THE CHARACTER TABLE OF A SHARPLY 5-TRANSITIVE SUBGROUP OF Alt(12)

NICK GILL AND SAM HUGHES

Abstract. In this paper we calculate the character table of a sharply 5-transitive subgroup of Alt(12), and of a sharply 4-transitive subgroup of Alt(11). Our presentation of these calculations is new because we make no reference to the sporadic simple Mathieu groups, and instead deduce the desired character tables using only the existence of the stated multiply transitive permutation representations.

1. Introduction

In this paper we prove the following theorem.

Theorem 1.

(1) If G is a sharply 5-transitive subgroup of Alt(12), then the character table is given by Table 8.

(2) If G is a sharply 4-transitive subgroup of Alt(11), then the character table is given by Table 7.

This theorem is not new – item (1) is a consequence of the fact that M_{12} is the unique sharply 5-transitive subgroup of Alt(12), and the fact that the character table of M_{12} is known; item (2) is a consequence of the two analogous statements for M_{11}.

Our proof of the theorem is new, however, because it makes no reference whatsoever to the groups M_{11} and M_{12} but, instead, deduces the character table using nothing more than the stated assumptions about sharp multiple transitivity.

Our interest in proving this theorem stems from our study of Frobenius’ famous 1904 paper in which (amongst many other things) he calculates the character table of M_{12} and M_{24} [Fro04]. We were curious to understand Frobenius’ methods because, in the late nineteenth century, there appeared to be some lingering doubt as to the “existence” of the Mathieu groups (that is to say, people questioned whether the permutations that Mathieu wrote down in his original paper [Mat73] generated alternating groups, rather than any genuinely “new” groups). As late as 1897, Miller published a paper claiming that M_{24} did not exist [Mil97], although he retracted this claim soon after [Mil00].

In fact, studying Frobenius’ 1904 paper, it seems that Frobenius was in no doubt as to the existence of the Mathieu groups and, indeed, he uses specific properties of these groups when he calculates their character tables (see the MathOverflow discussion on this subject for more detail [Lad]). Nonetheless, we were left wondering whether, in principle, Frobenius could have calculated the character table of M_{12} using nothing more than the property of sharp 5-transitivity – the main result of this paper confirms that the answer to this question is “yes”! In particular, note that the proof of Theorem 1 below uses little more than the basics of character theory, all of which would have been available to Frobenius in 1904 – the results that we make use of are summarized at the start of §3.

1.1. Further work. There are two natural avenues for extending the current work. First, it would be nice if the two occurrences of “Alt(12)” in the theorem could be replaced by “Sym(12)”. It seems quite plausible that such a theorem could be proved using the methods in the current paper, however it would likely make the conjugacy class calculations considerably more complicated and so we have not investigated this in detail.

Second, one wonders about proving the analogous theorem for M_{24}, something like: “If G is a 5-transitive subgroup of Alt(24) of order 244823040, then the character table of G is as follows...” It would be nice to prove such a theorem using only elementary character theory, although one might expect that the details would be rather onerous.

Date: April 30, 2020.

2000 Mathematics Subject Classification. 20C15; 20B20; 20C34.

Key words and phrases. Mathieu groups, sporadic groups, character table, permutation group, multiply transitive.

Type 1^{9} 2^{4} 3^{3} 4_{A}^{2} 4_{B}^{2} 4_{C}^{2}
Size 1 9 8 18 18 18

Table 1. Conjugacy class sizes in G_{9}

1.2. Structure, notation and thanks. In what follows we will consider groups G_{9}, G_{10}, G_{11} and G_{12}. For i = 9, …, 12, the group G_{i} denotes any sharply (i − 7)-transitive subgroup of Alt(i) on i points.

In §2, we deduce the conjugacy class structure of G_{9}, G_{10}, G_{11} and G_{12}; in §3, we deduce the character table of G_{9}, G_{10}, G_{11} and G_{12}.

All of our notation is entirely standard. We remind the reader that a character of G is called real if it takes real values for all elements of G; similarly, a conjugacy class C of G is called real if any element of C is conjugate to its inverse. The connection between these two concepts is given in Theorem 6, and will be exploited in §3.

In the course of this research, we asked a question on the website MathOverflow. We are particularly grateful to Frieder Ladisch, whose answer to our question clarified some of the history behind this area of research [Lad].

2. Conjugacy classes

In this section we calculate the conjugacy class structure of G_{9}, G_{10}, G_{11} and G_{12}. Note that, for i = 9, …, 12, G_{i} is any sharply (i − 7)-transitive subgroup of Alt(i) on i points.

We label conjugacy classes via the cycle structure of their elements. Where there is more than one conjugacy class with the same cycle structure, they are distinguished with subscript Roman letters – see, for instance, Table 1 which lists the conjugacy classes of G_{9}, three of which contain elements of type 4^{2}.

2.1. The conjugacy classes of G_{9}. Since |G_{9}| = 72, and is sharply 2-transitive on 9 points, it is straightforward to see that G_{9} has a normal regular subgroup N with N ≅ C_{3} × C_{3}. Now the stabilizer of a point, call it H, must act (by conjugation) regularly on the non-zero elements of N – then H is a group of order 8 that is isomorphic to a subgroup of GL_{2}(3), and one can check directly that H ≅ Q_{8}; thus G ≅ (C_{3} × C_{3}) ⋊ Q_{8}.

The conjugacy classes of G_{9} can now be written down in Table 1. Note that all classes are real.

2.2. The conjugacy classes of G_{10}. The group G = G_{10} has order 720, and elements that fix at least one point must have cycle structure in the list given in Table 1, with the possibility of fusion for the elements of cycle type 4^{2}.

When one considers the cycle type of fixed-point-free elements of G, one must exclude all elements that have powers that fix elements and that are not of a type listed in Table 1. One obtains immediately that the only possible new cycle types are 5^{2} and 8^{1}2^{1}.

There must be an element g ∈ G of type 5^{2}, since 5 divides |G|. What is more, since C_{Alt(10)}(g) = ⟨g⟩, we conclude that C_{G}(g) = ⟨g⟩ and so the conjugacy class containing g has size 144. Let P = ⟨g⟩, a Sylow 5-subgroup of G_{10}. Since C_{G}(P) = P, and N_{G}(P)/C_{G}(P) ≲ Aut(P) ≅ C_{5}. We conclude that N_{G}(P) has order dividing 20. Sylow’s theorems tell us that |G : N_{G}(P)| ≡ 1 (mod 5) and we conclude that |N_{G}(P)| = 20. Since all conjugates of P intersect trivially, we conclude that G contains 144 elements of order 5, hence there is precisely one conjugacy class of elements of type 5^{2}.

We now consider elements of types listed in Table 1. For elements of type 3^{3}, we observe first that an element of this type fixes a unique point and, since the stabilizer of a point contains a unique Sylow 3-subgroup of G, we conclude that each element of type 3^{3} lies in a unique Sylow 3-subgroup of G. Now we use the fact that the stabilizer of a point is maximal in G to conclude that there are precisely 10 Sylow 3-subgroups and, therefore, 80 elements of type 3^{3} in G.

All remaining elements, of which there are 450, must be of type 4^{2} or 2^{1}8^{1}. Now let P be a Sylow 2-subgroup of G and observe that P has two orbits, of size 2 and 8. Let {9, 10} be the smaller orbit; then the stabilizer of 9 is equal to the stabilizer of 10 and is equal to Q_{8}, a Sylow 2-subgroup of G_{9}. The points in the other orbit each have stabilizers in P of size 2, and there are four distinct stabilizers; this leaves four elements which must be fixed-point-free, and hence are of type 2^{1}8^{1}. What is more these elements cannot be central in P, otherwise P would be abelian. We conclude that there are at most two conjugacy classes in G of elements of type 2^{1}8^{1}, and they have size 90 (since they do not commute with any elements of odd order).

In fact, it is clear that any Sylow 2-subgroup of G_{10} is characterized by its orbit of size 2. This implies, first, that, since the same is true of elements of type 2^{1}8^{1}, each element of type 2^{1}8^{1} is in a unique Sylow 2-subgroup; it implies, second, that there are precisely 45 Sylow 2-subgroups of G_{10}, and so there are 180 elements of type 2^{1}8^{1}, split into two conjugacy classes of size 90.

Type 1^{10} 2^{4} 3^{3} 4_{A}^{2} 4_{B}^{2} 5^{2} 2^{1}8_{A}^{1}^{∗} 2^{1}8_{B}^{1}^{∗}
Size 1 45 80 90 180 144 90  90

Table 2. Conjugacy class sizes in G_{10}

Now there are at most three conjugacy classes in G of elements of type 4^{2} containing a total of 270 elements; since these elements are real in G_{9}, they are real in G_{10} and so the classes have even order. One of these classes must be squares of elements of order 8, and so at least one conjugacy class has size 90; the others have size 90 or 180. There are, therefore, two possibilities for elements of type 4^{2}: three conjugacy classes of size 90, or two of size 90 and 180. Note that, since there are 180 elements of type 2^{1}8^{1}, there are only 90 elements that are squares of these. Thus if there are three conjugacy classes of size 90, then two of these must have a centralizer isomorphic to C_{4} × C_{2}. It is easy to check, though, that this is not possible, given that all involutions are of type 2^{4}.

The conjugacy classes of G_{10} are summarized in Table 2. Note that we have asterisked the two conjugacy classes that are not real – it is clear that they are the only conjugacy classes that have a chance of being non-real; to see that they are not real, simply observe that there are no elements of type 4^{2} in Alt(8) that send an 8-cycle to its inverse.^{1}

2.3. The conjugacy classes of G_{11}. The group G = G_{11} has order 7920, and elements that fix at least one point must have cycle structure given in Table 2, with the possibility, once again, of fusion for the elements of cycle type 4^{2} or 2^{1}8^{1}.

Taking into account the same considerations as before, we obtain that any fixed-point-free elements of G must have cycle type 11^{1} or 2^{1}3^{1}6^{1}.

An element, g, of type 11^{1} is self-centralizing in Alt(11) and hence, also in G_{11}. We conclude that a conjugacy class of this type has size 720. On the other hand a Sylow 11-subgroup of Alt(11) has a normalizer of size 55, hence N_{G}(⟨g⟩) has size 11 or 55. If the former, then one immediately concludes that there are 10 conjugacy classes of this type; this means that there are a total of 720 elements in G that are not of type 11^{1}. Since a point-stabilizer of G_{11} has size 720 and does not contain any elements of type 11^{1}, we conclude that a point-stabilizer of G_{11} is normal, a contradiction. Thus |N_{G}(g)| = 55 and we conclude, furthermore, that there are two conjugacy classes in G of type 11^{1}.

We know that there is a unique conjugacy class of elements of type 5^{2} in G_{10}, hence the same is true in G_{11}. What is more these elements are self-centralizing, hence this class has size 1584.

Now consider an element, g, of type 3^{3}. It is clear that C_{G}(g) contains an involution if and only if there is an element of type 2^{1}3^{1}6^{1}. If this is not the case, then the conjugacy class of type 3^{3} has size 7920/9 = 880, and it is the only non-trivial conjugacy class that does not have size divisible by 3. But now 7920 − 880 − 1 = 7039 is not divisible by 3, and we have a contradiction. Hence we conclude that |C_{G}(g)| is even, and contains an element of type 2^{1}3^{1}6^{1}. Since the Sylow 2-subgroup of C_{Alt(11)}(g) is of size 2, we conclude that |C_{G}(g)| = 18 and the conjugacy class of type 3^{3} has size 440.

Let h be an element of type 2^{4}. Then g is central in a Sylow 2-subgroup of G, and is also centralized by an element of type 3^{3}; it is easy to check that it is not centralized by a Sylow 3-subgroup of G, hence the conjugacy class of involutions has size 7920/48 = 165.

Elements of type 2^{1}8^{1} are self-centralizing, so these conjugacy classes have size 990; any fusion of the two conjugacy classes would have to take place in a Sylow 2-subgroup and, since this Sylow 2-subgroup is the same as for G_{10}, we know that there are two such conjugacy classes.

The remaining elements are of type 4^{2} and of type 2^{1}3^{1}6^{1} and they make up the remaining 2310 elements. The size of the respective conjugacy classes is 7920/8 and 7920/6 and since these two numbers sum to 2310 we know that there is a unique conjugacy class of each type.

The conjugacy classes of G_{11} are now written down in Table 3. Note that we have asterisked the four conjugacy classes that are not real. It is clear that all the other classes are real, and it clear that the classes of type 11^{1} are non-real, since they are non-real in Alt(11); similarly, the classes of type 2^{1}8^{1} are non-real since they are non-real in G_{10} and any putative “reversing element” in G_{11} would have to lie in G_{10}.

2.4. The conjugacy classes of G_{12}. The group G = G_{12} has order 95040, and elements that fix at least one point must have cycle structure given in Table 3, with the possibility of fusion for the elements of order 8 and 11.

1It is well-known that the group G_{10} is, in fact, M_{10}, the unique non-split extension of Alt(6). One can compare our enumeration of the conjugacy classes of G_{10} with the enumeration of conjugacy classes of M_{10} that appears in the ATLAS [CCN^{+}85]; note that our class 4_{B}^{2} is labeled 4C in the ATLAS; similarly the two classes of elements of order 8 are labeled 8C and D** in the ATLAS.

Type 1^{11} 2^{4} 3^{3} 4^{2} 5^{2} 2^{1}3^{1}6^{1} 2^{1}8_{A}^{1}^{∗} 2^{1}8_{B}^{1}^{∗} 11_{A}^{1}^{∗} 11_{B}^{1}^{∗}
Size 1 165 440 990 1584 1320 990  990 720 720

Table 3. Conjugacy class sizes in G_{11}

Taking into account the same considerations as before, we obtain that any fixed-point-free elements of G must have cycle type 2^{1}10^{1}, 3^{1}9^{1}, 4^{1}8^{1}, 6^{2}, 2^{2}4^{2}, 3^{4}, 2^{6}. In order to exclude elements of type 3^{1}9^{1}, we need a lemma:

Lemma 2. Let P be a Sylow 3-subgroup of G. Then P ≅ He(3).

Here He(3) is the Heisenberg group over the field of order 3, a group of order 27. Note that all non-trivial elements of He(3) have order 3, hence there are no elements of order 9 in G.

Proof. Observe that |P | = 27. There are five groups of order 27, three abelian and two non-abelian.

Let us suppose first that P is abelian. Suppose that Λ is an orbit of P in its action on Ω, and let λ ∈ Λ. Then, since P is abelian, any elements that fixes λ must fix every element in Λ. Since no element of G fixes more than 4 points, this means that |Λ| = 1 or 3. The group P must not fix more than 4 points, thus there are at least two orbits of P of size 3, call these Λ_{1} and Λ_{2}. Now the orbit-stabilizer theorem asserts that P has a subgroup P_{1}, of order 9, that fixes every element of Λ_{1}; similarly P has a subgroup P_{2}, of order 9, that fixes every element of Λ_{2}. But now P_{1} ∩ P_{2} is non-trivial (by order considerations) and an element in the intersection fixes at least the 6 points of Λ_{1} ∪ Λ_{2}. This is a contradiction.

Suppose, then, that P is the non-abelian group of order 27 that is not He(3). This is the extraspecial group of exponent 9; it has center, Z, of order 3; it has a normal elementary-abelian subgroup, P_{0}, of order 9; and all the elements in P \ P_{0} are of order 9. Since all of the elements of order 9 are fixed-point-free, and since a stabilizer of a point in G has order divisible by 9, we conclude that P_{0} stabilizes a point, indeed it must fix 3 points. But, since the elements of order 3 in P have cycle type 3^{3} this implies that all elements of P of order 3 fix the same 3 points. This is impossible. □

The same argument as before gives two conjugacy classes of self-centralizing elements of type 11^{1}. Similarly there is a single conjugacy class of self-centralizing elements of type 2^{1}3^{1}6^{1}. In addition there is an easy counting argument that says that the number of elements that fix exactly four points is \binom{12}{4} × 7 = 3465. There are only two conjugacy classes that do this – of type 2^{4} and 4^{2}; what is more, using the fact that the stabilizer of 4 points is Q_{8}, we see that there are six times as many elements of type 4^{2} as there are of type 2^{4}. This leaves us with 495 of type 2^{4} and 2970 of type 4^{2}.

Let g be an element of type 5^{2}. Then |C_{G}(g)| is even if and only if there is an element of type 2^{1}10^{1}. Suppose that this is not the case – then C_{G}(g) is of order 5 and all other non-trivial conjugacy classes have order divisible by 5. But 95040 − 1 − 95040/5 is not divisible by 5, a contradiction. We conclude that |C_{G}(g)| = 10, and there exist elements of type 2^{1}10^{1}. What is more N_{G}(⟨g⟩) must be a group of size 40 with C_{G}(g) a normal cyclic subgroup of order 10. If h is a generator of C_{G}(g), then N_{G}(⟨g⟩) acts by conjugation on C_{G}(g) = ⟨h⟩ and is transitive on the elements of order 10 in ⟨h⟩; we conclude that there is a unique conjugacy class of elements of type 2^{1}10^{1}, and this conjugacy class has size 9504.

Let us examine the Sylow 3-subgroup of G in more detail. The orbits of P must be of size 3 and 9, and a count of elements in the stabilizers immediately yields that P contains 14 elements of type 3^{3} and 12 that are fixed-point-free, i.e. of type 3^{4}. Since 14 ≢ 0 (mod 3), we conclude that the non-trivial central elements of P are of type 3^{3}, and the centralizer of an element of type 3^{3} is divisible by 27. In addition, for g of type 3^{3}, |C_{Alt(12)}(g)| is not divisible by 4, and so we conclude that |C_{G_{12}}(g)| = 54.

A straightforward counting argument tells us that there are 35310 fixed-point-free elements in G; we currently have 47190 elements unaccounted for, of which the only type that is not fixed-point-free are those of type 2^{1}8^{1}. We conclude that there are 11880 of these; on the other hand, consulting Table 2, there are 90 elements in conjugacy class 2^{1}8_{A}^{1} for the stabilizer of any two letters. Thus each conjugacy class of this type has size \binom{12}{2} × 90 = 11880, and we conclude that there is a unique conjugacy class of this type.

We are left with the fixed-point-free elements of G – there are 35310 of these, and they are of types 2^{6}, 3^{4}, 2^{2}4^{2}, 6^{2}, 4^{1}8^{1} and 2^{1}10^{1} (although we do not yet know if all of these occur). We already know that there are 9504 elements of type 2^{1}10^{1} and this leave us 25806 elements for the rest.

Our earlier calculations imply that the centralizer of an element of type 3^{4} is divisible by 9 but not 27; we conclude that the centralizer is of size 9, 18 or 36. Suppose that g is of type 3^{4}, and that C_{G}(g) is not of size 9 – then there is an element of type 2^{6} that centralizes g and we conclude that G contains an element h of type 6^{2}. Now C_{G}(h) is of size at most 12. But now, since \frac{1}{9} = \frac{1}{12} + \frac{1}{36}, we conclude that in any case there are at least |G|/9 = 10560 elements of type 3^{4} or 6^{2}.

Type 1^{12} 2^{4} 2^{6}  3^{3}  3^{4}  4^{2} 2^{2}4^{2} 5^{2} 2^{1}3^{1}6^{1} 6^{2}  2^{1}8^{1} 4^{1}8^{1} 2^{1}10^{1} 11_{A}^{1}^{∗} 11_{B}^{1}^{∗}
Size 1 495 396 1760 2640 2970 2970 9504 15840 7920 11880 11880 9504 8640 8640

Table 4. Conjugacy class sizes in G_{12}

The remaining fixed-point-free elements of G all have order a power of 2. Let us, therefore, examine P, a Sylow 2-subgroup of G. Since P is of size 64, we know that P does not fix any points; thus the orbit structure of P is either 8−4 or 8−2−2 (since a Sylow 2-subgroup of G_{9} has orbit structure 8−1).

Suppose that the orbit structure is 8−2−2. If α is a point in an orbit of size 2, then P_{α} is a normal index 2 subgroup of P, and so P_{α} fixes all points in the orbit. If β is in the other orbit, then P_{β} does likewise, and so P_{α} ∩ P_{β} has size at least 16 and fixes 4 points, a contradiction.

Thus the orbit structure is 8−4. Let Λ be the orbit of size 4, let Γ be the orbit of size 8, and let K be the kernel of the action on Λ; then K ≅ Q_{8}, the Sylow 2-subgroup of G_{9}. Since N = N_{Sym(8)}(K) has size 192 = 64×3, we conclude that P is isomorphic to a Sylow 2-subgroup of N. What is more the natural map P → Sym(4), given by considering the action of P on Λ, yields that P/K ≅ D_{4}, the dihedral group of order 8.

Next observe that C = C_{Sym(8)}(K) is isomorphic to Q_{8}, and Z(C) = Z(K); indeed all non-central elements of C and K act on Γ as elements of type 4−4 that all square to the same element. Then the non-central elements of C must be of type 4−4− 2−2 in the action on Ω; in particular C induces the normal Klein 4-subgroup of Sym(4) in the action on Λ. But now one can easily check that the four cosets of K in N \ C all contain an 8-cycle; since two of these cosets must induce a 4-cycle on Λ, we conclude that G contains elements of type 8−4. These elements are self-centralizing in Alt(12), and so likewise in G; what is more there are not enough elements left unaccounted for to allow for more than one such conjugacy class. Thus there is a unique conjugacy class of type 8 − 4 and it has size |G|/8 = 11880.

There are, in addition, elements of type 2^{2}4^{2}, and these have a centralizer of size at most 32; again, a count of remaining elements leads us to conclude that there is a unique conjugacy class of elements of this type and it has size |G|/32 = 2970.

At this stage, then, we have \frac{83}{720}|G| elements unaccounted for; these are of type 3^{4}, 6^{2} and/or 2^{6}, and we know that at least \frac{80}{720}|G| of them are of type 3^{4} or 6^{2}. Thus there are at most |G|/240 elements of type 2^{6}. If there are no elements of type 2^{6} centralized by an element of order 3, then we conclude that there is a unique conjugacy class of elements of type 2^{6}, and it must have size |G|/320. But this also means that there are no elements of type 6^{2}, and that the elements of type 3^{4} all have centralizers of size 9. This does not yield the correct number of elements.

We conclude that there are elements of type 2^{6} centralized by elements of order 3. By counting remaining elements, we conclude that there is a unique conjugacy class of elements of type 6^{2}, and it has size |G|/12; similarly, there is a unique conjugacy class of elements of type 3^{4}, and it has size |G|/36.

There are, therefore, |G|/240 elements of type 2^{6}; let g be one such. Notice, first, that a Sylow 3-subgroup of G does not have a subgroup of order 9 for which all elements are of type 3^{4}; we conclude that |C_{G}(g)| is not divisible by 9. Next, notice by our arguments above, that we can take g ∉ Z(KC); it is now an easy matter to check that |C_{P} (g)| ≤ 16, where P is a Sylow 2-subgroup of G. Thus |C_{G}(g)| has size at most 240 and we conclude that there is exactly one conjugacy class of elements of type 2^{6}.

We summarise what we have worked out in Table 4. Note that we have asterisked the two conjugacy classes that are not real; it is clear that these are the only possible conjugacy classes that have a chance of being non-real, and it is equally clear that they are non-real, since they are non-real in Alt(12).

3. Character Tables

In this section we work out the character tables of G_{9}, G_{10}, G_{11} and G_{12}. To do this we will need nothing more than the basics of ordinary character theory, along with enough information about the ordinary characters of the symmetric group to calculate irreducible characters for G_{12}.

As a reminder we note down five results that will be particularly useful in what follows.

Theorem 3. [JL01, p. 342] If G acts 2-transitively on a finite set Ω, then the character given by χ(g) = |fix(g)| − 1 is irreducible.

Theorem 4. [JL01, p. 196] Let G be a finite group and let g ∈ G. Let χ be a faithful irreducible character of G, then χ^{2}(g) decomposes as the direct sum of a symmetric part \frac{1}{2}(χ^{2}(g) + χ(g^{2})) and an antisymmetric part \frac{1}{2}(χ^{2}(g) − χ(g^{2})).

Theorem 5. [JL01, p. 236] Let G be a finite group, x ∈ G, H ≤ G, and suppose that H ∩ x^{G} breaks up into l conjugacy classes of H. If χ is a character of H, then:

(1)  (χ ↑ G)(x) = |C_{G}(x)| (\frac{χ(x_{1})}{|C_{H}(x_{1})|} + ⋯ + \frac{χ(x_{l})}{|C_{H}(x_{l})|})

where x_{1}, …, x_{l} ∈ H are representative elements of the l classes of H.

Theorem 6. [JL01, p. 264] The number of real characters of G is the same as the number of real conjugacy classes of G.

Next we recall the definition of the inner product on characters: if χ and ϕ are characters of G, then we define

⟨χ, ϕ⟩ = ∑_{i=1}^{k} \frac{χ(g_{i})\overline{ϕ}(g_{i})}{|C_{G}(g_{i})|}.

The next theorem is a reminder of Schur’s Orthogonality Relations; in particular it tells us how to use this inner product to recognise irreducible characters.

Theorem 7. [JL01, p. 161] Let χ_{1}, …, χ_{k} be the irreducible characters of G and let g_{1}, …, g_{k} be representative elements of the conjugacy classes of G, then for every r, s ∈ {1, …, k} we have:

(2)  ⟨χ_{r}, χ_{s}⟩ = δ_{rs};
(3)  ∑_{i=1}^{k} χ_{i}(g_{r})\overline{χ_{i}}(g_{s}) = δ_{rs}|C_{G}(g_{r})|.

3.1. The Character Table of G_{9}. We know that G_{9} has 6 conjugacy classes and hence 6 irreducible characters. We also know that G_{9} has a normal subgroup N ≅ C_{3} × C_{3}, with G/N ≅ Q_{8}. We can lift 3 linear characters and a 2 dimensional character from Q_{8}. The final character χ_{5} is given by the 2-transitive action of M_{9} on 9 points (Theorem 3). The character table is given in Table 5.

\begin{matrix}G_{9} & 1^{9} & 2^{4} & 3^{3} & 4_{A}^{2} & 4_{B}^{2} & 4_{C}^{2} \\ χ_{0} & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{1} & 1 & 1 & 1 & −1 & 1 & −1 \\ χ_{2} & 1 & 1 & 1 & 1 & −1 & −1 \\ χ_{3} & 1 & 1 & 1 & −1 & −1 & 1 \\ χ_{4} & 2 & −2 & 2 & 0 & 0 & 0 \\ χ_{5} & 8 & 0 & −1 & 0 & 0 & 0 \end{matrix}

Table 5. The character table of G_{9}

3.2. The Character table of G_{10}. Note that G_{10} has 8 conjugacy classes, hence 8 irreducible characters. Let χ_{0} denote the trivial character of G_{10} and let χ_{2} denote the irreducible character obtained from the 3-transitive action of G_{10}.

\begin{matrix}G_{10} & 1^{10} & 2^{4} & 3^{3} & 4_{A}^{2} & 4_{B}^{2} & 5^{2} & 2^{1}8_{A}^{1} & 2^{1}8_{B}^{1} \\ χ_{0} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 9 & 1 & 0 & 1 & 1 & −1 & −1 & −1 \end{matrix}

We will now try inducing characters from the point stabilizer G_{9}. Let χ be a character of G_{9}, then using the centralizer orders and Theorem 5 we have:

(χ ↑ G_{10})(g) = \begin{cases}10χ(g) & if g ∈ 1^{10}; \\ 2χ(g) & if g ∈ 2^{4}, 4_{A}^{2}; \\ χ(g_{B}) + χ(g_{C}) & if g ∈ 4_{B}^{2}; \\ χ(g) & if g ∈ 3^{3}; \\ 0 & otherwise. \end{cases}

(Here g_{A} and g_{B} are elements from the G_{9}-classes 4_{A}^{2} and 4_{B}^{2} respectively.) Let χ_{A}, χ_{B} and χ_{C} denote the lifts of the characters χ_{1}, χ_{2} and χ_{4} for G_{9} respectively (see Table 5).

\begin{matrix}χ_{A} & 10 & 2 & 1 & −2 & 0 & 0 & 0 & 0 \\ χ_{B} & 10 & 2 & 1 & 2 & −2 & 0 & 0 & 0 \\ χ_{C} & 20 & −4 & 2 & 0 & 0 & 0 & 0 & 0 \end{matrix}

Taking inner products of the characters we find that χ_{A} is irreducible, ⟨χ_{B}, χ_{B}⟩ = 2 and ⟨χ_{C}, χ_{C}⟩ = 2. To keep with our naming convention, relabel χ_{A} to χ_{4}. Our next strategy will be to construct the anti-symmetric parts of both χ_{2}^{2} and χ_{4}^{2}, we will skip over the symmetric decomposition of these characters because they do not yield information that is useful to our endeavours. Let χ_{D} and χ_{E} denote the antisymmetric components of χ_{2}^{2} and χ_{4}^{2} respectively.

\begin{matrix}χ_{D} & 36 & −4 & 0 & 0 & 0 & 1 & 0 & 0 \\ χ_{E} & 45 & −3 & 0 & 1 & −1 & 0 & 0 & 0 \end{matrix}

Taking inner products of χ_{D} with known irreducible and compound characters we find that ⟨χ_{D}, χ_{D}⟩ = 3 and ⟨χ_{C}, χ_{D}⟩ = 2. Let χ_{7} = χ_{D} − χ_{C}; then ⟨χ_{7}, χ_{7}⟩ = 1 and, since χ_{7}(1) > 0, we conclude that χ_{7} is an irreducible character. Repeating the process with χ_{E} we find that ⟨χ_{E}, χ_{E}⟩ = 4, ⟨χ_{7}, χ_{E}⟩ = 1, ⟨χ_{B}, χ_{E}⟩ = 1 and ⟨χ_{C}, χ_{E}⟩ = 2. Let χ_{3} = χ_{E} − χ_{7} − χ_{C} = χ_{E} − χ_{D}, then ⟨χ_{3}, χ_{3}⟩ = 1 and, since χ_{3}(1) > 0, we conclude that χ_{3} is an irreducible character

\begin{matrix}χ_{3} & 9 & 1 & 0 & 1 & −1 & −1 & 1 & 1 \\ χ_{7} & 16 & 0 & −2 & 0 & 0 & 1 & 0 & 0 \end{matrix}

Finally, observe that ⟨χ_{B}, χ_{3}⟩ = 1. Then χ_{1} = χ_{B} − χ_{3} is irreducible. Using Theorem 6 and the fact that G_{10} has two non-real classes, we conclude that the remaining two characters occur as a complex conjugate pair. These can, then, be calculated using the Schur orthogonality relations.

\begin{matrix}G_{10} & 1^{10} & 2^{4} & 3^{3} & 4_{A}^{2} & 4_{B}^{2} & 5^{2} & 2^{1}8_{A}^{1} & 2^{1}8_{B}^{1} \\ χ_{0} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{1} & 1 & 1 & 1 & 1 & −1 & 1 & −1 & −1 \\ χ_{2} & 9 & 1 & 0 & 1 & 1 & −1 & −1 & −1 \\ χ_{3} & 9 & 1 & 0 & 1 & −1 & −1 & 1 & 1 \\ χ_{4} & 10 & 2 & 1 & −2 & 0 & 0 & 0 & 0 \\ χ_{5} & 10 & −2 & 1 & 0 & 0 & 0 & ω & \overline{ω} \\ χ_{6} & 10 & −2 & 1 & 0 & 0 & 0 & \overline{ω} & ω \\ χ_{7} & 16 & 0 & −2 & 0 & 0 & 1 & 0 & 0 \end{matrix}

Table 6. The character table of G_{10}, where ω = \sqrt{−2}.

3.2.1. The Structure of G_{10}. Using the character table, the following result can be easily derived about the structure of G_{10}.

Theorem 8. The group G_{10} is a non split extension of C_{2} by a normal subgroup K = ker(χ_{1}).

Proof. As the kernels of characters are normal subgroups we see that K ◁ G_{10} and moreover this is the only non trivial proper normal subgroup of G_{10}. Observe that G_{10}/K ≅ C_{2}. Since there are no involutions in G_{10} \ K, we conclude that we have a non-split extension. □

3.3. The Character Table of G_{11}. By assumption G_{11} acts 4-transitively on a set of size 11. By considering the number of fixed points of each conjugacy class we obtain a 10 dimensional irreducible character χ_{1} (Theorem 3).

\begin{matrix}G_{11} & 1^{11} & 2^{4} & 3^{3} & 4^{2} & 5^{2} & 2^{1}3^{1}6^{1} & 2^{1}8_{A}^{1} & 2^{1}8_{B}^{1} & 11_{A}^{1} & 11_{B}^{1} \\ χ_{0} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{1} & 10 & 2 & 1 & 2 & 0 & −1 & 0 & 0 & −1 & −1 \end{matrix}

Now, let χ_{S} and χ_{A} be the symmetric and antisymmetric decomposition of χ_{1}^{2}.

\begin{matrix}χ_{S} & 55 & 7 & 1 & 3 & 0 & 1 & 1 & 1 & 0 & 0 \\ χ_{A} & 45 & -3 & 0 & 1 & 0 & 0 & -1 & -1 & 1 & 1 \end{matrix}

A quick calculation gives ⟨χ_{S}, χ_{S}⟩ = 3 and ⟨χ_{A}, χ_{A}⟩ = 1. Moreover, ⟨χ_{S}, χ_{0}⟩ = 1 and ⟨χ_{S}, χ_{1}⟩ = 1. Define χ_{8} = χ_{A} and χ_{7} = χ_{S} − χ_{0} − χ_{1} and note that ⟨χ_{7}, χ_{7}⟩ = 1. Hence we have found two new irreducible characters of G_{11}.

\begin{matrix}χ_{7} & 44 & 4 & −1 & 0 & −1 & 1 & 0 & 0 & 0 & 0 \\ χ_{8} & 45 & −3 & 0 & 1 & 0 & 0 & −1 & −1 & 1 & 1 \end{matrix}

3.3.1. Induction from G_{10}. We will now try inducing characters from the 1-point stabilizer subgroup G_{10}. Let χ be a character of G_{10}; then, using the centralizer orders and Theorem 5, we have:

(χ ↑ G_{11})(g) = \begin{cases}11χ(g) & if g ∈ 1^{10}; \\ 3χ(g) & if g ∈ 2^{4}; \\ 2χ(g) & if g ∈ 3^{3}; \\ χ(g_{A}) + 2χ(g_{B}) & if g ∈ 4^{2}; \\ χ(g) & if g ∈ 5^{2}, 2^{1}8_{A}^{1}, 2^{1}8_{B}^{1}; \\ 0 & otherwise. \end{cases}

(Here g_{A} and g_{B} are elements from the G_{10}-classes 4_{A}^{2} and 4_{B}^{2} respectively.) Let χ_{4} be the induced character of the non trivial linear character of G_{10} and let χ_{9} be the the anti-symmetric decomposition of χ_{4}^{2}.

\begin{matrix}χ_{4} & 11 & 3 & 2 & −1 & 1 & 0 & −1 & −1 & 0 & 0 \\ χ_{9} & 55 & −1 & 1 & −1 & 0 & −1 & 1 & 1 & 0 & 0 \end{matrix}

We find that ⟨χ_{4}, χ_{4}⟩ = 1 and ⟨χ_{9}, χ_{9}⟩ = 1, hence, they are both irreducible characters of G_{11}.

3.3.2. Schur Orthogonality. The remaining four characters come in complex conjugate pairs. We can deduce this by using the fact that elements of cycle type 2^{1}8^{1} and 11^{1} are not real and applying Theorem 6. Given that each pair will have the same dimension, we can attempt to calculate the dimension of these 4 remaining characters. Let the dimension of the first pair be d_{1} and the dimension of the second pair be d_{2}.

We have d_{1}^{2} +d_{1}^{2} +d_{2}^{2} +d_{2}^{2} = 712, hence d_{1}^{2} +d_{2}^{2} = 356. By an exhaustive search we find that 356 can be expressed as the sum of two squares in exactly one way, that is 356 = 10^{2} + 16^{2}. It immediately follows that d_{1} = 10 and d_{2} = 16.

Let χ_{2}(1) = χ_{3}(1) = 10 and χ_{5}(1) = χ_{6}(1) = 16. We will now use the column relations to calculate the character values for the remaining conjugacy classes. We use the fact that χ(g) = \overline{χ(g^{−1})}, and let the characters take the following values:

\begin{matrix}χ_{2} & 11 & x_{1} & x_{2} & x_{3} & x_{4} & x_{5} & x_{6} & \overline{x_{6}} & x_{7} & \overline{x_{7}} \\ χ_{3} & 11 & x_{1} & x_{2} & x_{3} & x_{4} & x_{5} & \overline{x_{6}} & x_{6} & \overline{x_{7}} & x_{7} \\ χ_{5} & 16 & y_{1} & y_{2} & y_{3} & y_{4} & y_{5} & y_{6} & \overline{y_{6}} & y_{7} & \overline{y_{7}} \\ χ_{6} & 16 & y_{1} & y_{2} & y_{3} & y_{4} & y_{5} & \overline{y_{6}} & y_{6} & \overline{y_{7}} & y_{7} \end{matrix}

Substituting the column containing x_{i} for i = 1, … 5 and the first column of the character table into (3) we can obtain values for x_{i} and y_{i}. We shall demonstrate this with x_{1}:

1 + 20 + 2 × 10x_{1} + 33 + 2 × 16y_{1} + 176 − 135 − 55 = 0
(4)  ⟹  5x_{1} + 8y_{1} = −10.

Substituting the column containing x_{1} into (3) twice gives:

1 + 4 + 2x_{1}^{2} + 9 + 2y_{1}^{2} + 16 + 9 + 1 = 48
(5)  ⟹  x_{1}^{2} + y_{1}^{2} = 4.

Solving (4) and (5), we obtain two solutions x_{1} = −2, y_{1} = 0 and x_{1} = \frac{78}{89}, y_{1} = \frac{−160}{89}. The second set of these cannot be expressed as sum of 2nd roots of unity. Hence, x_{1} = −2 and y_{1} = 0. Continuing in this manner we obtain:

\begin{matrix}χ_{2} & 10 & −2 & 1 & 0 & 0 & 1 & x_{6} & \overline{x_{6}} & x_{7} & \overline{x_{7}} \\ χ_{5} & 16 & 0 & −2 & 0 & 1 & 0 & y_{6} & \overline{y_{6}} & y_{7} & \overline{y_{7}} \end{matrix}

The remaining values x_{6}, x_{7}, y_{6}, and y_{7} can be calculated by repeat applications of the row and column relations. We find that x_{6} = \sqrt{−2}, x_{7} = −1, y_{6} = 0 and y_{7} = \frac{1}{2}(−1 + \sqrt{−11}). Table 7 shows the complete character table.

\begin{matrix}G_{11} & 1^{11} & 2^{4} & 3^{3} & 4^{2} & 5^{2} & 2^{1}3^{1}6^{1} & 2^{1}8_{A}^{1} & 2^{1}8_{B}^{1} & 11_{A}^{1} & 11_{B}^{1} \\ χ_{0} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{1} & 10 & 2 & 1 & 2 & 0 & −1 & 0 & 0 & −1 & −1 \\ χ_{2} & 10 & −2 & 1 & 0 & 0 & 1 & α & \overline{α} & −1 & −1 \\ χ_{3} & 10 & −2 & 1 & 0 & 0 & 1 & \overline{α} & α & −1 & −1 \\ χ_{4} & 11 & 3 & 2 & −1 & 1 & 0 & −1 & −1 & 0 & 0 \\ χ_{5} & 16 & 0 & −2 & 0 & 1 & 0 & 0 & 0 & β & \overline{β} \\ χ_{6} & 16 & 0 & −2 & 0 & 1 & 0 & 0 & 0 & \overline{β} & β \\ χ_{7} & 44 & 4 & −1 & 0 & −1 & 1 & 0 & 0 & 0 & 0 \\ χ_{8} & 45 & −3 & 0 & 1 & 0 & 0 & −1 & −1 & 1 & 1 \\ χ_{9} & 55 & −1 & 1 & −1 & 0 & −1 & 1 & 1 & 0 & 0 \end{matrix}

Table 7. The character table of G_{11}, where α = \sqrt{−2} and β = \frac{1}{2}(−1 + \sqrt{−11}).

3.4. The Character Table of G_{12}. We begin by noting that G_{12} has 15 conjugacy classes and 15 irreducible characters, one of which is the trivial character χ_{0}.

3.4.1. The Permutation Character and Tensor Products. By assumption, G_{12} acts 5-transitively on a set of size 12. Hence, by Theorem 3, we get the permutation character χ_{1}.

\begin{matrix}G_{12} & 1^{12} & 2^{4} & 2^{6} & 3^{3} & 3^{4} & 4^{2} & 2^{2}4^{2} & 5^{2} & 2^{1}3^{1}6^{1} & 6^{2} & 2^{1}8^{1} & 4^{1}8^{1} & 2^{1}10^{1} & 11_{A}^{1} & 11_{B}^{1} \\ χ_{0} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{1} & 11 & −1 & 3 & 2 & −1 & 3 & −1 & 1 & 0 & −1 & −1 & 1 & −1 & 0 & 0 \end{matrix}

Now, let χ_{S} and χ_{A} be the symmetric and antisymmetric decomposition of χ_{1}^{2}.

\begin{matrix}χ_{S} & 66 & 10 & 6 & 3 & 0 & 6 & 2 & 1 & 1 & 0 & 2 & 0 & 1 & 0 & 0 \\ χ_{A} & 55 & −1 & −5 & 1 & 1 & 3 & −1 & 0 & −1 & 1 & −1 & 1 & 0 & 0 & 0 \end{matrix}

A quick calculation gives ⟨χ_{S}, χ_{S}⟩ = 3 and ⟨χ_{A}, χ_{A}⟩ = 1. Moreover, ⟨χ_{S}, χ_{0}⟩ = 1 and ⟨χ_{S}, χ_{1}⟩ = 1. Define χ_{8} = χ_{A} and χ_{6} = χ_{S} − χ_{0} − χ_{1}. We have found two new irreducible characters of G_{12}. Squaring these characters, however, is not a viable plan; the characters obtained have dimensions 2916 and 3025.

\begin{matrix}χ_{6} & 54 & 6 & 6 & 0 & 0 & 2 & 2 & −1 & 0 & 0 & 0 & 0 & 1 & −1 & −1 \\ χ_{8} & 55 & −1 & −5 & 1 & 1 & 3 & −1 & 0 & −1 & 1 & −1 & 1 & 0 & 0 & 0 \end{matrix}

3.4.2. Induction from G_{11}. We will now try inducing characters from the subgroup G_{11}. Let χ be a character of G_{11}; then using the centralizer orders and Theorem 5 we have:

(χ ↑ G_{12})(g) = \begin{cases}12χ(g) & if g ∈ 1^{12}; \\ 4χ(g) & if g ∈ 2^{4}, 4^{2}; \\ 3χ(g) & if g ∈ 3^{3}; \\ 2χ(g) & if g ∈ 5^{2}; \\ χ(g) & if g ∈ 2^{1}3^{1}6^{1}, 11_{A}^{1}, 11_{B}^{1}; \\ χ(g_{A}) + χ(g_{B}) & if g ∈ 2^{1}8^{1}; \\ 0 & otherwise. \end{cases}

(Here g_{A} and g_{B} are elements from the G_{11}-classes 2^{1}8_{A}^{1} and 2^{1}8_{B}^{1} respectively.) Inducing the trivial character of G_{11} gives a character equal to χ_{0} +χ_{1}. Inducing the integer valued 10-dimensional character gives a character equal to χ_{1} + χ_{6} + χ_{8}. Now, let χ_{12} be the induced character of a complex valued 10-dimensional character of G_{11}. We see that ⟨χ_{12}, χ_{12}⟩ = 1, therefore, χ_{12} is irreducible.

Let χ_{V} be the character obtained by inducing the 55-dimensional character of G_{11}. Finally, let χ_{B} be the induced complex valued 16-dimensional character of G_{11} and let ω = \frac{1}{2}(−1 + \sqrt{−11}), then we have:

\begin{matrix}χ_{12} & 120 & −8 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & −1 & −1 \\ χ_{V} & 660 & −4 & 0 & 3 & 0 & 0 & −4 & 0 & 0 & −1 & 2 & 0 & 0 & 0 & 0 \\ χ_{B} & 192 & 0 & 0 & −6 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & ω & \overline{ω} \end{matrix}

Note that ⟨χ_{B}, χ_{B}⟩ = 2 and ⟨χ_{V} , χ_{V} ⟩ = 6, but the inner product of χ_{B} with any known irreducible is 0. We have ⟨χ_{V} , χ_{12}⟩ = 1, but the inner product is 0 for any other known irreducibles. These do not give us any new irreducible characters, but we will use these characters later.

3.4.3. Restriction from Sym(12). Using the Frobenius character formula, it is possible to construct low dimensional characters of Sym(12) evaluated over the conjugacy classes of G_{12}; recall that these characters are labeled with partitions of 12. In this section we consider the restriction of some of these characters to the group G_{12}; for instance we note that (χ_{(11,1)} ↓ G_{12}) = χ_{1}, (χ_{(10,2)} ↓ G_{12}) = χ_{2} and (χ_{(10,1,1)} ↓ G_{12}) = χ_{3}.

The table below gives six new characters that we have constructed in this way. Note that we abuse notation here: given a partition, λ, of 12, we would normally write χ_{λ} for the associated character of Sym(12) whereas here we write χ_{λ} for the restriction of the associated character to G_{12}. Note too that, for ease of notation, we let λ_{A} = (9, 1, 1, 1) and λ_{B} = (8, 1, 1, 1, 1).^{2}

\begin{matrix}G_{12} & 1^{12} & 2^{4} & 2^{6} & 3^{3} & 3^{4} & 4^{2} & 2^{2}4^{2} & 5^{2} & 2^{1}3^{1}6^{1} & 6^{2} & 2^{1}8^{1} & 4^{1}8^{1} & 2^{1}10^{1} & 11_{A}^{1} & 11_{B}^{1} \\ χ_{(9,3)} & 154 & 10 & −6 & 1 & 4 & −2 & −2 & −1 & 1 & 0 & 0 & 0 & −1 & 0 & 0 \\ χ_{λ_{A}} & 165 & −11 & 5 & 3 & 3 & 1 & 1 & 0 & 1 & −1 & −1 & −1 & 0 & 0 & 0 \\ χ_{(8,4)} & 275 & 11 & 15 & 5 & −4 & −1 & 3 & 0 & −1 & 0 & −1 & 1 & 0 & 0 & 0 \\ χ_{(7,5)} & 297 & 9 & −15 & 0 & 0 & 5 & −3 & 2 & 0 & 0 & −1 & −1 & 0 & 0 & 0 \\ χ_{(3,2,1)} & 320 & 0 & 0 & −4 & −4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ χ_{λ_{B}} & 330 & −6 & 10 & 6 & −3 & −2 & −2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}

We first check the inner product of each character with itself and then with each of the known irreducibles. We find that the ⟨χ_{λ_{A}}, χ_{λ_{A}}⟩ = 2 and ⟨χ_{λ_{A}}, χ_{12}⟩ = 1, define χ_{5} = χ_{λ_{A}} − χ_{12} and note that ⟨χ_{5}, χ_{5}⟩ = 1. Hence, we have found a new irreducible character of G_{12}.

χ_{5}  45 −3 5 0 3  1  1  0  0  −1 −1 −1  0  1  1

We find that ⟨χ_{λ_{B}}, χ_{λ_{B}}⟩ = 3 and that ⟨χ_{λ_{B}}, χ_{12}⟩ = 1, hence we define χ_{X} = χ_{λ_{B}} − χ_{12}. Similarly we find that ⟨χ_{(8,4)}, χ_{(8,4)}⟩ = 4 and that ⟨χ_{(8,4)}, χ_{6}⟩ = 1, hence we define χ_{Y} = χ_{(8,4)} − χ_{6}. Checking the inner products of χ_{X} and χ_{Y} with themselves and each other we obtain ⟨χ_{X}, χ_{X}⟩ = 2, ⟨χ_{Y} , χ_{Y} ⟩ = 3 and ⟨χ_{X}, χ_{Y} ⟩ = 2. Define χ_{2} = χ_{Y} − χ_{X}, we find that ⟨χ_{2}, χ_{2}⟩ = 1; we have found a new irreducible character of G_{12}.

\begin{matrix}χ_{Y} & 221 & 5 & 9 & 5 & −4 & −3 & 1 & 1 & −1 & 0 & −1 & 1 & −1 & 1 & 1 \\ χ_{X} & 210 & 2 & 10 & 3 & −3 & −2 & −2 & 0 & 1 & −1 & 0 & 0 & 0 & 1 & 1 \\ χ_{2} & 11 & 3 & −1 & 2 & −1 & −1 & 3 & 1 & 0 & −1 & −1 & 1 & −1 & 0 & 0 \end{matrix}

Let χ_{S} and χ_{A} be the symmetric and antisymmetric decomposition of χ_{2}^{2}. We find that χ_{S} = χ_{0} + χ_{2} + χ_{6} and χ_{9} = χ_{A} is a new irreducible.

χ_{9}  55 −1 −5 1 1  3 −1 0  −1  1  1 −1  0  0  0

2The history of this sort of restriction is worth a note: a classical result of Frobenius asserts that if t is a natural number with t < n/2, then a subgroup G < Sym(n) is 2t-transitive if and only if every character of Sym(n) labeled by a partition (λ_{1}, λ_{2}, …, λ_{a})

with λ_{2} + ⋯ + λ_{a} ≤ t remains irreducible when restricted to G. This result appears in [Fro04], the same paper in which Frobenius calculates the character tables of M_{12} and M_{24}; indeed Frobenius makes use of this result in his calculation of these tables. Note that Theorem 3 is a special case of Frobenius’ result. A modern version of Frobenius result, making use of the Classification of Finite Simple Groups, was given in a beautiful paper of Saxl [Sax87] – his theorem considers G, a subgroup of Sym(n) or of Alt(n) and χ, an ordinary character of Sym(n) or of Alt(n), and he describes all pairs (χ, G) where the restriction of χ to G is irreducible. In our proof we do not make use of Frobenius’ (or Saxl’s) result – it is enough for us to be able to calculate the restriction of various characters of Sym(12)

directly and naively.

We will now check the inner product of every restricted character and χ_{V} with our known irreducible characters of G_{12}.

Character χ_{0} χ_{1} χ_{2} χ_{5} χ_{6} χ_{8} χ_{9} χ_{12}
χ_{(9,3)}  0 0 0 0 0 0 0 0
χ_{λ_{A}}  0 0 0 1 0 0 0 1
χ_{(8,4)}  0 0 1 0 1 0 0 0
χ_{(7,5)}  0 1 0 0 1 0 0 0
χ_{(3,2,1)}  0 0 0 0 0 0 0 0
χ_{λ_{B}}  0 0 0 0 0 0 0 1
χ_{V}  0 0 0 1 0 0 0 1

Define the following characters:

χ_{C} = χ_{(9,3)}
χ_{D} = χ_{(7,5)} − χ_{1} − χ_{6}
χ_{E} = χ_{(3,2,1)}
χ_{F} = χ_{λ_{B}} − χ_{12}
χ_{W} = χ_{V} − χ_{5} − χ_{12}

The values of these are as follows.

\begin{matrix}G_{12} & 1^{12} & 2^{4} & 2^{6} & 3^{3} & 3^{4} & 4^{2} & 2^{2}4^{2} & 5^{2} & 2^{1}3^{1}6^{1} & 6^{2} & 2^{1}8^{1} & 4^{1}8^{1} & 2^{1}10^{1} & 11_{A}^{1} & 11_{B}^{1} \\ χ_{C} & 154 & 10 & −6 & 1 & 4 & −2 & −2 & −1 & 1 & 0 & 0 & 0 & −1 & 0 & 0 \\ χ_{D} & 231 & 7 & −9 & −1 & 0 & −1 & −1 & 1 & 1 & 0 & −1 & −1 & 1 & 0 & 0 \\ χ_{E} & 320 & 0 & 0 & −4 & −4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ χ_{F} & 210 & 2 & 10 & 3 & −3 & −2 & −2 & 0 & 1 & −1 & 0 & 0 & 0 & 1 & 1 \\ χ_{W} & 485 & 5 & 5 & −1 & −1 & −3 & −3 & 0 & −1 & −1 & 1 & 1 & 0 & 1 & 1 \end{matrix}

Taking the inner products of each of these new characters with each other gives the following:

χ_{C} χ_{D} χ_{E} χ_{F} χ_{W}
χ_{C} 2 1 0 0 1
χ_{D} 1 2 1 0 1
χ_{E} 0 1 2 1 2
χ_{F} 0 0 1 2 2
χ_{W} 1 1 2 2 4

Note, first, that this table of values implies that each of these characters is the sum of distinct irreducibles. Writing these irreducibles as α, β, γ, and so on, we see immediately that we can write

χ_{C} = α + β;
χ_{D} = β + γ;
χ_{E} = γ + δ;
χ_{F} = γ + ϵ;
χ_{W} = α + γ + δ + ϵ.

Now one obtains that α = \frac{1}{2}(χ_{W} − χ_{F} + χ_{C} − χ_{D}). Once we have α it is an easy matter to obtain the other four irreducibles using the equalities just given. We therefore have five new irreducibles which we label as follows:

α = χ_{11}, β = χ_{7}, γ = χ_{14}, δ = χ_{13}, ϵ = χ_{10}.

Finally, we return to the character χ_{B} from earlier. We find that ⟨χ_{B}, χ_{14}⟩ = 1, and so we define χ_{3} = χ_{B} − χ_{14}. Letting χ_{4} be the complex conjugate of χ_{3}, we obtain our final two irreducibles. The full character table is given in Table 8.

\begin{matrix}G_{12} & 1^{12} & 2^{4} & 2^{6} & 3^{3} & 3^{4} & 4^{2} & 2^{2}4^{2} & 5^{2} & 2^{1}3^{1}6^{1} & 6^{2} & 2^{1}8^{1} & 4^{1}8^{1} & 2^{1}10^{1} & 11_{A}^{1} & 11_{B}^{1} \\ χ_{0} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{1} & 11 & 3 & −1 & 2 & −1 & 3 & −1 & 1 & 0 & −1 & 1 & −1 & −1 & 0 & 0 \\ χ_{2} & 11 & 3 & −1 & 2 & −1 & −1 & 3 & 1 & 0 & −1 & −1 & 1 & −1 & 0 & 0 \\ χ_{3} & 16 & 0 & 4 & −2 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & −1 & ω & \overline{ω} \\ χ_{4} & 16 & 0 & 4 & −2 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & −1 & \overline{ω} & ω \\ χ_{5} & 45 & −3 & 5 & 0 & 3 & 1 & 1 & 0 & 0 & −1 & −1 & −1 & 0 & 1 & 1 \\ χ_{6} & 54 & 6 & 6 & 0 & 0 & 2 & 2 & −1 & 0 & 0 & 0 & 0 & 1 & −1 & −1 \\ χ_{7} & 55 & 7 & −5 & 1 & 1 & −1 & −1 & 0 & 1 & 1 & −1 & −1 & 0 & 0 & 0 \\ χ_{8} & 55 & −1 & −5 & 1 & 1 & −1 & 3 & 0 & −1 & 1 & −1 & 1 & 0 & 0 & 0 \\ χ_{9} & 55 & −1 & −5 & 1 & 1 & 3 & −1 & 0 & −1 & 1 & 1 & −1 & 0 & 0 & 0 \\ χ_{10} & 66 & 2 & 6 & 3 & 0 & −2 & −2 & 1 & −1 & 0 & 0 & 0 & 1 & 0 & 0 \\ χ_{11} & 99 & 3 & −1 & 0 & 3 & −1 & −1 & −1 & 0 & −1 & 1 & 1 & −1 & 0 & 0 \\ χ_{12} & 120 & −8 & 0 & 3 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & −1 & −1 \\ χ_{13} & 144 & 0 & 4 & 0 & −3 & 0 & 0 & −1 & 0 & 1 & 0 & 0 & −1 & 1 & 1 \\ χ_{14} & 176 & 0 & −4 & −4 & −1 & 0 & 0 & 1 & 0 & −1 & 0 & 0 & 1 & 0 & 0 \end{matrix}

Table 8. The character table of G_{12}, where ω = \frac{1}{2}(−1 + \sqrt{−11}).

4. Final remarks

One can read off many properties of the groups G_{11} and G_{12} by looking at the character tables that we have constructed. Note, for instance, that all of the irreducibles of the two groups have trivial kernel; one concludes immediately that G_{11} and G_{12} are simple.

We saw above, in §2.1, that G_{9} ≅ (C_{3} × C_{3}) ⋊ Q_{8}. We should note that, although we have not deduced the isomorphism types of G_{10}, G_{11} and G_{12}, in each case it is well-known that they are unique up to group isomorphism. Indeed G_{10} ≅ M_{10}, the unique non-split degree 2 extension of Alt(6), while G_{11} ≅ M_{11} and G_{12} ≅ M_{12}, the two smallest sporadic simple groups of Mathieu.

References

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Department of Mathematics,, University of South Wales,, Treforest, CF37 1DL, U.K., nick.gill@southwales.ac.uk

Department of Mathematics,, University of South Wales,, Treforest, CF37 1DL, U.K., just.sam.hughes@gmail.com