian 3-group, so that S is isomorphic to a 5-subgroup of GL(5, 3), a contradiction, so |GL(5, 3)| is not divisible by 25. Hence S is elementary Abelian.

Some Properties of Local Subgroups

Lemma 2.7.1: P normalizes no non-trivial {2, 5, 7}-subgroup of G.

Proof: Othereise, P normalizes an elementary Abelian q-group E for q ∈ {2, 5, 7}. Then N_{G}(E) is irreducible so q ≠ 5 by Lemma 2.6.1. If q ∈ {2, 7}, then |E| = q^{10} and E contains an element of order q with an eigenspace of codimension 1, contrary to Lemma 2.1.6.

Lemma 2.7.2: Let y ∈ G be an element of order 7. Then y centralizes any {3, 5}-group which it normalizes.

Proof: Since the Sylow 5-subgroups of G are elementary Abelian of order at most 125, we only need to consider the case that y normalizes a 3-subgroup E minimal subject to E = [E, y] ≠ 1. Then E has class at most 2, and has exponent 3.

If E is Abelian, then |E| = 3^{6}, and Res_{E⟨y⟩}^{G}(χ) has one irreducible (faithful) constituent θ of degree 7 and a further four linear constituents, each containing E = [E, y] in their kernel. Since |E| = 729, there is an element u ∈ E with θ(u) = 5+e^{\frac{2πi}{3}} +e^{\frac{−2πi}{3}} . Then u has the eigenvalue 1 with multiplicity 9, contrary to Lemma 2.1.6.

Hence E is non-Abelian. Now, using Lemma 2.1.6 again, Res_{E}^{G}(χ) has no irreducible constituent of degree 9. Hence Res_{E}^{G}(χ) has an irreducible constituent of degree 3. By Clifford’s theorem, θ is y-stable, as χ(1) = 11. But then [E, y] ≤ kerθ by Lemma 1.2.10, contrary to the fact that E = [E, y].

Lemma 2.7.3: Let A be a non-cyclic 5-subgroup of G and E be an A-invariant {2, 3, 7}-subgroup of G with an Abelian Sylow 2-subgroup. Then [E, A] = 1. Furthermore, if A ∈ Syl_{5}(G), then A centralizes every 2-subgroup of G which it normalizes.

Proof: Suppose otherwise, and that E is chosen minimal subject to [E, A] ≠ 1. If E is a non-Abelian 3-group, then Res_{E}^{G}(χ) has an irreducible constituent of degree 3 which must be A-invariant. But once more [E, A] ≤ kerθ by Lemma 1.2.10, contrary to the fact that E = [E, A]. Suppose then that E is an Abelian 3-group. Then |E| = 81, and [E, a] = 1 for some element a ∈ A^{#}. Let L = EA. Then Res_{L}^{G}(χ) has (counting multiplicity) at least two non-linear irreducible constituents, again using Lemma 2.1.6. Each of these has degree 5. The remaining constituent of Res_{L}^{G}(χ) must be linear, and have E = [E, A] in its kernel. Hence all irreducible constituents of Res_{E}^{G}(χ) occur with multiplicity at most 2, so a acts with quadratic minimum polynomial in any reduction (mod 5) of G, contrary to Lemma 2.6.2. A similar argument applies if E is a (necessarily Abelian) 7-group or if E is an elementary Abelian 2-group.

Suppose then that A ∈ Syl_{5}(G) but that E is a non-Abelian special 2-group. Then [Z(E), A] = 1 and [E : Z(E)] = 16. Choose an element x ∈ A which normalizes some conjugate of P. Then by Lemma 2.1.5, |C_{G}(x)| is not divisible by 4 so Z(E) has order 2 and E is extraspecial of order 32. Let z be the central involution of E. Now each linear constituent of Res_{E}^{G}(χ) has z in its kernel, and the unique non-linear irreducible character of E has degree 4. Hence the multiplicity of −1 as an eigenvalue of t is divisible by 4. But by Lemma 2.1.5, since z ∈ C_{G}(x), z has the eigenvalue −1 with multiplicity 6, a contradiction.

Section Eight

In this section, we prove that G contains no element of order 35. We then use this to prove that if 7 divides |G|, then 25 does not divide |G|. We note that A_{12} has order divisible by 175 and contains an element of order 35.

Lemma 2.8.1: G contains no element of order 35.

Proof: Let us choose S so that S ∩ N_{G}(P) contains an element x of order 5. Suppose that s ∈ G is an element of order 5 and that u ∈ C_{G}(s) has order 7. Let C = C_{G}(s) and let S ∈ Syl_{5}(C_{G}(s)), so that also S ∈ Syl_{5}(G). By Lemma 2.4.1, S ≠ ⟨s⟩, so that S is not cyclic. By Lemma 2.7.3, [S, O_{5′}(C)] = 1. By Lemma 2.4.1 again, O_{5′}(C) has order 2, 3 or 6 ( and is not cyclic if it has order 6). Hence u centralizes O^{5′}(C). Also, by Lemma 2.7.2, u centralizes O_{5}(C). Hence u centralizes F (C). Thus C has a component, L say. Let E = E(C).

We claim that 7 divides |E|. Otherwise, E is a {2, 3, 5}-group; However, E has a u-invariant Sylow 5-subgroup A and a u-invariant Sylow 3-subgroup T. By Lemma 2.7.2, we have [A, u] = [T, u] = 1. Let Q ∈ Syl_{2}(E) be u-invariant. Then E = QC_{E}(u) and [E, u] = [Q, u] ≤ Q. But then E, ⟨u⟩, E] = [⟨u⟩, E, E] = 1, so E, E, ⟨u⟩] = 1 by the 3-subgroups Lemma. Hence E, ⟨u⟩] = 1 and [F^{∗}(C), u] = 1, a contradiction as 7 does not divide |F (C)|. Hence we may suppose that 7||L|, and we do so.

Since O_{5′}(C) is solvable, we also have 5||L|. Hence 35 divides |L|. Now L has Abelian Sylow 5-subgroups and Abelian Sylow 7-subgroups, so that Z(L) has order at most 3. There must be a non-trivial linear constituent θ of Res_{L}^{G}(χ). Now θ(1) ≥ 4 by Lemmas 1.3.1 and Lemma 1.3.2 . Also, θ(1) ≤ 8 by Lemma 2.1.6 since [L, s] = 1.

Case 1: θ(1) = 4: In this case, L/Z(L) ≅ A_{7} and |Z(L)| = 2. Let t be an involution of Z(L). Now S centralizes O_{5′}(C), so that t ∈ C_{G}(S). By Lemma 2.4.1, χ(t) = −1. Now the −1-eigenspace of t has dimension 6, and L acts on this eigenspace. Since L has no irreducible character of degree less than 4, the character of L afforded by its action on the −1-eigenspace of t is θ + 21_{L}, where λ_{0} is the trivial character. But t acts with trace −6 on this eigenspace, a contradiction.

Case 2: θ(1) = 5: This case does not occur by Lemma 1.3.4, since 35 divides [L : Z(L)].

Case 3: θ(1) = 7: In this case, θ is faithful since Z(L) ≤ C_{G}(S) and no non-identity element of C_{G}(S) has an eigenvalue of multiplicity greater than 6. Hence Z(L) = 1 since L = L′ and |Z(L)| ≤ 3. Also Res_{L}^{G}(θ) = θ + 4.1_{L}. Since 35||L|, we have L ≅ A_{8} or L ≅ Sp(6, 2) by Lemma 1.3.6. In each case, there is an involution t ∈ L with θ(t) = 3. But then χ(t) = 7 and t has the eigenvalue 1 with multiplicity 9, contrary to Lemma 2.1.6.

Case 4: θ(1) = 8: In this case, as in Case 3, we have Z(L) = 1 and θ is faithful. By Lemma 1.3.7, L ≅ A_{9} and contains an involution with θ(t) = 4. Once more, χ(t) = 7 and t has the eigenvalue 1 with multiplicity 9, contrary to Lemma 2.1.6.

Case 5: θ(1) = 6: In this case, since 35|[L : Z(L)], by Lemma 1.3.5, we have L/Z(L) ≅ A_{7} or L/Z(L) ≅ HJ, the Hall-Janko sporadic group of order 604800. In the latter case, we have |Z(L)| = 2. If |Z(L)| = 2, then we claim that L has generalized quaternion Sylow 2-subgroups, which eliminates the case L/Z(L) ≅ HJ. For let z be an involution of Z(L). If L contains another involution t, then either θ(t) = 2 or θ(zt) = 2. Hence we either have χ(t) = 7 or χ(zt) = 7, and either t or zt has a fixed-point space of dimension 9, contrary to Lemma 2.1.6.

Hence L/Z(L) ≅ A_{7} in this case. Now S normalizes L since χ(1) = 11. Since S is elementary Abelian, and Out(L) has order prime to 5, we have LS = L×C_{S}(L). Any element of order 5 in S which centralizes a non-central involution of L lies in C_{S}(L) since no element of order 5 of L centralizes a non-central involution of S. Likewise, any element of order 5 in S which centralizes a element of order 3 in L lies in C_{S}(L). Recall that there is an element x be an element of order 5 in S which normalizes P, a Sylow 11-subgroup of G. Since x has no eigenvalue of multiplicity greater than 3, x does not lie in C_{S}(L). Now θ extends to an irreducible character of LS, which we still call θ. For u ∈ L of order 5, we have θ(u) = 1 since u is conjugate within L to all its non-identity powers. We may suppose that x = uv for some v ∈ C_{S}(L). Hence θ(x) = ω for some 5-th root of unity ω such that θ(v) = 6ω.

Now if Z(L) = 1, then for any involution t ∈ L, we have θ(t) = 2 and χ(t) = 7, contrary to Lemma 2.1.6. Hence we may suppose that there is a nonidentity element z ∈ Z(L), of order 2 or 3. Then x ∈ C_{G}(z) and (using Lemma 2.4.1) x has 5 different eigenvalues on the 1-eigenspace of z, so has trace 0 in its action on this eigenspace. Thus θ(x) = 1, since C acts on the other eigenspace of z, which is 6-dimensional, and we have χ(x) = 1 by Lemma 2.4.1. Hence ω = 1.

Now v acts trivially on the 3-dimensional 1-eigenspace of x. Now v ∉ ⟨x⟩ since v centralizes an element of order 7, but x does not. Hence ⟨P, x, v⟩ acts trivially on the 1-eigenspace of P. However, this subgroup has order divisible by 275, and is not irreducible, contrary to Lemma 2.3.2.

Corollary 2.8.2: If 7 divides |G|, then 25 does not divide |G|.

Proof: Suppose that 175 divides |G|. Let T ∈ Syl_{7}(G), and let S ∈ Syl_{5}(G). By Lemmas 2,5,4, neither Res_{S}^{G}(χ) nor Res_{T}^{G}(χ) is rational-valued. By Lemma 1.2.12, G contains an element of order 35, contrary to Lemma 2.8.1.

Chapter Three

From now on (for the rest of this work, not just this chapter) to avoid lengthy analyses of small groups which are (at least implicitly) “known,” we suppose that 2^{11} divides |G| and that G has sectional 2-rank greater than 4. For if 2^{11} does not divide |G|, then G is classified by the results of B. Beisiegel [3] and V. Stingl, [29]. If G has sectional 2-rank at most 4, then G is known by the results of Gorenstein-Harada in [16]. In particular, by work of A. MacWilliams, this implies that we have SCN_{3}(2) ≠ ∅ in G. Hence if the centralizer of each involution in G is 2-constrained, then O_{2′}(C_{G}(t)) = 1 for all involutions t ∈ G, using now standard signalizer functor theory.

In this chapter, we will prove that 25 does not divide |G|. Using 2.8.2, we assume for the remainder of the chapter that 7 does not divide |G|. Let S be a Sylow 5-subgroup of G with S ∩ N_{G}(P) = ⟨x⟩ ≠ 1. By Lemmas 2.3.2 and 2.4.1, we have ⟨Res_{S}^{G}(χ), λ⟩ ≤ 2 for each linear character λ of S. We know already that S is elementary Abelian of order at most 125. We eliminate the cases |S| and |S| = 125 separately.

Section One

In this section we will prove that |S| ≠ 25. For the remainder of the section, we assume that |S| = 25.

Lemma 3.1.1*: N_{G}(S)/C_{G}(S) has Abelian Sylow 2-subgroups of order at most 8 and of rank at most 2.

Proof: Let A be a Sylow 2-subgroup of N_{G}(S)/C_{G}(S). We claim that A contains no element of order 4 whose square acts without non-trivial fixed point on S. If there were such an element, then the orbits of A on the non-trivial linear characters of S would all have length divisible by 4. Hence ⟨Res_{S}^{G}(χ), 1⟩ ≥ 3, contrary to Lemma 2.3.2. Since A is isomorphic to a subgroup of C_{4} ≀ C_{2}, we see that |A′| ≤ 2 and |A| ≤ 8. Also, A has exponent dividing 4. Each element of order 4 in A must be central (if there are any at all), so A must be Abelian (of rank at most two because it has a faithful two dimensional representation over GF(5)).

Lemma 3.1.2: [N_{G}(S) : C_{G}(S)] is not divisible by 3.

Proof: Suppose that there is an element y ∈ N_{G}(S) which induces an automorphism of order 3 on S via its conjugation action. Then y has two orbits in its conjugation action on maximal subgroups of S, each of length 3, and the nontrivial linear characters of S fall into orbits of length 3. Hence ⟨Res_{S}^{G}(χ), 1⟩ = 2, making use of Lemma 2.3.2. Let A = ⟨x⟩, and let B be a cyclic subgroup of order 5 of S which is not y-conjugate to A. Now using the proofs of Lemma 2.4.1 and Lemma 2.4.2 , we have

50 = 11 + 12 + 3( ∑_{v∈B^{#}} χ(b)),

so that ⟨Res_{B}^{G}(χ), 1⟩ = 4.

Similarly, we have

25⟨Res_{S}^{G}(χ), Res_{S}^{G}(χ)⟩ = 121 + 12 + 3( ∑_{v∈B^{#}} |χ(b)|^{2}).

This is an odd integer ( for note that the algebraic integers |χ(b)|^{2} and |χ(b^{−1}|^{2} are equal), and is congruent to 1 (mod 3). On the other hand, ⟨Res_{S}^{G}(χ), Res_{S}^{G}(χ)⟩ is an integer between 13 and 21, since χ(1) = 11, and Res_{S}^{G}(χ) contains the trivial character with multiplicity 2 and no linear character with multiplicity greater than 2. Thus ⟨Res_{S}^{G}(χ), Res_{S}^{G}(χ)⟩in{13, 19}. But the only way that 19 could arise would be with four linear constituents of multiplicty 2 and one with multiplicity 3, a contradiction. Hence ⟨Res_{S}^{G}(χ), Res_{S}^{G}(χ)⟩ = 13, and Res_{S}^{G}(χ) is the sum of 9 distinct non-trivial linear characters and twice the trivial character.

Hence if u ∈ C_{G}(S) has prime order p ≠ 5, then u acts as a transvection in any reduction (mod p) of G, contrary to Lemma 2.1.7. Hence C_{G}(S) = S.

Now we claim that S ∩ S^{g} = 1 for all g ∈ G\N_{G}(S). Suppose otherwise. Then there is an element s ∈ S^{#} such that S is not normal in C = C_{G}(s). Now [S, O_{5′}(C)] = 1 so that [S, F (C)] = 1 and O_{5′}(C) = 1. Hence C has a component E. Then E is simple and E is isomorphic to one of A_{5}, A_{6} or PSU(4, 2) by Lemma 1.2.17, since |E| has the form 2^{a}3^{b}5. Now S ⊲ C_{G}(x), so that s is not conjugate to any power of x. Hence s is conjugate to an element of B, and s has the eigenvalue 1 with multiplicity 4 from the calculation above. Let v be any element (E ∩ S)^{#}. Now S = ⟨s, x⟩ = ⟨s, v⟩. Now E acts non-trivially on the 1-eigenspace of s, since ⟨Res_{S}^{G}(χ), 1⟩ = 2. Since Z(E) = 1 and E has a faithful complex character of degree 4, we have E ≅ A_{5}.

Now we have see that ∑_{b∈B^{#}} χ(b) = 9, and we must also have

∑_{b∈B^{#}} |χ(b)|^{2} = 64

since

133 + 3( ∑_{b∈B^{#}} |χ(b)|^{2}) = 13 × 25 = 325.

Hence we have ⟨Res_{B}^{G}(χ), 1⟩ = 4 and ⟨Res_{B}^{G}(χ), Res_{B}^{G}(χ)⟩ = 37. Since a conjugate of s lies in B, we see that s has the eigenvalue 1 with multiplicity 4, one primitive 5-th root of unity with multiplicity 4, one primitive 5-th root of unity with multiplicity 2 and one primitive 5-th root of unity with multiplicity 1. Now E acts trivially on the eigenspaces of s of dimension less than 4 since E ≅ A_{5}. Since S = ⟨v, s⟩ and the trivial character of S the only constituent of Res_{S}^{G}(χ) of multiplicity greater than 1, this is a contradiction (considering the 2-dimensional eigenspace of s).

We have now established that S ∩ S^{g} = 1 for all g ∈ G\N_{G}(S), and we established earlier that C_{G}(S) = S. Since Res_{S}^{G}(χ) contains the trivial character with multiplicity 2 and 9 distinct linear characters with multiplicity 9 each, it follows that Res_{S}^{G}(χ) is not real-valued. Hence no involution of N_{G}(S) can invert every non-identity element of S. Thus N_{G}(S) has a cyclic Sylow 2-subgroup and a normal 2-complement. Furthermore, from the structure of N_{G}(S)/S, we see that its Sylow 3-subgroup must now be self-centralizing. Hence [N_{G}(S) : S] ∈ {3, 6}. However, since χ(1) = 11, Lemma 1.2.7 excludes the possibility that N_{G}(S) is a Frobenius group of order 25. Thus N_{G}(S)/S ≅ S_{3}.

Now we note that C_{G}(x) = S. For there are 6 maximal subgroups of S, permuted in two cycles of length 3 by the element y ∈ N_{G}(S) of order 3. The three involutions of N_{G}(S)/S each normalize two of the maximal subgroups (centralizing one of them and inverting the elements of the other). Hence N_{G}(S) still has two orbits on maximal subgroups of S. Now a generator b of the maximal subgroup B (which does not contain a conjugate of x) has the eigenvalue 1 with multiplicity 4, and three further eigenvalues with respective multiplicities 1,2 and 4. Hence b is not conjugate to its inverse. Hence b is centralized by an involution of N_{G}(S) and hence the element x must be inverted by an involution of N_{G}(S). Since S ⊲ C_{G}(x), we must have C_{G}(x) = S, as claimed. Hence we have |N_{G}(P)| = 55.

We also note that χ(b) has four different algebraic conjugates, so that χ requires the 5-th roots of unity. We have seen that G contains no element of order 20 under the current hypotheses, and that G contains no element of order 15. Hence χ takes rational values on a Sylow 2-subgroup of G and on a Sylow 3-subgroup of G, by Lemma 1.2.12. Hence [G : N_{G}(S)] divides 3^{4}.2^{14}.11, using Lemma 2.5.1. Also we have |G| ≡ 55 ( mod 121) by Sylow’s Theorem, so that [G : N_{G}(S)] ≡ 77 (mod 121) and \frac{[G:N_{G}(S)]}{11} ≡ 216 (mod 275). Since \frac{[G:N_{G}(S)]}{11} divides 2^{14}.3^{4}, this yields [G : N_{G}(S)] = 2376 and |G| = 2^{4}.3^{4}.5^{2}.11, contrary to our current hypotheses.

Lemma 3.1.3*: [N_{G}(S) : C_{G}(S)] ∈ {4, 8}.

Proof: It suffices to prove that [N_{G}(S) : C_{G}(S)] ≠ 2. Suppose otherwise. Then since G = G′, see that for all u ∈ S^{#}, we have C_{G}(u) ∩ N_{G}(S) = C_{G}(S), so that C_{G}(u) has a normal 5-complement, which is centralized by S by Lemma 2.7.3. Hence C_{G}(u) ≤ C_{G}(S) for all u ∈ S^{#}. Thus S ∩S^{g} = 1 for all g ∈ G\N_{G}(S). Set M = N_{G}(S), and let V be an R_{π}G-module affording χ, where R_{π} is the usual localization and π is a prime ideal containing 5. Then Res_{M}^{G}(V) is indecomposable, and is not induced from a proper subgroup. But then O_{5′}(M) ≤ Z(M), since Res_{O_{5′} (M)}^{G}(χ) lies over a single irreducible character of O_{5′}(M), which must be linear as χ(1) = 11. By unimodularity, we obtain O_{5′}(M) = 1, so that C_{G}(s) = S for all s ∈ S^{#}, contrary to Lemma 1.2.7.

Lemma 3.1.4: Let u be an element of order 5 which normalizes, but does not centralize, a 3-subgroup B of G. Then C_{G}(u) = QH, where Q is a normal Sylow 5-subgroup and H has order dividing 6, but contains no element of order 6. Furthermore, B is Abelian Proof: Let T = [B, u] and U = ⟨u⟩. Suppose first that T is non-Abelian. Then Res_{T}^{G}(χ) contains a non-linear irreducible constituent, say θ. Then θ(1) = 3, otherwise an element of order 3 in Z(T) contradicts Lemma 2.1.6. Since χ(1) = 11, θ must be U-stable by Clifford’s theorem. Then T = [T, u] ≤ kerθ by Lemma 1.2.10, a contradiction. Hence T is Abelian and u acts without non-trivial fixed points on T. Now Res_{T U}^{G}(χ) must have two non-linear irreducible constituents ( possibly equal), otherwise T contains an element of order 3 with a fixed-point space of dimension 8, contrary to Lemma 2.1.6. Hence ⟨Res_{T}^{G}(χ), 1⟩ = 1. But the argument above also shows that T ≤ kerψ) whenever ψ is a non-linear irreducible constituent of Res_{B}^{G}(χ), so there can be no such constituent, and B is Abelian.

Now in any reduction (mod 5) of G, we see that u must have two Jordan blocks of size 5. Hence the arguments of Lemma 2.4.1 may be applied to u.

Lemma 3.1.5*: |C_{G}(S)| is odd.

Proof: By Lemma 2.4.1, |O_{5′}(C_{G}(S))| divides 6 and C_{G}(S) contains no element of order 6. Suppose that C_{G}(S) contains an involution t. By Lemma 2.4.1, χ(t) = −1 and t has the eigenvalue −1 with multiplicity 6, the eigenvalue 1 with multiplicity 5.

We first claim that O_{5}(C_{G}(t)) ≠ 1. Suppose otherwise. By Lemma 2.7.3, we have [S, O_{5′}(C_{G}(t))] = 1, so certainly [S, F (C_{G}(t))] = 1, since S is Abelian. Hence E(C_{G}(t)) ≠ 1 and C_{G}(t) has a component L. Also, O_{5′}(C_{G}(t)) = ⟨t⟩. Now L has an irreducible character ψ of degree 6 or less with kerψ ≤ Z(L). Since 7 |̸|L|, by Lemma 1.3.5, we have 25 |̸|L|.

Suppose that L is the unique component of C_{G}(t). By 1.2.17, L/Z(L) ≅ A_{5}, A_{6} or PSU(4, 2). Hence Out(L) is a 5′-group in all cases. Since LS > L, we have C_{S}(L) ≠ 1. But then [C_{S}(L), F^{∗}(C_{G}(t))] = 1, so that C_{S}(L) ≤ Z(F (L)), a contradiction.

Hence C_{G}(t) has two distinct components L_{1} and L_{2} (and these are necessarily mutually centralizing). Let ψ_{1} be the character of C_{G}(t) afforded by the action on the −1 eigenspace of t and ψ_{2} be the character of C_{G}(t) afforded by the action of t on the 1-eigenspace of t. If (say) L_{1} acts trivially on the −1 eigenspace of t, then L_{1} contains an involution u with ψ_{2}(u) = 1, and then u has the eigenvalue 1 with multiplicity 9, contrary to Lemma 2.1.6. Hence L_{1} and L_{2} each act faithfully on the −1-eigenspace of t, and neither acts irreducibly on that eigenspace. Also, either L_{1} or L_{2} must act trivially on the 1-eigenspace of t, since L_{1}/⟨t⟩ and L_{2}/⟨t⟩ each have trivial center. Let us label so that L_{1} acts trivially on the 1-eigenspace of t.

If Res_{L_{1}}^{C_{G}(t)}(ψ_{1}) = α_{1} + α_{2} + α_{3}, where each α_{i} is irreducible of degree 2 ( but the α_{i} are not necessarily distinct), then L_{1} ≅ SL(2, 5) and L_{1} contains an element of order 6 with eigenvalues 1, e^{\frac{πi}{3}} , e^{− \frac{πi}{3}} , contrary to Lemma 1.3.10. Hence Res_{L_{1}}^{C_{G}(t)}(ψ_{1}) = α_{1} + α_{2}, where α_{1} is irreducible of degree 2 and α_{2} is irreducible of degree 4. But since Res_{L_{2}}^{C_{G}(t)}(ψ_{1}) is faithful, we see that L_{2} is Abelian, as [L_{1}, L_{2}] = 1, a contradiction.

Thus we have now established that O_{5}(C_{G}(t)) ≠ 1, as claimed.

Now let T ∈ Syl_{2}(G) with t ∈ T. Under current hypotheses, there is an an elementary Abelian A ⊲ T of order 8. We set θ(u) = O_{5}(C_{G}(u)) for each involution u ∈ G. We will prove that θ is balanced, and that θ is an A-signalizer functor.

Let a, b be distinct commuting involutions of G. Let C = C_{G}(b) and C̄ = C/O_{2′}(C). Let u ∈ (θ(a)∩C_{G}(b))^{#}. Note that C_{G}(u) contains a Klein 4-group, so by Lemma 3.1.4, u centralizes every 3-subgroup of G which it normalizes. Since S ∈ Syl_{5}(G) is Abelian, we see that u centralizes F (⟨u⟩O_{2′}(C_{G}(b))), as C_{G}(b) is a {2, 3, 5}-group. Hence u ∈ Z(F (⟨u⟩O_{2′}(C_{G}(b)))). Thus [O_{2′}(C_{G}(b)), u] is a 5-group.

Note that ū ∈ O_{2′}(CC̄(ā)), so that a now familiar argument due to J.G.Thompson yields [O_{2}(C̄), ū] = 1. Now we have that [O_{2}(C), u, u] is a 5-group, so that [O_{2}(C), u] = 1 and now we have [F (C), ⟨u⟩] = 1. Hence our claim is established unless C has a component, L say, so suppose there is such a component. If |C| is divisible by 25, then we have seen above that O_{5}(C) ≠ 1. Since S is Abelian, we see in any case that 25 |̸|L|. Hence L/Z(L) ≅ A_{5}, A_{6} or PSU(4, 2). Since now L ⊲ C, and none of these groups admits an outer automorphism of order 5, we have u ∈ F^{∗}(C) = LF (C), so that u ∈ L × O_{5}(C). Now, working in C/O_{5}(C), we see that [L, a] ≤ O_{5}(C) if L ≅ A_{5} or A_{6}, (since no outer automorphism of L centralizes a Sylow 5-subgroup in these cases), a contradiction. Hence we have L/Z(L) ≅ PSU(4, 2).

Since a centralizes a Sylow 5-subgroup of L, there must either be some ainvariant irreducible constituent ψ of Res_{L}^{G}(χ) of degree greater than 5, or else there must be an irreducible constituent ψ of Res_{L}^{G}(χ) which induces irreducibly to ⟨a⟩L. In the former case, we observe that L = [L, a] is in the kernel of all other constituents of Res_{L⟨a⟩}^{G}(χ). If ψ(1) = 6, then Z(L) = 1, otherwise there is an involution of L\Z(L) which contradicts Lemma 2.1.6. Also, because of Lemma 2.1.6, we must have ψ(v) = −2 for each involution v ∈ L. If b has the eigenvalue −1 with multiplicity 6 in this case, we obtain a contradiction to Lemma 2.1.6 by considering bv. However, PSU(4, 2) has a unique irreducible character of degree 6, and there are involutions y, z ∈ L with ψ(y) = 2 = −ψ(z), a contradiction. We can’t have ψ(1) = 8, since neither PSU(4, 2), nor its double cover, have an irreducible character of degree 8. By Lemma 2.1.6, we can’t have ψ(1) = 10, since b centralizes ⟨a⟩L. Hence the first alternative does not occur.

In the latter case, ψ(1) ≥ 4, and b must have an eigenvalue of multiplicity 8 (using Lemma 2.1.6). Hence b has the eigenvalue −1 with multiplicity 8, ψ(1) = 4 and Z(L) has order 2. In fact, ⟨b⟩ = Z(L), for otherwise, if z ∈ Z(L) is an involution, then bz has the eigenvalue 1 with multiplicity at least 8. Then unimodularity leads to a contradiction, again using Lemma 2.1.6. Furthermore, Res_{L}^{G}(χ) = ψ + \overline{ψ} + 3λ_{0}, where λ_{0} is the trivial character, since L has no nontrivial irreducible character of degree less than 4 and two complex conjugate irreducible characters of degree 4.

Recall that u ∈ O_{5}(C_{G}(a)) ∩ C_{G}(b). Let us assume (as we may without loss of generality) that S ∩ L = ⟨v⟩ ≠ 1. Now χ(v) = 1, and v has the eigenvalue 1 with multiplicity 3. Let W be the 1-eigenspace of v. Then W is an S-invariant subspace. Also C_{G}(b) acts trivially on W. Thus ⟨S, C_{G}(b)⟩ is not an irreducible subgroup of G, so H = ⟨S, C_{G}(b)⟩ < G. By Maschke’s theorem, there is an H-invariant complementary subspace to W. This must be the −1-eigenspace of b, since C_{G}(b) already acts irreducibly on that eigenspace. Hence H = C_{G}(b), since [H, b] acts trivially on both W and its complement. Thus S ≤ C_{G}(b). But now let s ∈ S be an element which normalizes a Sylow 11-subgroup of G. Then by Lemma 2.4.1, we have χ(s) = 1 and χ(sb) = χ(b) = −1, contrary to the fact that χ(b) = −5.

We have now established that θ is balanced. We now derive a contradiction by fairly standard signalizer functor analysis, but the proof in our special situation is somewhat simpler. Recall that T was a Sylow 2-subgroup of G with A ⊲ T.

Firstly, θ is complete by D. Goldschmidt’s signalizer functor theorem. Set X = θ(A) = ⟨θ(a) : a ∈ A^{#}⟩(≠ 1) and M = N_{G}(X). Then the arguments in the article by Gorenstein in [24] show that whenever 1 ≠ U ≤ T contains a Klein 4-subgroup, then we have N_{G}(U) ≤ M. By Lemma 1.6.2, we have T ≤ M′.

Using Lemma 3.1.2 if X is a Sylow 5-subgroup of G, then we know that M/C_{G}(X) is a 2-group. Hence T ≤ C_{G}(X), and then T is cyclic by Lemma 2.4.1, a contradiction.

Lemma 3.1.6: C_{G}(S) = S and S normalizes no non-trivial 5′-subgroup of G. Proof: By Lemma 2.7.3, it suffices to prove that C_{G}(S) = S. Suppose otherwise. Let x be an element of order 3 in C_{G}(S). By Lemma 2.4.2, we may suppose that χ(x) = −4 or χ(x) = 5 + 6e^{\frac{2πi}{3}} . We treat these cases separately. In the second, notice that x is not conjugate to its inverse in G.

Case 1: χ(x) = −4: We claim first that |C_{G}(x)| is not divisible by 9, from which it easily follows that |G| is not divisible by 9. In that case, x must be conjugate to its inverse in G. Since x ∈ C_{G}(S) and S ∈ Syl_{5}(G), x is already conjugate to x^{−1} within C_{G}(S), contrary to Lemma 3.1.5.

Now [O_{5′}(C_{G}(x)), S] = 1 by Lemma 2.7.3, so that [S, F (C_{G}(x))] = 1. Also, O_{5′}(C_{G}(x)) = ⟨x⟩. If E(C_{G}(x)) = 1, then C_{G}(x) ≤ N_{G}(S) and our claim is valid. Suppose then that C_{G}(x) has a component L of order divisible by 5. Now x ∉ C_{G}(x)′ since its action on either of its 5-dimensional eigenspaces is not unimodular. Since L has Abelian Sylow 5-subgroups, we may now conclude (since O_{5′(C_{G}(x))} = ⟨x⟩ that Z(L) = 1 and that L is simple. Also, L has a faithful complex irreducible character of degree at most 5, so that L ≅ A_{5} or A_{6}. We claim that C_{G}(x) has no other components. If M were another component of L, then M must act trivially on the 1-eigenspace of x and on one of the other eigenspaces of x ( for neither L nor M has an irreducible character of degree 2.) In that case, M contains an involution with a fixed-point space of codimension 2, contrary to Lemma 2.6.1. Hence L is the unique component of C_{G}(x). Suppose that O_{5}(C_{G}(x)) = 1. Then F (C_{G}(x)) ≤ Z(C_{G}(x)), and 25 |̸ |C_{G}(x)| since Out(L) is a 5′-group. Thus O_{5}(C_{G}(x)) ≠ 1. Choose u ∈ O_{5}(C_{G}(x))^{#}. Notice that if L acts irreducibly on both 5-dimensional eigenspaces of x, then Res_{S}^{G}(χ) is multiplicity-free, and x acts trivially in any reduction (mod 3) of G, a contradiction.

Hence L has a non-trivial complex irreducible character of degree at most 4. Since L is a simple {2, 3, 5}-group, we must have L ≅ A_{5}. Let V = ⟨x⟩W, where W is a subgroup of L isomorphic to A_{4}. Since W acts faithfully on each 5-dimensional eigenspace of x, we see that the character afforded by the action of W on either eigenspace has an irreducible character of degree 3 as a constituent. Thus no irreducible consituent of Res_{V}^{G}(χ) occurs with multiplicity three or more. Then u acts with minimum polynomial of degree at most 2 in any reduction (mod 5) of G, contrary to Lemma 2.6.2.

Case 2: χ(x) = 5 + 6e^{\frac{2πi}{3}} : Now C_{G}(x) ≰ N_{G}(S), for otherwise (as in the previous case) ⟨x⟩ ∈ Syl_{3}(G), but in this case G has a normal 3-complement, since x is not conjugate to its inverse in G.

As before, we have [F (C_{G}(x)), S] = 1 and E(C_{G}(x)) ≠ 1. Thus C_{G}(x) has a component L and Z(L) ≤ ⟨x⟩. Suppose that L has another component M. We claim that either L or M acts trivially on the 5-dimensional eigenspace of x. For otherwise, each of them acts as a simple group on that eigenspace, and has a 3-dimensional irreducible constituent, a contradiction, as they centralize each other. Suppose then that L acts trivially on the 5-dimensional eigenspace of x.

Now M must also act non-trivially on the 6-dimensional eigenspace of x, otherwise it contains an involution violating Lemma 2.1.6. Let H = LM and let ψ be the character of C_{G}(x) afforded by its action on the 6-dimensional eigenspace. Then since C_{G}(x) contains no central element of order 6, the only possibility is that Res_{LM}^{C_{G}(x)}(ψ) = α + β, where α and β are irreducible of degree 3 and L ≤ kerα. But then L contains an involution contradicting Lemma 2.1.6.

Hence L = E(C_{G}(x)). If L acts trivially on the 1-eigenspace of x, then ψ(t) = −2 for every involution t ∈ L (otherwise, Lemma 2.1.6 is contradicted). Let A be a Klein 4-subgroup of L (there is such a subgroup, since Z(L) has odd order). Then ⟨Res_{A}^{G}(χ), 1⟩ = 5 and each non-trivial linear character of A occurs with multiplicity 2 in Res_{A}^{G}(χ). Since x acts trivially on the fixed-point space of A and x has only one eigenvalue on V e_{λ} where V is the underlying R_{π}-module ( π a prime ideal containing 3 of a suitable ring of algebraic integers, and e_{λ} being the primitive idempotent of Z(R_{π}A) associated to the linear character λ of A), we deduce that x acts trivially in any reduction (mod 3) of G, a contradiction. Hence L acts non-trivially on the 5-dimensional 1-eigenspace of x.

Now Z(L) has odd order coprime to 5, so that L/Z(L) ≅ A_{5}, A_{6} or PSU(4, 2). In fact, either L acts irreducibly on the 1-eigenspace of x, or ψ(1) < 5 and L ≅ A_{5}. As above, since Out(L) is a 5′-group, we see that Q = O_{5}(C_{G}(x)) ≠ 1.

Suppose that L is not simple. Then Z(L) ≤ O_{5′}(C_{G}(x)) = ⟨x⟩, so that Z(L) = ⟨x⟩, and L/Z(L) does not have cyclic Sylow 3-subgroups. in that case, ψ restricts irreducibly to L, since L/Z(L) ≅ A̸_{5}.

Since Q acts on the 1-eigenspace of x and centralizes the subgroup L which acts irreducibly on that space, each generator of Q has only one eigenvalue on the 1-eigenspace of x. On the 6-dimensional eigenspace of x, we see that L either acts irreducibly, or else the eigenspace if the direct sum of two irreducible modules for L. Hence each element of Q has at most two eigenvalues on either of the eigenspaces of x, so since x has order prime to 5, each element of Q acts with minimum polynomial of degree at most two in any reduction (mod 5) of G, contrary to Lemma 2.6.2. Thus L is simple.

The same argument works if L ≅ A̸_{5}, since neither PSU(4, 2) nor A_{6} has a faithful irreducible character of degree less than 5. Hence we must have L ≅ A_{5}. Now F^{∗}(C_{G}(x)) ≅ C_{3} × C_{5} × A_{5} and C_{G}(x)/F^{∗}(C_{G}(x)) is isomorphic to a subgroup of Out(F^{∗}(C_{G}(x))), so is a 2-group.

Now we claim that x commutes with none of its other G-conjugates. For let u ∈ L be an element of order 3. Then u^{j}x^{k} has three different eigenvalues on the 1-eigenspace of x for 1 ≤ j ≤ 2 and 0 ≤ k ≤ 2. Hence none of these elements is conjugate to x, since x has only two eigenvalues in total. Since x is not conjugate to x^{−1} in G, the claim is established. Let R be a Sylow 3subgroup of G containing x. Then x ∈ Z(R), otherwise x centralizes at least 3 of its R-conjugates.

By the weak closure of {x} in R, we see easily that N_{G}(Y) ≤ C_{G}(x) whenever x ∈ Y ≤ R, so that C_{G}(x) controls fusion in R with respect to G. Thus

R = R ∩ G′ = R ∩ C_{G}(x)′.

However,

⟨x⟩ = R ∩ C_{G}(x)′ ∩ Z(C_{G}(x)) ≤ R′ = 1,

a contradiction.

Lemma 3.1.7: For some g ∈ G\N_{G}(S), we have S ∩ S^{g} ≠ 1.

Proof: Suppose that no such g. Then [G : N_{G}(S)] ≡ 1 (mod 25) and S ⊲ C_{G}(s) for all s ∈ S^{#}. By Corollary 3.1.3 and Lemma 3.1.6, we see that G contains no element of order 15. By Lemma 2.5.3, Res_{S}^{G}(χ) is not rational-valued, so by Lemma 1.2.12, χ is 3-rational. By Lemma 2.5.1, 3^{6} |̸|G|. We recall that we are assuming that 2^{11}||G|.

Since [N_{G}(S) : S] ∈ {4, 8}, we either have [G : S] ≡ 4 or 8 (mod 25). Hence |G| is of the form 2^{a}.3^{b}5^{2}.11, where 11 ≤ a ≤ 17, 0 ≤ b ≤ 5, 11.2^{a}.3^{b} ≡ 4 or 8 (mod 25) and 5^{2}.2^{a}.3^{b} ≡ 5 or 10 (mod 11).

Then we have 2^{a−6}.3^{b} ≡ 1 or 2 (mod 25) and 2^{a}.3^{b−2} ≡ 1 or 2 (mod 11). There are sol solutions to these congruences in the given ranges for a and b.

Lemma 3.1.8: |G| is not divisible by 729.

Proof: Suppose otherwise. Then by Lemma 2.51, χ is not 3-rational, while by Lemma 2.5.3, χ is not 5-rational. Then G contains an element u of order divisible by 15 by Lemma 1.2.12 (in fact, we need a slightly sharper form than is stated in the Lemma: χ(u) is moved by non-trivial Galois automorphisms σ and τ such that σ fixes 3′-roots of unity τ fixes 5′-roots of unity. Let z be the 5-part of u. Since |N_{G}(S)| is not divisible by 3, using by Lemmas 3.1.3 and 3.1.6, we conclude that S ⋪C_{G}(z). Now, as in several previous arguments, [S, F (C_{G}(z))] = 1, so that E = E(C_{G}(z)) ≠ 1. Since S is Abelian and E = E′, O_{5}(E) = 1. Hence |E| has the form 2^{a}3^{b}5. Since O_{5′}(E) ≤ C_{G}(S) = S, we deduce that E is simple. We claim that E ≅ A_{5}. Suppose otherwise. Then E ≅ A_{6} or PSU(4, 2).

In either case, E has no faithful complex irreducible character of degree less than 5. We claim that one of the irreducible constituents of Res_{E}^{G}(χ) does have degree 5. Suppose otherwise. If E ≅ A_{6} then (using 2.1.6, since E acts on each eigenspace of z), we see that Res_{E}^{G}(χ) = ψ + 3.1_{E}, where ψ is irreducible of degree 8, and that z has an 8-dimensional eigenspace. Then z has only 4 eigenvalues, so z^{5} acts trivially in any reduction (mod 5) of G and z^{5} = 1.

Now by the Baer-Suzuki theorem, there is a conjugate of z, say y, such that T = ⟨z, y⟩ is not a 5-group. Now Res_{T}^{G}(χ) has a linear constituent of multiplicity 5 or more, so that |T | is not divisible by 25, using Lemma 2.3.2. Now z is not conjugate to its inverse within T, for otherwise, 1 is the only eigenvalue of z which can occur with multiplicity greater than 5, but 1 occurs with odd multiplicity as an eigenvalue of z since χ(1) = 11 and all eigenvalues of z other than 1 occur in complex conjugate pairs. Then 1 occurs with multiplicity 9 as an eigenvalue of z, contrary to Lemma 2.1.6.

Since ⟨z⟩ ∈ Syl_{5}(T) and z is not conjugate to its inverse within T, we see that T has a normal 5-complement, N say. Let Now z ∉ O_{5}(T), so that [N, z] ≠ 1. Let U be a z-invariant subgroup of N minimal subject to [U, z] ≠ 1. Since z has only 4-eigenvalues in the given representation of G, by Lemma 1.2.10 (and the Hall-Higman theorem), we see that U is a non-Abelian special 2-group. Furthermore, since U = [U, z] is in the kernel of all linear constutuents of Res_{U⟨z⟩}^{G}(χ), we see that Res_{U}^{G}(χ) = θ + 7.1_{U}, where θ is irreducible of degree 4. Hence U is extraspecial of order 32. But then U contains an involution t with θ(t) = 0, so that χ(t) = 7, contrary to Lemma 2.1.6. Hence E ≅ A̸_{6}.

Suppose then that E ≅ PSU(4, 2), but still that Res_{E}^{G}(χ) has no irreducible constituent of degree 5. Then Res_{E}^{G}(χ) = ψ + 5.1_{E}, where ψ is irreducible of degree 6 ( Lemma 2.1.6 excludes the possibility ψ(1) = 10). There is not involution t ∈ E with χ(t) = 7, by Lemma 2.1.6, so that ψ(t) = −2 for each involution t ∈ E. Now E contains an elementary Abelian subgroup A of order 16, so that ⟨Res_{A}^{E}(ψ), 1⟩ < 0, a contradiction. Hence E ≅ P̸SU(4, 2) if there is no irreducible constituent of Res_{E}^{G}(χ) of degree 5.

Suppose then that Res_{E}^{G}(χ) has an irreducible constituent, say ψ, of degree 5. Now Res_{C_{G}(z)}^{G}(χ) has no irreducible constituent of degree 10 or more by Lemma 2.1.6. Since E ⊲ ⊲C_{G}(z), it follows that ψ extends to an irreducible character of degree 5 of C_{G}(z), which lies in a 5-block B of defect 1 of C_{G}(z). Since χ(1) = 11 and C_{G}(S) = S, χ lies in the principal 5-block of G.

By Brauer’s second and third main theorems, the character

∑_{μ∈B}⟨Res_{C_{G}(z)}^{G}(χ), μ⟩μ

vanishes identically on the 5-section of z in C_{G}(z). However, each irreducible character μ ∈ B has degree divisible by 5, so this character does not vanish at z, a contradiction.

Now we have established that E ≅ A_{5}, and that Res_{C_{G}(z)}^{G}(χ) has no irreducible constituent of degree 5. Note that every element of order 3 in E is conjugate to its inverse within E.

Since C_{G}(z) has self-centralizing elementary Abelian Sylow 5-subgroups of order 25, we may write C_{G}(z) = ⟨z⟩×H for some subgroup H. Now F^{∗}(H) = E and Out(E) is a 2-group, so we conclude that |C_{G}(z)| is not divisible by 9.

Now we recall that our element u of order divisible by 15 (and with 5-part z) was chosen so that χ(u)^{σ} ≠ χ(u), where σ is a non-trivial Galois automorphism which fixes 3′-roots of unity. However, let u = zw = wz where w ∈ E has order 3. Then χ(zw) ∈ ℤ[e^{\frac{2πi}{5}} ], since E acts on the eigenspaces of z and w is conjugate to its inverse within w. This final contradiction shows that 729 does not divide |G|.

Lemma 3.1.9*: [G : N_{G}(S)] ≡ 6 or 11 (mod 25), and there are two weakly closed cyclic subgroups W and Y of order 5 of S such that C_{G}(s) = S for all s ∈ S\(W ∪ Y). If |N_{G}(P)| = 110 for a Sylow 11-subgroup P, then we have [G : N_{G}(S)] ≡ 6 (mod 25).

Proof: By Lemma 3.1.2, N_{G}(S)/S is Abelian of order 4 or 8, and exponent 4. Let A be a complement to S in N_{G}(S). By Maschke’s theorem, we may write S = W × Y for cyclic subgroups W and Y of order 5, each normal in N_{G}(S). These subgroups are both weakly closed in S with respect to G, since N_{G}(S) controls the fusion in G of all subgroups of S. Since G = G′, neither W nor Y is central in N_{G}(S).

For any element s ∈ S\(W ∪ Y), we have A ∩ C_{G}(s) = 1, so that C_{G}(s) has a normal 5-complement. By Lemma 3.1.5, we have C_{G}(s) = S.

By Lemma 3.1.7, there is an element z ∈ S^{#} such that S is not normal in C_{G}(z). We know now that z ∈ W ∪ Y. Suppose for the sake of definiteness that z ∈ Y. Arguing as in the proof of Lemma 3.1.8, E = E(C_{G}(z)) ≅ A_{5}, and Res_{E}^{G}(χ) has no irreducible constituent of degree 5. In particular, N_{G}(Y) has 6 Sylow 5-subgroups.

If W is not a Sylow intersection, then we see that S permutes the Sylow 5subgroups of G not contained in Y in cycles of length 25. Hence [G : N_{G}(S)] ≡ 6 (mod 25). If W is a Sylow intersection, we find similarly that N_{G}(W) also has 6 Sylow 5-subgroups and [G : N_{G}(S)] ≡ 11 (mod 25).

If |N_{G}(P)| = 110, then there is an element s ∈ S with |C_{G}(s)| = 10 and S ⊲C_{G}(s), using Lemma 2.4.2 and the fact that C_{G}(S) = S. Then we must have ⟨s⟩ = W, since N_{G}(Y) has a component. Hence W is not a Sylow intersection in this case.

Lemma 3.1.10: N_{G}(S)/S is a Klein 4-group.

Proof: We have |G| = 2^{a}3^{b}5^{2}11, where 11 ≤ a ≤ 17 and 1 ≤ b ≤ 5. Also, from Sylow’s theorem we have 2^{a}3^{b}5 ≡ 1 or 2 (mod 11). We know also that 2^{a}3^{b}11 ≡ −1, −6, −2 or −12 (mod 25) since [N_{G}(S) : S] ∈ {4, 8} and [G : N_{G}(S)] ≡ 6 or 11 (mod 25).

Now 2 is a quadratic non-residue (mod 11), while 3 and 5 are residues. Hence if a is odd, then 2^{a}3^{b−2} ≡ 2 (mod 11), so that 2^{a+3−2b} ≡ 1 (mod 11) and a ≡ 2b − 3 (mod 10). If a is even, then a ≡ 2b + 4 (mod 10).

Also, (−1)^{b}.2^{a−3b} ≡ 9, 4, −7 or 8 (mod 25). If b is even, this gives a − 3b ≡ 14, 2, 15 or 3 (mod 20). If b is odd, it gives a − 3b ≡ 4, 12, 5 or 13 (mod 20).

The only possible solutions for these congruences in the given range are |G| = 2^{13}.3^{3}.5^{2}.11 or |G| = 2^{15}.3^{4}.5^{2}.11. However, in the first case, we have [G : N_{G}(P)] ≡ 1 (mod 11), so that |N_{G}(P)| = 110 and [G : N_{G}(S)] ≡ 6 (mod 25) by Lemma 3.1.9. Now [G : S] = 2^{13}.3^{3}.11 ≡ 2^{9}.3^{3} ≡ 24 (mod 25). In that case, we have [N_{G}(S) : S] = 4. In the second case, we also have |N_{G}(P)| = 110 by Sylow’s Theorem. Then [G : S] = 2^{15}.3^{4}.11 ≡ 2^{11}.3^{4} ≡ 13, (mod 25). However this congruence only arises in the case [N_{G}(S) : S] = 8 and [G : N_{G}(S)] ≡ 11 (mod 25), which we have excluded.

Let us examine the case |G| = 2^{13}.3^{3}.5^{2}.11 more closely. If N_{G}(S)/S is cyclic of order 4, then a generator of A acts as an automorphism of order 4 on one of W, Y and inverts the other. Suppose that A acts an an automorphism of order 4 on Z, where Z ∈ {W, Y }. Then C_{G}(Z) ∩ A = 1 and C_{G}(Z) has a normal 5-complement. Hence C_{G}(Z) = S, since we know that S normalizes no non-trivial 5′-group. Hence Z is not a Sylow intersection, so Z = W since N_{G}(Y) has a component. Since |N_{G}(P)| = 110, we know that the element s ∈ S which normalizes a Sylow 11-subgroup has C_{G}(s) ≠ S, so ⟨s⟩ ∈ {W, Y }. However S ⊲ C_{G}(s) so ⟨s⟩ ≠ Y, since N_{G}(Y) has a component, while we have seen that C_{G}(W) = S, a contradiction.

Lemma 3.1.11: |S| ≠ 25.

Proof: We have reached the following situation: C_{G}(S) = S and S normalizes no non-trivial 5′-subgroup of G. There is an element y ∈ S^{#} such that E = E(C_{G}(y)) ≅ A_{5}. We have N_{G}(S) = SA, where A is a Klein 4-group. There is an element x ∈ S^{#} which normalizes a Sylow 11-subgroup of G, and we have S ⊲ C_{G}(x). Also, C_{G}(y) = ⟨y⟩ × E. There is another weakly closed subgroup W of order 5 of S such that C_{G}(s) = S for all s ∈ S\(W ∪ Y), where Y = ⟨y⟩.

Furthermore, we have |G| = 2^{13}3^{3}5^{2}11, and |N_{G}(P)| = 110. We also have [G : N_{G}(S)] ≡ 6 (mod 25). We claim that M = N_{G}(Y) is strongly 5-embedded in G. We have [N_{G}(Y) : C_{G}(Y)] = 2 and so N_{G}(S) ≤ N_{G}(Y) by a Frattini argument. Since S ⊲ C_{G}(s) for all s ∈ S\Y, we now have C_{G}(s) ≤ M for all s ∈ S, so M is strongly 5-embedded in G as claimed. Notice also that M ≅ D_{10} × A_{5}. Also, now ⟨x⟩ = W ∈ Syl_{5}(E) is the other weakly closed cyclic subgroup of S of order 5, for S ⊲ C_{G}(x) so x ∉ Y, while we have C_{G}(s) = S for all s ∈ S\(W ∪ Y) and |C_{G}(x)| = 50.

Let us note that G has more than one conjugacy class of involutions. For we know that C_{G}(S) = S. Now there is an involution t_{1} ∈ C_{G}(Y) and an involution t_{2} ∈ C_{G}(W). Then Y ∈ Syl_{5}(C_{G}(t_{1})) and W ∈ Syl_{5}(C_{G}(t_{2}). Since W and Y are not G-conjugate, it follows from Sylow’s theorem that t_{1} and t_{2} are not G-conjugate.

Now we are in a position to calculate the character Res_{E}^{G}(χ). Notice that N_{E}(W) is dihedral of order 10, and that we now know that χ(w) = 1 for a generator w of W. If t is an involution of E that inverts w, then unimodularity forces χ(t) = −1, and we note that Res_{W ⟨t⟩}^{G}(χ) = λ + ρ, where λ is the nontrivial linear character and ρ is the regular character. We have made use of Lemmas 2.4.1 and 2.4.2.

In particular, we have ⟨Res_{E}^{G}(χ), 1⟩ ≤ 1. We have also seen in the course of earlier proofs that Res_{E}^{G}(χ) does not have an irreducible constituent of degree 5. The only possibilities to get the restriction of the character correct on W ⟨t⟩ are either Res_{E}^{G}(χ) = 1 + ψ_{1} + ψ_{2} + θ, where ψ_{1} and ψ_{2} are distinct irreducible characters of degree 3 and θ is irreducible of degree 4, or else Res_{E}^{G}(χ) = ψ_{1} + ψ_{2} + γ, where ψ_{1} and ψ_{2} are distinct irreducible characters of degree 3 and γ is irreducible of degree 5. In either case, since this character is multiplicity free, we deduce that C_{G}(E) is Abelian. However C_{G}(E) contains a dihedral group of order 10, a contradiction.

Section Two

In this section, we will complete the proof that |S| = 5 for S ∈ Syl_{5}(G). We suppose for the remainder of the section that |S| = 125. By Lemmas 1.4.2 and 2.1.5, there is an an element g ∈ G\N_{G}(S) such that S ∩ S^{g} ≠ 1.

Lemma 3.2.1: i) C_{G}(S) = S. ii) |S| ≠ 125.

Proof: i) By Lemma 2.7.3, we know that S centralizes any 5′-group it normalizes. By the results of section 2.4, C_{G}(S) has order at most 6, has a normal Sylow 3-subgroup of order at most 3, and contains no element of order 6.

Suppose that C_{G}(S) ≠ S, and let T = O_{5′}(C_{G}(S)). If |T | is divisible by 3, let W = O_{3}(T), otherwise, let W = T and M = N_{G}(W). Notice that N_{G}(S) ≤ M.

We claim that S ∩ S^{g} = 1 for all g ∈ G\M. Since |S| = 125 and M is not irreducible, this contradicts Lemma 1.4.2, as N_{G}(S) ≤ M.

Suppose that Q = S ∩ S^{g} ≠ 1 for some g ∈ G\M and choose Q of maximal possible order. Set H = N_{G}(Q). Then ⟨W, S, S^{g}⟩ ≤ H. If W ≤ F (H), then [W, S] = [W, S^{g}] = 1, so that W ≤ T^{g}, and W = W^{g} since W^{g} is a normal Sylow subgroup of T^{g} in any case. This contradicts the fact that g ∉ M. Hence we may suppose that W is not contained in F (H).

Now [W O_{5′}(H), S] = 1. If O_{5′}(H) ≠ 1, then it either contains W or else it has order 2 and normalizes W (using the structure of C_{G}(S)). In the latter case, we have [W, O_{5′}(H)] = 1, a contradiction, since C_{G}(S) contains no element of order 6 and W is not contained in F (H). Thus O_{5′}(H) = 1. Now S is not normal in H.

Each component of H has order divisible by 5, using Burnside’s p^{a}q^{b}-theorem, since C_{G}(Q) is a {2, 3, 5}-group. However E = E(H) has TI-Sylow 5-subgroups, so has a single component (for if L_{1}, L_{2} are distinct components, then L_{1} ≤ C_{E}(L_{2}) ≤ N_{E}(S) and likewise for L_{2}). If L is any component of H, then L is simple as O_{5′}(H) = 1 and L has Abelian Sylow 5-subgroups. If 25 divides |L|, then Lemma 1.2.7 is contradicted. Hence |S ∩ L| = 5. Now L ≅ A_{5}, A_{6} or PSU(4, 2). Hence W is not contained in L, since each of these possibilities has a self-centralizing Sylow 5-subgroup. Thus W induces outer automorphisms on L. This excludes the possibility that L ≅ PSU(4, 2), since that group has trivial outer automorphism group. But in the other cases, it is also the case that no automorphism of L of order prime to 5 acts trivially on the Sylow 5-subgroup, a contradiction.

ii) We argue similarly to part i) to show that there is a subgroup Q ≤ S of order 25 such that C_{G}(Q) ≰ N_{G}(S). For again, using Lemma 1.4.2, we choose Q ≠ 1 of such that Q = S ∩ S^{g} for some g ∈ G\N_{G}(S) and |Q| is maximal. We set H = N_{G}(Q). Now O_{5′}(H) = 1 by part i) and Lemma 2.7.3. Hence [S, F (H)] = 1, so H has a unique component L as in part i). Also |L| is not divisible by 25 by the argument of part i). Hence L ≅ A_{5}, A_{6} or PSU(4, 2). Since Out(L) is a 5′-group in each case, we have S ≤ F^{∗}(H) and |Q| = 25.

We claim that L ≅ A_{5}. Otherwise, no non-trivial irreducible character of L has degree less than 5. In fact, Res_{L}^{G}(χ) has no irreducible constituent of degree 5. For χ lies in the principal 5-block since χ(1) = 11 and C_{G}(S) = S. If ψ is an irreducible constituent of Res_{L}^{G}(χ) of degree 5, then there must be another non-trivial irreducible constituent, otherwise Lemma 2.1.6 is contradicted by an involution of L. Hence Res_{Q}^{G}(χ) has at most three distinct linear constituents. Since Q is not cyclic, there is a non-identity element of Q which has at most two eigenvalues, so acts with (at most) quadratic minimum polynomial in any reduction (mod 5) of G, contrary to Lemma 2.6.2.

Now, apart from irreducible characters of degree 5, PSU(4, 2) has no other non-trivial irreducible character of degree less than 10. But since no element of Q^{#} contains an element with an eigenspace of codimension 1, Res_{L}^{G}(χ) has no irreducible constituent of degree 10. Hence L ≅ A_{6}. If Res_{L}^{G}(χ) has an irreducible constituent ψ of degree 8, then let U be a subgroup of L which is isomorphic to (a natural copy of) A_{4}. Then Res_{U}^{G}(χ) = 2α + λ + μ + 3.1_{U}, where α is irreducible of degree 3 and λ and μ are non-trivial linear characters of U.

Since Q is not cyclic, there is an element x ∈ Q^{#} which has only two eigenvalues on the fixed-point space of U. Hence, by the arguments of Section 1.5, x acts with quadratic minimum polynomial in any reduction (mod 5) of G, again contrary to Lemma 2.6.2.

By Lemma 2.1.6, Res_{L}^{G}(χ) has no irreducible constituent of degree 9 or 10, otherwise each element of Q has an eigenspace of codimension 1 or 2. Thus L ≅ A̸_{6}, so we must have L ≅ A_{5}.

Let U be a Sylow 2-normalizer of L, a subgroup isomorphic to A_{4}. Suppose that Res_{U}^{G}(χ) has no irreducible constituent with multiplicity 4 and only one irreducible constituent with multiplicity 3. Then since Q is not cyclic, there is an element of Q^{#} which has at most two eigenvalues on each homogeneous component of the restriction of the underlying module to U. Such an element acts with quadratic minimum polynomial in any reduction (mod 5) of G, contrary to Lemma 2.6.2.

Hence Res_{L}^{G}(χ) must contain the trivial character with multiplicity 4 or more. Since Res_{L}^{G}(χ) must contain more than one non-linear irreducible constituent by Lemma 2.1.6.

It now follows that Res_{U}^{G}(χ) = 2ψ+5.1_{U}, where ψ is the irreducible character of degree 3 of L. Now let us note that C_{G}(Q) ≅ Q × N where E(N) ≅ A_{5} and N ≅ A_{5} or S_{5}, since C_{N}(S ∩ N) = (S ∩ N).

Then there are now four possibilities for Res_{L}^{G}(χ), where L is a component of H. Possibly Res_{L}^{G}(χ) = ψ_{1} +ψ_{2} +5.1_{L}, where ψ_{1} and ψ_{2} are different irreducible characters of degree 3. Or Res_{L}^{G}(χ) = 2ψ_{1} +5.1_{L}, where ψ_{1} and ψ_{2} is irreducible of degree 3. Or Res_{L}^{G}(χ) = ψ_{1} +θ_{1} +4.1_{L}, where ψ_{1} is irreducible of degree 3 and θ_{1} is irreducible of degree 4. Or Res_{L}^{G}(χ) = 2θ_{1} + 3.1_{L}, where θ_{1} is irreducible of degree 4.

Suppose that H/C_{G}(Q) contains an element of order 3. Then no non-trivial linear character of Q can occur with multiplicity greater than 3 in Res_{Q}^{G}(χ) since if α is an automorphism of order 3 of Q induced by the action of H, whenever λ is a non-trivial linear constituent of Res_{Q}^{G}(χ), both λ^{α} and λ^{α^{2}} occur with the same multiplicity as λ. We claim that this eliminates the last case. For in the last case, note that Res_{Q}^{G}(χ) must contain the trivial character with multiplicity 8 or more (since L acts on the homogeneous components for Q of the underlying module). But then Res_{Q}^{G}(χ) must have the form λ + λ^{α} + λ^{α^{2}} + 8.1_{Q}, where λ is a non-trivial linear character of Q. Then Q contains an element of order 3 with a fixed-point space of dimension 9, contrary to Lemma 2.1.6. Likewise, in case Res_{L}^{G}(χ) = ψ_{1} + θ_{1} + 4.1_{L}, where θ_{1} has degree 4 and ψ_{1} has degree 3, then Res_{Q}^{G}(χ) must contain the trivial character with multiplicity at least 7. However, since α acts without non-trivial fixed-points on Q and χ(1) = 11, the multiplicity of the trivial character in Res_{Q}^{G}(χ) must be congruent to 2 (mod 3), so still must be 8, yielding the same contradiction. Even in the case Res_{L}^{G}(χ) = ψ_{1} + ψ_{2} + 5.1_{L}, where ψ_{1} and ψ_{2} are (possibly equal) irreducible characters of degree 3, we still see that the multiplicity of the trivial character in Res_{Q}^{G}(χ) must be at at least 6, so in fact at least 8, and the same contradiction ensues. Hence we conclude in any case that [N_{G}(Q) : C_{G}(Q)] is not divisible by 3.

Now let v ∈ L be an element of order 3 and set Y = C_{G}(v). Then χ(v) = 5, and for ω a primitive complex cube root of unity, v has the eigenvalue 1 with multiplicity 7, and the eigenvalues ω and ω^{2} with multiplicity 2 each. Now Q ∈ Syl_{5}(C_{G}(v)) and certainly Q centralizes O_{5}(Y). We claim that [Q, X] = 1, where X = O_{5′}(Y). Now QX leaves each eigenspace of v invariant. By Lemma 1.2.10, [Q, X] acts trivially on the ω-eigenspace and the ω^{2}-eigenspace of v. Let V be a Q-invariant Sylow 3-subgroup of X and Z be a Q-invariant Sylow 3-subgroup of X. If [V, Q] ≠ 1, then [V, Q] has a 6-dimensional fixed-point subspace, and Res_{[V,Q]Q}^{G}(χ) = θ + 6λ_{0}, where θ is irreducible of degree 5 and λ_{0} is trivial. Then [V, Q] contains an element of order 3 which has the eigenvalue 1 with multiplicity 9 and the eigenvalues ω and ω^{2} with multiplicity 1 each, contrary to Lemma 2.1.6. Hence [V, Q] = 1. A similar argument applies to [Z, Q], with the additional possible configuration that Res_{[Z,Q]Q}^{G}(χ) = β + 7λ_{0}, where β is irreducible of degree 4 (and [Z, Q] is extra-special of order 32). This configuration still exhibits an involution which contradicts Lemma 2.1.6.

We will now prove that ⟨v⟩ ∈ Syl_{3}(G). To do this, it is enough to prove that Y is not divisible by 9. To do this, it is enough to show that Q ⊲ Y, since [N_{G}(Q) : C_{G}(Q)] is not divisible by 3 and C_{G}(Q) ≅ Q × N where F^{∗}(N) ≅ A_{5}.

Suppose otherwise. Then, since we have seen that [Q, F (Y)] = 1, there must be a component, say E, of Y. Also, we note that Q ∈ Syl_{5}(Y) since C_{G}(S) = S. Thus E ∩ Q ≠ 1 and Q normalizes E. If |E| is divisible by 25, then Q ≤ E. Otherwise, |Out(E)| is not divisible by 5, (since E/Z(E) ≅ A_{5}, A_{6} or PSU(4, 2), still yielding Q ≤ E, since [F (Y), Q] = 1 and Q/Q ∩ E is isomorphic to a subgroup of Out(E). Hence |E| is divisible by 25 in all cases. On the other hand, E has a complex irreducible character of degree 7 or less (and |E| is not divisible by 7, since |G| is not), since E acts trivially on the two 2-dimensional eigenspaces of v. However, Hence E is isomorphic to a quasi-primitive complex linear group of degree 6 or less with |E| divisible by 25, and |E| is not divisible by 7. There is no such group, by the classification of low-dimensional complex linear groups, as outlined in Section 1.3. Hence |G| is not divisible by 9.

Now we have enough information to derive a contradiction from block theoretic information. We recall that a Sylow 11-subgroup of G, say P, has order 11, and that [N_{G}(P) : C_{G}(P)] ∈ {5, 10}. Hence every irreducible character of G has degree congruent to one of {0, 1, −1, 5, −5} (mod 11). Since |G| is not divisible by 9 and G has no normal 3-complement, the degree of the irreducible characters of G are 1, x and x + 1 for some integer x congruent to 1 (mod 3). By Lemma 1.2.14, one of x and x + 1 is not divisible by 11. Since the principal 3-block is closed under algebraic conjugation nd G is simple neither x nor x + 1 is congruent to ±5 (mod 11). Hence we have x ≡ 0 (mod 11) or x ≡ −1 (mod 11).

If x ≡ 0 (mod 11), and 5 is a quadratic residue (mod 11), while x + 1 has the form 2^{a}5^{b}, we see that a is even, since 2 is not a quadratic residue (mod 11). Since x + 1 ≡ −1 (mod 3), x + 1 is divisible by 5. Hence a = 0, otherwise x = 1, since x is coprime to each of 2, 3 and 5, while x ≡ 1 (mod 3). This contradicts the simplicity of G.

If x ≡ −1 (mod 11), then x is even, since 5 is a quadratic residue (mod 11). If x is a power of 2, then that power must be odd, as 5 is a quadratic residue (mod 11). But x ≡ 1 (mod 3) precludes that possibility. Hence x is divisible by 5. Hence x + 1 = 11 since x + 1 is coprime to each of 2,3, and 5. Hence x = 10 and x + 1 = 11. Now G has a subgroup of index 11 or 12 by Lemma 1.3.9. Since G is simple, we see that |G| divides 12!, and 125 does not divide |G|.

Chapter Four

In this chapter, we consider the case that a Sylow 5-subgroup, S = ⟨s⟩, of G has order 5. In Section One, we dispose of the case [N_{G}(S) : C_{G}(S)] = 2. In Section Two, we will prove that C_{G}(S) ≠ S. After Sections 3,4 and 5, we will have proved that |G| is divisible by 7. In Section Six, we prove that 49 divides |G|, and then in Section Seven, we derive a final contradiction.

Section One

We assume in this section that [N_{G}(S) : C_{G}(S)] = 2. Then the principal 5block consists of two exceptional characters, the trivial character, and one more non-exceptional character, which is rational valued, of degree x, say. There is a sign ε such that x ≡ ε (mod 5) and x(x + ε) divides |G|. By Lemma 1.2.14, the principal 5-block and principal 11-block contain a common irreducible character other than the trivial character.

Notice that the exceptional characters in the principal 5-block are rational valued on 5-regular elements, so on all 11-singular elements. Hence they must be non-exceptional characters in the principal 11-block (we consider all irreducible characters of the principal 11-block as non-exceptional if there are 11 of them).

Lemma 4.1.1: 11 divides x(x + ε).

Proof: Otherwise, the irreducible characters of degree x and x + ε must both lie in the principal 11-block, and they must be non-exceptional in that block, since they both take rational vales on 11-singular elements. But then they both have degrees congruent to ±1 (mod 11), a contradiction.

Lemma 4.1.2: x ≠ 11.

Proof: Suppose that x = 11. Then G has a rational valued character of degree 11, so that G has a subgroup of index 11 or 12 by Lemma 1.3.9.

Corollary 4.1.3: i) |C_{G}(S)| ∈ {10, 15, 30} and C_{G}(S) contains no element of order 6. ii) If |C_{G}(S)| = 15, then no element of order 3 in C_{G}(S) is conjugate to its inverse in G.

Proof: i) χ does not lie in the principal 5-block as x ≠ 11, so that G must have more than one 5-block of maximal defect. By Lemma 2.4.1, C_{G}(S) has one of the listed orders. If C_{G}(S) has order 15, and u is an element of order 3 in C_{G}(S), then χ is not constant on the 5-section of S by Brauer’s second main theorem. Hence χ(su) ≠ χ(s). Using Lemma 2.4.1, there is a primitive cube root of unity ω such that χ(su) = ω and χ(u) = 6ω + 5. In that case, u is certainly not conjugate to its inverse in G.

Lemma 4.1.4: If ε = −1, then x = 99. If ε = 1, then x = 21.

Proof: Case i): Suppose first that 11 divides x. a) If x = 11.2^{a} and x + ε = 3^{b}7^{c}, then a is even as 2 is a quadratic non-residue (mod 5), while ±1 are both residues. Now x ≠ 44, since |G| is not divisible by 43. Hence a ≥ 4.

If c ≠ 0, then 2^{a+2} ≡ ε (mod 7), so that ε = 1 as a is even and −1 is not a quadratic residue (mod 7). Thus 3^{b}7^{c} ≡ 1 (mod 11), so that c is even as 3 is a quadratic residue (mod 11) but 7 is not. Thus 7^{c} ≡ 1 (mod 16). Since 3^{b}7^{c} ≡ x + 1 ≡ 1 (mod 16) (for x is divisible by 16), we deduce that b is also even. Now, however, 3^{b}7^{c} ≡ ±1 (mod 5), whereas x + 1 ≡ 2 (mod 5). Hence c = 0.

Now 3^{b} ≡ 2ε (mod 5), a quadratic non residue (mod 5), so b is odd. As a is even, 11.2^{a} ≡ 2 (mod 3), so ε = 1 as 11.2^{a} + ε = 3^{b}. Now 3^{b} ≡ 1 (mod 4), as a is even, so that b is also even, a contradiction.

b) If x = 11.3^{a} and x + ε = 2^{b}7^{c}, then a is even as x ≡ ε (mod 5), which is a quadratic residue. In particular, x ≡ 3 (mod 4). If b ≥ 2, then x + ε ≡ 0 (mod 4), so ε = 1. Thus 2^{b} ≡ 1 (mod 3) as a ≠ 0, and hence b is even. As 2^{b}7^{c} ≡ 2ε (mod 5), a quadratic non-residue, c is odd. Now, however, x ≡ −1 (mod 7), which contradicts the fact that a is even and x = 11.3^{a}, which is a quadratic residue (mod 7). Thus b ≤ 1, so that b = 1 as x is odd and x + ε = 2^{b}7^{c}.

Now c ≠ 0 as 11.3^{a} + ε = 2.7^{c}. Now ε ≡ 2 (mod 3), so that ε = −1. Hence 2.7^{c} ≡ −1 (mod 11), so c ≡ 2 (mod 10). By Lemma 2.1.4,—G— is not divisible by 7^{12}, so that c = 2 and x = 99.

c) If x = 11.7^{a} and x + ε = 2^{b}3^{c}, then ab ≠ 0 and 7^{a} ≡ ε (mod 5), so a is even. Hence x ≡ 3 (mod 8), so that x + ε ≢ 0 (mod 8). Thus b ≤ 2. If b = 1, then ε = −1 as x ≡ 3 (mod 4).Hence 2.3^{c} ≡ −1 (mod 7), so that 3^{c} is a quadratic non-residue (mod 7) and c is odd. However, now x − 1 ≡ −2 (mod 5), so that 3^{c} ≡ −1 (mod 5), a contradiction as c is odd.

If b = 2, then ε = 1 as x ≡ ±3 (mod 8). Hence 4.3^{c} ≡ 1 (mod 7) Hence 3^{c} is a quadratic residue (mod 7), so that c is even. Now x + 1 ≡ 2 (mod 5), a contradiction as c is even and x = 4.3^{c} − 1.

d) If x = 11.2^{a}.3^{b}, x + ε = 7^{c}. Now c is odd, as 7^{c} ≡ 2ε, a non-residue (mod 5). Also, the quadratic non-residue 7^{c} ≡ ε (mod 11), so ε = −1. Now x = 7^{c} +1 ≡ 8 (mod 16) as c is even. Thus a = 3. Also, x = 7^{c} + 1 ≡ 2 (mod 3), so b = 0. Hence x = 88, a contradiction, as 29 does not divide |G|.

e) If x = 11.2^{a}.7^{b}, and x+ε = 3^{c}, then ε = 1, since 3^{c} is a quadratic residue (mod 11). Also, c is odd, as 3^{c} ≡ 2 ( a non-residue) (mod 5). Hence x = 3^{c} − 1 ≡ 2 (mod 4), so that a = 1. Now 2.7^{b} ≡ 1 (mod 5), so that b ≡ 3 (mod 4). Now 3^{c} ≡ 1 (mod 7), contrary to the fact that c is odd and 3 is a quadratic nonresidue (mod 7).

f) If x = 11.3^{a}7^{b} and x + ε = 2^{c}, then 2^{c} ≡ 2 (mod 5), so that c ≡ 1 (mod 4). Since c is odd and 2 is a quadratic non-residue (mod 11), then ε = −1. Now b = 0, for otherwise 2^{c} ≡ −1 (mod 7), whereas −1 is a quadratic non-residue and 2 is a quadratic residue (mod 7). Now 11.3^{a} ≡ −1 (mod 5), so a is even and x ≡ 3 (mod 8). Now 2^{c} = x − 1 ≡ 2 (mod 8), so that c = 1, a contradiction.

Case ii): x + ε ≡ 0 (mod 11).

a) If x = 2^{a} and x+ε = 11.3^{b}.7^{c}, then as 2^{a} ≡ ε (mod 5), a is even. As 2^{a} ≡ −ε (mod 11) and a is even, we have ε = −1. Hence a ≡ 0 (mod 10). However, x − 1 is then divisible by 2^{10} − 1, so by 31, a contradiction.

b) If x = 3^{a}, x + ε = 11.2^{b}.7^{c}, then a is even as x ≡ ε (mod 5). Then 3^{a} ≡ −ε (mod 11), so ε = −1 as 3 is a quadratic residue (mod 11). Now a ≡ 0 (mod 5), so a ≡ 0 (mod 10). Now, however, 3^{5} − 1 divides x − 1, a contradiction, as 3^{5} − 1 = 242.

c) If x = 7^{a} and x + ε = 11.2^{b}3^{c}, then as above a is even, ε = −1 and a ≡ 0 (mod 10). This contradicts the fact that 7^{10} does not divide |G| by Lemma 2.1.4.

d) If x = 2^{a}3^{b} and x + ε = 11.7^{c}, then as 11.7^{c} ≡ 2ε (mod 5), c is odd. Thus x + ε ≡ 5 (mod 8) so a ≤ 2. Hence b ≠ 0. Now x + ε = 11.7^{c} ≡ 2 (mod 3), so that ε = −1, as x ≡ 0 (mod 3). Now x ≡ 1 (mod 11), so that a is even as 3 is a quadratic residue (mod 11), but 2 is not. However, x = 11.7^{c} + 1 ≡ 6 (mod 8), as c is even. This is a contradiction as a is even.

e) If x = 2^{a}7^{b} and x+ε = 11.3^{c}, then c is odd as 3^{c} ≡ 2ε (mod 5). Now 2^{a} ≡ −ε (mod 3). If a is even, then ε = −1. As c is odd, we have x = 11.3^{c} + 1 ≡ 2 (mod 8), as c is odd. Thus a =1, contrary to the fact that a is even.

Thus a is odd and ε = 1. Now b = 0, otherwise 11.3^{c} ≡ 1 (mod 7), contrary to the fact that c is odd and 3 is a quadratic non-residue (mod 7). Now we have 2^{a} + 1 = 11.3^{c}. Also, 11.3^{c} ≡ 2 (mod 5), so that c ≡ 3 (mod 4). then 11.3^{c} ≡ 9 (mod 16), so a ≤ 3, a contradiction, since a ≡ 5 (mod 10).

f) If x = 3^{a}7^{b} and x + ε = 11.2^{c}, then as 11.2^{c} ≡ 2ε (mod 5), c is odd, as 2^{c} is a quadratic non-residue (mod 5). If ε = −1, then x ≡ 2 (mod 3),so that a = 0. Now if b ≠ 0, then 2^{c+2} ≡ −1 (mod 7), a contradiction, since 2 is a quadratic residue (mod 7) but −1 is not. Hence b = 0, so x = 1, contrary to the simplicity of G. Thus ε = 1.

Now x ≡ −1 (mod 11), so that b is odd, since 3 is a quadratic residue (mod 11) but 7 is not. Since x ≡ 1 (mod 5) and b is odd, a must also be odd. Now x ≡ 5 (mod 8), so that x + 1 ≡ 6 (mod 8) and c = 1. Hence x = 21, x + ε = 22 in this case.

Lemma 4.1.5: x ≠ 21.

Proof: Suppose that x = 21. Let θ be the irreducible character of degree 21 in the principal 5-block of G, which we know to be rational valued. Let T ∈ Syl_{7}(G) and choose u ∈ T^{#}.

Then as 7 does not divide [G : C_{G}(u)], and \frac{[G:C_{G}(u)]θ(u)}{θ(1)} is a rational algebraic integer, we see that θ(u) = 7z for some integer z (with |z| < 3 as G is simple).

\frac{[G:C_{G}(u)]θ(u)}{θ(1)} ≢ [G : C_{G}(u)] (mod 7), so that θ does not belong to the principal 7-block of G. Since G contains no element of order 35, Lemma 1.2.14 shows that there is a non-trivial irreducible character in both the principal 7-block and the principal 5-block.

As the two exceptional characters from the principal 5-block agree on 5regular elements, they agree on 7-singular elements, so both lie in the same 7-block (since they have degree prime to 7, so do not vanish on all 6-singular elements). Hence both exceptional characters of degree 22 lie in the principal 7-block. Lemma 1.2.14 may now be applied with the 7-block which contains θ; we conclude that θ lies in a 7-block of defect 0 of G and that |T | = 7.

Now Res_{T}^{G}(χ) is not rational valued by Lemma 2.5.4, we have [N_{G}(T) : C_{G}(T)] ≠ 6. Now both exceptional characters (of degree 22) in the principal 5-block are non-exceptional characters in the principal 7-block. Hence there are at least 3 non-exceptional characters in the principal 7-block and [N_{G}(T) : C_{G}(T)] = 3. Now, however, the two non-trivial non-exceptional characters in the principal 7-block are not rational valued, as they are both exceptional characters in the principal 5-block. This contradicts Lemma 1.6.1, as G is simple.

Lemma 4.1.7: G has a self-centralizing Sylow 7-subgroup T of order 49.

Proof: By Lemma 2.1.8, T is elementary Abelian. Let ψ_{1} and ψ_{2} be the exceptional characters of the principal 5-block of G, each rational-valued on 5regular elements, and each of degree 98. Choose u ∈ T^{#}. Then ψ_{1}(u) = 49z for some integer z with |z| < 2. Since z ≢ 2 (mod 7), it follows that neither ψ_{i} is in the principal 7-block. By Lemma 1.2.14, we see that |T | = 49. We claim that C_{G}(u) = T. Since C_{G}(u) contains no element of order 5 (as |C_{G}(S)| divides 30), we see that Res_{C_{G}(u)}^{G}(ψ_{1}) is rational-valued. For any element x ∈ C_{G}(u), we deduce that ψ_{1}(x0 = 49w(x) for some integer w(x) ∈ {−1, 0, 1}. If x is an involution, then as G = G′, unimodularity forces ψ_{1}(x) ≡ 2 (mod 4), a contradiction.

Suppose that x ∈ C_{G}(T) has order 3. Then 49w(x) ≡ 98 (mod 3), so that w(x) = −1 and ψ_{1}(x) = −49.

Now let s be a generator of S, and set U = O_{5′}(C_{G}(s)). Let C_{y} denote the class sum of y ∈ G in the group algebra ℂG. We count the number of elements with 5-part s in the product of class sums C_{s}C_{x}. This is the sum over u ∈ U of the coefficient of C_{su} in C_{s}C_{x}, so is certainly a nonnegative integer, h say. On the other hand, this may be calculated using the formula of Burnside for the product of class sums. Using the fact that, by block orthogonality relations following from Brauer’s Second Main Theorem (and using Brauer’s Third Main Theorem), we have ∑_{u∈U} θ(su) = 0 for each irreducible character θ of G which does not lie in the principal 5-block, we obtain:

h = \frac{|U||G|}{|C_{G}(x)||C_{G}(s)|} (1 − \frac{48}{99} − \frac{3.49}{98}) .

However, this is certainly negative, a contradiction. Hence C_{G}(T) has order prime to 3. Since C_{G}(P) = P and C_{G}(S) has order prime to 7, the Lemma now follows.

Lemma 4.1.8: |C_{G}(S)| ≠ 15. Proof: Suppose otherwise, and let u ∈ C_{G}(s) be an element of order 3. Then |N_{G}(P)| = 55, as G contains no element of order 10. Now |G| = 2^{a}.3^{b}.5.7^{2}.11 for some integers a and b. Now [G : N_{G}(P)] = 2^{a}3.^{b}7^{2}, so that a is even as 3 is a quadratic residue (mod 11), but 2 is not.

Now let t be an involution of G, and let us count the number of elements with 5-part s which appear in the product of class sums C_{t}C_{t}. Since u is not conjugate to its inverse in G, either is su, so that s itself is the only element with 5-part s which is expressible as a product of two involutions.

Now every involution which inverts s belongs to N_{G}(S) and we know that |N_{G}(S)| = 30. Also, N_{G}(S) has a central Sylow 3-subgroup. Hence N_{G}(S) contains 5 involutions.

Let θ be the non-trivial non-exceptional irreducible character in the principal 5-block. Then the usual formula for the product of class sums and the block orthogonality relations yield, by an argument similar to that of the previous result:

\frac{3|G|}{|C_{G}(t)|^{2}} \frac{(99 − θ(t))^{2}}{99.98} = 5.

This contradicts the fact that 2^{a} is the exact power of 2 dividing |G| and a is even.

Lemma 4.1.9: All involutions of G are conjugate.

Proof: Let B be the 5-block which contains χ. By Lemma 2.5.4, Res_{T}^{G}(χ) is not rational-valued, and by Lemma 2.4.2, B is invariant under Galois automorphisms which fix all 7′-roots of unity. Hence B contains a Galois conjugate χ̃ ≠ χ which agrees with χ on 7-regular elements, so on 5-singular elements. Both χ and χ̃ are non-exceptional characters of B. Thus B contains two algebraically conjugate exceptional irreducible characters μ_{1} and μ_{2}, which have each degree 22.

Let t and u be any involutions of G with χ(1) = a and χ(u) = b. Then χ̃(t) = a and μ_{1}(t) = μ_{2}(t) = 2a. Similarly, χ̃(u) = b and μ_{1}(u) = μ_{2}(u) = 2b.

Hence

∑_{θ∈B} \frac{θ(t)θ(u)θ(y)}{θ(1)} = 0

for any 5-singular element y ∈ G. We already know that |C_{G}(s)| ∈ {10, 30}, and we consider these cases separately.

Case i): |C_{G}(s)| = 10. In this case, G has only two 5-blocks of full defect, which must be B and the principal 5-block.

Let t, u be involutions of G, as above. We wish to compute the coefficient of s in the class sum C_{t}C_{u}. By the computation above, only the principal 5block makes a non-zero contribution to the character theoretic formula for the computation of this expression. The contribution from the principal 5-block is

\frac{|G|}{|C_{G}(t)|.|C_{G}(u)|} \frac{(99 − ψ(t))(99 − ψ(u))}{99.98},

where ψ is the non-trivial, non-exceptional character in the principal 5-block. Since G is not simple, this expression is non-zero. Hence t and u must be G-conjugate as s has odd order.

Case ii) |C_{G}(s)| = 30. In this case, C_{G}(s) ≅ S × U where U ≅ S_{3}. Also, each involution of G is conjugate to an involution which inverts s, so that N_{G}(S) ≅ L×U, where L ≅ D_{10}. Now N_{G}(S) has three 5-blocks of full defect, and hence so has G (by Brauer’s First Main Theorem). The three 5-blocks of G correspond to the three irreducible characters of U, all of which are N_{G}(S)-stable when extended to irreducible characters of C_{G}(s).

Let x be an element of order 3 in U and z be an involution. Let B̃ be the 5block of G corresponding to the irreducible character of degree 2 of U. Then for each irreducible character μ of B̃ we have μ(s) = −2μ(sx) and μ(sz) = 0. Then B̃ ≠ B, the 5-block of G containing χ since by 2.4.2, we have 1 = χ(s) = χ(sx) = −χ(sz), so B is the block corresponding to the non-trivial linear character of U.

Let h_{1} be the coefficient of s in the product of class sums C_{t}C_{u} and h_{2} be the coefficient of sx in the product. Then the block B̃ makes zero contribution to the character theoretic expression for h_{1} + 2h_{2}. As above, the block B makes no contribution to this expression either.

The contribution from the principal 5-block is

\frac{3|G|}{|C_{G}(t)|.|C_{G}(u)|} \frac{(99 − ψ(t))(99 − ψ(u))}{99.98},

where ψ is the non-trivial, non-exceptional character in the principal 5-block. Since G is not simple, this expression is non-zero. Hence t and u must be G-conjugate as s and sx both have odd order.

Lemma 4.1.10: 2^{9} does not divide |G|.

Proof: Let t be an involution of G, and let us calculate the number of elements, h say, with 5-part s in the product of class sums C_{t}C_{t}.

A now familiar computation shows that this is is

\frac{|U||G|}{|C_{G}(t)|^{2}} \frac{(99 − ψ(t))^{2}}{99.98} ,

where ψ is the non-trivial, non-exceptional character in the principal 5-block. Hence we have

[G : C_{G}(t)] = \frac{h.|C_{G}(t)|.99.98}{|U|(99 − ψ(t))^{2}} .

We now claim that ψ(t) ∈ {−1, 19}. We may suppose that t inverts s, and we set L = ⟨t⟩S. Let λ be the non-trivial linear character of L. Then we have:

0 ≤ 10⟨Res_{L}^{G}(ψ), λ = 99 − 4 − 5ψ(t),

so that ψ(t) ≤ 19. A similar calculation using the trivial character of L yields ψ(t) ≥ −19.

Now t is conjugate to an involution u which centralizes s. Now we have ψ(s) = ψ(su) = −1, so that ψ(u) ≡ −1 (mod 5). Since G = G′, unimodularity yields ψ(u) ≡ ψ(1) (mod 4), so that ψ(u) = ψ(t) ≡ −1 (mod 20). Thus ψ(t) ∈ {−1, 19}, as claimed.

Now t is conjugate to an involution central in a Sylow 2-subgroup of G, so that [G : C_{G}(t)] is odd. Since |U| is divisible by 2, but not by 4, and since (99 − ψ(t))^{2} is not divisible by 2^{9}, we see from the equation above that 2^{9} does not divide |C_{G}(t)|, and thus 2^{9} does not divide |G|.

Section Two

In this section (assuming, as we may from now on. that [N_{G}(S) : C_{G}(S)] = 4) that C_{G}(S) ≠ S. In fact, we prove that the principal 5-block contains at most one irreducible character of degree 11, so that any such character must be rational-valued. By Lemma 1.3.9, this possibility may be excluded, since G would be isomorphic to a subgroup of A_{12} in that case.

Lemma 4.2.1: χ does not lie in the principal 5-block.

Proof: Suppose otherwise. Then as χ is not rational-valued, there is an another algebraic conjugate of χ in the principal 5-block. Under the current assumptions, the principal 5-block contains five irreducible characters, all of degree congruent to ±1 (mod 5). At least one of these has degree, x say, congruent to -1 (mod 5).

Then the possible degrees (including multiplicities) for the irreducible characters of the principal 5-block are either 1, 11, 11, x, x − 23 or 1, 11, 11, x, 23 − x. In fact, the second case can’t occur, for since x ≡ −1 (mod 5) and x(23 − x) divides |G|, we have x = 9 or x = 14. However, any irreducible character of G of degree prime to 11 lies in the principal 11-block, so has degree congruent to ±1 or ±5 (mod 11).

In the first case, the congruences for the principal 11-block yield x ≡ 0 or 1 (mod 11) since the character of degree x can’t be exceptional in the principal 11-block.

Case i): x ≡ 0 (mod 11)

a) If x = 11.2^{a} and x − 23 = 3^{b}7^{c}, then 3^{b}7^{c} ≡ −1 (mod 11), so that c is odd as 3 is a quadratic residue (mod 11) and 7 and −1 are not. Also, 3^{b}7^{c} ≡ 1 (mod 5), so b is also odd. Now 3^{b}7^{c} ≡ 5 (mod 8), so thta x ≡ 4 (mod 8) and a = 2. Hence x = 44 and x − 23 = 21.

Let ψ be the irreducible character of degree 21 in the principal 5-block. Then ψ is rational-valued. let u ∈ G be an element of order 7. Since G has Abelian Sylow 7-subgroups, [G : C_{G}(u)] ≢ 0 (mod 7). Hence ψ(u) is divisible by 7. Since G is simple, we have ψ(u) ∈ {−14, −7, 0, 7, 14}. Then \frac{[G:C_{G}(u)]ψ(u)}{ψ(1)} ≢ [G : C_{G}(u)] (mod 7) and ψ does not lie in the principal 7block. Let 1, α, β, γ be the other irreducible characters in the principal 5-block, of respective degrees 1,11,11,44. Since 1 + α + β − γ + ψ vanishes on 5-regular elements, it vanishes on 7-singular elements, as C_{G}(S) has order prime to 7, so G contains no element of order 35. Now not all of 1, α, β, γ, ψ are in the principal 7-block, and as α and β are algebraically conjugate, they are either both in the principal 7-block or both in non-principal 7-blocks. By Lemma 1.2.14, we either have 1 + α(1) + β(1) − γ(1) ≡ 0 ( mod |T |) or 1 − γ(1) ≡ 0 ( mod |T |), where T ∈ Syl_{7}(G). Clearly the second case does not occur, so we have |T | = 7, and ψ lies in a 7-block of defect 0 of G. Then γ is the unique irreducible character of degree 44 in the principal 5-block of G, so is rational-valued. Hence γ is a non-exceptional character in the principal 7-block, a contradiction since all such characters have degrees congruent to ±1 (mod 7).

b) If x = 11.3^{a} and x − 23 = 2^{b}7^{c}. As x ≡ 1 (mod 5), a is even. Now c is also even as x − 23 ≡ 1 (mod 5). Also, x ≡ 3 (mod 8) so that x − 23 ≡ 4 (mod 8) and b = 2. Then x − 23 ≢ −1 (mod 11), since 2^{b}7^{c} is a quadratic residue (mod 11). This is a contradiction, since x ≡ 0 (mod 11).

c) If x = 11.7^{a} and x − 23 = 2^{b}.3^{c}, then a is even as x ≡ 1 (mod 5). Again x − 23 ≡ 4 (mod 8) and b = 2. Then 4.3^{c} is a quadratic reside (mod 11), so 4.3^{c} ≢ −1 (mod 11).

d) If x = 11.2^{a}.3^{b} and x − 23 = 7^{c}, then c ≡ 0 (mod 4) as x − 23 ≡ 1 (mod 5). Thus 7^{c} ≢ −1 (mod 11), again a contradiction.

e) If x = 11.2^{a}.7^{b} and x − 23 = 3^{c}, then we obtain a contradiction as 3^{c} ≢ −1 (mod 11), since 3 is a quadratic residue (mod 11).

f) If x = 11.3^{a}.7^{b} and x − 23 = 2^{c}, then c is even as x − 23 ≡ 1 (mod 5). Then, however x − 23 ≢ −1 (mod 11), a contradiction.

Case ii) x ≅ 1 (mod 11).

a) If x = 2^{a} and x − 23 = 11.3^{b}.7^{c} then a is even, so x ≅ 1 (mod 3) and b = 0. Now c is even as 7^{c} ≡ 1 (mod 5). Then 7^{c} ≅ 1 (mod 16) and x ≡ 2 (mod 16). Hence a = 1, a contradiction.

b) If x = 3^{a} and x − 23 = 11.2^{b}.7^{c}, then as x ≅ −1 (mod 5), a ≡ 2 (mod 4). Thus x ≡ 1 (mod 4) and b = 1. Hence 22.7^{c} ≅ 1 (mod 5) and c is odd. Hence 22.7^{c} ≡ 10 (mod 16). However, as a ≡ 2 (mod 4), it follows that 3^{a} ≡ 9 (mod 16) and x − 23 ≡ 2 (mod 16), a contradiction.

c) If x = 7^{a} and x − 23 = 11.2^{b}.3^{c}, then x ≅ 1 (mod 3) so c = 0. Now a is even as 7^{a} ≡ −1 (mod 5), so that x ≡ 1 (mod 4) and x − 23 ≡ 2 (mod 4) and b = 1. Then x − 23 = 22, a contradiction as 5 does not divide x. d) If x = 2^{a}.3^{b} and x − 23 = 11.7^{c}, then c is even as x − 23 ≡ 1 (mod 5). Then x − 23 ≡ 3 (mod 8) and a = 1. Also 2.3^{b} ≡ −1 (mod 5) and b is odd. Now c ≠ 0 as x ≠ 34, so that 2.3^{b} ≡ 2 (mod 7). Then 3^{b} ≡ 1 (mod 7) and b ≡ 0 (mod 6), contrary to the fact that b is odd.

e) If x = 2^{a}.7^{b} and x − 23 = 11.3^{c}, then c is even, as x − 23 ≡ 1 (mod 5). Now x − 23 ≡ 3 (mod 8) so that a = 1. As 2.7^{b} ≡ −1 (mod 5), b is also odd. Now, however, 11.3^{c} ≡ 5 (mod 7), contrary to the fact that c is even and 5 is a quadratic non-residue (mod 7).

f) If x = 3^{a}.7^{b} and x − 23 = 11.2^{c}, then c is even as 11.2^{c} ≡ 1 (mod 5). Then x − 23 ≡ 2 (mod 3), so a = 0. Hence b is even as x ≡ −1 (mod 5). Thus x ≡ 1 (mod 4), so that c = 1, a contradiction.

Now all possibilities for the prime factorizations of x and x − 23 have been eliminated, so the proof of the Lemma is complete.

Corollary 4.2.2: i) C_{G}(S) ≠ S. ii) If |C_{G}(S)| = 15 and x ∈ C_{G}(S) has order 3, then for some primitive cube root of unity ω, we have χ(sx) = ω and χ(x) = 6ω + 5.

Proof: i) If C_{G}(S) = S, then G has only one 5-block of full defect, and χ must lie in the principal 5-block, contrary to Lemma 4.2.1.

ii) If there is an element x ∈ C_{G}(S) of order 3 such that C_{G}(S) = S⟨x⟩, but the values are not stated, then χ(s) = χ(sx) = χ(sx^{2}) = 1 by Lemma 2.4.2, and then χ is constant (but non-zero) on 5-singular elements, so lies in the principal 5-block by Brauer’s Second and Third Main Theorem, again contrary to Lemma 4.2.1.

Corollary 4.2.3: i) If |C_{G}(S)| = 15, then |G| is divisible by 27 and, if 81 does not divide |G|, then the Sylow 3-subgroups of G are elementary Abelian. ii) If |C_{G}(S)| = 15 and |G| is divisible by 7, then G contains an element of order 21.

Proof: Let R ∈ Syl_{3}(G) and suppose that |C_{G}(S)| = 15.

Let x ∈ C_{G}(S) have order 3. Then χ(x) = 6ω + 5 for some primitive cube root of unity. In particular, x is not conjugate to its inverse in G.

Suppose that R is Abelian of rank at most 2. Then R ≤ N_{G}(R)′ as G = G′. Now N_{G}(R) = C_{G}(R)K for some 2-group K. Also, R = [R, K] and C_{R}(K) = 1. Let T = Ω_{1}(R). Then N_{G}(R)/C_{G}(R) acts faithfully on T. Since x is not conjugate to its inverse in G, viewing N_{G}(R)/C_{G}(R) as a subgroup of GL(2, 3), (for R is not cyclic), we see that N_{G}(R)/C_{G}(R) does not contain a central involution of SL(2, 3). Then K/C_{K}(R) is cyclic of order 2, and [R, K] ≠ R, a contradiction.

Suppose that R is non-Abelian of order 27 and that x ∈ R. Then Res_{R}^{G}(χ) contains an irreducible constituent, say ψ, of degree 3, which is faithful. Now ψ(y) = 0 for all y ∈ R\Z(R). Since x has only two eigenvalues in total, we have ψ(x) ≠ 0, so that x is conjugate to an element of Z(R) and x is not conjugate in G to any element of R\Z(R). Thus x is not conjugate to any other element of R.

Hence {x} is weakly closed in any Sylow 3-subgroup containing x. Set H = C_{G}(x). Note that there is no G-conjugate of x in H other than x itself. Hence any two elements of H which are G-conjugate are already H-conjugate. For any element y ∈ R\Z(R), we see that C_{G}(y) has a Sylow 3-subgroup of order 9, and that this is ⟨x, y⟩. Hence (since no conjugate of y has a centralizer of order divisible by 27) C_{H}(y) has central Sylow 3-subgroup and a normal 3complement, and then C_{G}(y) has a normal 3-complement, since C_{H}(y) controls the fusion of 3-elements of C_{G}(y).

We claim that H/⟨x⟩ ≅ S_{6}. Note that N_{G}(S) ≤ C_{G}(x) = H. From 2.4.1 and 2.4.2, we know that the 1-eigenspace of x is 5-dimensional, and that S acts faithfully on this eigenspace. Let N be the kernel of the action of H on this eigenspace. Then x ∈ N, and we claim that N = ⟨x⟩. Now S normalizes the 5′- group N and C_{N}(s) = ⟨x⟩, so that N is nilpotent. Also, ⟨x⟩ since C_{N}(s) = ⟨x⟩ and 3^{4} does not divide |N|. Let θ be the character of H afforded by its action on the 6-dimensional ω-eigenspace of x. Let A = O_{2}(N) and V = O_{7}(N).

If V ≠ 1, then V has an S-invariant elementary Abelian subgroup E of order 7^{4}. Then Res_{SE}^{N}(θ) = α + λ_{1} + λ_{2}, where α is irreducible of degree and λ_{i} are (not necessarily different) linear characters which each contain E = [E, S] in their kernel. Then E contains an element of order 7 with an eigenspace of codimension 2, contrary to Lemma 2.1.6. A similar argument applies with the prime 2 (for C_{G}(S) has odd order under current assumptions, so every minimal S-invariant 2-subgroup of N is Abelian). Hence N = ⟨x⟩, and the action on H/⟨x⟩ on the 1-eigenspace of x is faithful.

Let ψ be the character of H afforded by its action on the 1-eigenspace of x. Let H̄ = H/⟨x⟩. We know that ψ(s) = 0, so ⟨Res_{S}^{H}(ψ), 1⟩ = 1. Since O_{5′}(H)) = ⟨x⟩, we see that Z(H̄) = 1. From the lists of low-dimensional linear groups, we can now conclude that ψ is irreducible of degree 5. Since 9 (but no higher power) divides |H̄| we must have E = E(H̄) ≅ A_{6}. Since the given degree 5 irreducible character of E extends to H̄ (and since N_{G}(S) ≤ H and [N_{G}(S) : C_{G}(S)] = 4), we conclude that H̄ ≅ S_{6}, as only one outer automorphism of A_{6} stabilizes one of its irreducible characters of degree 5).

Let y be an element of R\⟨x⟩. We have seen that C_{H}(y) has a normal 3complement. Now if z ∈ H is 3-regular with z̄ ∈ CH̄(ȳ). Then [y, z] ∈ ⟨x⟩ so that [y, z, z] = 1 (so z normalizes W = ⟨x, y⟩, and then W, z] = 1 as [W, z, z] = 1). Hence CH̄(ȳ) also has a normal 3-complement. However H̄ contains an element ȳ of order 3 with CH̄(ȳ) ≅ C_{3} × S_{3}, a contradiction. ii) Suppose that |C_{G}(S)| = 15 and that |G| is divisible by 7. For x ∈ C_{G}(S) of order 3, we know that χ(x) is not rational. Also, for T ∈ Syl_{7}(G), Res_{T}^{G}(χ) is not rational-valued by Lemma 2.5.4, so by Lemma 1.2.12, G contains an element of order 21.

We close this section with some remarks about integrality which will be useful later. Recall that s denotes a generator of S ∈ Syl_{5}(G).

Lemma 4.2.4: Let B_{1} denote the principal 5-block of G and B denote the 5-block containing the character χ. Then for any elements a, b ∈ G, both

\frac{6|G|}{|C_{G}(a)||C_{G}(b)|} ( ∑_{μ∈B_{1}} \frac{μ(a)μ(b)μ(s)}{μ(1)})

and

\frac{6|G|}{|C_{G}(a)||C_{G}(b)|} (∑_{μ∈B} \frac{μ(a)μ(b)μ(s)}{μ(1)})

are algebraic integers

Proof: In case |C_{G}(S)| = 10, we choose an involution t ∈ C_{G}(S) and we add the coefficient of s in the product of class sums C_{a}C_{b} to the coefficient of st in the product of class sums C_{a}C_{b} for the first claim. We subtract the coefficient of st in the product of class sums C_{a}C_{b} from the coefficient of st in the product of class sums C_{a}C_{b} for the second claim.

In case |C_{G}(S)| = 15, we choose an element x ∈ C_{G}(S) of order 3 and a primitive complex cube root of unity ω with χ(sx) = ω. We add the coefficients of s, sx and sx^{2} in the product of class sums C_{a}C_{b} for the first claim. We add the coefficient of sx in the product of class sums C_{a}C_{b} to ōmega times the coefficient of sx in the product of class sums C_{a}C_{b} and ω times the coefficient of sx^{2} in the product of class sums C_{a}C_{b} for the second claim.

In case |C_{G}(S)| = 30, we choose an element x ∈ C_{G}(S) of order 3 and an involution t ∈ C_{G}(S). We have seen in earlier proofs that in this case, there is a third 5-block of G, say B_{2} such that ψ(s) = −2ψ(sx) and ψ(st) = 0 for all irreducible character ψ ∈ B_{2}.

In this case, we add the coefficient of s in C_{a}C_{b} to twice the coefficient of sx in C_{a}C_{b} and then to three times the coefficient of st in C_{a}C_{b} for the first claim. We add the coefficient of s in C_{a}C_{b} to twice the coefficient of sx in C_{a}C_{b} and then subtract three times the coefficient of st in C_{a}C_{b} for the first claim.

Section Three

Until stated otherwise, we assume that |G| is not divisible by 7. In this section, we suppose that |C_{G}(S)| = 15. Recall that we have assumed already that 2^{11}||G|.

Lemma 4.3.1: |G| = 2^{16}.3^{3}.5.11. or |G| = 2^{12}.3^{11}.5.11

Proof: In this case, we have |N_{G}(P)| = 55 for P ∈ Syl_{11}(G). Now |G| has the form 2^{a}3^{b}.5.11 where 11 ≤ a ≤ 17 and 3 ≤ b ≤ 13. Also 2^{a}3^{b} ≡ 1 (mod 11) and 2^{a−2}3^{b−1} ≡ 1 (mod 5). Now 2^{2b}3^{b} ≡ 1 (mod 11), so we have a ≡ 2b (mod 10) and a is even. Now the congruence (mod 5) yields 2^{a−b−1} ≡ 1 (mod 5), so a ≡ b + 1 (mod 4) and b is odd. Then b = 1 leads to no feasible value of a and b = 3 yields a = 16. No feasible value of a arises from the cases b = 5, b = 7, b = 9 or b = 13. If b = 11, then a = 12.

Lemma 4.3.2: |G| ≠ 2^{16}.3^{3}.5.11.

Proof: Suppose otherwise. Then by Corollary 4.3.2, G has elementary Abelian Sylow 3-subgroups. Let x ∈ C_{G}(S) e an element of order 3. Let C = C_{G}(x) and K = O_{5′}(C). Now S normalizes K and has no non-trivial fixed-point on K/⟨x⟩, so that K is nilpotent. Now |O_{3}(K)| ≤ 3^{3}, so that S centralizes O_{3}(C). Hence O_{3}(C) = ⟨x⟩ as |C_{G}(S)| is not divisible by 3. Let M = O_{2}(C) and suppose that M ≠ 1.

Now as in Lemma 4.2.3, we may argue that C acts faithfully on the 5dimensional eigenspace of x and C/⟨x⟩ is irreducible in its action on that eigenspace. Since |C| is divisible by 27, but by no higher power of 3, we see that C = ⟨x⟩ × H, where E = E(H) ≅ A_{6}, since G has elementary Abelian Sylow 3-subgroups. Now let Q be a Sylow 3-subgroup of C. Each element of Q\⟨x⟩ has three different eigenvalues in its action on the 1-eigenspace of x. Since x only has two eigenvalues in the given representation of G, x is not conjugate to any element of Q\⟨x⟩. Since x is not conjugate to its inverse on G, we deduce that {x} is weakly closed in Q with respect to G and that C controls fusion in Q with respect to G. Hence Q ∩ G′ = Q ∩ H′ < Q, contrary to the fact that G = G′.

Lemma 4.3.3: |G| ≠ 2^{12}.3^{11}.5.11.

Proof: Let x ∈ C_{G}(S) be an element of order 3 and let C = C_{G}(x). Let R ∈ Syl_{3}(C_{G}(x)), and suppose (without loss of generality) that R consists of monomial matrices. Then the diagonal subgroup D of R has index at most 27. Furthermore, since x has only two eigenvalues, we see that xinD and D ≤ C. Hence 3^{8}||C|.

Now K = O_{5′}(C) is nilpotent as in earlier arguments, and C acts faithfully on the 1-eigenspace of x. Let V = O_{3}(C). Since |C_{V} (S)| = 3, we see that |V | has the form 3^{4h+1} for some integer h. Also V ≠ ⟨x⟩ since C has a faithfuk complex irreducible character of degree 4 or 5. By Lemma 3.1.4, V is Abelian. Now [V, S] has rank 4, for otherwise it would have rank 8, so V would have rank 9, and there would be a non-identity element of V with an eigenspace of codimension 2, contrary to Lemma 2.1.6.

Now Res_{S[S,V]}^{G}(χ) has the form λ + α + β where λ is linear and α and β are each irreducible of degree 5, β being afforded by the action on the 1-eigenspace of x. Hence Res_{S}^{G}(χ) has 11 distinct constituents, and C_{G}(V) is an Abelian 3-group by Lemma 1.5.1. Hence C_{G}(V) = V and O_{2}(C) = 1.

Since C acts irreducibly on the 1-eigenspace of x and V ≠ ⟨x⟩, we see that the 1-eigenspace of x is an primitive as a C-module. Hence [C : V] divides 5!. If |V | = 3^{5}, then a Sylow 3-subgroup of C has order at most 3^{6}, whereas we know that |C| is divisible by 3^{8}. Hence [V, S] is homocyclic of order 3^{8} and 3^{10} is the exact power of 3 dividing |C|.

Now let W be the weak closure of ⟨x⟩ in R with respect to G. Then W ≤ D and W is Abelian. Let u and v be G-conjugates of x other than x which lie in C. Suppose that [u, v] ≠ 1. Then ⟨u, v⟩ is not a 3-group, since otherwise it may be assumed to be monomial and u and v must both be diagonal, since u and v each have only two eigenvalues. For the same reason, the 1-eigenspace of x is a direct sum of 1- and 2- dimensional irreducible ℂ⟨u, v⟩-modules, and there must be at least one which is 2-dimensional, otherwise [u, v] ∈ ⟨x⟩. Then ⟨u, v⟩ has a subgroup isomorphic to SL(2, 3) which embeds isomorphically as a section of C/V, a contradiction as C/V is isomorphic to a subgroup of S_{5}, which has a dihedral Sylow 2-subgroup.

Hence the G-conjugates of x which lie in C generate an elementary Abelian normal subgroup of C. If this is ⟨x⟩, then x commutes with none of its other conjugates and is central in a Sylow 3-subgroup of G, contrary to the fact that 3^{11} divides |G| but 3^{11} does not divide |C|. Otherwise, since it is Sinvariant, it must be Ω_{1}(V).

Let M = N_{G}(Ω_{1}(V)). Then C ≤ M. If M leaves the 1-eigenspace of x invariant, then it also leaves the ω-eigenspace of x-invariant, so [M, x] = 1 and M = C. However, let Q be a Sylow 3-subgroup of G containing R. Then Ω_{1}(V)⊲N_{Q}(R), so N_{Q}(R) ≤ M and R = Q, a contradiction. Since S normalizes Ω_{1}(V), we see that either M has an orbit of length 10 on linear constituents of Res_{Ω_{1}(V)}^{G}(χ), or else Res_{M}^{G}(χ) is irreducible.

In the latter case, |M| is divisible by 11, but M has no normal 11-complement, since |M| is divisible by 11 and by 5, while no 5-local subgroup of G is irreducible by Lemma 2.6.1. Hence M contains a conjugate of N_{G}(P), which has order 55. Now [G : M] divides 2^{10} and is congruent to 1 (mod 55) since M contains a Sylow 5-normalizer of G and a Sylow 11-normalizer of G. Thus is a contradiction since M ≠ G.

In the former case, there is a linear character λ of M with λ(x) = ω ≠ 1. However, since G = G′ and Ω_{1}(V) is a weakly closed Abelian subgroup of the Sylow 3-subgroup Q of G, we have Q ∩ G′ = Q ∩ M′, a contradiction.

Corollary 4.3.4: If |C_{G}(S)| = 15, then 7 divides |G|.

Corollary 4.3.5: 3^{12} does not divide |G| if 7 does not divide |G|.

Proof: When 7 does not divide |G|, we know that |C_{G}(S)| is even. Hence the 5-block B which contains χ is invariant under algebraic conjugation by Brauer’s Second Main Theorem and the block orthogonality relations which follow from it. For whenever τ is a Galois automorphism, χ and χ^{τ} agree on the 5-section of s in G.

This 5-block contains only 5 irreducible characters. If a Sylow 3-subgroup R of G has exponent 3^{h}, then a Galois automorphim σ which all fixes roots of unity of 3′-order must permute the irreducible characters in orbits of length 1 and 2, or in orbits of length 1 and 3, or else in one orbit of length 2 and one of length 3. Hence either Res_{R}^{G}(χ) is real-valued, or else Res_{R}^{G}(χ + χ̄) is rational-valued, and the result follows by Lemma 2.5.2.

Section Four

In this section, we continue to assume that |G| is not divisible by 7, so that |C_{G}(S)| is even. We also suppose that G has no rational-valued irreducible character of degree 11.

Lemma 4.4.1: The principal 5-block of G contains an irreducible character of degree 16 or 66.

Proof: We know that χ is not rational-valued. If χ ≠ χ̄, set χ^{∗} = χ and if χ = χ̄ choose χ^{∗} be an algebraic conjugate of χ different from χ.

Now χχ^{∗} assumes the constant value 1 on 5-singular elements, and since χ^{∗} ≠ χ̄, ⟨χχ^{∗}, 1_{G}⟩ = 0. Hence χχ^{∗} has a non-trivial irreducible constituent θ in the principal 5-block of G, and with χ(1) ≡ 1 (mod 5). Also θ(1) is congunet to 0, ±1 or ±5 (mod 11). The only possibilities are 1,6,11, 16 or 66. As G is simple and θ is non-trivial θ(1) ≠ 1. Since G ≅ P̸SL(2, 11), we have θ(1) ≠ 6 by Lemma 1.2.3. If θ(1) = 11, then θ is rational-valued by Lemma 4.2.1. Then G has a subgroup of index 11 or 12 by Lemma 1.3.9. Hence θ(1) = 16 or 66.

Lemma 4.4.2: The possible sets of irreducible characters in the principal 5block are:

i) 1, 16, 16, 66, 99. ii) 1, 66, 66, 891, 1024. iii) 1, 66, 99, 1024, 1056. iv) 1, 66, 99, 144, 176.

v) 1, 66, 176, 7776, 8019.

Proof: Let θ be the character given by the previous Lemma. Suppose first that θ(1) = 16. Then θ is an exceptional character in the principal 11-block, and the other exceptional character from that block is also in the principal 5block. At least one irreducible character in the principal 5-block has degree, x say, congruent to −1 (mod 5). Then for some sign ε, the irreducible character degrees for the principal 5-block are 1, 16, 16, x, ε(x − 33). Also x ≡ 0 or ±1 (mod 11). If ε = −1, then x ≤ 33 and both x and 33 − x divide |G|. There are no possibilities for x, given that x = 11 has been excluded. Hence ε = 1.

Now x ≡ −1 (mod 5). Either x or x − 33 is odd. If x is odd, then the power of 3 dividing x is even. Since x ≠ 11, 9 divides x, so that 3 divides x − 33, but 9 does not. Thus the power of 2 dividing x − 33 is odd as x − 33 ≡ 1 (mod 5). If 4 divides x − 33, then x ≡ 1 (mod 4), so that 11 does not divide x (otherwise x = 11.3^{2m} for some integer m and x ≡ 3 (mod 4)). Hence x is the degree of a non-exceptional character in the principal 11-block, so x ≡ 1 (mod 11) as x is a power of 3 and 3 is a quadratic residue (mod 11). Also, x − 33 ≡ 1 (mod 11), and x − 33 = 3.2^{r} for some odd integer r, a contradiction as 2 is a quadratic non-residue (mod 11) and 3 is a residue. Thus x − 33 is not divisible by 4, so x = 99 and x − 33 = 66 in this case.

If x − 33 is odd, then 3 divides x − 33 to an even power since x − 33 ≡ ±1 (mod 5), and x − 33 ≠ 11, then 3 divides x but 9 does not. Now 4 divides x and x − 33 ≡ 3 (mod 4). Thus 11 divides x − 33, since otherwise x − 33 = 3^{2d} for some integer d, which is congruent to 1 (mod 4). Now x − 33 = 11.3^{2d} and x ≡ 4 (mod 8). Then x = 132, a contradiction, as this should be congruent to ±1 (mod 5).

If θ(1) = 66, suppose that the degrees of the irreducible characters in the principal 5-block are 1, 66, x, y and z. None of x, y or z can be 11. For example, if x = 11, then y and z are both prime to 11, so the irreducible character of degree 11 is rational valued.

We suppose, as we may, that x ≡ −1 (mod 5). We claim that we may suppose also that y ≡ 1 (mod 5). For if x, y and z are all congruent to −1 (mod 5), then x + y + z = 67. There is no choice of x, y or z which will satisfy the required congruences.

Now we have 1 + 66 − x + y + εz = 0 for some sign ε with z ≡ ε (mod 5).

Suppose first that the principal 5-block contains an exceptional character from the principal 11-block. Then it contains two such characters, of equal degree. Hence either x = z or y = z (for x ≠ z). Suppose first that x = z, and that these are degrees of exceptional characters.

Now 1 + 66 − 2x + y = 0, so that y is odd. If x is divisible by 3, then y = 11, a contradiction. Hence y is odd and x is prime to 3. Now x ≡ −5(1 + y) (mod 11), so that (1 + y) ≡ δ (mod 11) for some sign δ. Hence y is divisible by 11, so x is an odd power of 2, since −5 is a quadratic non-residue (mod 11). However, this contradicts the fact that x ≡ ±1 (mod 5).

If y = z, and these are degrees of exceptional characters, then we have 1 + 66 + 2y − x = 0, so that y ≡ 5 (mod 11) and x ≡ 0 (mod 11). Now x is odd, so x = 11.3^{s} for some integer s ≡ 2 (mod 4), since x ≡ −1 (mod 5). Now y = 2^{a} for some integer a and we have a ≡ 4 (mod 10) since y ≡ 16 (mod 11). Also, 2^{a} ≡ 1 (mod 5) so that a ≡ 0 (mod 4). Thus a ≡ 4 (mod 20), so that a = 4 as a ≤ 17. In this case, the degrees of the irreducible characters in the principal 5-block are 1, 16, 16, 66, 99.

Hence we may suppose that all irreducible characters in the principal 5-block of G of degree prime to 11 are non-exceptional characters in the principal 11block. We claim that an odd number of irreducible characters in the principal 5-block have degree divisible by 11. For let s be an element of order 5 in G and t be an element of order 11. Let the irreducible characters in the principal 5-block be 1, μ_{2}, μ_{3}, μ_{4}, μ_{5}, where μ_{2} has degree 66. Let ε_{i} = μ_{i}(s) for i = 3, 4, 5.

Then 1 + μ_{2} + ε_{3}μ_{3} + ε_{4}μ_{4} + ε_{5}μ_{5} vanishes on 5-regular elements of G, so on elements of order 11. Hence we have 1 + ε_{3}μ_{3}(t) + ε_{4}μ_{4}(t) + ε_{5}μ_{5}(t) = 0. Since μ_{i}(t) ∈ {−1, 0, 1} for each i, we deduce that (possibly after relabelling), we either have μ_{3}(t) = −ε_{3} and μ_{4}(t) = μ_{5}(t) = 0, or else μ_{3}(t) = ε_{3} and μ_{i}(t) = −ε_{i} for i = 4, 5.

Suppose that x is a power of 3. In this case, x = 3^{b}, where b ≠ 0. As x ≡ −1 (mod 5), b ≡ 2 (mod 4). Also, x ≡ 1 (mod 11) as 3 is a quadratic residue (mod 11), but -1 is not. Hence b ≡ 0 (mod 5) so b = 10, as 3^{20} does not divide |G|.

Now we have 1 + 66 − x + y + εz = 0, so that y + εz ≡ 14 (mod 81). Also, y + εz ≡ 0 (mod 11) as x ≡ 1 (mod 11). Suppose that yz ≢ 0 (mod 11). Then either y or z is a power of 2.

If y = 2^{c}, then c ≡ 0 (mod 4), as y ≡ 1 (mod 5). Hence 2^{c} ≡ ±1 (mod 11), so c ≡ 0 (mod 5). Thus c ≡ 0 (mod 20), a contradiction as c ≠ 0 by the simplicity of G, yet 2^{20} does not divide |G|.

If z is a power of 2, we must have z ≡ −1 (mod 5), or the same argument applies. In this case, we then obtain z = 2^{10}. Now y = 3^{10} + 2^{10} − 67. Now y ≡ 1 (mod 11), y ≡ 3 (mod 9) and y ≡ 2 (mod 4), so that y does not divide |G|, a contradiction.

Thus yz ≡ 0 (mod 11) and y +εz ≡ 0 (mod 11), so both y and z are divisible by 11. Now y and z are not both divisible by 3. If y = 11.2^{r}, then r ≡ 0 (mod 4), as y ≡ 1 (mod 5). Thus y ≡ 2 (mod 3), so z ≡ 0 (mod 3). Also εz ≡ 6 (mod 8) as r ≠ 0. Now z = 22.3^{k} for some odd integer k, as z ≡ ±1 (mod 5). Thus z ≡ 2 (mod 8), so ε = −1 and k ≡ 3 (mod 4). Now 11.2^{r} ≡ 14 (mod 27), so 2^{r} ≡ 16 (mod 27) and r − 4 ≡ 0 (mod 18).Thus r = 4 as 2^{22} does not divide |G|. Now y = 176 and we have 1 + 66 + 176 = 3^{10} + z, a contradiction.

Thus z = 11.2^{r} and a similar argument applies unless z ≡ −1 (mod 5), so suppose that z ≡ −1 (mod 5). Then r is even, so 11.2^{r} ≡ 2 (mod 3) and y ≡ 1 (mod 3). But then y = 11.2^{s} for some integer s, a case we have discounted.

This disposes of the case that x is a power of 3. A similar argument discounts the possibility that z is a power of 3 when z ≡ −1 (mod 5).

Suppose that y = 3^{b} for some integer b. Then b ≡ 0 (mod 4) as y ≡ 1 (mod 5). Also, y ≡ 1 (mod 11) as y is a quadratic residue (mod 11). Hence b ≡ 0 (mod 5), so that b ≡ 0 (mod 20), a contradiction as y ≠ 1 and 3^{20} does not divide |G|. Hence y is not a power of 3, and a similar argument shows that z is not a power of 3 when z ≡ 1 (mod 5). Hence none of x, y, z is a power of 3.

A similar argument shows that 2^{20} divides y if y is a power of 2, a contradiction. Likewise, a contradiction arises if z is a power of 2 and z ≡ 1 (mod 5).

Suppose that x = 2^{a} for some integer a. Then a ≡ 2 (mod 4) as x ≡ −1 (mod 5). Thus 2^{a} ≡ 1 (mod 11). Hence a ≡ 0 (mod 10), so that x = 2^{10} as 2^{20} does not divide |G|. Thus we now have 1+66−1024+y +εz = 0. Now 3 divides both y and z (otherwise neither is divisible by 3, but one of them is odd, so one of them must be 11, which has been excluded). However, y and z aren’t both divisible by 9. Hence one of x, y has the form 33.2^{r} for some odd integer r and the other has the form 11.3^{s} for some even integer s. If r = 1, then the degrees are 1, 66, 66, 891, 1024. Suppose then that r ≥ 3.

Now y + εz ≡ −3 (mod 8) and 11.3^{s} ≡ 3 (mod 8) so ε = −1 and z = 11.3^{s}. Hence s ≡ 2 (mod 4). Now we have 1 + 66 − 1024 + 33.2^{r} − 11.3^{s} = 0. Thus 2^{r} − 3^{s−1} = 29. If s ≥ 3, we have 2^{r} ≡ 2 (mod 27) as s ≥ 6 in that case. Hence r ≡ 1 (mod 18). But r > 1 and 2^{19} does not divide |G|, a contradiction. Hence s = 2, and the character degrees are 1, 66, 99, 1024, 1056.

A similar argument applies of z is a power of 2 and z ≡ −1 (mod 5). Thus we now assume that none of x, y, z is a power of 2. Hence one of x, y, z has the form 11.2^{a}, another has the form 11.3^{b} and another has the form 2^{c}.3^{d}. Furthermore, abcd ≠ 0, and a and b are both even.

Since 1 + 66 − x + y + εz = 0, it follows that there is a sign δ such that 2^{c}.3^{d} ≡ δ (mod 5) and 2^{c}.3^{d} ≡ −δ (mod 11).

If δ = −1, we may suppose that x = 2^{c}.3^{d}. As x ≡ −1 (mod 5), we have c − d ≡ 2 (mod 4). Also, c is even since 2 is a quadratic non-residue (mod 11) and 3 is a residue. Now d is also even, and since 67 − x + (y + εz) = 0 we have y + εz ≡ 1 (mod 4) and y + εz ≡ 5 (mod 9). Since a and b above are both even and non-zero, 11.2^{a} ≡ 0 (mod 4) and 11.3^{b} ≡ 3 (mod 4). Thus ε = −1 and z = 11.3^{b}. Hence b ≡ 2 (mod 4). Also, 11.2^{a} ≡ 5 (mod 9) and a ≡ 4 (mod 6). Thus a ≡ 4 (mod 12), since 11.2^{a} ≡ 1 (mod 4). As a ≤ 17, we have a = 4 or a = 16.

If a = 4, then 1 + 66 + 176 − 2^{c}3^{d} − 11.3^{b} = 0. Hence 11.3^{b} < 243, so b = 2 and the character degrees are 1, 66, 99, 144, 176.

If a = 16, then 11.2^{16} ≡ −4 (mod 27). Thus b = 2 or d = 2. If d = 2, then c ≡ 0 (mod 4) (since c − d ≡ 2 (mod 4) ), and c ≡ 4 (mod 10) (as 2^{c}.3^{d} ≡ 1 (mod 11) ). Hence c = 4, as c ≤ 17. Now the character degrees are 1, 66, 144, 11.2^{16}, 11.3^{b}, where 67 − 144 + 11.2^{16} = 11.3^{b}. Thus 2^{16} ≡ 7 (mod 3^{b}). Now 2^{8} ≡ 13 (mod 243), so 2^{16} − 7 ≡ 162 (mod 243) and b = 4, a contradiction.

Suppose then that δ = 1, and that y = 2^{c}.3^{d}. Then 2^{c}.3^{d} ≡ 1 (mod 5) and 2^{c}.3^{d} ≡ −1 (mod 11). Hence c is odd as 3 is a quadratic residue (mod 11) and 2 is not. Also c ≡ d (mod 4) from the congruence (mod 5). In particular, c and d are both odd. If d = 1, then 3.2^{c} ≡ −1 (mod 11),so c ≡ 7 (mod 10). Thus c = 17, since c ≤ 17. However, let θ be the irreducible character of degree 3.2^{17} in the principal 5-block. Then θ lies in a 2-block of defect zero, and vanishes on all 2-singular elements of G.

However, as |C_{G}(S)| is even and θ lies in the principal 5-block, we have θ(st) = θ(s) = −1 for a non-identity element s of order 5 in G and involution t ∈ C_{G}(S), a contradiction. Hence d ≠ 1.

If c = 1, then d ≡ 3 (mod 5), as 2.3^{d} ≡ −1 (mod 11). Also, d ≡ 1 (mod 4). Thus d ≡ 13 (mod 20), a contradiction, as 3^{13} does not divide |G|. Hence c ≠ 1.

Now c ≥ 3 and d ≥ 3. We have 67 − 2^{c}.3^{d} − (x − εz) = 0. Thus x − εz ≡ 13 (mod 27) and x − εz ≡ 3 (mod 8). Now one of x, z has the form 11.2^{a} and the other has the form 11.3^{b} (where a and b are even positive integers). From the congruences above, we must have x = 11.3^{b} and ε = 1, z = 11.2^{a}. Now z ≡ 5 (mod 9) and so a ≡ 4 (mod 6). As z ≡ 1 (mod 4), we also have a ≅ 0 (mod 4). Thus a = 4 or a = 16.

Now 1 + 66 + 11.2^{16} − 11.3^{b} + 2^{c}.3^{d} = 0, so as d ≥ 3, we have 11.3^{b} ≡ 9 (mod 27), a contradiction. Hence a ≠ 16, so a = 4.

Now we have 1 + 66 + 176 + 2^{c}.3^{d} − 11.3^{b} = 0, so that 11.3^{b} − 2^{c}3^{d} = 243. Since b ≡ 2 (mod 4), we have 11.3^{b} ≡ 3 (mod 16). Hence c ≥ 5 as c is odd.Thus 11.3^{b} ≡ 19 (mod 32), so that b ≡ 6 (mod 8). As b ≤ 13, we have b = 6. Hence the character degrees are 1, 66, 176, 7776, 8019 and the proof of the Lemma is complete.

Corollary 4.4.3: If |C_{G}(S)| = 30, then 27 divides |G|.

Proof: Each of the sets of character degrees in Lemma 4.4,2 include an irreducible character, say ψ, of degree divisible by 9. If C_{G}(S) has order 30, then C_{G}(S) contains an element u of order 3 with ψ(su) = ψ(s) = ±1, where s is a generator of S. Thus ψ does not lie in a 3-block of defect 0.

Lemma 4.4.4: The irreducible character degrees 1, 16, 16, 66, 99 do not occur for the principal 5-block.

Proof: Suppose that these degrees occur. Let χ_{2} be one of the irreducible characters of degree 16 and let t be an involution central in some Sylow 2subgroup R of order 2^{a} of G. Then gcd([G : C_{G}(t)], χ_{2}(1)) = 1, and by the usual argument (originating with Burnside), we have χ_{2}(t) = 0. Let χ_{3} be the other irreducible character of degree 16, and χ_{4} and χ_{5} be the remaining irreducible characters of degree 66 and 99.

For some integer a we have χ_{4}(t) = a and χ_{5}(t) = 1 + a, using the block orthogonality relations and the fact that χ_{4}(s) = 1 and χ_{5}(s) = −1 for a generator s of S.

Let u be an involution of C_{G}(S). Such an involution exists by the assumptions at the beginning of this section. Then χ_{2}(su) = χ_{2}(s) = 1, so that χ_{2}(u) ≡ 1 (mod 5). Also, since G = G′, we have χ_{2}(u) ≡ 0 (mod 4). Thus χ_{2}(t) ≡ 16 (mod 20). Since G is simple and χ_{2}(1) = 16, we have χ_{2}(u) = −4. In particular, u and t are not conjugate.

As the principal 11-block contains irreducible characters of degree 16, we have |N_{G}(P)| = 55 for P a Sylow 11-subgroup of G. Then [G : N_{G}(P)] = 2^{a}3^{b} for some integers a and b, so that a is even, as 3 is a quadratic residue (mod 11) but 2 is not.

Now we claim that χ_{2} and χ_{3} do not lie in the principal 2-block, while χ_{4} and χ_{5} do. For the involution t above we have

\frac{[G : C_{G}(t)]χ_{2}(t)}{χ_{2}(1)} ≢ [G : C_{G}(t)]

(mod 2), so χ_{2} does not lie in the principal 2-block (and neither does its algebraic conjugate χ_{3}). We note, incidentally, that χ_{2} and χ_{3} agree on 11-regular elements, so on 2-singular elements. Since neither of them vanishes on all 2singular elements, they are in the same 2-block.

Let z ∈ R be any element which is not conjugate to any element of C_{G}(S). By Lemma 1.2.14, χ_{2}(s)χ_{2}(z) + χ_{3}(s)χ_{3}(z) = 0. Hence χ_{2}(z) = χ_{3}(z) = 0 since χ_{2}(s) = χ_{3}(s) = 1.

Now χ_{2}(1) = 16, χ_{2}(y) = −4 if y is an element of R^{#} which is G-conjugate to an element of C_{G}(S) and χ_{2}(z) = 0 for all other elements of R. Suppose that there are h(≠ 0) elements of R which are G-conjugate to an element of C_{G}(S). Then

0 ≤ |R|⟨Res_{R}^{G}(χ_{2}), 1_{R}⟩ = 16 − 4h.

Since 16 divides |R|, we also have h ≡ 0 (mod 4). Hence h = 4. But then

|R|⟨Res_{R}^{G}(χ_{2}), Res_{R}^{G}(χ_{2})⟩ = 256 + 64,

so that |R| = 64, contrary to our standing assumption that 2^{11} divides |G|.

Lemma 4.4.5: The irreducible character degrees in the principal 5 block are not 1, 66, 99, 1024, 1056.

Proof: Suppose otherwise. Let R be a Sylow 2-subgroup of G and let t ∈ Z(R) be an involution. Let 1, chi_{2}, χ_{3}, χ_{4}, χ_{5} be the irreducible characters of respective degrees 1, 1024, 1056, 66, 99.

By an argument similar to that of the previous Lemma, χ_{2}(t) = 0. Hence there are integers a and b such that chi_{3}(t) = a, χ_{4}(t) = b and χ_{5}(t) = 1 + a + b. If t inverts no 5-singular element of G, then we obtain

(1 + \frac{a^{2}}{1056} + \frac{b^{2}}{66} − \frac{(1 + a + b)^{2}}{99} ) = 0.

Multiplication by 99 shows that 3 divides (1 + a + b)^{2}. Set (1 + a + b)^{2} = 9u. Then we see that 1056 + a^{2} + 16b^{2} = 96u. Hence 3 divides a^{2} + b^{2}. Thus both a and b are divisible by 3, contrary to the fact that 3 divides 1 + a + b. Hence t inverts some 5-singular element of G, so must invert an element of order 5, say s, which we may suppose lies in S.

Now 10⟨Res_{⟨t,S⟩}^{G}(χ_{4}, 1⟩ = 66+4+5χ_{4}(t), so that χ_{4}(t) ≥ −14. Similarly, using the multiplicity of the non-trivial linear character of ⟨S, t⟩, we have χ_{4}(t) ≤ 14. Likewise, we obtain −17 ≤ χ_{5}(t) ≤ 19, also making use of the fact that χ_{5}(t) ≡ 3 (mod 4) as G = G′. Now [G : C_{G}(t)] is odd, so that 32 divides χ_{3}(t).

We also have χ_{3}(t) = χ_{5}(t) − (1 + χ_{4}(t)). There are three possible configurations:

i) χ_{3}(t) = 0, χ_{5}(t) = 1 + χ_{4}(t). ii) χ_{3}(t) = 32, χ_{5}(t) = 19 and χ_{4}(t) = −14. iii) χ_{3}(t) = −32, χ_{5}(t) = −17 and χ_{4}(t) = 14.

By Lemma 4.2.4, we have:

z = \frac{6|G|}{|C_{G}(t)|^{2}} (1 + \frac{a^{2}}{1056} + \frac{b^{2}}{66} − \frac{(1 + a + b)^{2}}{99} )

is an integer. Let u = \frac{z|C_{G}(t)|^{2}}{6|G|} . As [G : C_{G}(t)] is odd and 2^{11}||G|, we see that

u = \frac{2^{10}r}{s} for integers r and s with s odd. We also recall that a is divisible by 32 and that b ≡ 2 (mod 4), say b = 4c + 2.

Working in the 2-adic integers, we have :

(1 + \frac{2(2c + 1)^{2}}{33} − \frac{(4c + 3)^{2}}{99} ) ≡ 0(mod32),

which may be rewritten as −8c^{2} ≡ 16c^{2} (mod 32). Thus c is even and b ≡ 2 (mod 8). This eliminates case iii) above.

Suppose that case ii) occurs. Then we obtain

u = (1 + \frac{32}{33} + \frac{98}{33} − \frac{361}{99} = \frac{128}{99}) ,

contrary to the fact that u is divisible by 1024 as a 2-adic integer. Hence case occurs, and a = 0. Set b = 8d + 2. Then u = 1 + \frac{2(4d+1)^{2}}{33} − \frac{(8d+3)^{2}}{99} , which simplifies to 99u = 96 + 32d^{2}. Since u is divisible by 1024 as a 2-adic integer, we have 3 + d^{2} ≡ 0 (mod 32), a contradiction.

Lemma 4.4.6: The possible orders of G are:

i) 2^{11}.3^{5}.5.11. ii) 2^{12}.3^{6}.5.11. iii) 2^{13}.3^{6}.5.11. iv) 2^{13}.3^{11}.5.11.

v) 2^{17}.3^{3}.5.11. vi) 2^{14}.3^{7}.5.11.

Proof: Let P be a Sylow 11-subgroup of G and S be a Sylow 5-subgroup of G. We have [N_{G}(S) : C_{G}(S)] = 4, and we know that |C_{G}(S)| = 10 or 30.

If |N_{G}(S)| = 40 and |N_{G}(P)| = 55, then |G| = 2^{a}.3^{b}.5.11. where 11 ≤ a ≤ 17 and 2 ≤ b ≤ 11, and we have 2^{a}3^{b} ≡ 1 (mod 11) and 2^{a−3}.3^{b} ≡ 1 (mod 5). Then a ≡ 2b (mod 10) and a ≡ b + 3 (mod 4), so a is even and b is odd. If a = 12, then b ≡ 6 (mod 20), so b = 13, a contradiction. If a = 14, then b ≡ 7 (mod 20), so a = 7. If a = 16, then b ≡ 13 (mod 20), a contradiction.

If |N_{G}(P)| = 55 and |N_{G}(S)| = 120, then |G| = 2^{a}.3^{b}.5.11. where 11 ≤ a ≤ 17 and 3 ≤ b ≤ 11, and we have 2^{a}3^{b} ≡ 1 (mod 11) and 2^{a−3}.3^{b−1} ≡ 1 (mod 5). Then a ≡ 2b (mod 10) and a ≡ b+2 (mod 4), so a and b are both even. If a = 12, then b ≡ 6 (mod 20), so b = 6 and |G| = 2^{12}.3^{6}.5.11. If a = 14, then b ≡ 12 (mod 20), a contradiction.If a = 16, then b ≡ 18 (mod 20), a contradiction.

If |N_{G}(P)| = 110 and |N_{G}(S)| = 40, then |G| = 2^{a}.3^{b}.5.11. where 11 ≤ a ≤ 17 and 3 ≤ b ≤ 11, and we have 2^{a−1}3^{b} ≡ 1 (mod 11) and 2^{a−3}.3^{b} ≡ 1 (mod 5). Then a ≡ 2b + 1 (mod 10) and a ≡ b + 3 (mod 4), so a is odd and b is even. If a = 11, then b ≡ 0 (mod 20), a contradiction. If a = 13, then b ≡ 6 (mod 20), so b = 6 and |G| = 2^{13}.3^{6}.5.11. If a = 15, then b ≡ 12 (mod 20), a contradiction. If a = 17, then b ≡ 18 (mod 20), a contradiction.

If |N_{G}(P)| = 110 and |N_{G}(S)| = 120, then |G| = 2^{a}3^{b}.5.11. where 11 ≤ a ≤ 17 and 3 ≤ b ≤ 11, and we have 2^{a−1}3^{b} ≡ 1 (mod 11) and 2^{a−3}.3^{b−1} ≡ 1 (mod 5). Then a ≡ 2b + 1 (mod 10) and a ≡ b + 2 (mod 4), so a and b are odd. If a = 11, then b ≡ 5 (mod 20) and |G| = 2^{11}.3^{5}.5.11. If a = 13, then b ≡ 11 (mod 20), so b = 11 and |G| = 2^{13}.3^{11}.5.11. If a = 15, then b ≡ 17 (mod 20), a contradiction. If a = 17, then b ≡ 3 (mod 20), so |G| = 2^{17}.3^{3}.5.11.

Lemma 4.4.7: |G| ≠ 2^{17}.3^{3}.5.11.

Proof: Suppose otherwise. Then |C_{G}(S)| = 30 by Lemma 4.4.6. Let x ∈ C_{G}(S) be an element of order 3. Let C = C_{G}(x). Then [S, O_{3}(C)] = 1, so that, by Lemma 2.4.1, ⟨x⟩ = O_{3}(C). Since x is inverted by all involutions of C_{G}(S), we see that S has no non-trivial fixed points on O_{5′}(C)/⟨x⟩, so that O_{5′}(C) is nilpotent and O_{5′}(C) = ⟨x⟩ × O_{2}(C). By Lemma 2.4.1, the eigenvalues of x are 1 (with multiplicity 1), ω and ω^{−1} (with multiplicity five each), where ω is a complex primitive cube root of unity. Hence x ∉ C′, since C acts on each eigenspace of x.

Suppose that C has a component L. Then as L = L′, the Sylow subgroups of L are Abelian and |Z(L)| is not divisible by 3. Now |L| is divisible by 5, but |Z(L)| is not.Thus S centralizes ⟨x⟩O_{2}(C), so that O_{2}(C) = 1 as C_{G}(S) contains no element of order 6 by Lemma 2.4.1. Hence L is simple, so that L ≅ A_{5} or A_{6}, since |G| is not divisible by 81. Suppose that L ≅ A_{5}. Since C = LN_{C}(S), we see that |C| is not divisible by 27,so that G has non-Abelian Sylow 3-subgroups. Let R be a Sylow 3-subgroup of G which contains x. Then whenever ψ is a nonlinear irreducible character of R, we have ψ(x) = 0. Since x has the eigenvalue 1 with multiplicity one, so that Res_{R}^{G}(χ) has exactly one non-linear irreducible constituent, and this has degree 3 and occurs with multiplicity one.

Let u ∈ Z(R) have order 3. Then without loss of generality, χ(u) = 8 + 3ω, possibly after replacing u by its inverse. Let α be the character of C afforded by its action on the ω-eigenspace of x. We note that L acts faithfully on the ω-eigenspace of x, using Lemma 2.1.6. Let y be an element of order 3 in L. Then ⟨x, u⟩ ∈ Syl_{3}(C), so we may suppose that u ∈ ⟨x, y⟩. However, y has three different eigenvalues on the ω-eigenspace of x, and so does every element of ⟨x, y⟩\⟨x⟩.

Since x and x^{−1} each have an eigenvalue other than 1 with multiplicity 5, we see that u ∉ ⟨x, y⟩, a contradiction. Hence L ≅ A_{6}.

We note then that the Sylow 3-subgroups of C/⟨x⟩ are TI sets, and that C now contains a Sylow 3-subgroup of G, say R, which is Abelian. Now using Lemma 2.1.6, we see that Res_{L}^{G}(χ) = 1+ψ_{1} +ψ_{2}, where the ψ_{i} are irreducible of degree 5, but need not be different. We claim that ⟨x⟩ is weakly closed in R with respect to G. Since χ(x) = −4, it suffices to prove that χ(y) ≠ −4 whenever y ∈ R\⟨x⟩.

We note that R = ⟨x⟩ × (R ∩ L). For any v ∈ R ∩ L, we have ψ_{i}(v) ∈ {−1, 2} for each i. If ψ_{1}(v) = ψ_{2}(v), then χ(v) ∈ {5, −1}. Otherwise χ(v) = 2. We see also that χ(xv) = χ(x^{−1}v) ∈ {−1, 2} in the former case, and that neither χ(xv) nor χ(x^{−1}v) is real in the latter case. Hence ⟨x⟩ is weakly closed in R with respect to G, as claimed. Then N_{G}(R) ≤ N_{G}(⟨x⟩). Set M = N_{G}(⟨x⟩). Also, ⟨x⟩ = O_{3}(N_{G}(S), so that N_{G}(S) ≤ M. Hence [G : M] ≡ 1 (mod 15). Now [G : M] has the form 11.2^{s} for some integer s. However, this is a contradiction since 2^{s+1} ≡ 1 (mod 3) yields that s is odd, while 2^{s} ≡ 1 (mod 5) yields that s is even. This eliminates the case L ≅ A_{6}, so that C has no component.

Now 9 divides |C| since 27 divides |G|. Hence N_{G}(S) ≰ C, since |N_{G}(S)| = 120. Hence O_{5}(C) = 1, so that F (C) = ⟨x⟩ × O_{2}(C) and K = O_{2}(C) ≠ 1. Since C_{G}(S) contains no element of order 6, S acts without non-trivial fixed points on K. Hence K is Abelian, since SK acts on the eigenspaces of x. Now K must be Abelian of rank 4, otherwise it has rank at least 8, and then contains an involution contradicting Lemma 2.1.6. For the same reason, K acts faithfully on each 5-dimensional eigenspace of x. Hence C acts faithfully (and irreducibly but imprimitively) on each 5-dimensional eigenspace of x. Hence C/F (C) is isomorphic to a subgroup of S_{5}, so that |C| is not divisible by 27.

Thus G has a non-Abelian Sylow 3-subgroup R (containing x) of order 27. As before, we may choose u ∈ Z(R) so that χ(u) = 8+3ω. Then x centralizes every Abelian 3′-group which it normalizes by the Hall-Higman-Shult theorems (more precise that Lemma 1.2.10). However, u ∈ C, since x ∈ R. Hence u centralizes O_{2}(C) and hence centralizes F^{∗}(C). Then u ∈ Z(F (C)), a contradiction, since ⟨x⟩ = O_{3}(C) and χ(x) = χ(x^{−1}) = −1.

Lemma 4.4.8: If the irreducible character degrees for the principal 5-block of G are 1, 66, 176, 7776, 8019, then |G| = 2^{13}.3^{11}.5.11.

Proof: Suppose otherwise. Since 8019 = 11 × 729, the possibilities for |G| are then 2^{12}.3^{6}.5.11, 2^{13}.3^{6}.5.11 or 2^{13}.3^{11}.5.11 or 2^{14}.3^{7}.5.11

If |G| = 2^{12}.3^{6}.5.11, then |C_{G}(S)| = 30 from the proof of Lemma 4.4.6. Let ψ be the irreducible character in the principal 5-block of degree 8019. Then ψ lies in a 3-block of defect zero, so vanishes on all 3-singular elements of G. However there is a generator s of S and an element x ∈ C_{G}(s) of order 3 such that ψ(sx) = ψ(s) = −1, a contradiction.

If |G| = 2^{13}.3^{6}.5.11, then |C_{G}(S)| = 10 from the proof of Lemma 4.4.6. Let ψ be the irreducible character of degree 7776 = 32×243 in the principal 5-block.

Then ψ is in a 3-block of defect 1, and is the only irreducible character from its 3-block which is also in the principal 5-block.

By Lemma 1.2.14, since G contains no element of order 15, we have ψ(1)ψ(s) ≡ 0 (mod 3^{6}), for any generator s ∈ S, a contradiction as ψ(s) = 1.

If |G| = 2^{14}.3^{7}.5.11, then we have |C_{G}(S)| = 10 by Lemma 4.4.6. Hence G contains no element of order 15. The above argument, this time using the character of degree 8019 = 11 × 729) yields a contradiction.

Lemma 4.4.9: If |G| is divisible by 2^{12}, then the degrees of the irreducible characters in the principal 5-block are not: 1, 66, 66, 891, 1024.

Proof: Suppose otherwise. Let ψ be the irreducible character of degree 1024 in the principal 5-block and let t be an involution which is central in a Sylow 2subgroup. By a now familiar argument, we have ψ(t) = 0 since [G : C_{G}(t)] and ψ(1) are coprime. We claim that t inverts an element of order 5. For otherwise, t inverts no 5-singular element, and there are integers a and b such that

1 + \frac{a^{2}}{66} + \frac{b^{2}}{66} + \frac{(1 + a + b)^{2}}{891} = 0,

a contradiction, as the expression is clearly positive.

As in earlier arguments, we have −14 ≤ a, b ≤ 14. By Lemma 4.2.4,

\frac{6|G|}{|C_{G}(t)|^{2}} (1 + \frac{a^{2}}{66} + \frac{b^{2}}{66} + \frac{(1 + a + b)^{2}}{891} )

is an integer, m say. Let r = \frac{m|C_{G}(t)|^{2}}{6|G|} . Since 2^{12} divides |G| and [G : C_{G}(t)] is odd, we see that (viewed as a 2-adic integer), r is divisible by 2^{11}. Since G = G′, we have a ≡ b ≡ 2 (mod 4). Let a = 2u and b = 2v for odd integers u and v. Then

r = (1 + \frac{2u^{2}}{33} + \frac{2v^{2}}{33} + \frac{(1 + 2u + 2v)^{2}}{891} ) .

Taking congruences (mod 128) and multiplying by -5 yields: 54(u^{2} + v^{2}) + (1 + 2u + 2v)^{2} ≡ 5 (mod 128). Hence 2[(u + v)^{2} + (u + v)] + 27(u^{2} + v^{2}) ≡ 2 (mod 64).

Then 29(u^{2} + v^{2}) + 2(u + v + 2uv) ≡ 2 (mod 64) and (u + v + 2uv) ≡ 0 (mod 4). Hence u + v ≡ 2 (mod 4), so u ≡ v (mod 4) as u and v are odd.

Now set u = 2z + 1 and v = 2w + 1. Then −4 ≤ z, w ≤ 3 and z − w is even. Now

29(4z^{2} + 4z + 1 + 4w^{2} + 4w + 1) + 2(2z + 2w + 2 + 4w + 4z + 8zw + 2) ≡ 2

(mod 64) so that

−3(4z^{2} + 4z + 4w^{2} + 4w) + 12z + 12w + 16zw ≡ 0

(mod 64). Hence 4zw − 3z^{2} − 3w^{2} ≡ 0 (mod 16). This forces z and w to each be divisible by 4.

If z = w = −4, then u = v = −7 and a = b = −14. Then

(1 + \frac{a^{2}}{66} + \frac{b^{2}}{66} + \frac{(1 + a + b)^{2}}{891} ) = 1 + \frac{196}{33} + \frac{27}{33} = \frac{256}{33},

contrary to the fact that r is divisible by 2^{11}.

If z = w = 0, then u = v = 1 and a = b = 2. Then

(1 + \frac{a^{2}}{66} + \frac{b^{2}}{66} + \frac{(1 + a + b)^{2}}{891} ) = 1 + \frac{4}{33} + \frac{25}{891} = \frac{1024}{891},

contrary to the fact that r is divisible by 2^{11}.

If z = 0 and w = −4, then u = 1, v = −7 and a = 2, b = −14.

(1 + \frac{a^{2}}{66} + \frac{b^{2}}{66} + \frac{(1 + a + b)^{2}}{891} ) = 1 + \frac{100}{33} + \frac{121}{891} = \frac{128 × 29}{891},

contrary to the fact that r is divisible by 2^{11}. The same argument applies if z = −4 and w = 0.

Corollary 4.4.10: |G| ≠ 2^{13}.3^{6}.5.11.

Proof: Suppose otherwise. Then the degrees of the irreducible characters in the principal 5-block are 1, 66, 99, 144, 176. Also, from the proof of Lemma 4.4.6, we know that |N_{G}(P)| = 110 and |N_{G}(S)| = 40 for a Sylow 11-subgroup P and Sylow 5-subgroup S of G (without loss of generality, we may assume that S ≤ N_{G}(P)). Let s be a generator of S.

Let t be an involution which inverts P and centralizes S. We count the number n of elements with 5-part s in the product of class sums C_{t}C_{t}. Now G has only two 5-blocks of positive defect by Brauer’s First Main Theorem. By Brauer’s second main theorem, whenever ψ is an irreducible character in the non-principal 5-block of positive defect, we have ψ(s) + ψ(st) = 0. Hence only irreducible characters from the principal 5-block can make a non-zero contribution to the character-theoretic formula for this integer.

Let 1, χ_{2}, χ_{3}, χ_{4}, χ_{5} be the irreducible characters of respective degrees 1, 66, 99, 144, 176. Let χ_{2}(t) = a,χ_{3}(t) = b,χ_{4}(t) = c and χ_{5}(t) = d. Then b + c = 1 + a + d.

Now a ≡ 2 (mod 4) as G = G′. Also, a ≡ 1 (mod 5) as χ_{2}(st) = 1. Hence a ≡ 6 (mod 20). However, ⟨Res_{⟨t⟩P}^{G} (χ_{2}), 1⟩ = 66 + 11χ_{2}(t) ≥ 0 so that a ≥ −6. Similarly, a ≤ 6, using the non-trivial linear character of P ⟨t⟩. Hence a = 6.

Similarly, we find that b ≡ 3 (mod 4) and b ≡ −1 (mod 5), so that b ≡ −1 (mod 20). Also, we find that |b| ≤ 9, so that b = −1. Likewise, c ≡ 4 (mod 20) and |c| ≤ 14, so that c = 4. Hence d = −4.

Now we have

n = \frac{2|G|}{|C_{G}(t)|^{2}} (1 + \frac{36}{66} − \frac{1}{99} − \frac{16}{144} + \frac{16}{176}) .

Hence n = \frac{100|G|}{33|C_{G}(t)|^{2}} .

In particular, as 2^{13} divides |G| but 2^{14} does not, the exact power of 2 dividing n is an odd power.

Now it t_{1}, t_{2} are two conjugates of t whose product is s or st, then t_{1} and t_{2} both lie in N_{G}(S). Hence n may be calculated within N_{G}(S).

We may write N_{G}(S) = ⟨t⟩ × F, where F is a Frobenius group of order 20. Let u be an involution in F. Then u and t are not conjugate within N_{G}(S) (though they might be in G). Any involution of N_{G}(S) which inverts s must be conjugate within N_{G}(S) to u or to tu. Hence t is G-conjugate to either u or to tu.

If u and tu are not G-conjugate in G, but t is G-conjugate to u, then st is not expressible as a product of two conjugates of t. For if t_{1}t_{2} = st, for conjugates t_{1} and t_{2} of t, then t_{1} and t_{2} both centralize t, and t_{1} and tt_{2} are conjugate via an element of S, since they are involutions whose product is in the group S, which has odd order. Hence n = 5 in this case. Likewise, in this case, n = 5 if t is conjugate in G to tu.

If u and tu are conjugate in G, we see that N_{G}(S) contains 10 conjugates of u, five of which are conjugate within N_{G}(S) to u, the other five being conjugate within N_{G}(S) to tu. Hence there are 10 expressions of s as a product of two conjugates of t. Likewise, there are ten expressions of st as a product of two conjugates of t, so that n = 20 in this case.

In either case, the exact power of 2 dividing n is an even power, a contradiction.

Lemma 4.4.11: |G| ≠ 2^{11}.3^{5}.5.11.

Proof: Suppose otherwise. Then |N_{G}(P)| = 110 and |C_{G}(S)| = 30 from the proof of Lemma 4.4.6.

Let x ∈ C_{G}(S) be an element of order 3 and let C = C_{G}(x). As in earlier proofs, we know that O_{5′}(C) is nilpotent, since it acted on by S with fixedpoint subgroup ⟨x⟩. As before, we know also that C acts on the 5-dimensional eigenspaces of x, and that x does not act with determinant 1 on these eigenspaces, so that x ∉ C′.

Suppose that O_{3}(C) ≠ ⟨x⟩. Then O_{3}(C)/⟨x⟩ is elementary Abelian of order 81. Since x ∉ C′, we see that O_{3}(C) is Abelian. Since O_{3}(C) is a direct product of homocyclic S-invariant subgroups, we see that O_{3}(C) is elementary Abelian of order 3^{5}. Hence C is solvable, since C/O_{3}(C) is a {2, 5}-group.

Since C_{G}(y) is a {2, 3}-group for each element y of order 3 in G which is not conjugate to x or x^{−1}, we conclude that C_{G}(u) is solvable for each u ∈ G of order 3. Let R be a Sylow 3-subgroup of G containing x.

Since R is elementary Abelian of order 3^{5}, the solvable signalizer functor theorem (Glauberman, [14]), implies that ⟨O_{3′}(C_{G}(u)) : u ∈ R^{#}⟩ is a solvable 3′-group, T say. If T ≠ 1 and set M = N_{G}(T). If T = 1, set M = N_{G}(R). Then standard signalizer functor analysis shows if T ≠ 1 (since R has rank 5), that M ∩ M^{g} has order prime to r for all g ∈ G\M. If T = 1, then R ⊲ C_{G}(u) for all u ∈ R^{#}, so that M ∩ M^{g} is an r′-group for all g ∈ G\M in this case too. Then [G : M] ≡ 1 (mod 3^{5}), a contradiction, since [G : M] divides 11264. Hence O_{3}(C) = ⟨x⟩.

Next, we prove that O_{2}(C) = 1. Suppose otherwise. As in the proof of Lemma 4.4.7, we conclude that |C| is not divisible by 27. Let R be a Sylow 3subgroup of G with x ∈ R. Then |C_{R}(x)| = 9. Since R has at least 9 irreducible characters, we conclude that θ(x) = 0 for any non-linear irreducible character of R. Since x only has the trivial character with multiplicity 1, we deduce that Res_{R}^{G}(χ) has just one non-linear irreducible constituent, including multiplicities.

We claim that [R, x] is cyclic. Since Res_{R}^{G}(χ) has eight linear constituents, each non-identity element of R has the eigenvalue 1 with multiplicity 8 or more. If [R, x] were not cyclic, it would contain an element with a fixed-point space of dimension 9, contrary to Lemma 2.1.6. Thus [R, x] is cyclic. Each nonidentity element y of [R, x] has at most 4 distinct eigenvalues, one of which has multiplicity 8 or more, and acts with at most one Jordan block of size 4 or more in any reduction (mod 3) of G. Hence y^{3} acts either trivially, or as a transvection, in any reduction (mod 3) of G. By Lemma 2.1.7, [R, x] has order 3. Thus [R : C_{R}(x)] ≤ 3, so that |R| ≤ 27, a contradiction. Hence O_{2}(C) = 1.

Now S acts with no nontrivial fixed-point on O_{5′}(C)/⟨x⟩, so that O_{5′}(C) is nilpotent, and hence O_{5′}(C) = ⟨x⟩. Now O_{5}(C) = 1, since |C| contains ⟨x, Z(R)⟩, so has order divisible by 9, while |N_{G}(S)| = 120. Hence C has a single component, L say, which is simple, as x ∉ C′.

Now L is isomorphic to one of A_{5}, A_{6} or PSU(4, 2). Now [C : L⟨x⟩] divides 4 (with equality unless L ≅ PSU(4, 2)), since C/F^{∗}(C) is isomorphic to a subgroup of Out(L). If L ≅ A_{5}, then 27 does not divide |C| and the argument above yields a contradiction. If L ≅ A_{6}, then |C| is not divisible by 81, then arguing as in the proof of Lemma 4.3.2, we see that ⟨x⟩ is weakly closed in W ∈ Syl_{3}(C) with respect to G, since there is no element w ∈ W \⟨x⟩ with χ(w) = −1 = χ(x). Let R ∈ Syl_{3}(G) with W ≤ R. Then N_{R}(W) ≤ C, so that R ≤ C, a contradiction as |R| = 3^{5}.

Now L ≅ PSU(4, 2). Then |L| = 2^{6}.3^{4}.5. Let M = N_{G}(⟨x⟩). Then M = LN_{G}(S) and |M| = 2^{7}.3^{5}.5 as |N_{L}(S)| = 20. Consequently, [G : M] = 176. Let ψ = Ind_{M}^{G}(1). Let t be an involution of N_{G}(P)∩C_{G}(S). where we have assumed with no loss of generality that [S, P] = P. Now ψ(t) ≥ 0 and furthermore ψ(t) ≡ 0 (mod 4) as G = G′. Now ψ(s) = ψ(st) = 1 since N_{G}(S) ≤ M. Hence ψ(t) ≡ 1 (mod 5), so that ψ(t) ≡ 16 (mod 20). Let λ be the non-trivial linear character of P ⟨t⟩. Then ⟨Res_{P ⟨t⟩}^{G}(ψ), λ⟩ = 176 − 11ψ(t) ≥ 0, so that ψ(t) ≤ 16. Hence ψ(t) = 16. Since |N_{G}(P)| = 110, we have ψ = ψ_{1} + ψ_{2}, where ψ_{1} and ψ_{2} are characters and ψ_{1} is irreducible with ψ_{1}(1) ≡ −1 (mod 11).

Since ψ_{1}(1) divides |G|, it must be the case that ψ_{1}(1) ∈ {10, 32, 54, 120}.

If ψ_{1}(1) = 10, then ψ_{1}(t) = 0 since t inverts P. This is not possible since G = G′, so ψ_{1}(t) ≡ ψ_{1}(1) (mod 4).

If ψ_{1}(1) = 54, then ψ_{1}(t) ≡ 2 (mod 4), while taking the inner product of Res_{P ⟨t⟩}^{G} with the two linear characters of P ⟨t⟩, we have (54 − 10) ± 11ψ_{1}(t) ≥ 0, so that −4 ≤ ψ(t) ≤ 4. Hence ψ_{1}(t) = ±2. A similar argument shows that −12 ≤ ψ_{2}(t) ≤ 12, contrary to the fact that ψ_{1}(t) + ψ_{2}(t) = 16.

If ψ_{1}(1) = 32, then taking the inner product with the linear characters of P ⟨t⟩, we have −2 ≤ ψ_{1}(t) ≤ 2, while we also need ψ_{1}(t) ≡ 0 (mod 4), so we have ψ_{1}(t) = 0. But we also see that −14 ≤ ψ_{2}(t) ≤ 14 using the linear characters of P ⟨t⟩, a contradiction.

If ψ_{1}(1) = 120, then ψ_{1}(t) ≡ 0 (mod 4), while also ψ_{1} vanishes on all 5singular elements of G and ψ_{1}(st) = 0, so that ψ_{1}(t) ≡ 0 (mod 5). Thus ψ_{1}(t) ≡ 0 (mod 20). Since ψ_{1} takes the constant value −1 on P^{#}, we find (again using the linear characters of P ⟨t⟩) that −10 ≤ ψ_{1}(t) ≤ 10, so that ψ_{1}(t) = 0. However, we see also that −6 ≤ ψ_{2}(t) ≤ 6, contrary to the fact that ψ_{1}(t) + ψ_{2}(t) = 16.

Lemma 4.4.12: |G| ≠ 2^{13}.3^{11}.5.11.

Proof: Suppose otherwise. Then by Lemma 4.4.6, we have |N_{G}(P)| = 110 and |C_{G}(S)| = 30. Let x ∈ O_{3}(C_{G}(S)) be an element of order 3 and set C = C_{G}(x). Then, as before, O_{5′}(C) is nilpotent. Set U = O_{5′}(C).

Now we claim that C acts faithfully on each 5-dimensional eigenspace of x. Since S ∈ Syl_{5}(C), and S acts faithfully on each of these eigenspaces, the kernel of the action on either eigenspace is contained in U. Suppose that there is a non-trivial kernel K in the action on one of the eigenspaces. Then K acts faithfully on the other 5-dimensional eigenspace, using Lemma 2.1.6. Also, [K, S] ≠ 1. If [K, S] has order divisibly by 3, then it contains an elementary Abelian subgroup of order 3^{4}, and contains an element of order 3 with a fixedpoint space of codimension at most 2. If [K, S] has even order, then it contains an involution with a fixed-point space of codimension at most 2. In either case, Lemma 2.1.6 is contradicted. Hence C does indeed act faithfully on each 5dimensional eigenspace of x. If T = O_{3}(U) > ⟨x⟩, then Res_{T}^{G}(χ) has 11 distinct constituents, and C_{G}(T) is an Abelian 3-group, so C_{G}(T) = C_{C}(T) = T. Also, [T, S] is Abelian of rank 4.

In that case, C has an Abelian normal (3)-subgroup A such that C/A is isomorphic to a subgroup of S_{5}. Let M = N_{G}(⟨x⟩). Then N_{G}(S) ≤ M as ⟨x⟩ = O_{3}(C_{G}(S)). There is an involution u ∈ C_{G}(S), and u normalizes [T, S].

However, S normalizes C_{[T,S]}(u), so that u inverts [T, S] since [T, S] admits S with an indecomposable action. In particular, Res_{[T,S]}^{G}(χ) is real-valued. Now [T, S] is homocyclic of rank 4. If [T, S] has exponent 9, then it contains an element y whose only eigenvalues are 1, η and η̄, where η is a primitive 9-the root of unity. Then y^{3} acts trivially in any reduction (mod 3) of G, a contradiction. Hence [T, S] is elementary Abelian of rank 4, and T is elementary Abelian of rank 5, since C_{T} (S) = ⟨x⟩. Hence the Sylow 3-subgroups of C have order at most 3^{6} in this case.

Suppose then that O_{3}(C) = ⟨x⟩, but that O_{2}(C) ≠ 1. Since C_{G}(S) contains no element of order 6, it follows that SO_{2}(C) is a Frobenius group with Abelian kernel, again of rank 4. Now C_{C}(O_{2}(C)) = O_{2}(C)⟨x⟩ since O_{3}(C) = ⟨x⟩. Now C/O_{5′}(C) is isomorphic to a subgroup of S_{5}, so that |C| divisible by 9, but not by 27, in this case.

Suppose then that F (C) = ⟨x⟩. Then C has a component L, which is unique by Burnside’s p^{a}q^{b}-theorem, as C is a {2, 3, 5}-group with a Sylow 5-subgroup of order 5. Also, L is simple, as x ∉ C′. Since L has a fathul irreducible representation of degree at most 5, we have L ≅ A_{5}, A_{6} or PSU(4, 2). Since C = LN_{C}(S), and the Sylow 3-subgroups of L has order at most 3^{4}, we see that C has a Sylow 3-subgroup of order at most 3^{5} in this case. In all cases, then, |R| is not divisible by 3^{7}.

Now let R ∈ Syl_{3}(G) with x ∈ R. Then R has a faithful (reducible, using Lemma 2.1.6) character of degree 9, and has an Abelian normal A subgroup of index at most 27, which consists of the diagonal matrices when R is expressed as a group of monomial matrices. Hence |A| ≥ 3^{8}. We claim that [A, x] = 1, which is a contradiction. For since x has the eigenvalue 1 with multiplicity 1, we see that the underlying permutation matrix for x has only one 3-cycle, so that [A, x] fixes (elementwise) a 9-dimensional space. Hence [A, x] is trivial by Lemma 2.1.6.

Remark 4.4.13: We have now reached the situation that |G| = 2^{12}.3^{6}.5.11 or 2^{14}.3^{7}.5.11. In the first case, by Lemma 4.4.6, we have |N_{G}(P)| = 55 and |N_{G}(S)| = 120 for P ∈ Syl_{11}(G) and S ∈ Syl_{5}(G). In the second case, we have |N_{G}(P)| = 55 and |N_{G}(S)| = 80.

Section Five

The principal 11-block contains 5 non-exceptional irreducible characters and 2 exceptional irreducible characters. The irreducible characters in the principal 5 block have degrees 1, 66, 99, 144 and 176. Hence the irreducible character of degree 144 in the principal 5-block is a non-exceptional character in the principal 11-block.

Hence there is an irreducible character ψ which is an irreducible constituent of χχ̄ and with ψ(1) ≡ −1 or 5 (mod 11). In the second case, ψ̄ is also an irreducible constituent of χχ̄. The possible values of ψ(1) are 10, 60, 120, 16, 27, 32, 54.

The case ψ(1) = 10 may be quickly eliminated, for we may choose an involution t ∈ C_{G}(S). Then if ψ(1) = 10, we have ψ(st) = 0 so that ψ(t) ≡ 0 (mod 5). Since G = G′, we have ψ(t) ≡ 2 (mod 4), so that ψ(t) ≡ 10 (mod 20), so ψ(t) = ±10, contrary to the simplicity of G. This argument shows that G has no irreducible character at all of degree 10.

Lemma 4.5.1: ψ(1) ≠ 16.

Proof: Suppose that ψ(1) = 16. Certainly ψ lies outside the principal 5-block. Since |C_{G}(S)| = 30, there are three 5-blocks of G. These are the principal block, the 5-block containing χ, and another 5-block B corresponding to the block of C_{G}(S) containing an irreducible character of degree 2. Each irreducible character in B has degree congruent to ±2 (mod 5). Hence ψ ∉ B. Thus ψ lies in the same 5-block. This 5-block is invariant under algebraic conjugation, so contains at least two irreducible characters of degree 11 as χ is not rational valued. It also contains two irreducible characters of degree 16 as ψ is exceptional. Hence the degrees of its irreducible characters are 11, 11, 16, 16 and 54.

The irreducible character of degree 54 also lies in the principal 11-block.Hence the degrees of the irreducible characters in the principal 11-block are 1, 144, 54, 16, 16, x, ε(x − 75), where ε = ±1 and x ≡ −1 (mod 11). If ε = −1, then x ≤ 75 and x(75 − x) divides |G|, a contradiction. Thus ε = 1. Now x is even, since 3 and 5 are quadratic residues (mod 11), but −1 is a non-residue. Thus x − 75 is odd, and must be divisible by 3. Since x − 75 ≡ 1 (mod 11) and x − 75 has the form 3^{a}.5^{b} with b ≤ 1, we have a ≥ 2. Hence 9 does not divide x. If x is not divisible by 5, then x has the form 3.2^{c} for some integer c and then c ≡ 7 (mod 10). Thus c = 7. But then x − 75 = 309, which does not divide |G|, a contradiction. Hence both x and x − 75 are divisible by 15.

Set x = 15.2^{d}. Then d ≡ 3 (mod 10) as x ≡ −1 (mod 11). If d = 13, then x^{2} > 2^{13}.3^{6}.5.11 = |G|, a contradiction. Hence d = 3, so that x = 120. The irreducible character degrees for the principal 11-block are 1, 45, 54, 120, 144, 16 and 16. Neither irreducible character of degree 16 is in the principal 2-block, since if z is a central involution in a Sylow 2-subgroup of G, we have α(z) = 0, where α is an irreducible character of degree 16. Let B̃ be the intersection of the principal 2-block with the principal 11-block. Let y be a generator of the Sylow 11-subgroup P. Then by Lemma 1.2.14, we have ∑_{μ∈B̃} μ(y)μ(1) ≡ 0 (mod |T |), where T is a Sylow 2-subgroup of G. Now

∑_{μ∈B̃} μ(y)μ(1) ≠ 0, while certainly |∑_{μ∈B̃} μ(y)μ(1)| ≤ 464 and 2^{13} divides |G|, a contradiction.

Lemma 4.5.2: ψ(1) ≠ 27.

Proof: Suppose that ψ(1) = 27. Then ψ does not lie in the principal 5-block of G, but lies in the 5-block B corresponding to the irreducible character of degree 2 of O_{5′}(C_{G}(S)). Now ψ is an irreducible constituent of χχ̄, and χχ̄ takes the constant value 1 on all 5-singular elements, so that the character μ = ∑ ϕ ∈ B⟨χχ̄, ϕ⟩ϕ vanishes on all 5-singular elements. Hence there is an irreducible constituent β of μ such that β(s) = −2 for all s ∈ S^{#}. Now β(1) ≡ 3 (mod 5) since ψ and ψ̄ are both constituents of μ. Hence β(1) ≤ 66. Also, β(1) ≡ 0, 1 or −1 (mod 11). The only possibility is β(1) = 33.

The degrees of the irreducible characters in B are therefore 27, 27, 33, x, ε(x − 21) where ε = ±1 and x ≡ −2 (mod 5). In fact, the case ε = −1 doesn’t occur, since it yields x < 21 and the given congruences are not satisfied for x in that range. Thus ε = 1. We note that 11 divides x(x − 21), for otherwise x ≡ 1 (mod 11) and x − 21 ≡ 2 (mod 11).

Since ∑_{ϕ∈B} ϕ(s)ϕ(1) = 0 and ϕ(s) = ±2 for all irreducible characters ϕ ∈ B, we see that x and x − 21 are both divisible by 3 and the irreducible character degrees are 27, 27, 33, 3h, 3(h − 7), where h is an integer congruent to 1 (mod 5).

Suppose that 11 divides h. If h is even, then h − 7 is a power of 3. Now 3(h − 7) ≡ 1 (mod 11) as 3 is a quadratic residue (mod 11) but −1 is not. Thus 3(h − 7) = 3^{5}, so h = 88 and x = 264 ≡ −1 (mod 5), a contradiction. Hence h is odd. If h is divisible by 3, then 3(h − 7) must have the form 3.2^{a} for some integer a. Now 3.2^{a} ≡ 1 (mod 11), so that a ≡ 2 (mod 10). Also, a ≡ 2 (mod 4) as x − 21 ≡ 2 (mod 5). Hence a = 2 and x = 33, a contradiction, since we assumed that h was divisible by 3. Hence x = 33 and x − 21 = 12 in this case.

Suppose that 11 does not divide h. Then if x − 21 is not even, for otherwise, x is a power of 3, while x ≡ −1 (mod 11), which is a quadratic non-residue (mod 11), while 3 is a quadratic residue. Thus x − 21 is odd. If 3 divides h − 7, then h = 2^{s} for some integer s. Since h ≡ 1 (mod 5), s is even, contrary to the fact that 3.2^{s} ≡ −1 (mod 11), which is a quadratic non-residue. Hence x − 21 = 33 in this case, so that h − 7 = 11, contrary to the fact that h ≡ 1 (mod 5).

Hence the irreducible character degrees for the block B are thus 27, 27, 33, 33 and 12.

Hence the degrees of the irreducible characters in the principal 11-block are 1, 12, 144, y, δ(130 − y), 27, 27, where y ≡ −1 (mod 11) and δ = ±1. If δ = 1, then y ≤ 130 and the only possibilities are y = 10 or y = 120 since y and 130−y both divide |G|. But we have seen earlier that G has no irreducible characters of degree 10. Thus δ = −1.

Suppose that y is not divisible by 5. Then neither is y − 130. If, additionally, y − 130 is not divisible by 3, then y − 130 is a power of 2 (and is greater than 2), so y is not divisible by 4. Then y = 2.3^{k} for some integer k, so that 2.3^{k} ≡ −1 (mod 11) and k ≡ 3 (mod 5). Thus k = 3 as |G| is not divisible by 3^{8}, so that y = 54, contrary to y > 130. Thus 3 divides y − 130, so that 3 does not divide y, and y is a power of 2, say 2^{r}. Then 2^{r} ≡ −1 (mod 11), so that r ≡ 5 (mod 10) and r = 5, since 2^{15} does not divide |G|. This again contradicts the fact that y > 130. Hence y and y − 130 are both divisible by 5.

Now one of y or y − 130 is not divisible by 3. Hence y and y − 130 are both even, as neither y nor y − 130 is equal to 5. However, y and y − 130 are not both divisible by 4, so one of them has the form 10.3^{j} for some integer j. Since 10.3^{j} is a quadratic non-residue (mod 11) and y ≡ −1 (mod 11), we have y = 10.3^{j} for some integer j > 0. Since y ≡ −1 (mod 11), we have j ≡ 0 (mod 5), so that j = 5 as 3^{10} does not divide |G|. Then y = 2430 and y − 130 = 2300, a contradiction as neither 23 nor 25 divides |G|.

Lemma 4.5.3: ψ(1) ≠ 54.

Proof: Suppose otherwise. Then ψ does not lie in the principal 5-block, and ψ(1) ≡ −1 (mod 5), so that ψ lies in the same 5-block B′ as χ, and this block is invariant under algebraic conjugation. Since χχ̄ takes the constant value 1 on 5-singular elements, there must be another irreducible character μ in B′ which is a constituent of χχ̄ and has μ(1) ≡ 1 (mod 5). Also μ(1) ≤ 66.Then μ(1) = 11, 16 or 66.

If μ(1) = 16, then the degrees of the irreducible characters in B′ are 11, 11, 16, 16, 54, a case which has been disposed of in Lemma 4.5.1. If μ(1) = 66, then the irreducible characters in B′ have degrees 11, 11, 54, 66 and 34, a contradiction is |G| is not divisible by 17.

If μ(1) = 11, then we claim that μ = χ = χ̄. For B′ only contains two irreducible characters of degree 11 ( recall that χ is not rational-valued, so there are at least 2). Otherwise, the irreducible character degrees for B′ are 11, 11, 11, 54, 21, whereas 7 does not divide |G|.

If χ is not real-valued, then χ and χ̄ occur with equal multiplicity in χχ̄. If they are different, then χχ̄ = 1 + a(χ + χ̄) + β for some positive integer a and character β. Now β has an irreducible constituent θ with θ(1) ≤ 44. But then the only possibility is that θ(1) = 44, so that the irreducible character degrees for B′ are 11, 11, 54, 44, 76 a contradiction as 19 does not divide |G|. Thus χ = χ̄. By Lemma 2.5.1, Res_{R}^{G}(χ) is not rational-valued for a Sylow 3subgroup, R, of G. Since χ is real-valued, the number of 3-conjugates of χ is at least 3, a contradiction, as B′ only contains two irreducible characters of degree 11, and is also invariant under algebraic conjugation.

Lemma 4.5.4: ψ(1) ≠ 32.

Proof: If ψ(1) = 32, then |G| = 2^{12}.3^{6}.5.11, for if |G| = 2^{14}.3^{7}.5.11 then |C_{G}(S)| = 10 and the degree of every irreducible character of G is congruent either to ±1 or to 0 (mod 5).

Then the degrees of the irreducible characters in the principal 11-block are 1, 32, 144, x, y, z, z. At least one of x, y or z is odd, so has the form 5^{a}.3^{b}, and is a quadratic residue (mod 11), so is congruent to 1 or 5 (mod 11). If a = 0, we have b = 3 or b = 5. If a = 1, then b = 2 or b = 5.

Suppose that a = 0 and b = 5. Let θ be the irreducible character of degree 243 in the principal 11-block. Then θ lies in the same 5-block as ψ, since θ(1) ≢ ±1 (mod 5). For a generator s of S, an element u ∈ P^{#}, we have ψ(s) = 2 and ψ(u) = −1 and θ(u) = 1 and θ(s) = −2. Then ψ(u)ψ(s)+θ(u)θ(s) = −4, so that by Lemma 1.2.14, there must be an irreducible character γ which lies both in the principal 11-block and in the 5-block B′. As γ(s) = ±2 and either γ(u) = ±1 or γ(u)+γ̄ = ±1, we deduce that x, y, z, z are each degrees of irreducible characters in B′, and that none of them is divisible by 5. One or more of x, y, z must be coprime to 3, so must be a power of 2, say 2^{k}. Now 2^{k} ≡ ±2 (mod 5), so that k is odd. Thus 2^{k} ≡ −1 or −5 (mod 11). If 2^{k} ≡ −1 (mod 11), then k ≡ 5 (mod 10). Thus k = 5, so the degrees are 1, 144, 32, 32, 243, 324, 324, but Lemma 1.2.14 is not satisfied for B′ and the principal 11-block.

If a = 0 and b = 3, the degrees for the irreducible characters in the principal 11-block are 1, 144, 32, x, ε(x − 86), 27 and 27, where x ≡ −1 (mod 11) and ε = ±1. If ε = 1, then as x − 86 ≡ 1 (mod 11), 2 must divide x − 86 to an even power, since 2 is a quadratic non-residue (mod 11) and 3 and 5 are quadratic residues. As x ≡ −1 (mod 11), 2 divides x to an odd power. Thus 4 does not divide x and 4 does divide x−86. If 5 does not divide x, then x = 2.3^{r} and r ≡ 3 (mod 5), so that r = 3 and x = 54, a contradiction as ε = 1 and x ≥ 86. Thus x has the form 10.3^{s}. Then as x ≡ −1 (mod 11), we have s ≡ 0 (mod 5) but s ≠ 0 as G has no irreducible character of degree 10. Thus s = 5 and x = 2430. Then x − 86 = 2344 = 8 × 293, which does not divide |G|, a contradiction. If ε = −1, then x ≤ 86 and the only possibility for the irreducible character degrees of the principal 11-block are 1, 144, 32, 32, 54, 27, 27. The irreducible character of degree 54 lies outside the principal 5-block, and lies outside the 5-block B′, since its degree is not congruent to ±2 (mod 5). Hence this character is the unique character from its

If a = 1 and b = 5, then the degrees for the irreducible characters of the principal 11-block are 1, 144, 32, x, ε(x−1102), 1215, 1215 where x ≡ 1 (mod 11) and ε = ±1. If ε = 1, then x − 1102 ≡ −1 (mod 11), so that 2 divides x − 1102 to an odd power. Similarly 2 divides x to and even power. Thus x − 1102 ≡ 2 (mod 4). As x − 1102 ∉ {2, 10}, x − 1102 must be divisible by 3, so x is not divisible by 3. Thus x = 5.2^{k}, so that k ≡ 6 (mod 10) as x ≡ 1 (mod 11). Hence k = 6. Hence x = 320, a contradiction as x > 1102. If ε = −1, then x ≤ 1102 and 1102 − x ≡ 1 (mod 11). Thus 2 divides both x and 1102 − x to an even power on consideration of quadratic residues (mod 11). As 1102 is not divisible by 4, we conclude that x and x − 1102 are both odd. Now either x or 1102 is coprime to 3. We may suppose that x is coprime to 3. Thus forces x = 5, a contradiction, as x ≡ 1 (mod 11).

If a = 1 and b = 2, then the degrees of the irreducible characters in the principal 11 block are 1, 144, 32, 45, y, z, z. If y ≡ 1 (mod 11) then z = ε(158+y) for some sign ε for some sign ε. Thus ε = 1 as z is positive. Hence z ≡ 5 (mod 11). Thus 2 divides both y and z to an even power, by the usual argument with quadratic residues. Since z = y+158, both y and z are odd.At least one of y or z is coprime to 3. Hence y = 5 or z = 5, a contradiction, since y ≡ 1 (mod 11) and z > 158. Thus y ≡ −1 (mod 11). The degrees of the irreducible characters in the principal 11-block are now, 1, 144, 32, 45, y, varepsilon(158 − y), ε(158 − y). If ε = 1, then y ≤ 158 and both y and 158−y divide |G|, a contradiction. Hence ε = −1. Since y ≡ −1 (mod 11) and y − 158 ≡ −5 (mod 11), 2 divides both y and y − 158 to an odd power. Thus either y or y − 158 is congruent to 2 (mod 4).

If 4 does not divide y, then 5 divides y, for otherwise y = 2.3^{k}, so that k ≡ 3 (mod 10) as y ≡ −1 (mod 11). Thus k = 3 as k ≤ 6. However, y > 158, a contradiction. Thus y = 10.3^{k} in this case, and k ≡ 0 (mod 5), so k = 5 as y ≠ 10. Thus y = 2430 and y − 158 = 2272 = 32 × 71, a contradiction. Thus 4 divides y, but does not divide y − 158.

Since y − 158 ∉ {2, 10}, y − 158 must be divisible by 3, so y is not. If 5 does not divide y then y = 2^{t} and t ≡ 5 (mod 10) as y ≡ −1 (mod 11), Thus t = 5 and y = 32, a contradiction as y > 158. Hence 5 divides y. Hence y −158 = 2.3^{r} and r ≡ 1 (mod 5) as 2.3^{r} ≡ 6 (mod 11). However, y − 158 ≡ 2 (mod 5), so that r ≡ 0 (mod 4). Thus r ≡ 16 (mod 20), a contradiction.

Lemma 4.5.5: ψ(1) ≠ 120.

Proof: Suppose that ψ(1) = 120. Then the degrees of the irreducible characters in the principal 11-block are 1, 120, 144, x, y, z where z ≡ ±5 (mod 11) and y and z are congruent to ±1 (mod 11). At least one of x, y and z is odd, and at least one of them is prime to 3.

If z = 5^{a}.3^{b}, then z is a quadratic residue (mod 11), so that z ≡ 5 (mod 11). If a = 0, then b ≡ 3 (mod 5), so b = 3. the irreducible character degrees for the principal 11-block are then 1, 144, 120, x, x + 2, 27, 27 where x ≡ −1 (mod 11). Hence 2 divides x to an odd power and 2 divides x + 2 to an even power. Thus x ≡ 2 (mod 4). Now x ∉ {2, 10}, so that x is divisible by 3. Hence 3 does not divide x + 2. If 5 does not divide x + 2, then x + 2 = 2^{c} for some integer c. As x + 2 ≡ 1 (mod 11), we have c ≡ 0 (mod 10), so that c = 10. Then x = 1022 which is coprime to 15 and does not divide |G|, a contradiction. Thus x + 2 is divisible by 5 but 5 does not divide x. Now x has the from 2.3^{k}. As x ≡ −1 (mod 11), so k ≡ 3 (mod 5). Thus k = 3 so x = 54 and x + 2 = 56, a contradiction as 7 does not divide |G|. Hence a = 1. Now 5.3^{b} ≡ 5 (mod 11), so that b ≡ 0 (mod 5) and b = 5. Thus z = 1215 and the irreducible character degrees for the principal 11-block are then 1, 144, x, 129, ε(1190 − x), 1215, 1215 where x ≡ 1 (mod 11) and ε = ±1. If ε = 1, then x ≤ 1190 and 1190 − x ≡ 1 (mod 11). Since x ≡ 1 (mod 11), both x and 1190 − x are divisible by 2 to some even power. Hence x and 1190 − x are both odd. One of these last two integers is coprime to 3. But then x = 5 or 1190 − x is 5 a contradiction, as x and 1190 − x are both congruent to 1 (mod 11). Thus ε = −1, so that x > 1190 and x−1190 ≡ −1 (mod 11). Now 2 divides x−1190 to an odd power and x divides x to an even power. Thus x − 1190 ≡ 2 (mod 4). Since x − 1190 ∉ {2, 10}, x − 1190 is divisible by 3 but x is not. Thus x has the form 5^{r}.2^{s} where r ≤ 1. If r = 0, then 2^{s} ≡ 1 (mod 11), so that s ≡ 0 (mod 10), and s = 10. But then x = 1024, whereas x > 1190. If r = 1, then 5.2^{s} ≡ 1 (mod 11), so that s ≡ 6 (mod 10) and s = 6, x = 320. This again contradicts the fact that x > 1190. Hence z is even.

Hence we may suppose that x is odd. Then x = 5^{a}.3^{b} where a ≤ 1. Now x ≡ 1 (mod 11), since 3 and 5 are both quadratic residues. If a = 0 then b ≡ 0 (mod 5), so that b = 5. Then the degrees of the irreducible characters in the principal 11-block are 1, 120, 144, 243, y, z, z. Since G has irreducible characters of degree congruent to ±2 (mod 5), we have |C_{G}(S)| = 30 and |G| = 2^{12}.3^{6}.5.11 in this case. If y ≡ −1 (mod 11), then 2 divides y to an odd power and z = ε(268 − y) for some sign ε. If ε = 1, then y ≤ 268 and we have a contradiction, as y and z both divide |G|. Thus ε = −1, so that z = y − 268 and z ≡ 6 (mod 11). Hence 2 divides z to an odd power on considering quadratic residues (mod 11). Now 268 is not divisible by 8, both y and y are congruent to 2 (mod 4). Thus 5 divides y, for otherwise y = 2.3^{s}, so that s ≡ 3 (mod 4) so s = 3 and y = 54, a contradiction as y > 268. Since 5 divides y, we have z = 2.3^{t} for some integer t. Since z ≡ 6 (mod 11), we have t ≡ 1 (mod 5). Now z ≡ 2 (mod 5), so that t ≡ 0 (mod 4), so that t ≡ 16 (mod 20), a contradiction. Thus y ≡ 1 (mod 11).

Now the irreducible character degrees for the principal 11-block are 1, 120, 144, 243, y, y+ 268, y + 268. If 3 does not divide z, then z has the form 5^{a}.2^{b}. As z ≡ 5 (mod 11), b is even on consideration of quadratic residues (mod 11). If a = 1, then b ≡ 0 (mod 10) so b = 10. Also, z ≡ 2 (mod 3), so that y ≡ 1 (mod 3). But then y is also a power of 2, and y = 2^{10} as y ≡ 1 (mod 11). This contradicts the fact that y + 268 = 5 × 2^{10}. Thus a = 0. But then z = 2^{d} for some integer d. Now z ≡ 5 (mod 11), so that d ≡ 4 (mod 10). Since 2^{14} does not divide |G| in this case, we have d = 4, contrary to z > 268. Thus 3 divides z in any case. Now 3 does not divide y. Thus 5 divides y, for otherwise y = 2^{r}, where r ≡ 0 (mod 10) as y ≡ 1 (mod 11). Then y ≡ 1 (mod 3) and y + 268 ≡ 2 (mod 3), a contradiction. Now y = 5.2^{s} ≡ 1 (mod 11), so that s ≡ 6 (mod 10). Thus s = 6 and y = 320, so that z = 588 = 12 × 49, which does not divide |G|, a contradiction.

Now we may suppose that a = 1. Then b ≡ 2 (mod 5). If b = 2, the irreducible character degrees for the principal 11-block are 1, 45, 120, 144, y, z, z. If y ≡ 1 (mod 11), then z = y + 70 and z ≡ 5 (mod 11). Thus 2 divides z to an even power and 2 divides y to an even power. Thus y and z are both odd, whereas we proved above that z is even. Thus y ≡ −1 (mod 11) and 2 divides y to an odd power. Also, z = δ(70 − y) for some sign δ. If δ = 1, then the only possibility is that y = 54 and z = 16, and we obtain a set of irreducible character degrees for the principal 11-block already excluded in Lemma 4.5.1. Thus δ = −1 and z ≡ 6 (mod 11). Now at least one of y, z is coprime to 3, and y, z can’t both be divisible by 4. Now y ≠ 10 and z ≠ 10, so that whichever of y, z is not divisible by 4 must be divisible by 3,and the other one is coprime to 3.

If z is coprime to 3, then z = 5^{r}2^{q}. If r = 0, then q ≡ 9 (mod 10) as z ≡ 6 (mod 11). Then q = 9 and z = 512, y = 582. But 582 = 6 × 97, which does not divide |G|. If r = 1, then q ≡ 5 (mod 10), so that q = 5. Then z = 160 and y = 230, a contradiction as 23 does not divide |G|. Thus 3 divides z, but 3 does not divide y. If y = 2^{t} for some integer t, then t = 5, since y ≡ −1 (mod 11).

However, y > 70, a contradiction. Thus y = 5.2^{d} for some integer d. Then d ≡ 1 (mod 10) as y ≡ −1 (mod 11). As y ≠ 10, we have d = 11 Then the irreducible character degrees for the principal 11-block are 1, 45, 120, 144, 10240, 10170, 10170. However 1017 = 9 × 113, which does not divide |G|, a contradiction.

Thus b = 7. Note that |G| = 2^{14}.3^{7}.5.11 in this case, and that |C_{G}(S)| = 10. Hence all irreducible characters of G have degree congruent to ±1 or 0 (mod 5). The degrees of the irreducible characters in the principal 11-block are now 1, 120, 144, 10935, y, z, z. If y ≡ −1 (mod 11), then z = ε(10960 − y) for some sign ε. If 3 does not divide y, then 5 does divide y, otherwise y = 2^{k} and k ≡ 5 (mod 10), so y = 32, a contradiction. Now y = 5.2^{t} for some integer t. As y ≡ −1 (mod 11), t ≡ 1 (mod 10). As y ≠ 10, we have t = 11 and then the irreducible character degrees for the principal 11-block are 1, 120, 144, 10240, 10935, 720, 720.

Since G contains no element of order 22, we deduce by Lemma 1.2.14 that all irreducible characters in the principal 11-block are contained in the principal 2-block. However, let θ be the irreducible character of degree 10240. We prove that θ does not lie in the principal 2-block, yielding a contradiction. Suppose otherwise. Let t be an involution central in a Sylow 2-subgroup of G. Then 2^{11} divides θ(t) since [G : C_{G}(t)] is odd. Since θ lies in the principal 2-block, we have θ(t) ≠ 0. Hence θ(t) ≥ 2^{11}. Then |C_{G}(t)| > 2^{22}. If 5 divides |C_{G}(t)|, then we also have θ(st) = 0 for some element s of order 5 in C_{G}(t). Hence θ(t) ≡ 0 (mod 5), a contradiction, for then 10240 divides θ(t), while θ(t) ≠ 0. Hence |C_{G}(t)| = 2^{14}.3^{h} for some integer h. Now 3^{h} > 2^{8}, so that h ≥ 6. Hence [G : C_{G}(t)] ∈ {55, 165}.

Let θ = Ind_{C_{G}(t)}^{G}(1). Then θ vanishes on all 5-singular elements and on all 11-singular elements. Since ⟨θ, 1⟩ = 1, we see θ has a non-trivial irreducible constituent β in the principal 5-block with β(1) ≡ −1 (mod 5). The degrees of the irreducible characters in the principal 5-block are 1, 66, 99, 144 and 176. Hence β(1) = 99 or 144. If β(1) = 99, then θ must have another irreducible constituent of degree congruent to 5 or −1 (mod 11), since it takes the constant value −1 on 11-singular elements. But there is no non-trivial irreducible character γ of degree less than 120 in the principal 11-block, a contradiction. If β(1) = 144, then as β takes the constant value 1 on 11-singular elements (recall that G has only one 11-block of positive defect), there still must be another irreducible constituent γ of degree congruent to 5 or −1 (mod 11) of θ, and this is a contradiction. Hence the degrees of the irreducible characters in the principal 11-block are not 1, 120, 144, 10240, 10935, 720 and 720.

Suppose then that y ≡ 1 (mod 11). Then z = 10960 + y. Suppose first that y and z are both coprime to 5. Then one of them is a power of 2. If y is a power of 2, then y = 2^{10}, and 32 does not divide z, while also z ≡ 2 (mod 3), so z ≤ 80, a contradiction. If z = 2^{s}, then s ≡ 4 (mod 10) since z ≡ 5 (mod 11). As z > 10960, we have s = 14. Then z ≡ 4 (mod 9), so that y ≡ 6 (mod 9). Also y ≡ 16 (mod 32), so y = 48, a contradiction. Hence both y and z are divisible by 5. Suppose then that z is coprime to 3. Then z = 5.2^{r} for some integer r. Thus r is even as z ≡ 5 (mod 11). Hence z ≡ 2 (mod 3), so y ≡ 1 (mod 3). Now y = 5.2^{k} and k is even, since y ≡ 1 (mod 11) and 5 is a quadratic residue (mod 11), while 2 is a non-residue. Thus y ≡ 2 (mod 3), a contradiction. Hence 3 divides z, but not y.

Now y has the form 5.2^{m} where m ≡ 6 (mod 10) as y ≡ 1 (mod 11). Thus m = 6 and z = 11280, which is divisible by 47, a contradiction.

Lemma 4.5.7: ψ(1) ≠ 60.

Proof: Suppose that ψ(1) = 60. Then the irreducible character degrees for the principal 11-block are 1, 144, x, y, z, 60, 60 where (without loss of generality) x is odd. Since 3 and 5 are quadratic residues (mod 11), then x ≡ 1 (mod 11). Hence x ∈ {45, 243, 10935}.

Suppose that x = 45. Since there are signs ε and δ such that 1 + 144 + 45 + εy + δz − 60 = 0, it is no loss of generality to suppose that y is not divisible by 4. If y is odd, then so is z. Since either y or z is prime to 3, then either z = 5 or y = 5, a contradiction, as y and z are congruent to ±1 (mod 11). Thus y ≡ 2 (mod 4), so that y ≡ −1 (mod 11) as y is a quadratic non-residue (mod 11). If 5 divides y, then y = 10.3^{s} where s ≡ 0 (mod 5). As y ≠ 10, we have y = 2430, leading to irreducible character degrees for the principal 11-block being 1, 144, 45, 2430, 2300, 60 and 60, a contradiction, as 2300 does not divide |G|. Hence y is not divisible by 5. Then t = 2.3^{r}, where r ≡ 3 (mod 5) as y ≡ −1 (mod 11). Thus r = 3 and y = 54. This leads to the irreducible character degrees for the principal 11-block being 1, 144, 45, 54, 76, 60, 60 a contradiction as |G| is not divisible by 19.

Suppose that x = 243. Then |G| = 2^{12}.3^{6}.5.11, for if |C_{G}(S)| = 10 then every irreducible character of G has degree congruent to 0 or ±1 (mod 5).Now the irreducible character, β say, of degree 243 in the principal 11-block belongs to a 3-block of defect 1 of G. Let S = ⟨s⟩ be a Sylow 5-subgroup of G and let w ∈ C_{G}(s) be an element of order 3. Then β(sw) = 1, so that ⟨w⟩ is a defect group for the 3-block containing β. By Lemma 1.2.14, since G contains no element of order 33, there is another irreducible character in the principal 11-block which belongs to the same 3-block as β. Hence we may suppose that y = 243t for some integer t not divisible by 3. Now z is not divisible by 3. Now z ≡ ±1 (mod 11) and z is either of the form 2^{a} or 5.2^{a}. Since z ≠ 10, a ≥ 4. Hence 8 divides t, but 16 does not, as 1 + 144 + 243 + ε_{1}243t + ε_{2}z − 60 for some signs ε_{1} and ε_{2}. Thus t ≡ −1 (mod 11) as 243t is a quadratic non-residue (mod 11). However, t ∈ {8, 40}, since 3 does not divide t, a contradiction.

Suppose that x = 5.3^{7} = 10935. Then for some signs ε and δ, we have 1 + 144 + 10935 + εy + δz − 60 = 0. Thus we may suppose that ε = −1 and y ≡ −1 (mod 11). Hence 2 divides y to an odd power, so that y and z are both even. If 8 divides y then z ≡ 4 (mod 8), so that 3 divides z, as z ∉ {4, 20}. Hence y is not divisible by 3, and y has the form 5^{a}.2^{b}. If a = 0 then 2^{b} ≡ −1 (mod 11), so that b ≡ 5 (mod 10) and then b = 5. But then y = 32 and z ≡ 9 (mod 27). Hence z ∈ {36, 180}, a contradiction. If a = 1, then b ≡ 1 (mod 10) since y ≡ −1 (mod 11). Since y ≠ 10, we have b = 11 and y = 10240. Then the irreducible characters in the principal 11-block have degrees 1, 144, 10935, 10240, 780, 60, 60, a contradiction, as 13 does not divide |G|. Hence y ≢ 0 (mod 8), so that y ≡ 2 (mod 4). Now δz ≡ y (mod 4), so that z ≡ 2 (mod 4). Now y and z are not both divisible by 3, so that either y = 10 or z = 10, a contradiction.

We have now eliminated every possibility for ψ(1), so that we must conclude that |G| is divisible by 7.

Section Six

Throughout this section, we will suppose that T, a Sylow 7-subgroup of G has order 7. As before, P will denote a Sylow 11-subgroup of G and S a Sylow 5-subgroup of G.

Lemma 4.6.1 [N_{G}(T) : C_{G}(T)] = 3.

Proof: By Lemma 2.5.4, Res_{T}^{G}(χ) is not rational-valued. Thus [N_{G}(T) : C_{G}(T)] ≠ 6. Since G has no normal 7-complement, we need to prove that [N_{G}(T) : C_{G}(T)] ≠ 2. Suppose then that [N_{G}(T) : C_{G}(T)] = 2. Then the principal 7-block consists of the trivial character, one other non-exceptional character of degree x congruent to ε = ±1 (mod 7), and three algebraically conjugate exceptional irreducible characters, all of degree x + ε. As G has no element of order 35, we see by Lemma 1.2.14 that some non-trivial irreducible character in the principal 7-block is also in the principal 5-block.

Since [N_{G}(S) : C_{G}(S)] = 4, each irreducible character in the principal 5block has degree congruent to ±1 (mod 5).Thus the principal 5-block does not contain both an irreducible character of degree x and one of x + ε. Hence some irreducible character in the principal 7-block does not lie in the principal 5-block.

The exceptional irreducible characters of the principal 7-block agree on 7regular elements, so on 5-singular elements. Hence they all either vanish identically on 5-singular elements, or else they all lie in the same 5-block. If they vanish identically on 5-singular elements, then x ≡ −ε (mod 5).

Suppose then that the exceptional irreducible characters of the principal 7block all lie in the same 5-block, say B. If this is the principal 5-block, then the irreducible character of degree x is the only irreducible character of the principal 7-block which lies outside the principal 5-block. Since this character assumes the constant value ε on 7-singular elements, we deduce (using Lemma 1.2.14) that εx ≡ 0 (mod 5), so 5 divides x.

If B is not the principal 5-block, then a similar argument shows that x+ε ≡ 0 (mod 5), so the exceptional irreducible characters all vanish on 5-singular elements, a contradiction, as they all lie in the same 5-block.

Since the principal 11-block does not contain irreducible characters both of degree x and of degree x + ε, we also have x(x + ε) ≡ 0 (mod 11). Hence x(x + ε ≡ 0 (mod 55).

If x = 55.2^{a} and x + ε = 3^{b}, then ε = 1 as 3 is a quadratic residue (mod 11) and −1 is not. Thus b ≡ 0 (mod 5). so that 3^{5} − 1 divides x = 3^{b} − 1. But 121 divides 3^{5} − 1 and x divides |G|, a contradiction.

If x = 55.3^{a} and x + ε = 2^{b}, then the exceptional irreducible characters from the principal 7-block all have degree congruent to ε (mod 5), so b is even. Then ε = 1, since 2^{b} is now a quadratic residue (mod 11). Thus b ≡ 0 (mod 10), so that 2^{5} − 1 divides x, a contradiction, as 31 does not divide |G|.

If x = 55, then ε = −1 and x − 1 = 54. Then the degrees of the irreducible characters in the principal 5-block are 1, 54, 54, 54 and 161, a contradiction as |G| is not divisible by 23.

If x = 11.2^{a} and x + ε = 5.3^{b}, then ε = 1, since 3 and 5 are quadratic residues (mod 11), but −1 is not. Thus 5.3^{b} ≡ 2 (mod 7) and b ≡ 3 (mod 6). Thus 5.3^{b} ≡ 7 (mod 8) and x ≡ 6 (mod 8). Hence a = 1, contrary to the fact that a is even, as x ≡ ±1 (mod 5).

If x = 11.3^{a} and x + ε = 5.2^{b}. Now 5.2^{b−1} is a quadratic non-residue (mod 7) and 5.2^{b−1} ≡ ε (mod 7), so that ε = −1. Now 11.3^{a} ≡ 1 (mod 5), so that a is even. However, 11.3^{a} ≡ −1 (mod 7), a contradiction.

If x = 5.2^{a} and x + ε = 11.3^{b}, then b is even as 11.3^{b} ≡ ε (mod 5). Thus 11.3^{b} is a quadratic residue (mod 7), so ε = 1 as x + ε ≡ 2ε (mod 7). Then we have 5.2^{a} ≡ 1 (mod 7), a contradiction, as 5 is a quadratic non-residue (mod 7) and 2 is a quadratic residue (mod 7).

If x = 5.3^{a} and x + ε = 11.2^{b} Then x ≡ −ε (mod 11), so that ε = −1 as 3 and 5 are quadratic residues (mod 11). However, 11.2^{b} ≡ −2 (mod 7), a contradiction as 2 and 11 are quadratic residues (mod 7), but −1 is not.

If x = 2^{a} and x + ε = 55.3^{b}, then 2^{a} ≡ ε (mod 7), and 2 is a quadratic residue (mod 7), so ε = 1. Also, a is even as 2^{a} ≡ −1 (mod 5). Now, however, 2^{a} ≡ −1 (mod 11), a contradiction as 2^{a} is a quadratic residue (mod 11), but −1 is not.

If x = 3^{a} and x + ε = 55.2^{b}, then a is even, as 3^{a} ≡ −ε (mod 5), a is even.Since 3^{a} ≡ ε (mod 7) we have ε = 1. Now, however, we have 3^{a} ≡ −1 (mod 11), a contradiction, as 3 is a quadratic residue (mod 11) and −1 is not.

The case x = 2^{a}.3^{b} and x + ε = 55 does not occur. For x + ε ≡ 2ε (mod 7), but 55 ≡ −1 (mod 7).

Lemma 4.6.2: C_{G}(T) ≠ T.

Proof: Suppose that C_{G}(T) = T. By Corollary 4.2.3, we know that |C_{G}(S)| ≠ 15. Since C_{G}(S) ≠ S by the results of Section 4.2, C_{G}(S) contains an involution t. By Lemma 2.4.2, we have χ(t) = −1. Also, we now have [N_{G}(T) : C_{G}(T)] = 3, so that t inverts no 7-singular element of G. Let u be a generator of t.

Since C_{G}(T) = T, the irreducible character χ must all lie in the principal 7-block. Furthermore, since χ(1) ≡ −3 (mod 7), χ is an exceptional character in the principal 11-block. Hence χ̄ is also exceptional in the principal 7-block. There must be an irreducible non-exceptional character in the principal 7-block whose degree y is congruent to −1 (mod 7).

Then the degrees of the irreducible characters in the principal 7-block are 1, y, 12 − y, 11, 11 or 1, y, y − 12, 11, 11. In the former case, we have y = 6, contrary to Lemma 1.2.3.

Suppose that the latter case holds. Let ψ_{2} be the irreducible character of degree y, ψ_{3} be the character of degree y − 12. Then the block orthogonality relations show (using the fact that χ(t) = −1 and χ̄(t) = −1), that there is an integer a such that ψ_{2}(t) = ψ_{3}(t) = a. Also ψ_{2}(u) = −1, ψ_{3}(u) = 1, while there is an algebraic integer b such that χ(u) = b and χ̄(u) = 1 − b.

Since C_{t}C_{t} involves no element with 7-part u, we have

(1 − \frac{a^{2}}{y} + \frac{a^{2}}{y − 12} + \frac{1}{11}) = 0.

This leads to \frac{12}{11} + \frac{12a^{2}}{y(y−12)} = 0, a contradiction. Thus C_{G}(T) ≠ T.

Lemma 4.6.3: C_{G}(T) is Abelian of order 14 or 21.

Proof: Since [N_{G}(T) : C_{G}(T)] = 3 and Res_{T}^{G}(χ) is not rational-valued by Lemma 2.5.4, χ is an exceptional irreducible character in its 7-block.

Consequently, there is an irreducible character ψ of degree 3 of N_{G}(T) such that χ(y) = −ψ(y) for each 7-singular y ∈ N_{G}(T). Then Res_{C_{G}(T)}^{G}(χ) = ∑_{i=1}^{8} ψ_{i}, where ψ_{i} is linear for 1 ≤ i ≤ 7, and where χ(y) = ψ_{8}(y) for each 7-singular y ∈ C_{G}(T). Then the ψ_{i} agree on O_{7′}(C_{G}(T)) for 1 ≤ i ≤ 7. Also, there is a linear character ϕ of O_{7′}(C_{G}(T)) and a primitive complex 7-th root of unity ω such that ψ_{8}(uy) = (1+ω^{3} +ω^{5} +ω^{6})ϕ(y) for all y ∈ O_{7′}(C_{G}(T)). Thus C_{G}(T) is Abelian. We also have χ(y) = 7ψ_{1}(y) + 4ϕ(y) for all y ∈ O_{7′}(C_{G}(T)).

Let z ∈ O_{7′}(C_{G}(T)) be an element of prime order p (where p ∈ {2, 3}). Now ψ_{1}(z) ≠ ϕ(z). Hence zu has 11-distinct eigenvalues, and C_{G}(uz) is an Abelian {7, p}-group. Since C_{G}(u) is Abelian, we see that C_{G}(u) is an Abelian {7, p}-group, as p and z were arbitrarily chosen.

Now u has no eigenvalue with multiplicity greater than 2. Hence every pelement of C_{G}(u) acts with (at most) quadratic minimum polynomial in any reduction (mod p)) of G. Hence the Sylow p-subgroup of C_{G}(T) is elementary Abelian. It must be cyclic, otherwise there would be an element of order p in ker(ψ_{1}ϕ̄).

Lemma 4.6.4: i) 2^{16} does not divide |G|. ii) 3^{12} does not divide |G|. iii) If |C_{G}(S)| = 15, then |C_{G}(T)| = 21.

Proof: Let B be the 5-block of G which contains χ. We claim that B only contains one other algebraic conjugate of χ.

Since there is no element of order 35 in G, the other exceptional character in the 7-block of χ (which agrees with χ on 5-singular elements) is also in B, since χ does not vanish on all 5-singular elements. The Galois group of an appropriate cyclotomic field is Abelian, so all irreducible characters in an orbit have the same stabilizer, which is contained in the stabilizer of B. Hence the number of algebraic conjugates of χ in B is even. If there are 4 or more, then the irreducible character degrees in B are 11, 11, 11, 11 and 44. Since there is no element of order 35 in G, these characters must all be in the same 7-block by Lemma 1.2.14, a contradiction, as the 7-block of χ only contains two exceptional characters.

i) If C_{G}(S) has even order, then B is invariant under all algebraic conjugation, since χ takes rational values on all 5-singular elements. so that χ is rationalvalued on 2-elements and on 3-elements. Part i) now follows by Lemma 2.5.1 (and 3^{6} does not divide |G| in this case). If y ∈ C_{G}(T) is an element of order 3, then χ(y) = 4α + 7β for two cube roots of unity α and β, a contradiction. Thus |C_{G}(T)| = 14 in this case.

ii) If |C_{G}(S)| = 15, then χ is not rational-valued on 3-elements by Corollary 4.2.2. Also, G contains an element of order 21 by Corollary 4.2.3. Now B is invariant under any Galois automorphism which fixes cube roots of unity, since such an automorphism does not affect the values of χ on 5-singular elements. Thus χ takes rational vales on 2-elements, and χ + χ̄ takes rational values on 3-elements. Part i) now follows from Lemmas 2.5.1 and 2.5.2.

Lemma 4.6.5: |G| = 2^{15}.3^{11}.5.7.11.

Proof: If |C_{G}(S)| = 15, then |G| = 2^{a}.3^{b}.5.7.11, where 11 ≤ a ≤ 15, 3 ≤ b ≤ 11. Also, we have |N_{G}(P)| = 55, |N_{G}(S)| = 60 and |N_{G}(T)| = 63. Now 2^{a−2}.3^{b−1}.7 ≡ 1 (mod 5), while 2^{a}.3^{b}.7 ≡ 1 (mod 11) and 2^{a}.3^{b−2}.5.11 ≡ 1 (mod 7).

This gives a ≡ b (mod 4), a ≡ 2b − 7 (mod 10) and 2a + b + 1 ≡ 0 (mod 6). Thus b is odd from the last equation, so a is also odd from the first. From the second equation b ≡ 3a+1 (mod 5). The last equation also gives b ≡ a+2 (mod 3), so b ≡ a + 2 (mod 6). Now we have a − b ≡ 4 (mod 12). Thus a ≡ 2b − 2 (mod 5) and a ≡ b + 4 (mod 12). If a = 11, there is no valid choice of b. If a = 13, there is no valid choice of b. If a = 15, then b = 11.

If |C_{G}(S)| = 10, then |G| = 2^{a}.3^{b}.5.7.11, where 11 ≤ a ≤ 15,1 ≤ b ≤ 5. Also, we have |N_{G}(S)| = 40 and |N_{G}(T)| = 42, using Lemma 4.6.4 and its proof. Now 2^{a−2}.3^{b} ≡ 1 (mod 5), while 2^{a}.3^{b} ≡ 1 (mod 7). Thus a−b ≡ 2 (mod 4) and 2a+b ≡ 0 (mod 6), so a ≡ b (mod 3). Now a and b are both even. Now we have a − b ≡ 6 (mod 12). Since 12 ≤ a ≤ 14 and 2 ≤ b ≤ 4, we have a contradiction.

If |C_{G}(S)| = 30, then |G| = 2^{a}.3^{b}.5.7.11, where 11 ≤ a ≤ 15,1 ≤ b ≤ 5. Also, we have |N_{G}(S)| = 120 and |N_{G}(T)| = 42, using Lemma 4.6.4 and its proof. Now 2^{a−2}.3^{b−1} ≡ 1 (mod 5), while 2^{a}.3^{b} ≡ 1 (mod 7). Thus a − b ≡ 1 (mod 4) and 2a + b ≡ 0 (mod 6), so a ≡ b (mod 3). Now a is odd and b is even. Now we have a − b ≡ 9 (mod 12). Since 11 ≤ a ≤ 15 and 2 ≤ b ≤ 4, we see that a − b = 9. Now we have |N_{G}(P)| = 55, for otherwise [G : N_{G}(P)] = 2^{a−1}3^{b}7, which is a quadratic non-residue (mod 11) as a is odd and 7 is a non-residue (mod 11), while 3 is a residue (mod 11).

Hence 2^{b+9}.3^{b}.7 ≡ 1 (mod 11). Then b − 1 + 8b + 7 ≡ 0 (mod 10) and b ≡ 6 (mod 10), a contradiction.

Lemma 4.6.6: |G| ≠ 2^{15}.3^{11}.5.7.11

Proof: Suppose that |G| = 2^{15}.3^{11}.5.7.11. In that case, we have |C_{G}(S)| = 15, |C_{G}(T)| = 21 and |N_{G}(P)| = 55.

By Lemma 4.6.3, and using unimodularity, C_{G}(T) contains an element x of order 3 such that χ(x) = 4ω + 7ω̄, where ω is a primitive complex cube root of unity. Let C = C_{G}(x). We claim that |C| is not divisible by 27. If T ⊲ C, this is clear, so suppose otherwise.

Let K = O_{7′}(C). Let θ be the character of C afforded by its action on the ωeigenspace of C. Then kerθ ≤ K, while by Lemma 1.2.10, we have [K, T] ≤ kerθ.

By Lemma 2.7.2, T centralizes a Sylow 3-subgroup of K, so ⟨x⟩ ∈ Syl_{3}(K). Thus ker(θ) is a 3′-group. It suffices to prove that 27 does not divide [C : ker(θ)]. Note that K = C_{K}(T)[K, T] = ⟨x⟩kerθ.

Let H = C/kerθ, and let bars denote images in H. Now x̄ ∉ H′, since the action of x on its 4-dimensional ω-eigenspace is not unimodular. Hence we now have O_{7′}(H) = ⟨x̄⟩ and O_{7′}(H′) = 1.

Now we have [T̄, F (H)] = 1. If T̄ ⊲ H, then |H| is not divisible by 27. Hence we may suppose that T̄ is not normal in H. Then H must have a component, L say, which must have order divisible by 7, and hence is the unique component of H. Since L has cyclic Sylow 7-subgroups, Z(L) is a 7′-group. Also, x̄ ∉ L since x̄ ∉ H′. Thus Z(L) = 1 and L is simple. Then L ≅ PSL(2, 7), since L has order divisible by 7 and has a faithful irreducible complex character of degree at most 4. Now H = LN_{H}(T̄). Since N_{L}(T̄) has order divisible by 3 and N_{H}(T̄) has order divisible by 9, but not by 27, we see that |H| is not divisible by 27, as claimed.

Let R ∈ Syl_{3}(G) with x ∈ R. Then |R| = 3^{11}, and by Lemma 2.5.1, R has an Abelian normal subgroup of index dividing 27. We may suppose that R is represented by monomial matrices. Let D be the (normal) subgroup of R consisting of diagonal matrices. Then as in the proof of Lemma 2.5.1, we have [R : D] ≤ 27. We now prove that D ≤ C_{G}(x), a contradiction.

As x has only two eigenvalues, we see by Mackey’s formula that whenever λ is a linear character of a proper subgroup Q of R such that β = Ind_{Q}^{R}(λ) is an irreducible constituent of Res_{R}^{G}(χ), then x is represented by diagonal matrices in the monomial representation affording character β. Hence x does lie in the diagonal subgroup D, as claimed. Hence D ≤ C, a contradiction as |D| ≥ 3^{8} and |C| is not divisible by 27.

Remark 4.6.7: We have now proved that 49 divides |G|.

Section Seven

In this section, we fix a Sylow 7-subgroup T of G of order at least 49.

Let B be the 5-block of G which contains χ. By Lemma 2.8.1, B is invariant under 7-conjugation, and by Lemma 2.5.4, Res_{T}^{G}(χ) is not rational-valued. The results of this section prove that B contains more than one other irreducible character which is 7-conjugate to χ. Lemma 2.5.4 guarantees the existence of at least one other irreducible character in B which is 7-conjugate to χ.

We assume for the duration of this section that B contains only one other irreducible character which is 7-conjugate to χ. We recall that the subgroup T is elementary Abelian. Now Res_{T}^{G}(χ + χ̄) is rational valued, and for any nontrivial linear constituent λ of Res_{T}^{G}(χ), a further two algebraic conjugates of λ occur as constituents of Res_{T}^{G}(χ).

As χ(1) = 11 and algebraically conjugate characters have the same kernel, we see that |T | ≤ 343 and ⟨Res_{T}^{G}(χ), 1⟩ ≡ 2 (mod 3). In fact, we can improve upon this.

Lemma 4.7.1: i) |T | = 49. ii) C_{G}(T) = T. iii) [N_{G}(T) : T] ∈ {3, 6, 9, 18}.

Proof: We first prove that T contains no non-identity element with a fixedpoint space of codimension 3. Suppose, on the contrary, that there is such an element, x say. As G is simple, there is a conjugate of x, say y, with [x, y] ≠ 1. Let K = ⟨x, y⟩. Since x and y each have a fixed-point space of dimension 8, we wee that K has a fixed-point space of dimension 5 or more. Since ⟨Res_{S}^{G}(χ), 1⟩ = 3, K is a 5′-group. As x and y do not commutes, K does not have a normal Sylow 7-subgroup. As x has (at most) four distinct eigenvalues, we have [O_{7′}(K), x] = 1 by Lemma 1.2.10. Since the same argument applies to y, we have O_{7′}(K) ≤ Z(K). Since K has Abelian Sylow 7-subgroups, we have [O_{7}(K), x] = [O_{7}(K), y] = 1 and F (K) ≤ Z(K). Hence K has a component, L say, and L has a faithful complex character of degree 6 or less. We write Res_{L}^{G}(χ) = ψ + 5.1_{L}, where ψ is a (possibly reducible)faithful complex character of L. Since |L| is not divisible by 5, we note that L has no complex irreducible character of degree 5 and also that L has no complex irreducible character of degree 5. There are several cases to consider, depending on the decomposition of ψ into irreducible constituents. We note, incidentally, that both x and y normalize L, since it is not possible for L to have 7 mutually centralizing conjugates.

Suppose that ψ = α + β, where α and β are both irreducible of degree 3. Since x and y have order 7, they both fix α and β. Now α and β are distinct characters of L, otherwise Res_{⟨x⟩}^{K}(ψ) contains the trivial character with even multiplicity, which is not the case. Hence they both extend to irreducible characters of K. Since α(1) = 3, we see that L ≅ PSL(2, 7). Now K/LZ(K) is isomorphic to a subgroup of Out(L), which is a 7′-group. Hence x has no repeated eigenvalue on either 3-dimensional K-invariant subspace, unless it acts by scalar multiplication on that subspace. But if [K, x] ≤ kerα (for example), we have a contradiction, as K/kerα is an irreducible linear group, yet [x, y] ∈ kerα, so that K/kerα is Abelian.

Suppose that ψ = α + t1, where α is irreducible and non-trivial, and t = 6 − α(1) > 0. If α(1) = 3, then any involution t ∈ L contradicts Lemma 2.1.6. If α(1) = 4, the same argument applies unless L has generalized quaternion Sylow 2-subgroups and every involution of L is central. But if t ∈ Z(L), then on each eigenspace except the 1-eigenspace of an element u ∈ L of order 7, t has only the eigenvalue −1. On the 1-eigenspace of u, t has the eigenvalue 1 with multiplicity 7 and the eigenvalue −1 with multiplicity 1. Hence t acts as a transvection in any reduction (mod 2) of G, contrary to Lemma 2.1.7.

Suppose that ψ is irreducible of degree 6. Then ψ extends to an irreducible character of K. By Lemma 1.2.6, [K : O_{7}(K)] ≤ 7. However, as O_{7}(K) ≤ Z(K), we have O_{7}(K) = 1, since every element of Z(K)^{#} has only two eigenvalues, contrary to Corollary 1.4.4. Hence x and y both lie in L, so K = L. However, in this case, the theory of blocks of defect 1 excludes this configuration, since in a representation affording ψ, x has the eigenvalue 1 with multiplicity 3 and three algebraically conjugate primitive 7-th roots of unity, each with multiplicity 1.

Hence our claim that no non-identity element of T has a fixed-point space of codimension 3 is established. If |T | ≥ 7^{3}, then we may write

Res_{T}^{G}(χ) = ψ_{1} + ψ_{2} + ψ_{3} + 21_{T} ,

where each ψ_{i} is the sum of three algebraically conjugate non-trivial linear characters. However, there is a non-identity element x ∈ kerψ_{1} ∩ kerψ_{2}, and then x has a fixed-point space of dimension 8, a contradiction. Thus |T | = 49.

We now claim that the trivial character is the only irreducible constituent of Res_{T}^{G}(χ) which occurs with multiplicity greater than 1, and that it occurs with multiplicity precisely two. If ⟨Res_{T}^{G}(χ), 1⟩ ≥ 5, then we may write

Res_{T}^{G}(χ) = ψ_{1} + ψ_{2} + 51_{T} ,

where each ψ_{i} is the sum of three algebraically conjugate non-trivial linear characters. Since T is not cyclic, there is a non-identity element x ∈ kerψ_{1}, and then x has the eigenvalue 1 with multiplicity at least 8, a contradiction. Hence we do have ⟨Res_{T}^{G}(χ), 1⟩ = 2.

If we write

Res_{T}^{G}(χ) = 2ψ_{1} + ψ_{2} + 21_{T} ,

where each ψ_{i} is the sum of three algebraically conjugate non-trivial linear characters, then kerψ_{1} contains a non-identity element as T is not cyclic. Again this yields a non-identity element with a fixed-point space of dimension 8 or more, a contradiction. Thus the trivial character is the only linear constituent of Res_{T}^{G}(χ) which occurs with multiplicity greater than 1, and it does indeed occur with multiplicity 2.

Hence for each prime p ≠ 7, any element of order p in C_{G}(T) acts as a transvection in any reduction (mod p) of G, contrary to Lemma 2.1.7. Hence C_{G}(T) = T.

Now N_{G}(T)/T is isomorphic to a 7′-subgroup of GL(2, 7). Furthermore, as Res_{T}^{G}(χ) is not real-valued ( for otherwise, that restriction would be rational valued, contrary to Lemma 2.5.4), not all elements of T can be inverted by an element of N_{G}(T). Thus [N_{G}(T) : T] ∈ {3, 6, 9, 18}, since N_{G}(T)/T (identified with a subgroup of GL(2, 7) ) intersects SL(2, 7) in a subgroup of odd order.

Lemma 4.7.2: T ∩ T^{g} = 1 for all g ∈ G\N_{G}(T). Furthermore, [N_{G}(T) : T] ≠ 3.

Proof: For the first claim, we need to prove that T ⊲ C_{G}(v) whenever v ∈ T^{#}. Choose such an element v. Let C = C_{G}(v) and K = O_{7′}(C). We first prove that [K, T] = 1. Suppose otherwise. Then there is an irreducible constituent ψ of Res_{KT}^{G} (χ) such that [K, T] ≰ kerψ. By Lemma 1.2.10, ψ(1) ≥ 7. As T K acts on each eigenspace of v, v has at least one eigenvalue with multiplicity 7 or more. Since Res_{T}^{G}(χ + χ̄) is rational-valued, v must have the eigenvalue 1 with multiplicity 7 or more (any other eigenvalue has another two algebraic conjugates which occur with the same multiplicity as it does). Since we have ⟨Res_{⟨v⟩}^{G}(χ), 1⟩ ≡ 2(mod3), v must have the eigenvalue 1 with multiplicity 8, which was excluded during the proof of Lemma 4.7.1.

Thus we do have [K, T] = 1 so that K = 1 by 4.7.1. Also, [F (C), T] = 1 as T is Abelian. Hence E = E(C) ≠ 1. Since v ∉ E (by a transfer argument) and |T | = 49, E is simple. Now v has no eigenvalue of multiplicity greater than 5, and |C| is not divisible by 5, so that L has a faithful complex irreducible character of degree 3 or 4. Since L is simple, we have L ≅ PSL(2, 7). Each non-trivial irreducible constituent of Res_{L}^{G}(χ) has degree 3. If there is only one such constituent, then L contains an involution contradicting Lemma 2.1.6. Hence there are at least two such constituents. Thus v has at least two different eigenvalues which occur with multiplicity 3. It follows that for some primitive 7-th root of unity ω, v has the eigenvalues ω, ω^{2} and ω^{4} with multiplicity 3 each, and the eigenvalue 1 with multiplicity 2. Furthermore, L acts trivially on the 1-eigenspace of v, and acts irreducibly on the other three eigenspaces (for if one of its elements of order 7 acts trivially on one 3-dimensional eigenspace, then it acts trivially on each of them, using algebraic conjugation).

Let D be a dihedral subgroup of L of order 6. Then D acts trivially on the 1-eigenspace of v. The character of D afforded by its action on any on the other eigenspace of v has the form α + λ, where α is the unique irreducible character of degree 2 of D, and λ is the non-trvial linear character of D. Hence Res_{D}^{G}(χ) contains no irreducible constituent of multiplicity greater than 3. Since v centralizes the 7′-group D, v acts with minimum polynomial of degree at most three in any reduction (mod 7) of G, by Lemma 1.5.1. This contradicts Lemma 2.1.5. Hence we do have T ∩ T^{g} = 1 for all g ∈ G\N_{G}(T).

Suppose then that [N_{G}(T) : T] = 3. Since G = G′, this implies that N_{G}(T) is a Frobenius group of order 147. Then C_{G}(u) has a normal 7-complement for all u ∈ T^{#}. Since T ⊲ C_{G}(u) for all such u, we have C_{G}(u) = C_{G}(T) for all u ∈ T^{#}. Since C_{G}(T) = T by Lemma 4.7.1, we have C_{G}(u) = T for all u ∈ T^{#}. Since χ(1) = 11, this contradicts Lemma 1.2.7.

Lemma 4.7.3: |G| is not divisible by 2^{16} and is not divisible by 3^{12}. Furthermore, if |C_{G}(S)| is even, then 3^{6} does not divide |G|.

Proof: Recall that B is the 5-block which contains χ. We prove that B only contains one other algebraic conjugate of χ. Suppose otherwise. Since B contains only one other 7-conjugate, say χ^{τ}, of χ, then whenever χ^{σ} ≠ χ^{τ} is another algebraic conjugate of χ in B, then χ^{στ} ∉ {χ^{τ} , χ, χ^{σ}} but χ^{στ} ∈ B. Since B only contains five irreducible characters, we conclude that B contains exactly four algebraic conjugates of χ (including χ itself), and one other irreducible character of degree 44. Let these be χ, χ_{1}, χ_{2}, χ_{3}, ψ where χ_{1} is 7-conjugate to χ, χ_{2} is conjugate to χ via some Galois automorphism fixing 7-th roots of unity, χ_{3} is 7-conjugate to χ_{2} and ψ is fixed by any Galois automorphism which stabilizes B. Let s be a generator of the Sylow 5-subgroup S of G. Note that χ_{i}(s) = 1 for each i and that ψ(s) = −1.

Let u be an element of T^{#} such that χ(u) is not rational, and set χ(u) = α. Then χ_{1}(u) = χ_{3}(u) = ᾱ, χ_{2}(u) = α and ψ(u) = 2(α + ᾱ). By Lemma 4.2.4, we see that

\frac{6|G|}{|C_{G}(u|^{2}} (\frac{2(α^{2} + ᾱ^{2})}{11} − \frac{4(α + ᾱ)^{2}}{44} )

is an integer. Since |T | divides |C_{G}(u)| and [G : C_{G}(u)] is not divisible by 7, we deduce that (α − ᾱ)^{2} is an integer multiple of 49.

Now χ(u) = α = q +r(ω +ω^{2} +ω^{4})+s(ω̄ +ω̄^{2} +ω̄^{4}) for a primitive 7-th root of unity ω and integers q, r, s with q+3(r+s) = 11. Then (α−ᾱ)^{2} = −7(r−s)^{2}. Hence r ≡ s (mod 7). Since q + 3(r + s) = 11 and q, r, s are all non-negative, we have r = s. But then χ(u) = q − r, a contradiction as χ(u) is not rational. Hence B contains only one other algebraic conjugate of χ.

We claim next that χ assumes rational values on all 2-elements. The only other algebraic conjugate of χ in B is a 7-conjugate, so that χ is fixed by any Galois automorphism which moves roots of unity of 2-power order and fixes 7-th roots of unity. Hence χ takes rational values on 2-elements. By Lemma 2.5.1, |G| is not divisible by 2^{16}.

By Lemma 2.4.2, B is invariant under any Galois automorphism which fixes cube roots of unity. By an argument like that above, χ + χ̄ takes rational values on all 3-elements, so that 3^{12} does not divide |G| by Lemma 2.5.2.

If |C_{G}(S)| is even, then B is invariant under all algebraic conjugation. Thus χ is fixed by any Galois automorphism which fixes 7-th roots of unity, so χ takes rational values on all 3-elements, and 3^{6} does not divide |G| by Lemma 2.5.1.

We now derive a contradiction from the assumptions of this section.

Lemma 4.7.4: B contains at least two other irreducible characters 7-conjugate to χ.

Proof: Suppose otherwise. Let P ∈ Syl_{11}(G), S ∈ Syl_{5}(G) and T ∈ Syl_{7}(G). We now know that |T | = 49, that [G : N_{G}(T)] ≡ 1 (mod 49) and that [N_{G}(T) : T] ∈ {6, 9, 18}. Also |N_{G}(P)| ∈ {55, 110} and |N_{G}(S)| ∈ {40, 60, 120}. Furthermore, |G| has the form 2^{a}3^{b}.5.7^{2}.11 where 11 ≤ a ≤ 15 and b ≤ 11. Furthermore, if |C_{G}(S)| ≠ 15, then b ≤ 5. Consideration of quadratic residues (mod 11) shows that |N_{G}(P)| = 55 if a is even and |N_{G}(P)| = 110 if a is odd.

If |N_{G}(S)| = 60, then G contains no element of order 10, so that |N_{G}(P)| = 55. Also, b ≥ 2, since C_{G}(S) contains an element of order 3 which is not conjugate to its inverse in G (as B is not the principal 5-block) and G is simple.

Now we have 2^{a}.3^{b}.7^{2} ≡ 1 (mod 11) and 2^{a−2}3^{b−1}7^{2} ≡ 1 (mod 5). Hence a − 2b + 4 ≡ 0 (mod 10) and a ≡ b − 1 (mod 4). Now a is even and b is odd. Also a ∈ {2b + 6, 2b − 4} as a ≥ 11 and b ≤ 11. The only possibility is b = 3 and a = 12, contrary to a ≡ b − 1 (mod 4).

If |N_{G}(S)| = 40, then 11 ≤ a ≤ 15 and 1 ≤ b ≤ 5. Then 2^{a−1}.3^{b} ≡ 1 (mod 5), so that a ≡ b + 1 (mod 4). If |N_{G}(P)| = 55, then a is even and b is odd. Then 2^{a+4−2b} ≡ 1 (mod 11), so that a ≡ 2b + 6 (mod 10). Again, we must have b = 3 and a = 12. But in that case [G : T] ≡ 2^{12}.3^{3}.6 (mod 49). But then [G : T] ≡ −6 (mod 49), a contradiction, as [G : T] ∈ {6, 9, 18}.

If |N_{G}(S)| = 120, then 11 ≤ a ≤ 15 and 1 ≤ b ≤ 5. Then 2^{a−1}.3^{b−1} ≡ 1 (mod 5), so that a ≡ b (mod 4). If |N_{G}(P)| = 55, then a and b are both even. Then 2^{a+4−2b} ≡ 1 (mod 11), so that a ≡ 2b + 6 (mod 10) and a = 2b + 6, a contradiction. If |N_{G}(P)| = 110, then a and b are both odd and 2^{a−1+4−2b} ≡ 1 (mod 11), and a ≡ 2b + 7 (mod 10). Then a = 2b + 7, a contradiction, as this forces b ≡ 1 (mod 4), so a ∈ {9, 17}.

Section Eight

In this section, we derive a final contradiction. We assume that |T | ≥ 49 and that B contains at least two other irreducible characters which are 7-conjugate to χ, where B is the 5-block which contains χ. Let ω be a primitive complex 7-th root of unity.

Lemma 4.8.1: |G| is divisible neither by 2^{16}, nor by 3^{12}. If |C_{G}(S)| is even, then |G| is not divisible by 3^{6}.

Proof: We first prove that B contains exactly three irreducible characters of degree 11. If not, then B contains 4 irreducible characters of degree 11 and one irreducible character of degree 44, considering the values at a generator s of S ∈ Syl_{5}(G). Since three of these are 7-conjugate, the other must be 7-rational. But then Lemma 2.5.4 yields a contradiction.

Now we may proceed as we did in Lemma 4.7.3, after noting that χ is invariant under any Galois automorphism which fixes 7-th roots of unity and cube roots of unity. When |C_{G}(S)| is even, B is invariant under all algebraic conjugation, and χ takes rational values on all 2-elements and all 3-elements.

Lemma 4.8.2: The only possible sets of degrees for the irreducible characters in B are {11, 11, 11, 66, 99}, {11, 11, 11, 231, 264} or {11, 11, 11, 891, 924}.

Proof: There is an integer x ≡ −1 (mod 5) such that the degrees of the irreducible characters in B are 11, 11, 11, x, ε(x − 33) for some sign ε. Since x and x − 33 both divide |G|, and since x ≡ ±1, ±5 or 0 (mod 11), we have ε = 1.

In fact, if x is not divisible by 11, then x must be non-exceptional in its 11-block, otherwise an 11-conjugate of x also lies in B, which is not possible. Hence, in that case, x ≡ ±1. Suppose then that x ≡ ±1 (mod 11).

If x = 2^{a}, then a > 0 and x − 33 is prime to 2, 3, 5 and 11, so x − 33 = 7^{b}. Now x ≡ −1 (mod 5), so a is even. Hence x is a quadratic residue (mod 11), so that x ≡ 1 (mod 11). Hence a ≡ 0 (mod 10), so x − 33 ≡ 7 (mod 8) and b is odd. However, x − 33 ≡ 1 (mod 5), so that b is even, a contradiction.

If x = 2^{a}3^{b} (and b > 0), then x − 33 = 3^{c}.7^{d}. If c > 1, then b = 1. Then a ≡ 3 (mod 4), as x ≡ −1 (mod 5). Hence x is a quadratic non-residue (mod 11), so that x ≡ −1 (mod 11) and a ≡ 7 (mod 10). Thus a ≡ 7 (mod 20), a contradiction. Hence c = 1. Now 3.7^{d} ≡ 1 (mod 5), so d is odd. Hence x − 33 ≡ 5 (mod 8), so that a = 1. Hence x is a quadratic non-residue (mod 11) and x ≡ −1 (mod 11). Now 1 − 2b ≡ 5 (mod 10), so b ≡ 3 (mod 5). Also b − 1 ≡ 2 (mod 4), so b ≡ 3 (mod 20). Hence b = 3 so x = 54 and x − 33 = 21.

Let ψ be the irreducible character of degree 21 in B. Then Res_{T}^{G}(χ) is rational-valued. Hence ψ(u) ≥ 0 for all u ∈ T, since all primitive 7-th roots of unity occur with equal multiplicity as eigenvalues of u in any representation of G affording ψ. We claim that ψ is the only irreducible character from its 7-block which belongs to the block B. In that case, however, Lemma 1.2.14 is violated, as G contains no element of order 35 by Lemma 2.8.1, a contradiction. For each element u ∈ T^{#}, we have ψ(u) ∈ {0, 7, 14} as G is simple.

Thus \frac{[G:C_{G}(u)]ψ(u)}{ψ(1)} ≡ 0, 5[G : C_{G}(u)] or 3[G : C_{G}(u)] (mod 7). However, any other irreducible character α ∈ B has degree prime to 7, and we find that \frac{[G:C_{G}(u)]α(u)}{α(1)} ≡ [G : C_{G}(u)] (mod π), where π is a prime ideal of ℤ[ω] containing 7. Since [G : C_{G}(u)] is not divisible by 7 when u ∈ T^{#}, we deduce that ψ is in a different 7-block from α for all irreducible characters α ≠ ψ ∈ B. Thus x ≠ 2^{a}.3^{b}.

If x = 2^{a}7^{b} and b > 0, then a = 0, for otherwise x − 33 is coprime to 2, 3, 5, 7 and 11, whereas x divides |G| and G is simple. Hence 7^{b} − 33 is a power of 2. Now b ≡ 2 (mod 4) as x ≡ −1 (mod 5). Thus x ≡ 1 (mod 11) as −1 is a quadratic non-residue (mod 11). Hence b ≡ 0 (mod 10), contrary to the fact that |G| is not divisible by 7^{10}.

If x = 2^{a}.3^{b}.7^{c} and a, b, c > 0, then x − 33 = 3^{d} and d is even as x − 33 ≡ 1 (mod 5). Thus b = 1. Also, x−33 is a quadratic residue (mod 11), so x−33 ≡ 1 (mod 11) and 5 divides d. Since 3^{d} ≡ 1 (mod 5), we also have d ≡ 0 (mod 4), so that 20 divides d, a contradiction.

If x = 3^{a}.7^{b} and a, b > 0, then x − 33 = 2^{c}.3^{d}. If a = 1, then b is odd, as x ≡ −1 (mod 5). Thus x ≡ −1 (mod 11), as x is a quadratic non-residue (mod 11). Then x − 33 ≡ −1 (mod 11), so that c is odd, as 3 is a quadratic residue (mod 11). Hence d is also odd However, as a and b are odd, we have x ≡ 5 (mod 8), so x − 33 ≡ 4 (mod 8) and c = 2, a contradiction. So we suppose now that 11 divides x.

Now set x = 11y. Then we need y ≡ −1 (mod 5) and y(y − 3) = 2^{a}.3^{b}.7^{c}.

If y = 2^{a}, then a ≡ 2 (mod 4). Now y − 3 is prime to 3, so y − 3 = 7^{b} and y − 3 ≡ 1 (mod 5) and b ≡ 0 (mod 4). Hence y ≡ 4 (mod 8) and x = 44. Then the degrees of the irreducible characters in B are 11, 11, 11, 11 and 44, a case which has already been excluded.

If y = 2^{a}.7^{b} where b > 0, then a = 0, for otherwise y − 3 is prime to 2, 3, 5, 7 and 11, so y = 1, a contradiction, as y ≡ −1 (mod 5). As a = 0 , b is even since y ≡ −1 (mod 5). Then y ≡ 1 (mod 8), so y − 3 = 2, a contradiction.

Suppose that y = 2^{a}.3^{b} where b > 0. If y − 3 is not divisible by 7, then y − 3 = 3^{c} and c ≡ 0 (mod 4) as y ≡ 1 (mod 5). Now c ≠ 0, otherwise x = 11 again. Thus b = 1 and 2^{a} − 1 is a power of 3. Now 3.2^{a} ≡ −1 (mod 5), so a ≡ 3 (mod 4) and 2^{a} ≡ 2 (mod 3), a contradiction. Thus we may suppose that 7 divides y − 3. Then b is odd, as 3 is a quadratic non-residue (mod 7) and 2 is a quadratic residue. Then a is also odd, as y ≡ −1 (mod 5). We claim that y − 3 is not divisible by 9. This is clear if b > 1, so suppose that b = 1. Then as a is odd, we see that y − 3 = 3(2^{a} − 1) is not divisible by 9, as claimed. Hence y − 3 = 3.7^{c} for some integer c. Then c ≡ 1 (mod 4) as y − 3 ≡ 1 (mod 5). Now y ≡ 8 (mod 16), so that a = 3. Now we have 8.3^{b} = 3.(7^{c} + 1) and 8.3^{b−1} = (7^{c} + 1). Now 3 does not divide 7^{c} + 1, so that b = 1 and y = 24. This yields x = 264 and x − 33 = 231.

Suppose that y = 3^{a}. Then a is even as y ≡ −1 (mod 5). Thus y − 3 ≡ 2 (mod 4). Also y − 3 is not divisible by 9. Thus y − 3 = 6.7^{b} for some integer b. However, since a is even, y = 3^{a} is a quadratic residue (mod 7), which 3 itself is not. Thus y − 3 is not divisible by 7, so y = 9. This leads to the irreducible character degrees for B being 11, 11, 11, 66 and 99.

Suppose that y = 3^{a}.7^{b} with b > 0. Then y − 3 = 3^{c}.2^{d}. Note that y − 3 is certainly even, so d > 0. Now we have 3^{c}.2^{d} ≡ −3 (mod 7), which is a quadratic residue, so c must be even. Now y − 3 = 3^{c}.2^{d} ≡ 1 (mod 5), so that c ≡ d (mod 4). As y ≡ −1 (mod 5), a ≡ b (mod 2). Thus y ≡ 1 (mod 4), so that 4 does not divide y − 3, contrary to the fact that d is even and positive.

Suppose that y = 2^{a}.3^{b}.7^{c}, where abc > 0. Then y − 3 = 3^{d} for some integer d, and d ≡ 0 (mod 4) as y − 3 ≡ 1 (mod 5). Thus y ≡ 4 (mod 16), so that a = 2. Now y has the form 12.7^{c}, and we have 4.7^{c} − 1 = 3^{d−1}. Thus 3^{d−1} ≡ −1 (mod 7) and d ≡ 4 (mod 6). Since we already have d ≡ 0 (mod 4), we have y ≡ 4 (mod 12). Since 3^{12} does not divide |G|, we have d = 4, so y = 84. Then x = 924 and x − 33 = 891.

Lemma 4.8.3: The irreducible character degrees for B are not 11, 11, 11, 231, 264.

Proof: Suppose that these degrees do occur. We claim that |T | ≥ 7^{3}. For if |T | = 49, then the irreducible character of degree 231 in B is in a 7-block of defect 1, and is the only irreducible character in its 7-block which is in B. Since G contains no element of order 35 by Lemma 2.8.1, Lemma 1.2.14 shows that this irreducible character of degree 231 lies in a 7-block of defect 0, a contradiction.

Now we see from Lemma 1.2.14 again that all irreducible characters of B lie in the same 7-block.

By Lemma 2.5.4, there is an element u ∈ T such that χ(u) is not rational. Let χ_{1} and χ_{2} be the other 7-conjugates of χ in B, labelled so that χ_{i}(u) = χ(u^{2^{i}}) for each i. Let θ_{1} be the irreducible character of degree 231 in B and θ_{2} be the irreducible character of degree 264 in B. Let s be a generator of the Sylow 5-subgroup S of G. Note that θ_{1} and θ_{2} are 7-rational, so that θ_{1}(u) and θ_{2}(u) are integers. We will apply Lemma 4.2.4 with a = u and b = u^{−1}. Let y = θ_{1}(u) and y + z = θ_{2}(u).

Then we have z = χ(u)+χ(u^{2})+χ(u^{4}). Notice that 11+2z = 7⟨Res_{⟨y⟩}^{G}(χ), 1⟩. Hence z ≡ 5 (mod 7). We also note that y is divisible by 7, as [G : C_{G}(u)] is not divisible by 7 and θ_{1}(1) is.

We set r = |χ(u)|^{2} + |χ(u^{2})|^{2} + |χ(u^{4})|^{2}, and we note that 121 + 2r = 7⟨Res_{⟨u⟩}^{G}(χ), Res_{⟨u⟩}^{G}(χ)⟩, so that r ≡ 6 (mod 7). Now set y = 7t. Since ψ_{1} lies in a 7-block of full defect (the same 7-block as χ), we have

\frac{[G : C_{G}(u)]θ_{1}(u)}{θ_{1}(1)} ≡ [G : C_{G}(u)]

(mod 7), so that θ_{1}(u) ≡ θ_{1}(1) (mod 49) and t ≡ 5 (mod 7).

By Lemma 4.2.4, we see that

\frac{6|G|}{|C_{G}(u)|^{2}} (\frac{r}{11} + \frac{y^{2}}{231} − \frac{(y + z)^{2}}{264} )

is an integer. Since [G : C_{G}(u)] is not divisible by 7, this quickly leads to

r + \frac{7t^{2}}{3} − \frac{(7t + z)^{2}}{24} = |T |m

for some 7-adic integer m.

In particular, this yields 24r+7t^{2} −14tz −z^{2} ≡ 0 (mod 49). We have already seen that t ≡ z ≡ 5 (mod 7). Hence t^{2} − 2tz ≡ 3 (mod 7) and 7t^{2} − 14tz ≡ 21 (mod 49). Thus z^{2} − 24r ≡ 21 (mod 49).

Let us write χ(u) = a + b(ω + ω̄) + c(ω^{2} + ω̄^{2}) + d(ω^{4} + ω̄^{4}), where a, b, c, d are non-negative integers with a + 2(b + c + d) = 11. Also we know that a = ⟨Res_{⟨y⟩}^{G}(χ), 1⟩ and

a^{2} + 2(b^{2} + c^{2} + d^{2}) = ⟨Res_{⟨u⟩}^{G}(χ), Res_{⟨u⟩}^{G}(χ)⟩

.

Thus 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). Then z^{2} ≡ 35a + 18 (mod 49) and 24r ≡ 35(a^{2} + 2(b^{2} + c^{2} + d^{2})) + 18 (mod 49).

Hence 21 ≡ z^{2} − 24r ≡ 35(a − a^{2} − 2(b^{2} + c^{2} + d^{2})) (mod 49), so that a^{2} − a + 2(b^{2} + c^{2} + d^{2}) ≡ 5 (mod 7). Thus (a − 4)^{2} + (3b)^{2} + (3c)^{2} + (3d)^{2} ≡ 0 (mod 7). We also recall that (a − 4) + 2(b + c + d) = 7.

Now a is odd, and u has at least 4 different eigenvalues by Corollary 2.1.5, so that at least two of b, c and d are non-zero. Also b, c and d are not all equal, since χ(u) is not rational.

The only cases we need to consider (up to symmetry in b, c and d) are: a = 3, b = 3, c = 1 or a = 1, b = 3 and c = 2. So far, we have only demonstrated a congruence (mod 49), which has eliminated all but two cases. It will take consideration of congruences (mod 343) to eliminate these cases.

Suppose then that a = 3, b = 3 and c = 1. In this case, we have z = 5 and r = 41, since 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). We originally set out to solve the congruence r + \frac{7t^{2}}{3} − \frac{(7t+z)^{2}}{24} ≡ 0 (mod 343).

Suppose that t satisfies 41 + \frac{7t^{2}}{3} − \frac{(7t+5)^{2}}{24} ≡ 0 (mod 343).

Then 7t^{2} − 70t + 959 ≡ 0 (mod 343). Then we have t^{2} − 10t + 137 ≡ 0 (mod 49). Thus (t − 5)^{2} + 112 ≡ 0 (mod 49), a contradiction, as t ≡ 5 (mod 7).

Suppose then that a = 1, b = 3 and c = 2. In this case, we have z = −2 and r = 34, since 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). We originally set out to solve the congruence r + \frac{7t^{2}}{3} − \frac{(7t+z)^{2}}{24} ≡ 0 (mod 343).

Suppose that t satisfies 34 + \frac{7t^{2}}{3} − \frac{(7t−2)^{2}}{24} ≡ 0 (mod 343).

Then 7t^{2} + 28t + 812 ≡ 0 (mod 343). Then we have t^{2} + 4t + 116 ≡ 0 (mod 49). Thus (t + 2)^{2} + 112 ≡ 0 (mod 49), again a contradiction, as t ≡ 5 (mod 7).

Lemma 4.8.4: The irreducible character degrees for B are not 11, 11, 11, 891, 924.

Proof: Suppose that these degrees do occur. We claim that |T | ≥ 7^{3}. For if |T | = 49, then the irreducible character of degree 924 in B is in a 7-block of defect 1, and is the only irreducible character in its 7-block which is in B. Since G contains no element of order 35 by Lemma 2.8.1, Lemma 1.2.14 shows that this irreducible character of degree 924 lies in a 7-block of defect 0, a contradiction.

Now we see from Lemma 1.2.14 again that all irreducible characters of B lie in the same 7-block.

By Lemma 2.5.4, there is an element u ∈ T such that χ(u) is not rational. Let χ_{1} and χ_{2} be the other 7-conjugates of χ in B, labelled so that χ_{i}(u) = χ(u^{2^{i}}) for each i. Let θ_{1} be the irreducible character of degree 891 in B and θ_{2} be the irreducible character of degree 924 in B. Let s be a generator of the Sylow 5-subgroup S of G. Note that θ_{1} and θ_{2} are 7-rational, so that θ_{1}(u) and θ_{2}(u) are integers. We will again apply Lemma 4.2.4 with a = u and b = u^{−1}. Let y = θ_{1}(u) and y + z = θ_{2}(u).

Then we have z = χ(u)+χ(u^{2})+χ(u^{4}). Notice that 11+2z = 7⟨Res_{⟨y⟩}^{G}(χ), 1⟩. Hence z ≡ 5 (mod 7). We also note that y + z is divisible by 7, as [G : C_{G}(u)] is not divisible by 7 and θ_{2}(1) is.

We set r = |χ(u)|^{2} + |χ(u^{2})|^{2} + |χ(u^{4})|^{2}, and we note as before that that 121 + 2r = 7⟨Res_{⟨u⟩}^{G}(χ), Res_{⟨u⟩}^{G}(χ)⟩, so that r ≡ 6 (mod 7). Now set y + z = 7t. Since θ_{2} lies in a 7-block of full defect (the same 7-block as χ), we have

\frac{[G : C_{G}(u)]θ_{2}(u)}{θ_{2}(1)} ≡ [G : C_{G}(u)]

(mod 7), so that θ_{2}(u) ≡ θ_{2}(1) (mod 49) and t ≡ 6 (mod 7).

By Lemma 4.2.4, we see that

\frac{6|G|}{|C_{G}(u)|^{2}} (\frac{r}{11} + \frac{y^{2}}{891} − \frac{(y + z)^{2}}{924} )

is an integer. Since [G : C_{G}(u)] is not divisible by 7, this quickly leads to

r + \frac{(7t − z)^{2}}{81} − \frac{7t^{2}}{12} = |T |m

for some 7-adic integer m.

In particular, this yields −2r − 7t^{2} + 7tz + 3z^{2} ≡ 0 (mod 49). We have already seen that t − 1 ≡ z ≡ 5 (mod 7). Hence 7tz − 7t^{2} ≡ 7 (mod 49). Thus 3z^{2} − 2r ≡ 42 (mod 49).

Let us again write χ(u) = a + b(ω + ω̄) + c(ω^{2} + ω̄^{2}) + d(ω^{4} + ω̄^{4}), where a, b, c, d are non-negative integers with a + 2(b + c + d) = 11. Also we know that a = ⟨Res_{⟨y⟩}^{G}(χ), 1⟩ and

a^{2} + 2(b^{2} + c^{2} + d^{2}) = ⟨Res_{⟨u⟩}^{G}(χ), Res_{⟨u⟩}^{G}(χ)⟩

.

Thus 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). Then 3z^{2} ≡ 7a + 5 (mod 49) and 2r ≡ 7(a^{2} + 2(b^{2} + c^{2} + d^{2})) − 23 (mod 49).

Hence 14 ≡ 3z^{2} − 2r ≡ 7(a − a^{2} − 2(b^{2} + c^{2} + d^{2})) (mod 49), so that again a^{2} − a + 2(b^{2} + c^{2} + d^{2}) ≡ 5 (mod 7). Thus (a − 4)^{2} + (3b)^{2} + (3c)^{2} + (3d)^{2} ≡ 0 (mod 7). We also recall that (a − 4) + 2(b + c + d) = 7.

Now again a is odd, and u has at least 4 different eigenvalues by Corollary 2.1.5, so that at least two of b, c and d are non-zero. Also b, c and d are not all equal, since χ(u) is not rational.

Once more, the only cases we need to consider (up to symmetry in b, c and d) are: a = 3, b = 3, c = 1 or a = 1, b = 3 and c = 2. So far, we have only demonstrated a congruence (mod 49), which has eliminated all but two cases. It will take consideration of congruences (mod 343) to eliminate these cases.

Suppose then that a = 3, b = 3 and c = 1. In this case, we again have z = 5 and r = 41, since 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). We originally set out to solve the congruence r + \frac{(7t−z)^{2}}{81} − \frac{7t^{2}}{12} ≡ 0 (mod 343).

Suppose that t satisfies 41 + \frac{(7t−5)^{2}}{81} − \frac{7t^{2}}{12} ≡ 0 (mod 343).

Then t^{2} − 40t + 1912 ≡ 0 (mod 49). Then we have t^{2} − 40t + 1 ≡ 0 (mod 49). However, t ≡ 6 (mod 7), so that (t − 20)^{2} ≡ 0 (mod 49), a contradiction, as 399 ≢ 0 (mod 49).

Suppose then that a = 1, b = 3 and c = 2. In this case, we again have z = −2 and r = 34, since 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). We originally set out to solve the congruence r + \frac{7t^{2}}{3} − \frac{(7t+z)^{2}}{24} ≡ 0 (mod 343).

Suppose that t satisfies 34 + \frac{(7t+2)^{2}}{81} − \frac{7t^{2}}{12} ≡ 0 (mod 343).

Then t^{2} +16t+8 ≡ 0 (mod 49). However, t ≡ 6 (mod 7), so t^{2} +16t+64 ≡ 0 (mod 49), a contradiction.

Lemma 4.8.4: The Sylow 7-subgroup T has order 49.

Proof: Suppose that |T | ≥ 343. We are left with the only remaining possibility for the irreducible characters in B being 11, 11, 11, 66, 99.

We see from Lemma 1.2.14 again that all irreducible characters of B lie in the same 7-block as G contains no element or order 35.

By Lemma 2.5.4, there is an element u ∈ T such that χ(u) is not rational. Let χ_{1} and χ_{2} be the other 7-conjugates of χ in B, labelled so that χ_{i}(u) = χ(u^{2^{i}}) for each i. Let θ_{1} be the irreducible character of degree 66 in B and θ_{2} be the irreducible character of degree 99 in B. Let s again be a generator of the Sylow 5-subgroup S of G. Note that θ_{1} and θ_{2} are 7-rational, so that θ_{1}(u) and θ_{2}(u) are integers. We will again apply Lemma 4.2.4 with a = u and b = u^{−1}. Let y = θ_{1}(u) and y + z = θ_{2}(u).

Then we have z = χ(u)+χ(u^{2})+χ(u^{4}). Once again, 11+2z = 7⟨Res_{⟨y⟩}^{G}(χ), 1⟩. Hence z ≡ 5 (mod 7).

We set r = |χ(u)|^{2} + |χ(u^{2})|^{2} + |χ(u^{4})|^{2}, and we note as before that that 121 + 2r = 7⟨Res_{⟨u⟩}^{G}(χ), Res_{⟨u⟩}^{G}(χ)⟩, so that r ≡ 6 (mod 7).

By Lemma 4.2.4, we see that

\frac{6|G|}{|C_{G}(u)|^{2}} (\frac{r}{11} + \frac{y^{2}}{66} − \frac{(y + z)^{2}}{99} )

is an integer. Since [G : C_{G}(u)] is not divisible by 7, this quickly leads to

r + \frac{y^{2}}{6} − \frac{(y + z)^{2}}{9} = |T |m

for some 7-adic integer m.

In particular, this yields 18r + y^{2} − 4yz − 2z^{2} ≡ 0 (mod 49). Now y ≡ 3 (mod 7) and z ≡ 5 (mod 7). Hence (y − 2z)^{2} ≡ 0 (mod 49). Thus z^{2} ≡ 3r (mod 49).

Let us one more write χ(u) = a + b(ω + ω̄) + c(ω^{2} + ω̄^{2}) + d(ω^{4} + ω̄^{4}), where a, b, c, d are non-negative integers with a + 2(b + c + d) = 11. Then a = ⟨Res_{⟨y⟩}^{G}(χ), 1⟩ and

a^{2} + 2(b^{2} + c^{2} + d^{2}) = ⟨Res_{⟨u⟩}^{G}(χ), Res_{⟨u⟩}^{G}(χ)⟩

.

Thus 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). Then z^{2} ≡ 35a + 18 (mod 49) and 3r ≡ 35(a^{2} + 2(b^{2} + c^{2} + d^{2})) − 10 (mod 49).

Hence 3r − z^{2} ≡ 14(a − a^{2} − 2(b^{2} + c^{2} + d^{2})) + 21 (mod 49), so that again a^{2} − a + 2(b^{2} + c^{2} + d^{2}) ≡ 5 (mod 7). Thus (a − 4)^{2} + (3b)^{2} + (3c)^{2} + (3d)^{2} ≡ 0 (mod 7). We also recall that (a − 4) + 2(b + c + d) = 7.

Now again a is odd, and u has at least 4 different eigenvalues by Corollary 2.1.5, so that at least two of b, c and d are non-zero. Also b, c and d are not all equal, since χ(u) is not rational.

Once more, the only cases we need to consider (up to symmetry in b, c and d) are: i) a = 3, b = 3, c = 1, or ii) a = 1, b = 3 and c = 2. So far, we have only demonstrated a congruence (mod 49), which has eliminated all but two cases. It will take consideration of congruences (mod 343) to eliminate these cases.

In case i) , we again have z = 5 and r = 41, since 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})). In case ii), we again have z = −2 and r = 34, since 11 + 2z = 7a and 121 + 2r = 7(a^{2} + 2(b^{2} + c^{2} + d^{2})).

We say that a non-trivial cyclic subgroup, A, of T is if type 1 if one of its generators falls into case i) above (up to the symmetry in b, c, d), of type 2 if one of its generators falls into case ii) (up to the same symmetry), and of type 3 if Res_{A}^{G}(χ) is rational valued. This is well-defined. Suppose that there are m cyclic subgroups of type 1, n of type 2 and p of type 3. Then

m + n + p = \frac{|T | − 1}{6} … 1.

121 + 82m + 68n + 96p = |T |⟨Res_{T}^{G}(χ), Res_{T}^{G}(χ)⟩ … 2.
11 + 10m − 4n + 24p = |T |⟨Res_{T}^{G}(χ), 1⟩ … 3.

Hence taking congruences (mod —T—), we have

6m + 6n + 6p ≡ −1.
121 + 82m + 68n + 96p ≡ 0.
121 + 110m − 44n + 264p ≡ 0.

Thus 28m − 112n + 168p ≡ 0 (mod |T |). Let |T | = 7^{k}.

We now have the following congruences (mod 7^{k−1}):

m − 4n + 6p ≡ 0
6m + 6n + 6p ≡ −1

and

10m − 4n + 24p ≡ −11.

Hence we obtain (subtracting 6 times the first equation from the second, and 10 times the first equation from the second) (mod 7^{k−1}):

m − 4n + 6p ≡ 0
30(n − p) ≡ −1

and

36(n − p) ≡ −11.

Subtracting 5 times the second equation from 6 times the first gives 49 ≡ 0 (mod 7^{k−1}). Hence k ≤ 3.

Suppose then that |T | = 343. We analyse this case yet further.

Suppose that there are no cyclic subgroups A of type 2, that is, n = 0. Then we have m + p = 57 and 30p ≡ 1 (mod 49). Then p ≡ 18 (mod 49), so that p = 18 as m and p are non-negative. Thus m = 39. Now 121 + 82m + 96p = 5047 = 49 × 103, a contradiction, since 121 + 82m + 68n + 96p is divisible by |T |.

Suppose then that n ≠ 0. Then T has a cyclic subgroup A such that ⟨Res_{A}^{G}(χ), 1⟩ = 1. Certainly, then, ⟨Res_{T}^{G}(χ), 1⟩ ≤ 1. On the other had, Res_{T}^{G}(χ) is real-valued, so that ⟨Res_{T}^{G}(χ), 1⟩ is odd, and we must have ⟨Res_{T}^{G}(χ), 1⟩ = 1.

Thus we have 11 + 10m − 4n + 24p = |T | = 343 and m + n + p = 57. Thus, subtracting 10 times the second equation from the first and rearranging, we have 14n−14p = 238, and then n−p = 17, a contradiction, since we saw earlier that 30(n − p) ≡ −1 (mod 49). Hence |T | ≠ 343, so that |T | = 49.

Lemma 4.8.5: |G| is divisible by 27, and |G| is divisible by 81 if |C_{G}(S)| = 10. Proof: As before, let θ_{2} denote the irreducible character of degree 99 in B. If C_{G}(S) contains an element x of order 3, and s is a generator of S, then by Lemma 2.4.2, we have θ_{2}(sx) neq0, so that θ_{2} does not lie in a 3-block of defect zero, and 27 divides |G|.

Suppose that |C_{G}(S)| = 10, but that 81 does not divide |G|. If 27 divides |G|, then θ_{2} lies in a 3-block of defect 1, and is the only character in that block which is also in the 5-block B, contrary to Lemma 1.2.14, since G contains no element of order 15 in this case. If 27 does not divide |G|, then 9 certainly does, as θ_{2}(1) = 99. Hence the irreducible character θ_{1} ∈ B, of degree 66, lies in a 3-block of defect 1, and is the only irreducible character in B which also lies in that 3-block. This again contradicts Lemma 1.2.14.

Lemma 4.8.6: The group G does not exist.

Proof: By Lemmas 4.8.1, 4.8.4 and 4.8.5, |G| is of the form 2^{a}.3^{b}.5.7^{2}.11, where 11 ≤ a ≤ 15 and 3 ≤ b ≤ 11. If C_{G}(S)| is even, then 3 ≤ b ≤ 5 and if |C_{G}(S)| = 10, then ≠ 3. For P ∈ Syl_{11}(G), we have |N_{G}(P)| ∈ {55, 110} and for S ∈ Syl_{5}(G),|N_{G}(S)| ∈ {40, 60 or 120. If N_{G}(P)| = 110, then evidently |C_{G}(S)| is even.

Suppose that |N_{G}(P)| = 110 and |C_{G}(S)| = 10. Then 2^{a−1}3^{b}.7^{2} ≡ 1 (mod 11), so that a is odd, as 3 is a quadratic residue (mod 11) and 2 is not. Also, 2^{a−3}.3^{b}.7^{2} ≡ 1 (mod 5), so that b is even as 2 is a quadratic non-residue (mod 5). Since 4 ≤ b ≤ 5 in this case, we must have b = 4. Since 2^{a−1}.3^{4}.7^{2} ≡ 1 (mod 11), we see that a ≡ 5 (mod 10). Since 2^{a−3}.3^{4}.7^{2} ≡ 1 (mod 5), we see that a ≡ 1 (mod 4). Hence a ≡ 5 (mod 20), a contradiction.

Suppose that |N_{G}(P)| = 55 and |C_{G}(S)| = 10. Since 2^{a}.3^{b}.7^{2} ≡ 1 (mod 11), a is even, as 3 is a quadratic residue (mod 11), but 2 is not. Also, b is odd, since 2^{a−3}.3^{b}.7^{2} ≡ 1 (mod 5) and a is odd, while 3 is a quadratic non-residue (mod 5). Since 4 ≤ b ≤ 5 in this case, we have b = 5. Since 2^{a}.3^{5}.7^{2} ≡ 1 (mod 11), we have ≡ 6 (mod 10), a contradiction.

Suppose that |C_{G}(S)| = 30 and |N_{G}(P)| = 55. In that case, a is again even, as 2^{a}.3^{b}.7^{2} ≡ 1 (mod 11), and b is even, since 2^{a−3}.3^{b−1} ≡ −1 (mod 5). Thus b = 4, since 3 ≤ b ≤ 5 in this case. Since 2^{a}.3^{4}.7^{2} ≡ 1 (mod 11), we have a ≡ 4 (mod 10). Also, a ≡ 0 (mod 4), as 2^{a−3}.3^{3}.7^{2} ≡ 1 (mod 5). Hence a ≡ 4 (mod 20), a contradiction.

Suppose that |N_{G}(P)| = 110 and |C_{G}(S)| = 30. Then a is odd, on consideration of quadratic residues (mod 11), as 2^{a−1}.3^{b}.7^{2} ≡ 1 (mod 11). Hence b is odd since 2^{a−3}.3^{b−1}.7^{2} ≡ 1 (mod 5). Now b ∈ {3, 5}. If b = 3, then a ≡ 3 (mod 10) since 2^{a−1}.3^{3}.7^{2} ≡ 1 (mod 11). However, a ≡ 3 (mod 4), since 2^{a−3}.3^{2}.7^{2} ≡ 1 (mod 5), so that a ≡ 3 (mod 20), a contradiction.

Suppose then that |C_{G}(S)| = 15. In that case, |N_{G}(P)| = 55, since G contains no element of order 10. We also have 2^{a}.3^{b}.7^{2} ≡ 1 (mod 11) and 2^{a−2}.3^{b−1}.7^{2} ≡ 1 (mod 5).

Hence a ≡ 2b + 6 (mod 10) and a ≡ b − 1 (mod 4). Thus a is even and b is odd. If a = 12, then b ≡ 3 (mod 5) and b ≡ 1 (mod 4), so b ≡ 13 (mod 20), a contradiction. If a = 14, then b ≡ 4 (mod 5) and b ≡ 3 (mod 4), so b ≡ 19 (mod 20), a contradiction. This contradiction completes the proof of the main theorem of this thesis.

References

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[2] Aschbacher, M.; Hall, M., Jr., Groups generated by a class of elements of order 3, J. Algebra 24 (1973), 591-612.

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