Blocks with trivial intersection defect groups ^{∗}

Jianbei An and Charles W. Eaton

May 31, 2002

This paper is dedicated to Jon Alperin on the occasion of his 65th birthday.

Abstract

We show that each block whose defect groups intersect pairwise trivially is Morita equivalent to a block with cyclic or generalised quaternion defect groups or to one of a given list of blocks of automorphism groups of quasisimple groups. In particular we classify all blocks of automorphism groups of non-abelian simple groups whose defect groups are non-cyclic and intersect pairwise trivially.

1 Introduction and notation

The classification given by Gorenstein and Lyons in [14] of finite simple groups possessing a strongly p-embedded subgroup, together with the classification given by Suzuki in [24], yields a classification of finite simple groups whose Sylow p-subgroups are TI (trivial intersection, meaning that each pair of distinct conjugates intersect trivially). Clifford theory then allows certain questions about groups with TI Sylow p-subgroups to be reduced to questions about finite simple groups (and their covering and automorphism groups). This approach has lead to the verification various conjectures in representation theory for this class of groups (see [5] and [7]).

By a well-known theorem of Green, when a finite group has TI Sylow p-subgroups each block must either have defect zero or maximal defect, and hence such blocks must have TI defect groups. A natural generalisation is to consider all blocks with TI defect groups (call these TI defect blocks). Here we classify all TI defect blocks of finite groups of the form M ≤ G ≤ Aut(M), where M is quasisimple. Clifford-theoretic arguments then allow us to establish Morita equivalences between any given TI defect block and a member of one of these Morita equivalence classes or a block with cyclic or generalised quaternion defect groups. The Clifford-theoretic arguments are essentially those of Fong, observing that similar arguments apply when the covered block is a block of defect zero, rather than just a block of a p′-group. As such the arguments may be considered elementary.

Behind the reduction lies the fact that TI defect blocks are precisely those with p-local rank one as defined in [3], and that the p-local rank is compatible with any reductions we make. Note that the class of TI defect blocks contains the class of TI blocks (see [4]).

∗This research was supported in part by the Marsden Fund of New Zealand via grant UOA 810.

2000 Mathematics Subject Classification. Primary 20C20

Our main result is the following:

Theorem 1.1 Let G be a finite group and B ∈ Blk(G) have non-normal TI defect groups. Then B is Morita equivalent to one of the following:

(a) a block with cyclic or generalised quaternion defect groups;

(b) a 2-block of A_{n} with Klein-four defect groups, where n = m^{2}/2 + m + 4 or n = m^{2}/2 + m + 6 for some integer m;

(c) the unique block of J_{2} or Ru with Klein-four defect groups;

(d) the unique block of O′N, Aut(O′N), 2.Suz, Aut(Suz) with defect groups of the form C_{3} × C_{3};

(e) the principal 3-block of M_{11};

(f) a 5-block of maximal defect of 3.McL or Aut(McL);

(g) the principal 11-block of J_{4};

(h) a block of Sp_{2m}(3) with defect groups of the form Q_{8}, where m ≥ 4;

(i) a p-block of maximal defect of a group X with N ≤ X ≤ Aut(N), where (p, [X : N]) = 1 and N is a p′-central extension of P SL_{2}(p^{m}) or P SU_{3}(p^{m}), where m > 1;

(j) a 2-block of maximal defect of a group X with N ≤ X ≤ Aut(N), where N ≅ 2B_{2}(2^{2m+1}), where m ≥ 1;

(k) a 3-block of maximal defect of a group X with N ≤ X ≤ Aut(N), where (3, [X : N]) = 1 and N ≅ ^{2}G_{2}(3^{2m+1}), where m ≥ 1;

(l) the principal 3-block of Aut( ^{2}G_{2}(3)′);

(m) the principal 5-block of ^{2}F_{4}(2)′, ^{2}F_{4}(2) or Aut( ^{2}B_{2}(2^{5})).

(n) a 3-block of maximal defect of a group X with N ≤ X ≤ Aut(N), where (3, [X : N]) = 1 and N is a 3′-central extension of P SL_{3}(4).

Further (b)-(n) comprise a complete list of non-cyclic TI defect blocks occuring in automorphism groups of quasisimple groups.

Note that we do not assert that the above are representatives of distinct Morita equivalence classes - this is certainly not the case. The main objective is to give a framework within which results about TI defect blocks may be proved.

Remark 1.2 Noting that Donovan’s conjecture holds for tame blocks (i.e., those with cyclic, dihedral, quasidihedral or quaternion defect), it follows immediately from Theorem 1.1 that for a given p-group D there are finitely many Morita equivalence classes of block with non-normal TI defect group D.

The paper is structured as follows. In what remains of this section we outline some general notation and definitions. Section 2 contains a collection of properties of the TI defect blocks and the p-local rank. In section 3 we give an elementary account of the correspondences used in the reduction step, and in section 4 we give the reduction itself. Sections 5 to 7 contain the analyses of the alternating groups, sporadic simple groups and groups of Lie type comprising the classification. Section 8 contains the proof of Theorem 1.1 itself.

Let G be a finite group and p a prime. Then H ≤ G is a TI subgroup if H^{g} ∩H = 1 for each g ∈ G − N_{G}(H). Here we also stipulate that G ≠ N_{G}(H).

Let R be a local complete discrete valuation ring containing a primitive |G|^{3} root of unity, whose residue field k = R/J(R) is algebraically closed of characteristic p and whose field of fractions K has characteristic zero. Denote by Blk(G) the set of blocks of G with respect to this system and let B ∈ Blk(G). If H ≤ G, then denote by Blk(H, B) the set of blocks of H with Brauer correspondent B. If H ◁ G, write Blk_{b}(G) for the set of blocks of G covering b ∈ Blk(H).

A radical p-subgroup of G is one with Q = O_{p}(N_{G}(Q)), where O_{p}(H) is the unique maximal normal p-subgroup of H. A radical p-chain σ of G is a chain Q_{0} < ⋯ < Q_{n} of p-subgroups of G, with strict inclusions, such that Q_{0} is radical in G and Q_{i+1} is a radical p-subgroup of N_{G}(Q_{0}) ∩ ⋯ ∩ N_{G}(Q_{i}) for each i with 0 ≤ i ≤ n − 1. Let |σ| = n be the length of σ and G_{σ} = N_{G}(Q_{0}) ∩ ⋯ ∩ N_{G}(Q_{n}). Denote by R(G) the set of radical p-chains of G. Following [3] and [23], define the p-local rank to be

plr(B) = max{|σ| : σ ∈ R(G, B)},

where R(G, B) ⊆ R(G) consists of those chains σ for which Blk(G_{σ}, B) ≠ ∅. By [3, 5.1] plr(B) = 1 if and only if B has a defect group D for which D/O_{p}(G) ≠ 1 and is a TI subgroup of G/O_{p}(G).

Write Irr(G) for the set of irreducible characters of G. If N ◁G and μ ∈ Irr(N), then write Irr(G, μ) for the set of irreducible characters of G covering μ. Denote by Irr(G, B) the set of irreducible characters belonging to B, and more generally, for H ≤ G write Irr(H, B) for the set of irreducible characters of H belonging to Brauer correspondents of B.

2 General properties of TI defect blocks and the p-local rank

Lemma 2.1 Let N be a normal subgroup of a finite group G and D a TI radical subgroup of G. If Q = D ∩ N ≠ 1, then N_{G}(Q) = N_{G}(D) and Q is a TI radical subgroup of N. In particular, if b ∈ Blk(N) is covered by a TI defect block B ∈ Blk(G), then b has a TI defect group. If further O^{p′}(G) ≤ N, then b and B have a defect group in common.

Proof. Suppose W is a defect group of b. Then W = N ∩ R for some defect group R of B (see for example [1, 15.1]). Thus either W = 1 or W ≠ 1. In both cases W is a TI subgroup. The last statement is trivial. □

We observe that the p-local rank of a block is compatible with the reductions of later sections:

Proposition 2.2 Let B ∈ Blk(G) have defect group D. Let H ≤ G and N ◁G. Then: (i) plr(B) = 0 if and only if D ◁ G; plr(B) ≤ 1 if and only if D/O_{p}(G) is TI; (ii) If b ∈ Blk(H, B), then plr(b) ≤ plr(G); (iii) Suppose μ ∈ Irr(N) is G-stable and extends to θ ∈ Irr(G). If ψ ∈ Irr(G/N)

lies in the block b_{N} of G/N and θψ ∈ Irr(G, B), then plr(b_{N}) ≤ plr(B). If further plr(B) = 1 and O_{p}(G) = 1, then b_{N} either has a normal defect group or TI defect groups;

(iv) If G = G_{1} × G_{2} and B = B_{1} × B_{2}, where B_{i} ∈ Blk(G_{i}), then plr(B) = plr(B_{1}) + plr(B_{2});

(v) If b ∈ Blk(N) is covered by B, with plr(B) = 1 and O_{p}(G) = 1, then plr(b) ≤ 1 and O_{p}(N) = 1;

(vi) If N ≤ Z(G), plr(B) = 1 and \overline{B} is the unique block of G/N corresponding to B under the natural epimorphism, then plr(B) = plr(\overline{B}).

Proof. (i) is [3, 5.1]; (ii) is [3, 3.2]; the first part of (iii) is [3, 4.1], and the second part follows from its proof; (iv) is [3, 3.5]; (v) follows immediately from (i) and Lemma 2.1; (vi) follows from (i), noting that O_{p}(Z(G)) ≤ O_{p}(G) is contained in every defect group. □

Lemma 2.3 Suppose that B ∈ Blk(G) has a TI defect group D. Let 1 ≠ x ∈ Z(D). Then D = O_{p}(L) whenever C_{G}(x) ≤ L ≤ N_{G}(⟨x⟩), and L possesses a block with Brauer correspondent B and defect group D.

Proof. Write Q = ⟨x⟩ and suppose C_{G}(Q) ≤ L ≤ N_{G}(Q). Then D ≤ L ≤ N_{G}(Q) ≤ N_{G}(D), so that D ◁ L. But there is a block b_{Q} of L with b_{Q}^{G} = B, which must have a defect group D(b_{Q}) satisfying D ≤ O_{p}(L) ≤ D(b_{Q}) ≤ D. □

Lemma 2.4 Let D be a p-subgroup of G, and N ≤ O_{p′}(Z(G)). If D ∉ Syl_{p}(G) and C_{G/N}(DN/N) ≤ DN/N, then there is no block with defect group D.

Proof. It’s well known that if D is a defect group, then D = P^{g} ∩ P for some P ∈ Syl_{p}(G) and some g ∈ C_{G}(D) (see, for example [1, 13.6]). If C_{G/N}(DN/N) ≤ DN/N, then P^{g} = P for every P ∈ Syl_{p}(G) containing D and every g ∈ C_{G}(D), so D ∈ Syl_{p}(G), a contradiction. □

Lemma 2.5 Let G be a finite group with O_{p}(G) = 1 and P a p-subgroup of G. Then P is a non-trivial TI radical subgroup of G if and only if M = N_{G}(P) is a maximal p-local subgroup of G with P = O_{p}(M) a non-trivial TI subgroup of G.

Proof. Let P be a non-trivial TI radical subgroup of G. Then N_{G}(P) is plocal and so N_{G}(P) ≤ M for some maximal p-local subgroup M of G. In particular, O_{p}(M) ≤ O_{p}(N_{G}(P)) = P. If x ∈ M, then O_{p}(M) = O_{p}(M)^{x} ≤ P ∩ P^{x}, so that x ∈ N_{G}(P) and M = N_{G}(P). Thus O_{p}(M) = P is TI.

Conversely, suppose M is maximal p-local and 1 ≠ P = O_{p}(M) is a TI subgroup. If P is non-radical, then M < N_{G}(P), so that N_{G}(P) = G and P ≤ O_{p}(G), which is impossible. □

Note that the following lemma applies in particular to defect groups.

Lemma 2.6 Let P be a radical p-subgroup of G, where G = G_{1}×G_{2}. Then P = P_{1}×P_{2} for uniquely defined radical p-subgroups P_{1}, P_{2} of G_{1} and G_{2} respectively. Conversely, P_{1} × P_{2} is a radical p-subgroup of G whenever P_{1}, P_{2} are radical subgroups of G_{1}, G_{2}.

Proof. See [21, 2.2]. □

In performing the reductions it may happen that we reduce to the case where the block has a normal defect group. We will see that, once we have reduced to the case where every normal p′-subgroup is central, this happens only when the defect groups intersect trivially with the generalized Fitting subgroup. The following two results show that this implies that the defect group cannot contain an elementary abelian subgroup of order p^{2}, i.e., that the defect groups are either cyclic or generalised quaternion (cf. [13, Theorem 5.4.10]).

Recall that a quasisimple group H is a perfect group where H/Z(H) is non-abelian simple. A component of G is a subnormal quasisimple subgroup. E(G) is the normal subgroup of G generated by the components and F^{∗}(G) = E(G)F (G) is the generalised Fitting subgroup.

Lemma 2.7 Let N ◁ G and D ≤ G be a TI defect group for some block B ∈ Blk(G) such that D ∩ N = 1. Then C_{N}(x) = C_{N}(D) for each x ∈ D − {1}. In particular for each n ∈ N, either C_{D}(n) = D or C_{D}(n) = 1.

Proof. Let x ∈ D − {1}. Then C_{N}(D) ≤ C_{N}(x). Suppose that n ∈ C_{N}(x) − C_{N}(D). Then there is y ∈ D such that n ∉ C_{N}(y). Now y^{n} = y[y, n] and [y, n] ∈ N, so y^{n} ∉ D, for otherwise 1 ≠ [y, n] ∈ D ∩ N = 1. Hence D^{n} ≠ D whilst 1 ≠ x ∈ D^{n} ∩ D, a contradiction. So C_{N}(x) = C_{N}(D). The last part follows immediately. □

Proposition 2.8 Suppose that D is a (non-normal) TI defect group for B ∈ Blk(G), where O_{p′}(G) ≤ Z(G). If D ∩ F^{∗}(G) = 1, then D is either cyclic or generalised quaternion.

Proof. Suppose that D possesses a subgroup Q = ⟨x, y : x^{p} = y^{p} = [x, y] = 1⟩ ≅ C_{p} × C_{p} (otherwise D is cyclic or generalised quaternion).

Note that C_{G}(F^{∗}(G)) ≤ F^{∗}(G), that E(G) ≠ 1 and that every component has order divisible by p. We claim first that Q fixes every component of G.

Write M_{1}, …, M_{n} for the components of G, and H = ∩_{i=1}^{n}N_{G}(M_{i}). Suppose that Q ≰ H. Consider the permutation action of Q on {1, …, n} induced by permutation of the components (by conjugation). Suppose that x ∈ H but y ∉ H. Let M_{i} lie in an orbit of length p of y. Since p| |M_{i}|, x must fix some non-trivial m ∈ M_{i}. Clearly y does not fix m, a contradiction by Lemma 2.7. Hence Q ∩ H = 1, so that Q acts faithfully. Without loss of generality we may assume that x moves the points {1, …, p} transitively but y doesn’t. Now y fixes one of {1, …, p}, say 1, or moves one of these points to the outside of the x-orbit. In the first case we may choose non-trivial m ∈ M_{1} fixed by y, since p| |M_{1}|, obtaining a contradiction by Lemma 2.7. In the latter case, x fixes some non-trivial m ∈ M_{1} ∗ ⋯ ∗ M_{p}. But y ∉ C_{G}(m), again a contradiction to Lemma 2.7. Hence Q ≤ H as claimed.

Since C_{G}(F^{∗}(G)) ≤ F^{∗}(G), there must be a component, say M, not centralized by Q. We examine the various possibilities for MQ = M ⋊ Q.

We may assume that M is simple, since otherwise we may just quotient by the centre of M (which is contained in the centre of G). Since C_{MQ}(M) = 1, no element of Q (which we may regard as a subgroup of Aut(M)) may act as an inner automorphism.

Since Q ≅ MQ/M is a subgroup of Out(M), it follows that Out(M) is non-cyclic, so M is a group of Lie type (we are making use of the isomorphism A_{6} ≅ P SL_{2}(9)). We may write x = x_{i}x_{o} and y = y_{i}y_{o}, where x_{i}, y_{i} are inner automorphisms, and x_{o} = x_{d}x_{f}x_{g}, y_{o} = y_{d}y_{f}y_{g}, where x_{d}, y_{d} are diagonal automorphisms, x_{f}, y_{f} are field automorphisms and x_{g}, y_{g} are graph automorphisms. Let r be the characteristic of the field of definition of M. Note that Lemma 2.7 tells us that ⟨C_{M}(F) : 1 ≠ F ≤ Q⟩ = C_{M}(Q) ≠ M.

Suppose first that p ≠ r. Then by [15, 7.3.4], the above discussion and the structure of Out(M) we must have ⟨C_{M}(F) : 1 ≠ F ≤ Q⟩ = M (see the proof Case (1) of Proposition 7.3), a contradiction.

Suppose that p = r. Then M possesses no diagonal automorphisms and we may choose x = x_{i}x_{f}, y = y_{i}y_{g}. By [15, 4.9.1.d] we may replace D by an M-conjugate if necessary to further assume that x = x_{f}. But then by [15, 7.3.8] we must have ⟨C_{M}(F) : 1 ≠ F ≤ Q⟩ = M (see the proof Case (2) of Proposition 7.3), again a contradiction, so we are done. □

3 Some correspondences like Fong’s

In [10] Fong studies modular representation theory with respect to a normal p′- subgroup, establishing the well-known Fong correspondences. These give a Morita equivalence between a block of the original group and a block of a p′-central extension of a certain section of that group. Now the same general methods may be applied for blocks of defect zero of any normal subgroup (thus generalising the results of [10]). This is nothing new, since this situation has been studied, among others, by Dade, and Külshammer and Puig (see [17], where strong results are proved concerning blocks covering nilpotent blocks). The results of [17] may be applied directly to establish the equivalences we require, but the authors feel that since the full strength of [17] is not being used, readers may benefit from a more elementary treatment. Our starting point is the treatment given in [8], where correspondences like Fong’s are given along with an elementary proof that the correspondences respect blocks. Our proof makes use of techniques from [10].

Let B ∈ Blk(G), where O_{p}(G) = 1. We suppose that B has positive defect. Let N ◁ G. Suppose that B covers b ∈ Blk(N), a block of defect zero. Let ζ be the unique irreducible character lying in b. It is well-known (see, for example [9, V.2.5]) that there is a unique block B_{I} of I_{G}(ζ), the inertia subgroup of ζ in G, such that induction gives a Morita equivalence between B and B_{I}. Hence it suffices to consider the case I_{G}(ζ) = G.

In the following paragraph we summarise section 1 of [8]:

We may choose a central extension Ĝ of G (with Ŵ ≤ Z(Ĝ) where Ĝ/Ŵ ≅ G and Ŵ ≤ [Ĝ, Ĝ]) such that there is an irreducible character θ̂ ∈ Irr(Ĝ) extending ζ.

All extensions Ĥ of H by A derived in this way that we consider in this paper satisfy A ≤ [Ĥ, Ĥ]. We therefore do not make further explicit reference to the fact. Let N̂ be the subgroup of Ĝ identified with N, so that N̂ ∩ Ŵ = 1. Set G̃ = Ĝ/N̂, a central extension of G/N by W̃ (where W̃ is the image of Ŵ under the natural epimorphism). Then θ̂ lies over a unique linear character μ̂ of Ŵ. Let μ̃ be the complex conjugate of μ̂, regarded as a character of W̃. There is a 1-1 correspondence between Irr(G, ζ) and Irr(G̃, μ̃), given by χ ↔ θ̂χ̃, where of course we are identifying χ with its inflation to Ĝ. Now there is a collection of blocks B̃_{1}, …, B̃_{r} of G̃ (the Dade correspondents of B) so that, writing B̃ = B̃_{1} + … + B̃_{r}, there is a correspondence between Irr(G, B, ζ) and Irr(G̃, B̃, μ̃), where implicitly we are using the 1-1 correspondence between blocks of G and blocks of Ĝ covering the principal block of O_{p′}(Ŵ). Note that the Dade correspondence respects the Brauer correspondence as described in [8]. Let S be an RN-module affording ζ and let T be an RĜ-module extending S. We observe that the functor M → (T^{∗} ⊗_{R} M)N̂ gives rise to a Morita equivalence between B and B̃ (for details see [18]).

We claim that we may choose W̃ to be a p′-group. For consider Ĝ/O_{p}(Ŵ). Let l ≠ p be a prime. Let P_{l} ≤ Ĝ contain N̂ × Ŵ so that P_{l}/(N̂ × Ŵ) is a Sylow l-subgroup of Ĝ/(N̂ × Ŵ). Now ζ extends to Ĝ so extends to ϕ ∈ Irr(P_{l}). Let α ∈ Irr(N̂ × Ŵ) be the canonical extension of the complex conjugate of the unique irreducible constituent of ϕ on restriction to O_{p}(Ŵ) (i.e., α is trivial on O_{p′}(Ŵ) × N̂). By [16, 8.16] α extends to, say ψ ∈ Irr(P_{l}). Then ϕψ is also an extension of ζ to P_{l} by standard Clifford theory. Note that O_{p}(Ŵ) ≤ Ker(ϕψ), so ζ (after appropriate identifications are made) extends to P_{l}/O_{p}(Ŵ). Now let P_{p} ≤ Ĝ contain N̂ × O_{p′}(Ŵ) so that P_{p}/(N̂ × Ŵ) is a Sylow p-subgroup of Ĝ/(N̂ × Ŵ). Identify b with the block of defect zero containing the canonical extension of ζ to N̂ × O_{p′}(Ŵ). By [9, V.3.5] there is a unique block of P_{p}/O_{p}(Ŵ) covering b, and so by [1, 15.1] this has a defect group, Q say, such that Q ∩ (N̂ × O_{p′}(Ŵ)) = 1 and P_{p}/O_{p}(Ŵ) = Q(N̂ × O_{p′}(Ŵ)), i.e., P_{p}/O_{p}(Ŵ) = (N̂ × O_{p′}(Ŵ)) ⋊ Q, and ζ extends to P_{p}/O_{p}(Ŵ). Hence ζ extends P_{q}/O_{p}(Ŵ) for every prime q, and so extends to Ĝ/O_{p}(Ŵ) by [16, 11.31] as claimed. In other words we may, and do, assume that O_{p}(Ŵ) = 1, so that Ŵ is a p′-group. A consequence of this is that Irr(G̃, B̃, μ̃) = Irr(G̃, B̃).

We now observe that in fact r = 1, so that B is Morita equivalent to a block B̃ of G̃. As with all results in this section, we make no claims to originality.

Proof. Let Λ be the subset of Irr(G, B, ζ) = Irr(G, B) of characters corresponding to characters of B̃_{1}. We show that Λ = Irr(G, B), which gives r = 1. For convenience, identify Λ with the set of inflations of its elements to Ĝ.

By block orthogonality (see for example [20, 3.7]), whenever x̃ ∈ G̃_{p′} (the set of p-regular elements of G̃) and ỹ ∈ G̃ − G̃_{p′} (i.e., ỹ is p-singular), we have

∑_{χ̃∈Irr(G̃,B̃_{1})} χ̃(x̃)χ̃(ỹ) = 0.

We use the converse (due to Osima) of the block orthogonality result to show that Λ is the set of irreducible characters belonging to a collection of blocks of Ĝ and hence is Irr(Ĝ, B̂), which would give the required result.

Let x̂ ∈ Ĝ_{p′} and ŷ ∈ Ĝ − Ĝ_{p′}. Recall that G̃ = Ĝ/N̂. Now x̂N̂ is p-regular, but we can’t guarantee that ŷN̂ is p-singular. We have

∑_{χ̂∈Λ} χ̂(x̂)χ̂(ŷ) = ∑_{χ̃∈Irr(G̃,B̃_{1})} χ̃(x̂N̂)θ̂(x̂)χ̃(ŷN̂)θ̂(ŷ) = θ̂(x̂)θ̂(ŷ) ∑_{χ̃∈Irr(G̃,B̃_{1})} χ̃(x̂N̂)χ̃(ŷN̂).

If ŷN̂ is p-singular, then this is zero by block orthogonality. If ŷN̂ is p-regular, and ŷ has order c, then ŷ^{c_{p′}} ∈ N̂, where c_{p′} is the p′-part of c. But ŷ^{c_{p′}} is the unique p-part of ŷ, and so is non-trivial. But then θ̂(ŷ^{c_{p′}}) = ζ(ŷ^{c_{p′}}) = 0 since ζ is in a block of defect zero. Hence ∑_{χ̂∈Λ} χ̂(x̂)χ̂(ŷ) = 0, and by [22, Theorem 3] Λ is the set of irreducible characters belonging to a collection of blocks as required. □

Write B̃ = B̃_{1} for the unique Dade correspondent. Note that at each stage, since we have a Morita equivalence, B, B_{I} and B̃ have isomorphic defect groups.

In summary, we have

Theorem 3.1 Let B ∈ Blk(G) be a block of positive defect and N ◁G, with O_{p}(G) = 1. Suppose that B covers a p-block b of defect zero of N. Write I = I_{G}(b). Then there is a central extension Ĩ of I/N by a cyclic p′-group W̃ and a block b̃ (of defect zero) of W̃ such that B is Morita equivalent to a block B̃ ∈ Blk(Ĩ) covering b̃. Further, if plr(B) = 1, then either plr(B̃) = 1 and O_{p}(G̃) = 1, or plr(B̃) = 0.

Proof. It remains to observe that each step we have made is compatible with the p-local rank of a block. Since B_{I} ∈ Blk(I_{G}(b)) has Brauer correspondent B, by part (ii) of Proposition 2.2 we have plr(B_{I}) ≤ plr(B) = 1. We must rule out the case that plr(B_{I}) = 0, i.e., that B_{I} has a normal defect group. Note that [O_{p}(I), N] ≤ O_{p}(I) ∩ N ≤ O_{p}(N) = 1. Hence O_{p}(I) ≤ C_{G}(N) ≤ I. But then O_{p}(I) ◁ C_{G}(N), and O_{p}(I) ≤ O_{p}(C_{G}(N)) ≤ O_{p}(G) = 1, since C_{G}(N) ◁ G. Hence B_{I} cannot have a normal defect group, since we are assuming that B (and so B_{I}) has positive defect.

By part (vi) of Proposition 2.2 plr(B̂) = plr(B_{I}) = 1, where B̂ is the unique block of Ĝ naturally corresponding to B_{I}. The rest follows from part (iii) of Proposition 2.2. □

4 Reduction to covers of automorphism groups of simple groups

Let G be a finite group and B ∈ Blk(G) with TI defect groups. We assume that B does not have a normal defect group, so that O_{p}(G) = 1.

Let E(G) be a central product M_{1} ∗ ⋯ ∗ M_{s} of normal subgroups of G, where each M_{i} is a central product M_{i1} ∗ ⋯ ∗ M_{it} of quasisimple groups. For each i, G acts transitively on M_{i1}, …, M_{it}.

Lemma 4.1 Consider B as above, and suppose that D possesses a subgroup of the form C_{p} × C_{p}. Let H ◁ G be a p′-group. Then DH/H is a non-normal TI subgroup of G/H.

Proof. Write \overline{G} for the quotient by H. Since

H = ⟨C_{H}(x) : 1 ≠ x ∈ D⟩ ≤ N_{G}(D)

(see [13, 6.2.4]), it follows that H ≤ N_{G}(D), so D ≤ C_{G}(H). Then N\overline{G}(\overline{D}) = \overline{N_{G}(D)} and the result follows. □

Theorem 4.2 B as above is Morita equivalent to a block with cyclic or generalised quaternion defect groups or a block of a finite group X where Z(X) is a p′-group and M ≤ X/Z(X) ≤ Aut(M), where M is non-abelian simple.

Proof. First observe that by Theorem 3.1 (with N = O_{p′}(G)) B is Morita equivalent to a block B̃ of G̃, where O_{p′}(G̃) ≤ Z(G̃) and either plr(B̃) = 0 or plr(B̃) = 1 and O_{p}(G̃) = 1. Now G̃ = Ĝ/N̂ in the notation of the previous section, where N = O_{p′}(G). If B (and hence B̃) does not have cyclic or generalised quaternion defect groups, then by Lemma 4.1 the image of D in G̃ (a defect group of B̃) is non-normal and trivial intersection. So we assume that O_{p′}(G) ≤ Z(G). Let D be a defect group for B. If E(G) = 1, then C_{G}(F^{∗}(G)) ≤ F^{∗}(G) = O_{p′}(G) ≤ Z(G), and G is abelian, contradicting out assumption that N_{G}(D) ≠ G. Hence we may assume that E(G) ≠ 1, and further that every component has order divisible by p.

If D∩E(G) = 1, then by Proposition 2.8 D is again cyclic or generalised quaternion. Hence we assume that D ∩ E(G) ≠ 1.

Consider M_{i} ◁ G, and let b_{i} ∈ Blk(M_{i}) be covered by B. By Proposition 2.2 (v) either b_{i} has positive defect and (non-normal) TI defect groups, or has defect zero. If b_{i} has defect zero, then by Theorem 3.1 B is Morita equivalent to a block B̃ of a central extension of I_{G}(b_{i})/M_{i} (where B̃ has TI defect groups or cyclic or generalised quaterion defect groups), and so we may assume that b_{i} has positive defect, and p-local rank one. We may also assume that b_{i} is G-stable.

Now M_{i} = M_{i1} ∗ ⋯ ∗ M_{it}, and b_{i} = b_{i1} × ⋯ × b_{it}. But b_{ij} = b_{ik} for all j, k since b_{i} is G-stable and G acts transitively on the components of M_{i}. By Proposition 2.2 (iv) we have 1 = plr(b_{i}) = t(plr(b_{i1})), so t = 1. Similarly the block b = b_{1} × ⋯ × b_{s} of E(G) has p-local rank one, and s = plr(b_{1}) + ⋯ + plr(b_{s}) = plr(b) = 1. Hence G has a unique component, say M. Note that M is the unique minimal non-central normal subgroup of G. Hence C_{G}(M) = Z(G), else G would have more than one component. Hence M ≤ G/Z(G) ≤ Aut(M) as required. □

For any χ ∈ Irr(G), denote by d(χ) and κ(χ) the nonnegative integers such that 1 ≤ κ(χ) ≤ (p − 1),

p^{d(χ)} = \frac{|G|_{p}}{χ(1)_{p}} and κ(χ) ≡ \frac{|G|_{p′}}{χ(1)_{p′}} (mod p).

If H ≤ G and ξ is a character of H, then

d(Ind_{H}^{G}(ξ)) = d(ξ) and κ(Ind_{H}^{G}(ξ)) = κ(ξ).

Remark 4.3 Let B and B̃ be the blocks given by Theorem 4.2 such that B is Morita equivalent to B̃. Suppose χ ∈ Irr(B) corresponds to χ̃ ∈ Irr(B̃). Then

d(χ) = d(χ̃) and κ(χ) ≡ κ_{0}κ(χ̃) (mod p)

for some integer κ_{0} ≢ 0 (mod p) dependent only on the block B.

Indeed, in the notation of Section 3, χ ↔ θ̂χ̃ with θ̂(1) = ζ(1), so

d(χ) = d(ζ) + d(χ̃) = d(χ̃) and κ(χ) ≡ κ_{0}κ(χ̃) (mod p),

where κ_{0} ≡ κ(ζ) (mod p). Thus the remark follows by the note above.

In the following sections we classify all TI defect blocks of p′-central extensions of groups M ≤ H ≤ Aut(M), where M is a non-abelian simple group. For the sake of space, and since blocks with cyclic defect groups are well understood, we only give a classification of TI defect blocks with non-cyclic defect groups.

5 Alternating groups

Let Z(G) = O_{p′}(G) ≤ M ≤ G where M/Z(G) ≤ G/Z(G) ≤ Aut(M/Z(G)) and M/Z(G) ≅ A_{n}, an alternating group, n ≥ 5. Write \overline{H} for Z(G)H/Z(G) whenever H ≤ G, and A^{π} for the preimage of A ≤ \overline{G} in G. We demonstrate that, but for a small list of exceptions, the only blocks B ∈ Blk(G) of p-local rank one are those which have defect groups of order p.

For the sake of exposition it is convenient to work first of all with the symmetric groups. For odd primes this is sufficient.

Proposition 5.1 Suppose that \overline{G} = S_{n}, where n ≥ 5, and suppose B ∈ Blk(G) has a TI defect group D ≠ 1. Then |D| = p unless (n, p) = (6, 3), in which case D ≅ C_{3}×C_{3}.

Proof. Let x ∈ Z(D) have order p, and write \overline{x} = xZ(G) ≠ Z(G). Write Q = ⟨x⟩. Note that C\overline{G}(\overline{x})^{π} ≤ N_{G}(Q). Suppose that \overline{x} consists of exactly k cycles of length p and n − kp of length one. Then C\overline{G}(\overline{x}) ≅ (C_{p} ≀ S_{k}) × S_{n−pk}.

Suppose first that p > 3, so that in particular O_{p}(S_{m}) = 1 for every m. By Lemma 2.3 and Lemma 2.6 D = O_{p}(C\overline{G}(\overline{x})^{π}) ≅ (C_{p})^{k}. [We will be implicitly using Lemma 2.6 in this way repeatedly throughout this proof, so we will no longer make explicit reference to it]. Suppose that k > 1. We may choose an element \overline{y} ∈ \overline{D} = Z(\overline{D}) consisting of precisely one cycle of length p. Then C\overline{G}(\overline{y}) ≅ C_{p} × S_{n−p}, so D = O_{p}(C\overline{G}(\overline{y})^{π}) ≅ C_{p}, a contradiction. Hence k = 1 and D ≅ C_{p} as required.

Now suppose that p = 3. If k > 3, then D = O_{p}(C\overline{G}(\overline{x})^{π}) ≅ (C_{p})^{k} × O_{3}(S_{n−3k}). As above, there is \overline{y} ∈ Z(\overline{D}) consisting of just one cycle length 3. We have C\overline{G}(\overline{y}) ≅ C_{p} × S_{n−3}, and so D = O_{p}(C\overline{G}(\overline{y})) ≅ C_{p} × O_{3}(S_{n−3}). Since k > 3 we have n > 9, so that O_{3}(S_{n−3}) = 1, giving a contradiction.

If k = 3, then D = O_{3}(C\overline{G}(\overline{x})^{π}) ≅ (C_{3} ≀ C_{3}) × O_{3}(S_{n−9}). Write D = D_{1} × D_{2}, where D_{1} ≅ C_{3} ≀C_{3} and D_{2} ≅ O_{3}(S_{n−9}). Note that D_{1} is isomorphic to a Sylow 3-subgroup of S_{9}. Now S_{9} does not have TI Sylow 3-subgroups (see [5] for example), so in particular D cannot be TI.

If k = 2, then D = O_{3}(C\overline{G}(\overline{x})^{π}) ≅ (C_{3})^{2} × O_{3}(S_{n−6}). If n ≠ 9, then D ≅ (C_{3})^{2}. Let \overline{y} ∈ Z(\overline{D}) consist of just one cycle of length 3. Then C\overline{G}(\overline{y}) ≅ C_{3} × S_{n−3}, so D = O_{p}(C\overline{G}(\overline{y})^{π}) ≅ C_{3} × O_{3}(S_{n−3}). If n ≠ 6, then |D| = 3 and we are done in this case. If n = 9, then again we may choose y ∈ Z(D) consisting of just one cycle of length 3. Then C\overline{G}(\overline{y}) ≅ C_{3} × S_{6}, so D ≅ C_{3}, a contradiction. If n = 6, then G has self-centralizing TI Sylow 3-subgroups, and so G has (two) blocks with TI defect groups isomorphic to C_{3} × C_{3}.

Finally suppose that p = 2. In this case we may take Z(G) = 1 unless n = 6 or n = 7 (since S_{n} has Schur multiplier of order two when n = 5 or n ≥ 8). We have D ≅ (C_{2} ≀ O_{2}(S_{k})) × O_{2}(S_{n−2k}). Suppose that O_{2}(S_{k}) = 1 (i.e., k = 3 or k ≥ 5). As usual we may choose a transposition \overline{y} ∈ Z(\overline{D}). Then C\overline{G}(\overline{y}) ≅ C_{2} × S_{n−2}, and D ≅ C_{2} × O_{2}(S_{n−2}). If n > 6, then O_{2}(S_{n−2}) = 1 so that |D| = 2. If n ≤ 6, then it is easy to find examples to show that D is not TI. If k = 2, then C\overline{G}(\overline{x}) ≅ (C_{2} ≀C_{2})×S_{n−4} and D ≅ (C_{2} ≀ C_{2}) × O_{2}(S_{n−4}). Hence C\overline{G}(\overline{D}) ≤ \overline{D}, and if n ≥ 7, then D ∉ Syl_{2}(G), so that by Lemma 2.4 D cannot be TI.

If k = 4, then n > 7 and so Z(G) = 1. We have D ≅ (C_{2} ≀ (C_{2} × C_{2})) × O_{2}(S_{n−8}). We have C_{G}(x) ≅ (C_{2} ≀ S_{4}) × S_{n−8}, and this must possess a block with defect group D by Lemma 2.3. Write C_{G}(x) = H × L, where H ≅ C_{2} ≀ S_{4} and L ≅ S_{n−8}. Then H must possess a block with defect group P ≅ C_{2} ≀ (C_{2} × C_{2}). Notice that P ∉ Syl_{2}(H). Hence since C_{H}(P) = Z(P) ≤ O_{p}(H), by Lemma 2.4 P cannot be a defect group in H, so D cannot be a defect group of C_{G}(x). □

Corollary 5.2 Suppose that A_{n} ≤ G/Z(G) ≤ Aut(A_{n}), where n ≥ 5, and suppose B ∈ Blk(G) has p-local rank one and is of positive defect. If D is a defect group of B, then |D| = p unless (n, p) = (6, 3) or p = 2. If (n, p) = (6, 3), then D ≅ C_{3} × C_{3}. If p = 2, then D is of order two or is elementary abelian of order 4. In particular A_{n} possesses a 2-block with Klein-four TI defect group precisely when n = m^{2}/2 + m + 4 or n = m^{2}/2 + m + 6 for some integer m ≥ 1.

Proof. Suppose that H is a covering group of A_{n}, where n ≥ 5, and G is a covering group of S_{n} containing H. Let b be a TI defect block of H with defect groups of order greater than p. Let D be a defect group of b.

First suppose p ≠ 2. If C_{G}(D) ≤ H, then the unique block B of G is a Brauer correspondent of b and so plr(B) = plr(b) = 1 by Proposition 2.2 (ii). Hence (n, p) = (6, 3) as required. If C_{G}(D) ≰ H, then it follows that [G : N_{G}(D)] = [H : N_{H}(D)], so that B and b have precisely the same defect groups. Hence again plr(B) = 1, so that (n, p) = (6, 3).

Now suppose that p = 2. We treat the cases n = 6, 7 separately, so that we may assume as before that Z(G) = 1. Let x ∈ Z(D) have order two. Suppose that x consists of precisely k cycles of length two, so that k is even. Then C_{A_{n}}(x) ≅ ((C_{2}≀A_{k})×A_{n−2k})⟨t⟩, where t is an involution, so that H = (C_{2} ≀ A_{k}) × A_{n−2k} ◁ C_{A_{n}}(x) ◁ (C_{2} ≀ S_{k}) × S_{n−2k} = C_{S_{n}}(x). Hence D = O_{2}(C_{A_{n}}(x)) ≤ O_{2}(C_{S_{n}}(x)) = (C_{2} ≀ O_{2}(S_{k})) × O_{2}(S_{n−2k}). We show that we may assume k = 2.

Suppose that k ≠ 2. If further n − 2k ≠ 2, then O_{2}(A_{k}) = O_{2}(S_{k}) and O_{2}(A_{n−2k}) = O_{2}(S_{n−2k}), so that D ≅ (C_{2} ≀ O_{2}(S_{k})) × O_{2}(S_{n−2k}). Now C_{H}(Q) must possess a block, b_{Q} say, with defect group D, where Q = ⟨x⟩. Let b̃ be a block of H covered by b_{Q}. This must also have defect group D. Write H = H_{1} × H_{2}, where H_{1} = C_{2} ≀ A_{k} and H_{2} = A_{n−2k}. Write b̃ = b̃_{1} × b̃_{2}, where b̃_{i} is a uniquely determined block of H_{i}. Then D = D_{1} × D_{2}, where D_{i} is a (the) defect group for b̃_{i}. We have C_{H_{1}}(D_{1}) ≤ D_{1}, so such a block can only exist when D_{1} ∈ Syl_{2}(H_{1}), i.e., when k = 4. Notice that D_{1} is isomorphic to a Sylow 2-subgroup of A_{8} when k = 4, but A_{8} does not have TI Sylow 2-subgroups, so in fact D cannot be TI in this case.

Suppose that n − 2k = 2. Then C_{2} ≀ A_{k} ≤ C_{A_{n}}(x) ≤ (C_{2} ≀ S_{k}) × C_{2}. As above, we may assume n ≠ 8, so that k ≥ 6, and D is elementary abelian of order 2^{k} or 2^{k+1}. Hence we may choose z ∈ Z(D) = D consisting of precisely two cycles of length two, i.e., we may as well have assumed that k = 2.

Now suppose k = 2. We need to show that, where it exists, B must have defect groups of the form C_{2} × C_{2}. We have (C_{2})^{2} × A_{n−4} ◁ C_{A_{n}}(x) ◁ (C_{2} ≀ C_{2}) × S_{n−4}, and (C_{2})^{2} × O_{2}(A_{n−4}) ≤ D ≤ (C_{2} ≀ C_{2}) × O_{2}(S_{n−4}). There is no block with non-cyclic TI defect groups for A_{8}. If n ≠ 8, then (C_{2})^{2} ≤ D ≤ C_{2} ≀ C_{2}. Note that in this instance, C_{2} ≀ C_{2} is isomorphic to a subgroup of S_{4}, but there is no such subgroup of A_{4}, so D ≅ (C_{2})^{2}. Now C_{A_{n}}(D) ≅ D × A_{n−4}, and so B does indeed exist whenever A_{n−4} possesses a p-block of defect zero. The precise conditions for the existence of such a block are well-known and easy to derive.

In the cases n = 6 and n = 7, each linear character of Z(G) extends to every normaliser of a Klein-four subgroup of H, so there is a defect-preserving 1-1 correspondence between blocks of H/Z(G) and blocks of H covering a given block of defect zero of Z(G). Hence it suffices to note that there are no TI defect blocks of A_{6} or A_{7}.

The remaining automorphism group we must consider is G/Z(G) ≅ A_{6}.2^{2}. It suffices to consider the case p = 2. By Lemma 2.1, a block B of G with TI defect groups must cover blocks of defect zero of A_{n}, since the only 2-defect groups for A_{6} are trivial and the Sylow 2-subgroups, which are not trivial intersection. Hence |D| = 2 or 4. But by [6] no involution has centralizer H with |O_{p}(H)| = 4, so |D| ≠ 4 by Lemma 2.3. □

6 Sporadic simple groups

Proposition 6.1 Suppose that N/Z(G) ≤ G/Z(G) ≤ Aut(N/Z(G)), where N/Z(G) is one of the 26 sporadic simple groups. Then B ∈ Blk(G) has non-cyclic TI defect group D if and only if one of the following occurs:

(i) D ≅ C_{2} × C_{2} and G/Z(G) is isomorphic to one of J_{2} or Ru;

(ii) D ≅ C_{3} × C_{3} and N/Z(G) ≅ O′N or Suz;

(iii) p = 3 and N/Z(G) ≅ M_{11};

(iv) p = 5 and N/Z(G) ≅ McL;

(v) p = 11 and N/Z(G) ≅ J_{4}.

Note that in cases (i) and (ii) D is not a Sylow p-subgroup and in cases (iii)-(v) D is a Sylow p-subgroup.

Throughout we use the notation of [15], including for the labelling of conjugacy classes of sporadic groups. So for instance xZ(G) ∈ 2A means that xZ(G) is a member of the conjugacy class 2A (of whichever group we are considering at the time) as labelled in [15].

As in the previous section write \overline{H} for the quotient by Z(G), and A^{π} for preimages. Of course we may take Z(G) = O_{p′}(G).

In this section let \overline{N} be a sporadic simple group and \overline{N} ≤ \overline{G} ≤ Aut(\overline{N}). Let B ∈ Blk(G) have non-cyclic TI defect group D. We use Lemma 2.3 and Lemma 2.4 to eliminate most of the possibilities for conjugacy classes containing non-trivial elements of Z(D). We first suppose that G = N.

Suppose that 1 ≠ x ∈ Z(D). First, by Lemma 2.3 D = O_{p}(N_{G}(Q)), where Q = ⟨x⟩. Hence we may eliminate the cases where O_{p}(N_{G}(Q)) is cyclic. We identify these from [15, Table 5.3], using the fact that N\overline{G}(\overline{Q})^{π} = N_{G}(Q), and list them in Table 1. Of course we consider only cases where p^{2}||G|.

\overline{G} Conjugacy classes  \overline{G} Conjugacy classes
M_{11}  none  M_{12}  2A, 3B
M_{22}  3A  M_{23}  3A
M_{24}  3A, 3B  J_{1}  2A
J_{2}  3A, 3B, 5A, 5B  J_{3}  3A
J_{4}  3A  Co_{3}  2A, 2B, 3C, 5B
Co_{2}  3B, 5B  Co_{1}  3A, 3D, 5A, 5B, 7A, 7B
HS  2B, 3A, 5B  McL 2A
Suz  3A, 5A, 5B  He 3A, 3B, 5A, 7A, 7B
Ly  2A, 3A  Ru 3A, 5B
O′N  2A  F i_{22}  2A, 3A, 5A
F i_{23}  2A, 3A, 5A  F i′_{24}  2A, 3A, 5A, 7A
HN  2A, 3A, 5A  T h 3A, 7A
F_{2} = BM  2A, 3A, 5A, 7A F_{1} = M  2A, 3A, 3C, 5A, 7A, 11A, 13A

Table 1: Conjugacy classes containing xZ(G) with O_{p}(N_{G}(⟨x⟩)) cyclic, p^{2}||G|

Of the remaining possibilities we may apply Lemma 2.4 to eliminate all but a small number of cases.

Lemma 6.2 Let \overline{G} be a sporadic simple group and xZ(G) in one of the conjugacy classes listed in Table 2. Then G cannot have a TI defect defect group D with x ∈ Z(D).

Proof. By Lemma 2.4 it suffices in each case to show that O_{p}(N_{G}(⟨x⟩)) is not a Sylow p-subgroup of G and that its quotient is self-centralizing. In all but one case this follows from consideration of the orders of centralizers of elements contained in N_{G}(⟨x⟩). The only different case is G = BM and x ∈ 3B. In this case N_{G}(⟨x⟩) ≅ 3_{+}^{1+8} : 2_{−}^{1+6} ^{·} U_{4}(2).2, and D ≅ 3_{+}^{1+8} ∉ Syl_{3}(G). The only involutions in G which can centralize a subgroup of G of order 3^{9} are those in 2A. Suppose z is an involution in C_{G}(D). Then by [6] D is a Sylow 3-subgroup of C_{G}(z), but Sylow 3-subgroups of C_{G}(z) have exponent 9 whilst D has exponent 3. Hence C_{G}(D) ≤ D. □

G  Conjugacy classes  G Conjugacy classes
M_{12}  2B  M_{11}, M_{22}  2A
M_{23}  2A  M_{24}  2A, 2B
J_{2}  2A  J_{3}  2A
J_{4}  2A, 2B, 11B  Co_{3}  3A, 3B
Co_{2}  2A, 2B, 2C, 3A  Co_{1}  2A, 2C, 3C, 5C
HS  2A, 5C  McL  3A, 3B, 5B
Suz  2A, 3B  He 2B, 7D, 7E
Ly  3B, 5A, 5B  Ru 2A
O′N  7B  F i_{22}  2B, 2C, 3B, 3C, 3D
F i_{23}  2C, 3B, 3C, 3D  F i′_{24}  2B, 3B, 3C, 3D
HN  2B, 3B, 5B, 5C, 5D, 5E  T h 2A, 3B, 3C
F_{2} = BM  2B, 2D, 3B, 5B  F_{1} = M  2B, 3B, 5B, 7B

Table 2: Conjugacy classes containing x with O_{p}(N_{G}(⟨x⟩)) ∉ Syl_{p}(G) and self-centralizing

We already know the classification of sporadic groups with TI Sylow p-subgroups (see [5]). This says that the only such cases are M_{11} with p = 3, McL with p = 5 and J_{4} with p = 11 (and their p′-covering groups). Note that in each of these cases we have one block of maximal defect for each block of defect zero of the centre. Hence we may now assume that O_{p}(N_{G}(⟨x⟩)) ∉ Syl_{p}(G).

We are thus left with, for p = 2: 2B in J_{2}; 2B in Co_{1}; 2B in Suz; 2A in He; 2B in Ru; 2B in F i_{23}; and 2C in BM; for p = 3: 3B in Co_{1}; 3C in Suz; 3A in O′N; and 3E in F i′_{24}.

In the remaining cases D ≅ C_{p} × C_{p}. Landrock in [19] gives a complete list of non-principal 2-blocks (but not for the covering groups). From this, we are able to eliminate Co_{1}, He, F i_{23} and BM for p = 2. We must eliminate Suz. If xZ(G) ∈ 2B, then \overline{N_{G}(D)} ≅ (A_{4} × L_{3}(4) : 2) : 2. In this case N_{G}(D) does not possess a block with defect group D, and so neither does G.

J_{2} and Ru each possess a single 2-block with defect group C_{2} × C_{2}. Further neither J_{2} nor Ru possesses a radical 2-subgroup of order two, so these blocks are indeed TI defect.

Suppose that \overline{G} ≅ Co_{1} and xZ(G) ∈ 3B lies in the centre of a non-cyclic TI defect group D. Then D ≅ C_{3} × C_{3}. But by [15, Table 5.3l] O_{3}(\overline{N_{G}(⟨x⟩)}) contains an element of 3A, so that D = ⟨x⟩, a contradiction.

Suppose that \overline{G} ≅ F i′_{24} and x with xZ(G) ∈ 3E lies in the centre of a non-cyclic TI defect group D. Then D ≅ C_{3} × C_{3}. But by [15, Table 5.3v] O_{3}(\overline{N_{G}(⟨x⟩)}) contains an element of 3A, so that D = ⟨x⟩, a contradiction.

If \overline{G} ≅ O′N and xZ(G) ∈ 3A, or if \overline{G} ≅ Suz and xZ(G) ∈ 3C, then by [15, Table 5.3s] and [15, Table 5.3o], \overline{N_{G}(⟨x⟩)} has the form ((C_{3} × C_{3}) · 2) × A_{6}, and the maximal normal 3-subgroup D of this is TI. By [6] N_{G}(⟨x⟩) is maximal, so is equal to N_{G}(D). This possesses a 3-block with defect group D and so by Brauer’s first main theorem a suitable B exists.

It remains to consider the automorphism groups of the sporadic groups. Since [Aut(N) : N]|2 for each sporadic simple group N, for odd primes, Aut(N) has noncyclic TI defect groups if and only if N does (by the same argument as in Corollary 5.2. It thus suffices to consider p = 2. Note that by Proposition 2.2 (v) and since blocks of Aut(N) covering blocks of defect zero of N have defect groups of order dividing two, if Aut(N) possesses a non-cyclic TI defect group, then N possesses a (possibly cyclic of order two) TI defect group. By [19] this leaves the cases Aut(J_{2}), Aut(McL) and Aut(F i_{22}).

We may eliminate Aut(J_{2}) (noting that here Z(G) = 1), since if x ∈ Z(D), where D is a non-cyclic TI defect group, is not contained in N, then D = O_{p}(N_{G}(⟨x⟩)) has order two. But by [6] the non-cyclic TI defect block of J_{2} is stabilised by the outer automorphisms, so by [1, 15.1] D must contain an element outsider of N.

We may eliminate Aut(McL) since if D is a TI defect group, then Z(D) must contain an involution outside of N. But then by [15, Table 5.3n] |D| = 2.

We may eliminate Aut(F i_{22}) since if D is a non-cyclic TI defect group, then |D| = 4, and D must contain an element x with xZ(G) ∈ 2A. But then D = ⟨x⟩. This completes the proof of Proposition 6.1.

7 Finite groups of Lie type

We will follow the notation of [15]. In particular, if K is a finite group of Lie type, then K_{u} and K_{a} are the corresponding universal and adjoint finite groups.

Lemma 7.1 Let K ∈ {SL_{m}^{ϵ}(q), E_{7}(q)_{u}, Sp_{2m}(q), Ω_{2m+1}(q), Ω_{2m}^{ϵ}(q)} with odd q and m ≥ 1, and let B be a 2-block of K with a defect group O_{2}(Z(K)), where ϵ = ± = ±1. Then one of the following holds.

(i) K = SL_{m}^{ϵ}(q) and there exists a 2-element τ ∈ GL_{m}^{ϵ}(q)\Z(GL_{m}^{ϵ}(q))K stabilizing B.

(ii) K = Ω_{2m}^{ϵ}(q) and there exists a 2-element τ ∈ SO_{2m}^{ϵ}(q)\Ω_{2m}^{ϵ}(q) stabilizing B.

(iii) There is a 2-element τ ∈ Outdiag(K) stabilizing B.

(iv) K = Sp_{2m}(q), B = E_{2}(K, (s)) for some semisimple 2′-element s in the dual group K^{∗} such that C_{K^{∗}}(s) is a Coxeter torus of K^{∗} and |C_{K^{∗}}(s)|_{2} = 2. Moreover, C_{K^{∗}}(s) ≅ GL_{1}(q^{m}) or GU_{1}(q^{m}) according as 4|q + 1 or 4|q − 1, and in the former case, m is odd. In addition, there exists a 2-element τ ∈ GSp_{2m}(q)\Z(GSp_{2m}(q))K stabilizing B.

Proof. Set ν(n) = log_{2}(n_{2}) for any integer n ≥ 1, where n_{2} is the 2-part of n.

Suppose K = E_{7}(q)_{u}, so that D(B) = Z(K) = 2. If χ is the canonical character of B, then χ can be viewed as an irreducible character of K_{a} = E_{7}(q)_{a}. Let H = Inndiag(K), χ′ an irreducible character H covering χ and B_{H} the 2-block of H containing χ′. Then the dual group H^{∗} is K. Since χ has defect 0 as a character of K_{a}, it follows that |D(B_{H})| = 1 or 2, and in the former case χ′(1) = 2χ(1) and in the later case χ′(1) = χ(1).

If χ′(1) = χ(1), then the inertia group I(B) is H and there exists a 2-element τ ∈ H\K_{a} stabilizing B. Suppose χ′(1) = 2χ(1), so that D(B_{H}) = 1. If (s, μ) is the pair of semisimple and unipotent labels of χ′, then

ν(χ′) = ν(K: C_{K}(s)) + ν(μ).

In particular, ν(μ) = ν(C_{K}(s): Z(C_{K}(s))) and |Z(C_{K}(s))|_{2} = 1, since μ(1) divides |C_{K}(s): Z(C_{K}(s))|. This is impossible as C_{2} = Z(K) ≤ C_{K}(s).

Suppose K is classical. Let V be the underlying space of K, I(V) = Isom(V) the set of all isometries on V defined in [15] and let I_{0}(V) be the subset of I(V) consisting of elements of determinant 1.

Suppose K ≠ Ω_{2m+1}(q) and Ω_{2m}^{ϵ}(q). Then K = I_{0}(V) and B = E_{2}(K, (s)) for some semisimple odd element s of the dual group K^{∗}. If K = SL_{m}^{ϵ}(q), then we may suppose s ∈ K and set s^{∗} = s. If K = Sp_{2m}(q), then let s^{∗} be the dual element of s defined in [2], so that s^{∗} ∈ K. Thus D(B) is a Sylow 2-subgroup of C_{K}(s^{∗}). Since D(B) is cyclic, it follows that C_{K}(s^{∗}) is a maximal torus T of K.

If K = Sp_{2m}(q), then T ≅ GL_{1}^{η}(q^{m}) is a Coxeter torus and |Z(K)| = 2 is the exact power of 2 dividing q^{m} − η (written 2∥q^{m} − η). If 4|q − 1, then 4|q^{m} − 1, so that η = −1 and m is arbitrary. If 4|q + 1, then 2∥q^{m} + 1 if and only if m is even. But if 2∥q^{m} + 1, then m must be odd since m is the degree of a monic irreducible polynomial Δ such that ω is a root of Δ if and only if ω^{−q} is its root (cf. [11, p.111]). In both cases, there is a 2-element τ ∈ GSp_{2m}(q)\Z(GSp_{2m}(q))K such that τ centralizes s^{∗}. It follows that τ stabilizes B.

Suppose K = SL_{m}^{ϵ}(q), so that I(V) = GL_{m}^{ϵ}(q). If T is non-cyclic, then C_{I(V)}(s) = GL_{1}^{ϵ_{1}}(q^{m_{1}})×GL_{1}^{ϵ_{2}}(q^{m_{2}}). Thus O_{2}(C_{I(V)}(s)) > O_{2}(Z(I(V)))D(B) and there exists some τ ∈ O_{2}(C_{I(V)}(s))\O_{2}(Z(I(V)))D(B) stabilizing B. If T is cyclic, then C_{I(V)}(s) = GL_{1}^{η}(q^{m′}) is a Coxeter torus of I(V), where (η, m′) = (1, m) when ϵ = 1, and (η, m′) = (1, \frac{m}{2}) or (−1, m) when ϵ = −1.

If m is odd, then D(B) = 1 and there exists some τ ∈ O_{2}(C_{I(V)}(s))\D(B) stabilizing B. If ϵ = 1 and α = ν(m), then α ≥ 1 and 2^{b+α}∥q^{m} − 1, where b ≥ 2 is an integer such that 2^{b+1}∥q^{2} − 1. In particular, 2^{b+α} > 2^{b} ≥ |D(B)|. Suppose ϵ = −1. Since m is even and since q + 1 is not a factor of q^{m} + 1, it follows that η = 1 and so m′ = m/2. But Z(I(V)) = q+1 ≤ C_{I(V)}(s), so q+1|q^{m′} −1 and m′ is even. Similarly, if α′ = ν(m′) ≥ 1, then 2^{b+α′}∥q^{m′} − 1 and 2^{b+α′} > 2^{b} ≥ |D(B)|. Thus in both cases, there exists some τ ∈ O_{2}(C_{I(V)}(s))\O_{2}(Z(I(V)))D(B) stabilizing B.

Suppose K = Ω_{2m+1}(q) or Ω_{2m}^{ϵ}(q), so that D(B) = 1 or 2. If K is abelian, then K = Ω_{2}^{ϵ}(q) = C_{\frac{q−ϵ}{2}} and the proof is trivial since SO_{2}^{ϵ}(q) = C_{q−ϵ}.

Let B̃ be the unique block of I_{0}(V) covering B, so that B̃ = E_{2}(I_{0}(V), (s)) for some semisimple 2′-element s of I_{0}(V)^{∗}. Thus D(B̃) ∩ K = D(B) and |D(B̃): D(B)| ≤ 2.

If |D(B̃): D(B)| = 2, then I(B) = I_{0}(V) and there exists a 2-element τ ∈ I_{0}(V)\K stabilizing B.

Suppose D(B̃) = D(B), and let s^{∗} ∈ I_{0}(V) be a dual element of s in I_{0}(V). Then D(B̃) is a Sylow 2-subgroup of C_{I_{0}(V)}(s^{∗}). Let C_{I_{0}(V)}(s^{∗}) = ∏_{Γ} L_{Γ}, where L_{Γ} = I_{0}(V_{0}) with V_{0} the fixed-point set of s^{∗} on V or L_{Γ} ≅ GL_{n_{Γ}}^{η_{Γ}}(q^{m_{Γ}}) for some integers n_{Γ}, m_{Γ} and sign η_{Γ}. Since D(B̃) is cyclic and K is non-abelian, it follows that C_{I_{0}(V)}(s^{∗}) ≅ GL_{1}^{ϵ}(q^{m}) is a Coxeter torus and D(B̃) = O_{2}(C_{I_{0}(V)}(s^{∗})) ≠ 1. Thus D(B) = D(B̃) = 2 and K = Ω_{2m}^{ϵ}(q).

Since −1_{V} ∈ K, it follows that \frac{m(q−1)}{2} is even or odd according as ϵ = 1 or −1. If ϵ = −1, then m is odd and 2∥q − 1, so that 2^{b}∥q^{m} + 1 and |O_{2}(C_{I_{0}(V)}(s^{∗}))| = 2^{b} ≥ 4. If ϵ = 1 and m is even, then 2^{b+ν(m)}∥q^{m} − 1 and 2^{b+ν(m)} ≥ 8. If ϵ = 1 and m is odd, then 4|q − 1 and 2^{b}∥q^{m} − 1. In all cases, |O_{2}(C_{I_{0}(V)}(s^{∗})| ≥ 2|D(B)|, which is impossible. □

Remark 7.2 (i) In the notation of Lemma 7.1 (i) and (ii), let K = SL_{m}^{ϵ}(q) or Ω_{2m}^{ϵ}(q). Then either τ = −1_{V} ∉ K or τ induces a non-trivial element of Outdiag(K).

(ii) If B̃ is a 2-block of I_{0}(V) = SO(V) such that D(B̃) = Z(SO(V)), then there exists a 2-element τ ∈ GSO(V)\Z(GSO(V))SO(V) stabilizing B̃.

Proof. Indeed, in the notation above, C_{GSO(V)}(s^{∗}) = ⟨σ, C_{GSO(V)}(s^{∗})⟩, where σ^{q−1} ∈ Z(C_{GSO(V)}(s^{∗})) and σ centralizes C_{SO(V)}(s^{∗}) (cf [12, (1A)]). Thus C_{GSO(V)}(s^{∗}) is abelian and |C_{GSO(V)}(s^{∗})|_{2} > |Z(GSO(V))C_{SO(V)}(s^{∗})|_{2}, so that there exists such a 2-element τ. □

Let Z(G) = O_{p′}(G) ≤ M ≤ G where M/Z(G) ≤ G/Z(G) ≤ Aut(M/Z(G)) and M/Z(G) is a finite non-abelian simple group of Lie type. If D is a TI defect group of a block of G, then

N_{G}(D)/Z(G) = N_{G/Z(G)}(D)

and D is a TI radical subgroup of G/Z(G). Thus we first classify TI radical subgroups of G/Z(G).

Proposition 7.3 Let K = K_{a} be a finite non-abelian simple group of Lie type and K ≤ G ≤ Aut(K). Suppose that D is a TI radical subgroup of G. Then one of the following holds.

(i) D is cyclic

(ii) C_{G}(D) ≤ D.

(iii) K = P SL_{3}(4) and D is a Sylow subgroup of K.

(iv) D is a generalized quaternion with D ∩ K = Q_{8} a quaternion, and

K ∈ {^{3}D_{4}(3), F_{4}(3), E_{6}^{ϵ}(3), E_{7}(3), E_{8}(3), P SL_{n}^{ϵ}(3), P Sp_{2n}(3)},

where n ≥ 3 and n ≠ 4 when K = P SL_{n}^{ϵ}(3).

Proof. Let H = Inndiag(K), R = H ∩ D and Q = K ∩ D.

Case (1) Suppose p and the characteristic r of the underlying field of K are distinct and D contains a non-cyclic abelian subgroup W and let E = Ω_{1}(W). Then E is a non-cyclic elementary abelian group acting on K and N_{G}(E) ≤ N_{G}(D). Let

Γ = Γ_{E,1}^{r′} = ⟨O^{r′}(C_{K}(F)) | 1 ≠ F ≤ E⟩

and Γ′ = Γ′_{E,1} = ⟨C_{K}(F) | 1 ≠ F ≤ E⟩. Then Γ′ ≤ Γ ≤ N_{G}(E) ∩ K.

Suppose Γ = K or Γ′ = K. If Q ≠ 1, then by Lemma 2.1, Q is a radical TI subgroup of K and N_{G}(Q) = N_{G}(D). But K ≤ N_{G}(D), so N_{K}(Q) = N_{G}(Q) ∩ K = K and Q ≤ O_{p}(K), which is impossible. Thus Q = 1 and KD = K × D as D = O_{p}(N_{G}(D)), which is also impossible. So Γ′ ≤ Γ < K.

Since Γ < K, it follows by [15, Theorem 7.3.3] that C_{E}(K) = 1. Applying [15, Theorem 7.3.4] we get a finite list of pairs (E, K). By Lemma 2.5, N_{G}(D) is a maximal p-local subgroup of G with E ≤ D ◁ N_{G}(D) and |N_{G}(D)|_{p} > |E| whenever E is not a Sylow subgroup of G. Note that if N_{G}(D) is contained in a maximal (non plocal) subgroup N of G, then N_{G}(D) is a maximal p-local subgroup of both N and G. Using the maximal subgroups of Aut(K) and K given in [6], we can get the possible N_{Aut(K)}(D). The possible triples (E, K, N_{Aut(K)}(D)) are given in Table 3.

\begin{matrix}|E| & K & N_{Aut(K)}(D) \\ 2^{2} & A_{1}(5) & S_{4} \\ 2^{2} & A_{1}(7) & D_{16}, S_{4} \\ 2^{2} & A_{1}(9) & S_{4}.2 × 2 \\ 2^{2} & B_{2}(3) & 2^{4}: S_{5}, 2.(A_{4} × A_{4}).2.2 \\ 2^{2}, 2^{3} & ^{2}G_{2}(3)′ & 2^{3}: 7: 3 \\ 3^{2} & A_{2}(4) & 3^{2}: 2S_{4} × 2 \\ 3^{2} & U_{4}(2) & 3_{+}^{1+2}: 2S_{4}, 3^{3}: (S_{4} × 2) \\ 3^{2} & B_{2}(2)′ & 3^{2}: D_{8} \\ 3^{2} & C_{3}(2) & 3_{+}^{1+2}: 2S_{4}, 3^{3}: (S_{4} × 2) \\ 3^{2} & C_{4}(2) & 3^{2}: D_{8} × S_{6}, S_{3} ≀ S_{4} \\ 3^{2} & G_{2}(2)′ & 3_{+}^{1+2}: 8: 2 \\ 3^{2} & ^{2}F_{4}(2)′ & 3^{2}: 2S_{4}, 3_{+}^{1+2}: SD_{16} \\ 3^{2} & L_{2}(8) & 9: 6 \\ 5^{2} & ^{2}F_{4}(2)′ & 5^{2}: 4S_{4} \\ 5^{2} & ^{2}B_{2}(2^{5}) & 25: 20 \end{matrix}

Table 3: The possible triples (E, K, N_{Aut(K)}(D))

If Aut(K) = G = K = C_{4}(2) and N_{G}(D) = 3^{2}: D_{8} × S_{6} < (S_{6} × S_{6}).2, then D = 3^{2} contains an element y of type 3A, so that C_{G}(y) = 3 × Sp_{6}(2) ≤ N_{G}(D), which is impossible. If (E, K) = (3^{2}, A_{2}(4)), then E ∈ Syl_{3}(K) and Aut(K) = A_{2}(4).D_{12}.

Thus D ∈ Syl_{3}(K) or D ∈ Syl_{3}(Aut(K)). In the latter case, N_{G}(E) = N_{G}(D) and so D = O_{3}(N_{G}(E)), which is impossible. If (E, K, N_{Aut(K)}(D)) ≠ (3^{2}, C_{4}(2), 3^{2}: D_{8} × S_{6}) and (3^{2}, A_{2}(4), 3^{2}: 2S_{4} × 2), then C_{G}(D) ≤ C_{N_{Aut(K)}(D)}(D) ≤ D.

Case (2) Suppose p = r and Q = K ∩ D ≠ 1. Then N_{G}(Q) = N_{G}(D) and by [15, Corollary 3.1.4],

F^{∗}(N_{G}(Q)) = O_{r}(N_{G}(Q)) = D.

Thus C_{G}(D) = C_{N_{G}(D)}(D) ≤ D.

Suppose Q = D∩K = 1. If x ∈ N_{G}(D)∩K and y ∈ D, then x^{−1}y^{−1}xy ∈ D∩K = 1, so that C_{K}(D) = N_{K}(D) ◁ N_{G}(D).

In the notation of [15, Theorem 2.5.12], Out(K) = Outdiag(K): Φ_{K}Γ_{K} and so D ≤ Φ_{K}Γ_{K}, since Outdiag(K) is a p′-group.

Suppose D is non-cyclic. Then |Γ_{K}|_{p} ≠ 1 and p = 2 or 3. If |Γ_{K}|_{2} ≠ 1, then |Γ_{K}|_{2} = 2, and K = A_{m}(2^{a}), B_{2}(2^{a}), D_{m}(2^{a}), F_{4}(2^{a}) or E_{6}(2^{a}), and moreover, Φ_{K}Γ_{K} is cyclic whenever K = B_{2}(2^{a}) and F_{4}(2^{a}). Thus K = A_{m}(2^{a}) with m ≥ 2, D_{m}(2^{a}) or E_{6}(2^{a}) with 2|a, and Φ_{K}Γ_{K} = Φ_{K} × Γ_{K}. If |Γ_{K}|_{3} ≠ 1, then |Γ_{K}|_{3} = 3 and K = D_{4}(3^{a}) with 3|a. So D is abelian of 3-rank 2.

Let ϕ be a field of graph-field automorphism of order p. Then K_{0} = C_{K}(ϕ) is simple and by [15, Theorem 7.1.4], C_{G}(ϕ) = ⟨ϕ⟩. It follows that ϕ ∉ D and D is cyclic, which is impossible.

Case (3) Suppose p ≠ r and every abelian subgroup of D is cyclic. Thus D is cyclic or a generalized quaternion group Q_{2^{α}} of order 2^{α} ≥ 8 (cf. [13, Theorem 5.4.10]). In the latter case, each subgroup of D is either cyclic or generalized quaternion, and D has a unique element z of order 2 with ⟨z⟩ = Ω_{1}(Z(D)).

Suppose D is non-cyclic, so that p = 2, D = Q_{2^{α}} and r is odd with q = r^{a}.

Case (3.1) Suppose, moreover that D ∩ K = 1, so that N_{K}(D) = C_{K}(D) and D is a quaternion subgroup of Out(K). In particular, a Sylow 2-subgroup of Out(K) is non-abelian, so that K is classical, K ≠ P Sp_{2m}(q), and m ≥ 3 when K = P SL_{m}^{ϵ}(q)

Since Out(P Ω_{m}(q)) is either abelian or Out(P Ω_{m}(q)) ∈ {D_{8}×C_{a}, S_{3}×C_{a}, S_{4}×C_{a}}, it follows that Out(P Ω_{m}(q)) contains no quaternion subgroup, so that K = P SL_{m}^{ϵ}(q).

Suppose D contains a field or graph-field automorphism ϕ of order 2. Then K_{0} = C_{K}(ϕ) ≤ N_{G}(D) and C_{K}(ϕ)D = C_{K}(D) × D. In particular, D ≤ C_{G}(K_{0}). Since r is odd, it follows that K_{0} is simple and by [15, Theorem 7.1.4], C_{G}(K_{0}) = ⟨ϕ⟩, so that D is cyclic, which is impossible.

Since Outdiag(K) is cyclic and since D has exactly one involution z, it follows that R = D ∩ H is cyclic. If moreover, R = 1, then D ≤ Φ_{K} × Γ_{K}, which is impossible. Thus R ≠ 1, z ∈ R, N_{G}(D) = C_{G}(z).

Let M = GL_{m}^{ϵ}(q), so that H = Inndiag(K) = M/Z(M). Then z ∈ H\K, R = O_{2}(C_{H}(z)) and C_{H}(z) is given by [15, Table 4.5.2]. If L = O^{r′}(C_{H}(z)), then

L ∈ \{SL_{m−1}^{ϵ}(q), SL_{i}^{ϵ}(q) ∗ SL_{m−i}^{ϵ}(q), SLϵ_{\frac{m}{2}} (q^{2})\},

where 2 ≤ i ≤ \frac{m}{2} and SL_{i}^{ϵ}(q) ∗ SL_{m−i}^{ϵ}(q) is a central product of SL_{i}^{ϵ}(q) and SL_{m−i}^{ϵ}(q). Let [D, L] = ⟨y^{−1}x^{−1}yx | y ∈ D, x ∈ L}, so that [D, L] ≤ D ∩ L = O_{2}(L). Since O_{2}(L) ≤ D ∩ H is cyclic, it follows that O_{2}(L) ≤ Z(L). But y^{−1}x^{−1}yx = 1 for any y ∈ D and 2′-element x ∈ L, so [D, L] = 1 and D ≤ C_{C_{G}(z)}(L).

Let ι be the inverse-transpose map or 1 according as M = GL_{m}(q) or M = U_{m}(q). We may suppose that

C = C_{Aut(K)}(z) = ⟨C_{H}(z), γ, ι⟩,

where γ is a field automorphism of M. Note that if xβ centralizes L for some x ∈ C_{H}(z) and β ∈ ⟨γ, ι⟩, then x ∈ C_{L}(β). It follows that C_{C}(L) = C_{C_{H}(z)}(L) and O_{2}(C_{C_{H}(z)}(L)) ≤ D ∩ H is cyclic. This is impossible as D ≤ C_{C}(z).

Case (3.2) Suppose Q = D ∩ K ≠ 1. Then Ω_{1}(Z(Q)) has order 2 and z ∈ Z(Q). So N_{G}(D) = C_{G}(z), N_{K}(Q) = C_{K}(z), D = O_{2}(C_{G}(z)) and Q = O_{2}(C_{K}(z)). The proof is similar to above.

The centralizer C_{H}(z) is given by [15, Table 4.5.1]. If L = O^{r′}(C_{H}(z)), then either L = L_{1} or L_{1} ∗ L_{2} for some finite groups L_{i} of Lie type. Since L ≤ C_{G}(z) = N_{G}(D), it follows that D ∩ L ◁ L and D ∩ L = O_{2}(L), so that [D, L] ≤ O_{2}(L).

Note that each O_{2}(L_{i}) is cyclic except when L_{i} ∈ {Ω_{3}(3) = A_{4}, SL_{2}(3) = SU_{2}(3) = Sp_{2}(3) = Q_{8}: 3, Ω_{4}(3) = SL_{2}(3) ∗ SL_{2}(3)}.

Since O_{2}(L) ≤ D is cyclic or generalized quaternion, it follows that either O_{2}(L_{i}) is cyclic for all i or L_{i} = SL_{2}(3) for a unique i. In the former case each L_{i} is quasisimple or cyclic, O_{2}(L_{i}) ≤ Z(L_{i}), and [D, L] ≤ Z(L), since L is generated by 2′-elements. In the later case q = 3 and O_{2}(L) = Q_{8}.

Suppose O_{2}(L) = Q_{8}, so that q = 3. If K is exceptional, then

K ∈ {^{3}D_{4}(3), G_{2}(3), F_{4}(3), E_{6}^{ϵ}(3), E_{7}(3), E_{8}(3)}.

If K is classical, then K ∈ {P SL_{n}^{ϵ}(3), P Sp_{2n}(3)}. By [15, Table 4.5.1], each K has exactly one possible conjugacy class of involutions z, and L = L_{1}∗L_{2} with L_{1} = SL_{2}(3) and

L_{2} = \{ \begin{matrix}SL_{2}(27) & if K = ^{3}D_{4}(3), \\ SL_{2}(3) & if K = G_{2}(3), \\ Sp_{6}(3) & if K = F_{4}(3), \\ SL_{6}^{ϵ}(3) & if K = E_{6}^{ϵ}(3), \\ Spin_{12}^{+}(3)/⟨z_{s}⟩ & if K = E_{7}(3), \\ E_{7}(3)_{u} & if K = E_{8}(3), \\ SL_{n−2}^{ϵ}(3) & if K = P SL_{n}^{ϵ}(3), \\ Sp_{2n−2}(3) & if K = P Sp_{2n}(3). \end{matrix}  (7.1)

If K = G_{2}(3), then Q ≥ O_{2}(L) = 2_{+}^{1+4}, which is impossible. Similarly, we may suppose K ≠ P SL_{4}^{ϵ}(3) and P Sp_{4}(3). Thus Q = O_{2}(L) = Q_{8} and C_{K}(Q) = C_{H}(Q) = L_{2}.

Suppose O_{2}(L) is cyclic and L is non-abelin, so that L is quasisimple. and D ≤ C_{C_{G}(z)}(L) ≤ C_{C_{Aut(K)}(z)}(L). In particular, Q ≤ R ≤ C_{C_{H}(z)}(L) and by [15, Theorem 4.5.1], R is cyclic.

In the notation of [15], Aut(K) = HΦ_{K}Γ_{K}. It follows by [15, Theorem 4.5.1] that we may suppose

C_{Aut(K)}(z) = C_{H}(z)Φ_{K}Γ_{K}.

If xβ ∈ C_{C_{Aut(K)}(z)}(L) for some x ∈ C_{H}(z) and β ∈ Φ_{K}Γ_{K}, then x ∈ C_{L}(β). It follows that C_{C_{Aut(K)}(z)}(L) ≤ C_{H}(z) and so D ≤ C_{C_{H}(z)}(L) is cyclic, which is impossible.

Suppose L is cyclic, so that K = P SL_{2}(q). Let G̃ be the general semilinear group on the underlying space V of K_{u} and A = G̃⟨ι⟩, where ι is the inverse-transpose map. In addition, let Z = {α1_{V} : 0 ≠ α ∈ 𝔽_{q}} be a subgroup of A, so that Aut(K) = A/Z and we may suppose z = tZ with O_{2}(Z) ≤ ⟨t⟩, since Q ≠ 1.

If xZ ∈ C_{G}(z), then x^{−1}tx ∈ tZ, so that x^{−1}tx ∈ ⟨t⟩ and x ∈ N_{A}(⟨t⟩). Thus C_{Aut(K)}(z) = N_{A}(⟨t⟩)/Z and C_{H}(z) = N_{GL_{2}(q)}(t)/Z.

Let D_{A} be a 2-subgroup of N_{A}(⟨t⟩) such that D_{A}/O_{2}(Z) = D. Now N_{A}(⟨t⟩) = ⟨GL_{1}(q^{2}), ρ, γ, ι⟩ and N_{GL_{2}(q)}(⟨t⟩) = ⟨GL_{1}(q^{2}), ρ⟩, where ρ and γ induce field automorphisms of order 2 and a, respectively and ι inverses each element of GL_{1}(q^{2}).

We may choose ρ such that ρ^{4} = 1, ρZ has order 2 in GL_{2}(q)/Z and [ρ, γ] = [ρ, ι] = 1. Suppose y ∈ D_{A} ∩ ⟨ρ, γ, ι⟩ such that 2^{w} = |y| ≥ 4. Then y^{2^{w−1}} = ϕ ∈ D_{A}, which is impossible. Here ϕ is the field automorphism of order 2. Thus D_{A} ∩ ⟨ρ, γ, ι⟩ ≤ ⟨ρ, ϕ, ι⟩ and

O_{2}(GL_{1}(q^{2})) < D_{A} ≤ ⟨O_{2}(GL_{1}(q^{2})), ρ, ϕ, ι⟩.

Since ⟨ρ, ϕ, ι⟩Z/Z is elementary abelian of order 2^{3}, it follows that m_{2}(D_{A}/O_{2}(Z)) ≥ 2, which is impossible. □

Proposition 7.4 Let Z(G) = O_{p′}(G) ≤ M ≤ G such that M/Z(G) is a finite nonabelian simple group of Lie type and M/Z(G) ≤ G/Z(G) ≤ Aut(M/Z(G)). Suppose B ∈ Blk(G) has a trivial intersection defect group D ≠ 1. Then one of the following holds:

(i) D is cyclic.

(ii) D is a non-cyclic Sylow p-subgroup of G. If D∩M is non-cyclic, then the possible pairs (M/Z(G), p) are listed in [5, Proposition 1.3] and moreover, [G: M]_{p} = 1, so that D ≤ M. If D ∩ M is cyclic, then Z(G) = 1, (M, p) = (^{2}B_{2}(2^{5}), 5) or (L_{2}(8), 3), in which case G = Aut(M) = M.p.

(iii) p = 2, Z(G) = 1, G = P Sp_{2m}(3) with even m ≥ 4 and D = Q_{8} is a quaternion group. In this case, let B_{u} be a block of K_{u} = Sp_{2m}(3) containing B. Then B_{u} = E_{2}(K_{u}, (s)) for some semisimple 2′-element s of the dual group K_{u}^{∗} = SO_{2m+1}(3), D(B_{u}) ≅ Q_{8} × C_{2} and C_{K_{u}^{∗}}(s) ≅ SO_{3}(3) × GL_{1}(3^{m−1}).

In particular, if D is non-cyclic, then G/Z(G) has a block with a defect group D = DZ(G)/Z(G).

Proof. Let K = M/Z(G). Since D = DZ(G)/Z(G) is a radical subgroup of G/Z(G), it follows by Proposition 7.3 that D is cyclic, C_{G/Z(G)}(D) ≤ D, K = A_{2}(4) with D = 3^{2} or D is generalized quaternion with D ∩ K = Q_{8} and K given by Proposition 7.3 (iv).

Suppose C_{G/Z(G)}(D) ≤ D. By Lemma 2.4 D is Sylow of both G and G/Z(G), so that by Lemma 2.1, Q = D ∩ K is a TI Sylow subgroup of K.

Suppose p ≠ r. If Q is non-cyclic, then by [5, Proposition 1.3],

(K, p) = {(P SL_{3}(4), 3), (^{2}F_{2}(2)′, 5)}

and if Q is cyclic, then by Table 3,

(K, p) ∈ {(L_{2}(8), 3), (^{2}B_{2}(2^{5}), 5)}.

If [G: M]_{p} ≠ 1, then (K, p) ≠ (^{2}F_{2}(2)′, 5), so that [G: M]_{p} = p.

If (K, p) = (L_{3}(4), 3), then Aut(K)/K = 2×S_{3}, N_{K}(Q) = 3^{2}: Q_{8} and N_{Aut(K)}(Q) = 3^{2}: 2S_{4} × 2. It follows that Q is a radical subgroup of G and D is not TI, which is impossible.

If (K, p) = (L_{2}(8), 3) or (^{2}B_{2}(2^{5}), 5), then Q is cyclic and G/Z(G) = Aut(K) = K.p. Since Aut(K)/K is cyclic and M(K) = 1, it follows by [7, Lemma 3.4] that M(G) = 1, so that Z(G) = 1 and G = Aut(K). If (K, p) = (L_{2}(8), 3), then K = ^{2}G_{2}(3)′, G = ^{2}G_{2}(3) and N_{G}(Q) = 9: 6. If (K, p) = (^{2}B_{2}(2^{5}), 5), then N_{G}(Q) = 25: 20. Thus Q is non-radical in G and so the Sylow subgroup D is a TI subgroup of G.

Suppose p = r. By [5, Proposition 1.3], K = A_{1}(p^{a}),^{2}A_{2}(p^{a}),^{2}B_{2}(2^{a}) or ^{2}G_{2}(3^{a}). Since p = r, it follows that a Sylow p-subgroup of Out(K) is cyclic generated by a field automorphism.

If |Out(K)|_{p} = 1, then D = Q is Sylow in G. In particular, we may suppose K ≠ ^{2}B_{2}(2^{a}) as in this case a is odd.

Suppose [G: M]_{p} ≠ 1, so that |Out(K)|_{p} ≠ 1. Since C_{K}(Q) ≤ Q and since Out(K) is cyclic, it follows that Q < D. Thus D contains a field automorphism ϕ of order p. By [15, Proposition 4.9.1], C_{K}(ϕ) = A_{1}(p^{a/p}),^{2}A_{2}(p^{a/p}) and ^{2}G_{2}(3^{a/3}), respectively. But C_{K}(ϕ) ≤ N_{Aut(K)}(D), so C_{K}(ϕ) is a subgroup of the Borel subgroup N_{K}(Q). In particular, C_{K}(ϕ) is soluble. Thus p = 2, C_{K}(ϕ) = A_{1}(2) or ^{2}A_{2}(2). In both cases, |Out(K)|_{p} ≤ p and D is Sylow in G.

If (K, p) = (A_{1}(4), 2), then Q = 2^{2}, N_{K}(Q) = A_{4} and N_{Aut(K)}(Q) = S_{4}. If (K, p) = (^{2}A_{2}(4), 2), then Q = 2^{2+4}, N_{K}(Q) = 2^{2+4}: 15 and N_{Aut(K)}(Q) = 2^{2+4}: (3 × D_{10}).2. In both cases, Q is radical in Aut(K) and a Sylow 2-subgroup D of Aut(K) is not a TI defect group.

Suppose D is generalized quaternion with D ∩ K = Q_{8} and

K ∈ {^{3}D_{4}(3), F_{4}(3), E_{6}^{ϵ}(3), E_{7}(3), E_{8}(3), P SL_{n}^{ϵ}(3), P Sp_{2n}(3)},

where n ≥ 3 and n ≠ 4 when K = P SL_{n}^{ϵ}(3). Thus

Aut(K) = \{ \begin{matrix}^{3}D_{4}(3).3 & if K = ^{3}D_{4}(3), \\ F_{4}(3) & if K = F_{4}(3), \\ E_{6}^{ϵ}(3).2 & if K = E_{6}^{ϵ}(3), \\ E_{7}(3).2 & if K = E_{7}(3), \\ E_{8}(3) & if K = E_{8}(3), \\ P SL_{n}^{ϵ}(3).(n, 3 − ϵ).2 & if K = P SL_{n}^{ϵ}(3), \\ P Sp_{2n}(3).2 & if K = P Sp_{2n}(3). \end{matrix}  (7.2)

In particular, Aut(K)/K is either cyclic or a 2-group. Since K ≠ P SU_{4}(3), it follows that M(K) is a 2-group, and by [7, Lemma 3.4], M(X) is a 2-group for any K ≤ X ≤ Aut(K). Since Z(G) is a 2′-group, it follows that we may suppose Z(G) = 1.

Let B_{K} be the block of K covered by B. Then we may suppose Q = K∩D = D(B_{K}) and in addition, suppose (Q, b_{Q}) is a Sylow B_{K}-subgroup.

If L = O^{3′}(C_{H}(Z(Q))), then L = L_{1} ∗ L_{2}, Q = O_{2}(L) and C_{K}(Q) = L_{2}, where L_{1} = SL_{2}(3) and L_{2} given by (7.1). Since (Q, b_{Q}) is a Sylow B_{K}-subgroup, it follows that b_{Q} ∈ Blk(L_{2}) with defect group Z(Q) = Z(L_{2}). Moreover, suppose K ≠ P Sp_{2n}(3). By [15, Table 4.5.1], N_{K}(Q) ≥ ⟨Q ∗ L_{2}, τ⟩ for some 2-element τ, and in addition, τ induces a non-trivial element of Outdiag(L_{2}), except when L_{2} = Spin_{12}^{+}(3)/⟨z_{s}⟩, in which case we can suppose τ ∈ SO_{12}^{+}(3)\Ω_{12}^{+}(3). By Lemma 7.1, τ stabilizes b_{Q}, so that τ ∈ N_{K}(Q, b_{Q}), which is impossible as (Q, b_{Q}) is maximal.

Suppose K = P Sp_{2n}(3) and G ≠ K. Then G = Aut(K) = K.2 = GSp_{2n}(3) and so C_{G}(Q) = L_{2}, N_{G}(Q) = ⟨Q ∗ L_{2}, τ⟩ for some 2-element τ ∈ G. By Lemma 7.1 again, τ stabilizes b_{Q} and so τ ∈ N_{G}(Q, b_{Q}), which is impossible. □

8 Proof of the main theorem

By Theorem 4.2 it suffices to find all TI defect blocks of groups X where M ≤ X/Z(X) ≤ Aut(M) for a non-abelian simple group M. Thus we have reduced to p′-central extension of groups of automorphisms of simple groups. Since it is not always the case that the Schur multiplier of a simple group and that of its automorphism group are the same, we show here that it suffices to consider automorphism groups of central extensions of simple groups. In our case, all the non-abelian simple groups we consider have outer automorphism group which is a product of at most two cyclic groups. We deal with each cyclic group in turn.

Proposition 8.1 Suppose that M is a non-abelian simple group and M ≤ X ≤ Aut(M). Suppose further that Out(M) is a product of at most two cyclic p′-groups. Let G be a central extension of X by a p′-group, and B ∈ Blk(G) a TI defect block. Then there is a central extension L of M by a p′-group and a finite group H with L ≤ H ≤ Aut(L) such that H possesses a TI defect block B_{H} Morita equivalent to B.

Proof. Let Z(G) ≤ N ◁ T ◁ G such that N/Z(G) = M, and T/N and G/T are cyclic p′-groups. Let W be the central extension of T/Z(G) by T′ ∩ Z(G) with W ≤ T. We repeat first the construction from [5, 3.6].

Choose u ∈ G such that uW Z(G) = uT generates the cyclic p′-group G/W Z(G) = G/T of order m. Hence u^{m} ∈ W Z(G), say u^{m} = wz where w ∈ W, z ∈ Z(G). Let V be a cyclic p′-group such that Z(G) ≤ V, and containing an element v ∈ V such that v^{m} = z. Write A = G ∗ V, where we identify Z(G) with G ∩ V.

Setting h = uv^{−1} ∈ A, define H = ⟨W, h⟩ ◁ A. Define L = N′. Following the argument in [7, 3.7], and using the fact that T/Z(G) has Schur multiplier contained in that of M (see [7, 3.5]) we see that Z(L) ≤ H′ and C_{H}(L) = Z(L) = Z(H) = O_{p′}(H), so that H is indeed an automorphism group of the perfect p′-central extension L of M.

Now there are Z_{G}, Z_{H} ≤ Z(A) such that A/Z_{G} ≅ G and A/Z_{H} ≅ H. By the constructions of section 3 there is a block B_{A} of A Morita equivalent to B, and a block B_{H} of H Morita equivalent to B_{A}. By Proposition 2.2 (vi) B_{A} and B_{H} have TI defect groups, and we are done. □

If X/M is cyclic, then by [7, 3.5] X has smaller Schur multiplier then M. Thus using this fact and Proposition 8.1, the result holds when M is a sporadic or alternating group by Corollary 5.2 and Proposition 6.1. Otherwise, if the defect group is not a Sylow p-subgroup, then the result holds by Proposition 7.4. If the defect group is a Sylow p-subgroup (and M is a group of Lie type), then the result holds by Proposition 7.4 and Proposition 8.1, but here we use the fact that G has TI Sylow p-subgroups if and only if G/Z(G) does (given that Z(G) is a p′-group), and that in each case Out(M) is either cyclic or a direct product of cyclic p′-groups.

Theorem 1.1 follows, noting that A_{6} ≅ P SL_{2}(9) and C_{3} × C_{3} is its Sylow subgroup.

ACKNOWLEDGEMENTS

We thank Paul Fong for his valuable suggestions, and Geoffrey Robinson for some useful discussions. We also thank Karin Erdmann for her suggestion regarding the generality of the results presented. The second author expresses his deep gratitude for the warm hospitality shown by the Department of Mathematics at the University of Auckland during his visit, when this research was done. Finally we thank the Marsden Fund for their support.

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