ALMOST IRREDUCIBLE TENSOR SQUARES

Gunter Malle

Universität Heidelberg

Let G be a covering group of a finite almost simple group. We determine those faithful irreducible complex characters χ of G for which χ ⊗ χ^{∗} − 1 is again irreducible. This gives a classification of the quasi-simple absolutely irreducible subgroups of GL_{n}(q) of order prime to q which act irreducibly on the Lie algebra of type A_{n−1} via the adjoint representation. The proof uses Lusztig’s description of the degrees of irreducible characters of reductive groups and the determination of Brauer trees by Fong and Srinivasan to handle the case of groups of Lie type. It turns out that the only infinite series of examples are characters of Weyl representations for SU_{n}(𝔽_{2}) and Sp_{2n}(𝔽_{3}).

1. Introduction and statement of the result

Let G be a finite group, M an irreducible ℂG-module and χ its character. Let χ^{∗} be the character of the dual of M. Then χ⊗χ^{∗} contains the trivial character precisely once and all its values are real and non-negative. Thus it resembles the character π of a transitive permutation representation for G. Such a permutation representation is doubly transitive if and only if the character π − 1 is irreducible. Here we study the analogue for χ⊗χ^{∗}. More precisely we determine those irreducible projective complex characters χ of finite almost simple groups such that χ ⊗ χ^{∗} − 1 is irreducible.

Theorem. Let G be a finite group such that G/Z(G) is almost simple (i.e., there exists a non-abelian simple group S such that S ≤ G/Z(G) ≤ Aut(S)), and χ an irreducible faithful complex character of G. Then χ ⊗ χ^{∗} − 1 is irreducible if and only if (G, χ(1)) are as in Table 7.1. (The notation for extensions and covering groups of simple groups follows [3].)

These characters play a rôle in the investigation of maximal subgroups of finite groups of Lie type. Namely, assume that G is an irreducible subgroup of GL_{n}(K), K an algebraically closed field of characteristic 0 or prime to |G|, with corresponding (Brauer) character χ. Then via the adjoint representation G also acts on the Lie algebra of type A_{n−1} over K, with character ψ := χ ⊗ χ^{∗} − 1. Thus for G/Z(G) almost simple the above theorem determines those cases where ψ is again irreducible. The case where the characteristic of K divides the order of G remains open.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 20C15.

The author gratefully acknowledges financial support by the Deutsche Forschungsgemeinschaft

Acknowledgement: I would like to thank Thomas Weigel for drawing my attention to this problem and for a very helpful conversation.

2. First reductions

First note that if χ is as in the theorem, then either χ(1) = 2 or χ is non-real. Indeed, if χ is real, then χ ⊗ χ splits into the symmetric and the anti-symmetric part, both of which have degree at least 2 if χ(1) > 2.

We will use the following simple but crucial observations.

Remark 2.1. Let ℓ be a prime dividing |G| such that the Sylow ℓ-subgroup of G is cyclic. Let χ ∈ Irr(G) and assume that ψ := χ ⊗ χ^{∗} − 1 is irreducible. Clearly not both χ, ψ can be ℓ-projective. If χ is projective, then so is χ ⊗ χ^{∗}, hence ψ lies in the principal ℓ-block of G and is connected to the trivial character on the Brauer tree. Since the trivial character can be connected to only one other node, this determines the degree of ψ (and hence of χ) uniquely. Furthermore, if the trivial character is not connected to the exceptional node, this leaves only one possibility for ψ.

Remark 2.2. Now assume that moreover S ≤ G/Z(G) ≤ Aut(S) for some non-abelian simple group S. Let χ ∈ Irr(G) be faithful such that ψ := χ ⊗ χ^{∗} − 1 is irreducible. Since χ is faithful the restriction ψ|_{G_{0}} of ψ to G_{0} = F^{∗}(G) does not contain the trivial character, so χ_{0} := χ|_{G_{0}} remains irreducible. Thus

χ_{0} ⊗ χ_{0}^{∗} = 1 + ψ_{1} + … + ψ_{s}

for certain non-trivial ψ_{i} ∈ Irr(G_{0}). If χ_{0} is of ℓ-defect zero, then one of the ψ_{i} is connected to the trivial character on the Brauer tree of the principal ℓ-block of G_{0}, hence all of them are. Thus in this situation, either ψ|_{G_{0}} is irreducible, or 1 is connected to the exceptional node. This allows to reduce the determination of almost irreducible tensor squares to covering groups of the simple group S.

The general strategy for the proof of our Theorem will be the following. For a simple group S, let G be its full covering group. We exhibit primes ℓ_{i}, i = 1, 2, …, such that the ℓ_{i}-Sylow subgroups of G are cyclic. For each ℓ_{i} Remark 2.1 applies and, knowing the real stem of the Brauer tree, we get one possibility for each ℓ_{i}. By Remark 2.2 this carries over to the automorphism group of S, unless the trivial character is connected to the exceptional node. Then, from a knowledge of the possible character degrees it will follow that very few irreducible characters have degree divisible by none of the ℓ_{i}. Mostly, these have large degrees and thus can again be excluded easily.

We now go through the list of finite simple groups.

3. The classical groups of Lie type

3A. The basic setup.

In order to treat the groups of Lie type we will always work in the following general setup. Let G be a connected reductive algebraic group over the algebraic closure of a finite field of characteristic p, F : G → G a Frobenius map and G := G^{F} the finite group of fixed points. Generically, the full covering groups of the finite simple groups of Lie type arise as groups G as above if G is taken to be simple and simply-connected. Their complex irreducible characters Irr(G) have been studied and classified by Lusztig. We will need in particular the following fact, which is Lusztig’s Jordan decomposition of characters. Let G, F be as before, let G^{∗} be a group dual to G with dual Frobenius map F^{∗}, and G^{∗} := G^{∗}^{F^{∗}}. Then there is a natural partition of Irr(G) as

Irr(G) = ⋃_{s∈G^{∗}/∼ semisimple} E(G, s)

where the union runs over G^{∗}-conjugacy classes of semisimple elements. Furthermore for each s there is a bijection

j_{s}: E(G, s) ⟶^{∼} E(C_{G^{∗}}(s), 1)

such that the degrees are related by

(3.1)  χ(1) = (G^{∗} : C_{G^{∗}}(s))_{p′} j_{s}(χ)(1)  for χ ∈ E(G, s)

(see [5, 13.23]). The characters in E(G, 1) are called unipotent. They were classified for all types by Lusztig; for classical groups they can be parametrized by suitable combinatorial objects (partitions and so-called symbols), for the exceptional groups there are explicit lists. A precise description can be found in [2].

We also need some information on the classes of maximal tori and their orders. If G is simple of type A_{n}, then the G-classes of F -stable maximal tori of G are parametrized by partitions α ⊢ n + 1 of n + 1. In the case of the linear group, the order of T_{α}^{F} is given by

(3.2)  |T_{α}^{F} | = (q − 1)^{−1} ∏_{α_{i}}(q^{α_{i}} − 1) ,

where α = (α_{1}, …, α_{r}). For the unitary group the order is obtained from the above formula by formally replacing q by −q (and adjusting the sign if necessary).

For a simple group G of type B_{n} or C_{n} the G-classes of F -stable maximal tori of G are parametrized by pairs of partitions (α, β) ⊢ n of n, i.e., α = (α_{1}, …, α_{r}), β = (β_{1}, …, β_{s}) and ∑ α_{i} + ∑ β_{j} = n. The order of T_{(α,β)}^{F} is given by

(3.3)  |T_{(α,β)}^{F}| = ∏_{α_{i}}(q^{α_{i}} − 1) ∏_{β_{j}}(q^{β_{j}} + 1) .

If G is of type D_{n}, then the G-classes of F -stable maximal tori are parametrized by pairs of partitions (α, β) ⊢ n such that β has an even number of parts if G^{F} is the split orthogonal group, respectively such that β has an odd number of parts if G^{F} is the non-split orthogonal group. The order of T_{(α,β)}^{F} is again given by (3.3).

In order to exhibit suitable cyclic Sylow subgroups we will make use of the existence of primitive prime divisors. More precisely let q be a prime power and n ≥ 3 an integer. If (n, q) ≠ (6, 2) then there exists a prime ℓ dividing q^{n} − 1 but not dividing q^{k} − 1 for 1 ≤ k < n (see [11, 5.2.14]). We will write l(n) for such a Zsigmondy prime ℓ (the relevant q should always be clear from the context). It is easily seen that l(n)|(q^{m} − 1) implies that n|m [11, 5.2.15]. Another way to express the property of Zsigmondy primes is the following. Let Φ_{k} denote the kth cyclotomic polynomial evaluated at q (where q should always be clear from the context), so

q^{n} − 1 = ∏_{k|n} Φ_{k} .

Then l(n)|Φ_{n} but l(n) |̸Φ_{k} for k < n.

Proposition 3.4. Table 7.1 is correct if S = F^{∗}(G/Z(G)) is a classical group of Lie type.

For the proof let G be a connected reductive simple simply-connected algebraic group of type A_{n}, B_{n}, C_{n} or D_{n} defined over 𝔽_{q}, and F : G → G the corresponding Frobenius map such that the finite group G := G^{F} of fixed points is a classical group. Then by (3.2), (3.3), G has F -stable maximal tori T_{1}, T_{2} such that T_{1} := T_{1}^{F} and T_{2} := T_{2}^{F} have orders as given in Table 3.5, depending on the type of (G, F). We denote by G^{∗} a group dual to G and by T_{i}^{∗} tori in G^{∗} dual to T_{i}, i = 1, 2. We will say that we are in the generic situation if

(1) both Zsigmondy primes ℓ_{1} and ℓ_{2} in Table 3.5 exist, and

(2) the group G is the full covering group of its simple quotient group. In this case it follows from the property of Zsigmondy primes that ℓ_{i} divides precisely one of the Φ_{k} occurring in the order formula for G and it can be verified using (3.2) and (3.3) that the corresponding cyclotomic polynomials Φ_{k} divide the order polynomial of G only once. Hence for example by [1, Cor. 3.13] the ℓ_{i}-Sylow subgroups of G are cyclic for i = 1, 2 in the generic case.

3.5. Tori and Zsigmondy primes for classical groups.

(G, F)  |T_{1}|  |T_{2}|  ℓ_{1}  ℓ_{2}
A_{n}  (q^{n+1} − 1)/(q − 1)  q^{n} − 1  l(n + 1)  l(n)
2A_{n}  (n ≡ 0(4)) (q^{n+1} + 1)/(q + 1)  q^{n} − 1  l(2n + 2)  l(n)
(n ≡ 1(4)) (q^{n+1} − 1)/(q + 1)  q^{n} + 1  l((n + 1)/2) l(2n)
(n ≡ 2(4)) (q^{n+1} + 1)/(q + 1)  q^{n} − 1  l(2n + 2) l(n/2)
(n ≡ 3(4)) (q^{n+1} − 1)/(q + 1)  q^{n} + 1  l(n + 1)  l(2n)
B_{n}, C_{n} (n odd)  q^{n} + 1  q^{n} − 1  l(2n)  l(n)
(n even)  q^{n} + 1  (q^{n−1} + 1)(q + 1)  l(2n)  l(2n − 2)
D_{n}  (n odd) (q^{n−1} + 1)(q + 1)  q^{n} − 1  l(2n − 2)  l(n)
(n even) (q^{n−1} + 1)(q + 1) (q^{n−1} − 1)(q − 1) l(2n − 2) l(n − 1)
2D_{n}  q^{n} + 1  (q^{n−1} + 1)(q − 1)  l(2n)  l(2n − 2)

We will consider the various classical types in the subsequent sections.

3B. The linear groups.

Let G be of type A_{n}, so G = SL_{n+1}(q), and let d := gcd(n + 1, q − 1). We first consider the generic situation, i.e., n ≥ 3 and (n, q) ≠ (3, 2), (5, 2), (6, 2). As remarked above the Sylow ℓ_{i}-subgroups of G are cyclic. If χ ∈ Irr(G) is ℓ_{i}-projective for i = 1 or 2, then by Remark 2.1 a possible irreducible ψ_{i} = χ ⊗ χ^{∗} − 1 must be connected to the trivial character on the ℓ_{i}-Brauer tree. The Brauer trees of general linear groups are described in [6, Th. C]. In the respective cases, the quantity e in loc. cit. is equal to n + 1, n respectively. Hence ψ_{i} is the unipotent character labeled by the partition (n)(1) respectively (n − 1)(2) (note that ψ_{i} is not exceptional since n ≥ 2). By [2, 13.8] the degrees of the ψ_{i} are then given by

(3.6)  ψ_{1}(1) = q \frac{q^{n} − 1}{q − 1},  ψ_{2}(1) = q^{2} \frac{(q^{n+1} − 1)(q^{n−2} − 1)}{(q^{2} − 1)(q − 1)}.

In the first case, χ(1)^{2} = (q^{n+1} − 1)/(q − 1), which divides |G|. This contradicts the assumption that χ is of ℓ_{1}-defect zero. To exclude ψ_{2} we use the fact that a faithful irreducible representation of G in characteristic 0 has degree at least q^{n} − 1 (see for example [11, 5.3.9]; this can also be obtained from Lusztig’s classification of characters by some elementary estimates, see [16]). But the inequality

(q^{n} − 1)^{2} − 1 ≤ q^{2} \frac{(q^{n+1} − 1)(q^{n−2} − 1)}{(q^{2} − 1)(q − 1)}

which is equivalent to

q^{2n+1}(q^{2} − q − 2) + q^{n}(q^{n} − q^{3} + 2q^{2} + 2q − 1) − q^{2} ≤ 0

does not have solutions for n ≥ 3 and q ≥ 2. In view of Remarks 2.1 and 2.2 this handles the generic situation if χ is ℓ_{1}- or ℓ_{2}-projective.

Now assume that χ is neither ℓ_{1}- nor ℓ_{2}-projective. The degree χ(1) is a product of cyclotomic polynomials Φ_{k}. By the defining property of Zsigmondy primes, ℓ_{1} |̸χ(1) implies that Φ_{n+1} does not occur in this product, and similarly Φ_{n} does not occur since ℓ_{2} |̸χ(1). So χ lies in a Lusztig series E(G, s) for an element s ∈ G^{∗} with T_{i}^{∗} ≤ C_{G^{∗}}(s) for i = 1, 2, i.e., for an element s contained in suitable conjugates of both T_{1}^{∗} and T_{2}^{∗}. The only such element is s = 1, so χ is unipotent. By the degree formula [2, 13.8] the fact that neither Φ_{n+1} nor Φ_{n} divides χ(1) implies that χ is parametrized by a partition λ ⊢ n + 1 having an n + 1- and an n-hook. The only such partitions are λ = (n+1) and λ = (1)^{n+1} which parametrize χ = 1 and the Steinberg character (see also the proof of [12, Th. 2.1]). But the first one is not faithful, while the second has degree q^{n(n+1)/2}, so χ(1)^{2} − 1 does not divide |G| by Zsigmondy’s theorem. Hence no examples arise.

We next consider the cases without Zsigmondy primes. Let first n = 1, so q ≥ 4. The smallest degree of a non-trivial irreducible character of SL_{2}(q) is (q − 1)/d. The largest degree of an irreducible character of the automorphism group of G is f(q + 1) where q = p^{f}. But (q − 1)^{2}/d^{2} − 1 ≤ f(q + 1) implies q ∈ {4, 5, 7, 9}. These can be checked individually. Next, let G = SL_{3}(q). There always exists a Zsigmondy prime ℓ_{1} dividing |T_{1}|. If χ is ℓ_{1}-projective, then ψ has to be the character connected to the trivial character on the ℓ_{1}-Brauer tree, which has degree q(q + 1) by (3.6). This leads to a contradiction as above. Thus χ(1) is not divisible by ℓ_{1}. From the known list of character degrees this implies that

χ(1) ∈ {q(q + 1), q^{3}, (q − 1)^{2}(q + 1)/d, (q − 1)^{2}(q + 1)} .

On the other hand, the largest possible degree of an irreducible character of the automorphism group is equal to 2f(q + 1)(q^{2} + q + 1). An easy estimate shows that necessarily q ∈ {2, 4}.

It remains to consider the groups

S ∈ {L_{3}(4), L_{4}(2), L_{6}(2), L_{7}(2)} .

The character tables of their covering groups are contained in the GAP library [14] and can easily be checked by computer.

3C. The unitary groups.

Next, let G = SU_{n+1}(q), d := gcd(n + 1, q + 1). The generic situation occurs for n ≥ 3, (n, q) ≠ (3, 2), (3, 3), (5, 2). The Brauer trees for unitary groups were determined in [7, (5A), (6A)]. Let ℓ_{i} be as in Table 3.5. It turns out that the characters ψ_{i} connected to the trivial character on the ℓ_{i}-Brauer tree are the unipotent characters parametrized by the partitions (n − 1)(1)^{2}, (n − 1)(2) respectively. The degrees of ψ_{1}, ψ_{2} are

(3.7) q^{3} \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{(q^{2} − 1)(q + 1)}, q^{2} \frac{(q^{n+1} − (−1)^{n+1})(q^{n−2} − (−1)^{n−2})}{(q^{2} − 1)(q + 1)}

[2, 13.8]. By [11, 5.3.9] the minimal degree of a faithful irreducible character of G is at least q(q^{n} − 1)/(q + 1) if n is even and (q^{n+1} − 1)/(q + 1) if n is odd. Assume first that n is even. Then the inequality

(\frac{q(q^{n} − 1)}{q + 1})^{2} − 1 ≤ q^{3} \frac{(q^{n} − 1)(q^{n−1} + 1)}{(q^{2} − 1)(q + 1)}

is equivalent to

\frac{q^{2}(q^{n} − 1)}{(q^{2} − 1)(q + 1)}(q^{n}(q − 2) − 2q + 1) ≤ 1

so χ ⊗ χ^{∗} − 1 = ψ_{1} can only occur if q = 2. A similar estimate rules out ψ_{2} for all q. Also, for odd n one finds again that the only possibility is q = 2 and ψ = ψ_{1}. So let q = 2. The dual group G^{∗} = PGU_{n+1}(2) contains a subgroup 3.PGU_{n}(2). Let s be the semisimple element of order 3 in its center. Then the character χ ∈ E(G, s) with j_{s}(χ) = 1 has degree

(2^{n+1} − (−1)^{n+1})/3.

(This is the character of a Weyl representation for SU_{n+1}(2).) In particular it is of ℓ_{1}-defect 0. Hence χ_{1} ⊗ χ_{1}^{∗} − 1 contains ψ_{1}. Comparison of degrees shows that ψ_{1} is the only constituent. This gives the first generic example in Table 7.1. If 3|(n + 1) then χ_{1} is invariant under the diagonal automorphism of order 3, thus leading to the example for GU_{n+1}(2) = SU_{n+1}(2).3.

If on the other hand χ is neither ℓ_{1}- nor ℓ_{2}-projective, then by the same argument as for SL_{n+1}(q) it follows that either χ = 1 or χ is the Steinberg character. Neither of them leads to a new example.

We next consider the cases without Zsigmondy primes. For G = SU_{3}(q), q ≠ 2, at least the Zsigmondy prime ℓ_{1} = l(6) exists. The corresponding character ψ on the ℓ_{1}-Brauer tree has degree q^{3} by (3.7), and the above estimates rule out this possibility. The degrees of non-trivial characters of G not of ℓ_{1}-defect zero are

χ(1) ∈ {q(q − 1), q^{3}, (q + 1)^{2}(q − 1)/d, (q + 1)^{2}(q − 1)} .

On the other hand, the largest degree of an irreducible character of the automorphism group of G is equal to 2f(q + 1)(q^{2} − 1), where q = p^{f}. Using that f ≤ q/4 for q ≠ 2, 3, 4, 8 we find that χ(1) is too large unless q ∈ {3, 4, 8} (note that U_{3}(2) is solvable). The character tables of these groups, as well as of the remaining exceptional cases

S ∈ {U_{4}(2), U_{4}(3), U_{6}(2)} .

are contained in the GAP-libraries and can be checked individually.

3D. The odd-dimensional orthogonal groups.

Next, let G = Spin_{2n+1}(q), n ≥ 2. The generic situation occurs for n ≥ 3, (n, q) ≠ (3, 2), (3, 3), (4, 2). The characters ψ_{i} connected to the trivial character on the ℓ_{i}- Brauer tree are the unipotent characters labeled by the symbols

(3.8)  \binom{0 n}{1}, and \binom{1 n}{0} (n odd) resp. \binom{0 n − 1}{2} (n even) ,

[7, (5A), (6A)], with degrees

(3.9) \frac{q}{2} \frac{(q^{n} − 1)(q^{n−1} + 1)}{q − 1}, \frac{q}{2} \frac{(q^{n} + 1)(q^{n−1} − 1)}{q − 1}, \frac{q^{2}}{2} \frac{(q^{2n} − 1)(q^{n−1} − 1)(q^{n−3} + 1)}{(q^{2} − 1)^{2}}

respectively. The minimal degree of a faithful projective character of SO_{2n+1}(q) is at least q^{n−1}(q^{n−1} − 1) for odd q [11, 5.3.9] and \frac{1}{2}(q^{n} − 1)(q^{n} − q)/(q + 1) for even q [16, Th. 5.5]. The standard estimate shows that for all (n, q) ≠ (3, 2) none of the above ψ_{i} has large enough degree.

Thus we may assume that χ(1) is neither divisible by ℓ_{1} nor ℓ_{2}. Then by (3.1) the character χ lies in a Lusztig series E(G, s) for some s ∈ G^{∗} such that |C_{G^{∗}}(s)| is divisible by both ℓ_{1} and ℓ_{2}. For this in particular C_{G^{∗}}(s) has to contain tori of order divisible by l(2) and l(n) (respectively l(2n − 2) if n is even). According to (3.3) only the tori conjugate to T_{1} have order divisible by l(2n). Now assume first that n is odd. Then only the conjugates of T_{2} have order divisible by l(n). The greatest common divisor of |T_{1}|, |T_{2}| divides 2, so s is of order at most 2. If n is even, then a maximal torus of order divisible by l(2n − 2) has order (q^{n−1} + 1)(q ± 1). The gcd of |T_{1}| with these orders is again at most 2, so in this case as well s has order divisible by 2. The centralizers of involutions in the dual group G^{∗} = PCSp_{2n}(q) are easily determined, and none of them has order divisible by ℓ_{1} and ℓ_{2}. Thus χ as above must lie in E(G, 1).

The unipotent characters of G are indexed by symbols of rank n and odd defect. The degree of the character parametrized by a symbol Λ is not divisible by ℓ_{1} if Λ has an n-cohook. It is not divisible by ℓ_{2} if Λ has an n-hook if n is odd, respectively an n − 1-cohook if n is even. The symbols having n-hooks and cohooks were determined in the proof of [12, Th. 2.3]: they are the ones indexing the trivial and the Steinberg character. By an argument as in loc. cit. one sees that the symbols of rank n and odd defect having an n- and an n − 1-cohook are the ones for the trivial and the Steinberg character as well as the symbols

(3.10)  \binom{0 1 n}{−}  and  \binom{0 1 … n − 2 n − 1 n}{1 … n − 2} .

The degrees of the unipotent characters χ_{i} corresponding to these latter symbols both are divisible by l(n − 1), so both are of ℓ_{3}-defect zero for a Zsigmondy prime ℓ_{3} = l(n − 1). Since n ≥ 4 is even, the ℓ_{3}-Sylow subgroups of G are cyclic. Thus χ_{i}(1)^{2} − 1 should be equal to the degree of the character ψ_{3} connected to 1 on the ℓ_{3}-Brauer tree, with symbol

\binom{2 n − 1}{0}.

This is not the case, and the treatment of the generic case for Spin_{2n+1}(q) is complete.

Now assume that G = Spin_{5}(q) = Sp_{4}(q). Then at least the Zsigmondy prime ℓ_{1} = l(4) exists. The degree of the corresponding character ψ_{1} is q(q + 1)^{2}/2. This only leads to a solutions for q = 3. The non-trivial character degrees of G not of ℓ_{1}-defect zero are

{(q^{2} − 1)/2, q(q − 1)^{2}/2, q(q + 1)^{2}/2, q^{2}(q^{2} − 1)/2, (q^{2} − 1)^{2}, q^{4}} .

The maximal degree of an irreducible character of Aut(G) equals 2f(q + 1)^{2}(q^{2} + 1). This forces either q ≤ 4, and these groups can be checked individually, or χ(1) = (q^{2} − 1)/2 (which only occurs for odd q). In the latter case we would have ψ(1) = (q^{2} + 1)(q^{2} − 3)/4. This divides |G| only if q = 3.

Finally, it remains to consider the groups

S ∈ {S_{4}(2), S_{4}(3), S_{6}(2), O_{7}(3), S_{8}(2)}

which can again be done in GAP.

3E. The symplectic groups.

Next, let G = Sp_{2n}(q), n ≥ 3, q odd. Here we are always in the generic case.

In order to treat the Weyl characters if n is even we need a slight strengthening of [16, Prop. 5.1 and Th. 5.2] which is almost immediate from the proof given there.

Lemma 3.11. Let G = Sp_{2n}(q), n ≥ 3, q odd. Then a non-trivial irreducible character χ of G either has degree larger than \frac{1}{4}q^{2n}, or χ is one of the four unipotent characters with labels

(3.12)  \binom{0 1 n}{−},  \binom{0 1}{n},  \binom{1 n}{0},  \binom{0 n}{1} ,

or χ is semisimple of degree

\frac{1}{2}(q^{n} ± 1), \frac{1}{2} \frac{q^{2n} − 1}{q ± 1}, \frac{q^{2n} − 1}{q ± 1} .

Proof. First assume that χ is unipotent. Then the statement is a consequence of the proof of [16, Prop. 5.1]. Indeed, a careful study of the arguments there shows that either χ(1) > q^{2n}/4 or χ is labeled by a symbol as above.

Otherwise χ ∈ E(G, s) for some non-central semisimple element s ∈ G^{∗}. From the proof of [16, Th. 5.2] it follows that the only possibilities are those listed in the Lemma. □

The characters ψ_{i} connected to the trivial character on the ℓ_{i}-Brauer tree have the same labels (3.8) and degrees (3.9) as for type B_{n} [7, (5A), (6A)]. A non-trivial irreducible character χ either has degree at least \frac{q}{2}(q^{n} − 1)(q^{n−1} − 1)/(q + 1) or it is one of the Weyl characters of G of degree

(3.13)  χ(1) = \frac{1}{2}(q^{n} ± 1)

by [16, Th. 5.2] (see also Lemma 3.11). The standard estimate shows that the degrees of the ψ_{i} are too small unless q = 3. For q = 3 it turns out that the characters χ in (3.13) lead to the examples given in Table 7.1.

Thus we may assume that χ(1) is neither divisible by ℓ_{1} nor ℓ_{2}. As in the case of Spin_{2n+1}(q) it follows that χ lies in a Lusztig series E(G, s) for an element s ∈ G^{∗} = SO_{2n+1}(q) of order at most 2. If n is odd, no element with centralizer order divisible by ℓ_{1} and ℓ_{2} exists. But if n is even, s might be an involution with centralizer of type Spin_{2n}^{−}(q). The corresponding characters χ in E(G, s) have degrees

χ(1) = \frac{1}{2}(q^{n} − 1) μ(1)

where μ ∈ E(Spin_{2n}^{−}(q), 1) is a unipotent character of a group of type ^{2}D_{n}. It will be shown in section 3F that the only unipotent characters of such a group with degree not of ℓ_{1}- or ℓ_{2}-defect 0 are the trivial and the Steinberg character. If μ is the Steinberg character, then the degree of χ is too large. If μ = 1 then χ is one of the Weyl characters in (3.13) and ψ has degree (q^{n} + 1)(q^{n} − 3)/4. The candidates for constituents are listed in Lemma 3.11. Since n ≥ 3 there exists a Zsigmondy prime ℓ_{3} = l(n). Now ψ(1) ≡ −1 (mod ℓ)_{3} and the only character in Lemma 3.11 with degree congruent to −1 mod ℓ_{3} is the unipotent character χ with symbol

\binom{1 n}{0} .

Comparison of degrees shows that ψ ≠ χ for q ≠ 3, so ψ is not irreducible in this case.

Now let q = 3. In the following table we have collected those characters from Lemma 3.11 whose degree is not bigger than ψ(1) for q = 3, together with values of these characters on elements of prime orders as indicated. (Each row may stand for several characters of the same degree; their values on the given elements are identical.)

1  l(2n)  l(2n − 2)
μ_{1}  (q^{n} − 1)/2  −1  −(q + 1)/2 = −2
μ_{2}  (q^{n} + 1)/2  0  −(q − 1)/2 = −1
μ_{3}  (q^{n} − 1)(q^{n} − q)/2(q + 1)  1  q  = 3
μ_{4}  (q^{2n} − 1)/2(q + 1)  0  (q − 1)/2  = 1
μ_{5}  (q^{n} + 1)(q^{n} + q)/2(q + 1)  0  0
μ_{6}  (q^{n} + 1)(q^{n} − q)/2(q − 1)  0  q  = 3
ψ  (q^{n} + 1)(q^{n} − q)/2(q − 1)  0  (q^{2} + 2q − 3)/4 = 3

Assume that ψ is not irreducible, so ψ ≠ μ_{6}. From the table we see that μ_{1} and μ_{3} have to have same multiplicity in ψ. Comparison of degrees shows that μ_{1} + μ_{3} can occur at most twice. But then the value on an element of order l(2n−2) shows that in fact their multiplicity must vanish. Then μ_{3} has to occur at least three times, which is not possible. This contradiction shows that ψ = μ_{6} is irreducible, which leads to the example in Table 7.1.

If on the other hand s = 1 then χ itself is unipotent, and this case can be ruled out precisely as for Spin_{2n+1}(q) since both the degrees and the Brauer trees for unipotent characters are the same in types B_{n} and C_{n}. This completes the treatment of the generic case and hence of Sp_{2n}(q).

3F. The even-dimensional non-split orthogonal groups.

Next, let G = Spin_{2n}^{−}(q), n ≥ 4. We are in the generic situation unless (n, q) = (4, 2). The characters ψ_{i} attached to the trivial character on the ℓ_{i}-Brauer tree are unipotent and parametrized by the symbols

(3.14)  \binom{0 1 n}{1} and \binom{0 1 n − 1}{2},

[7, (5A), (6A)] with degrees

(3.15) ψ_{1}(1) = q^{2} \frac{q^{2n−2} − 1}{q^{2} − 1}, ψ_{2}(1) = \frac{q^{3}}{2} \frac{(q^{n} + 1)(q^{n−1} − 1)(q^{n−2} − 1)(q^{n−3} + 1)}{(q^{2} − 1)^{2}}

[2, 13.8]. By [11, 5.3.9] the minimal degree of a faithful complex character of G is (q^{n−1} + 1)(q^{n−2} − 1) . The standard estimate shows that neither ψ_{1} nor ψ_{2} has large enough degree.

So we may assume that χ is not ℓ_{i}-projective for i = 1, 2. By (3.3) the only tori of order divisible by l(2n) or l(2n − 2) are those conjugate to T_{1}, T_{2} respectively. The gcd of their orders divides 2, so χ ∈ E(G, s) for some s ∈ G^{∗} of order at most 2. But no involution in G^{∗} = PCO_{2n}^{−}(q) has centralizer order divisible by both l(2n) and l(2n − 2). Thus χ must be unipotent. The unipotent characters of the G are parametrized by symbols Λ of rank n and defect congruent to 2 mod 4. If the degree of the character χ_{Λ} parametrized by the symbol Λ is not divisible by ℓ_{1} and ℓ_{2}, then Λ has to have n- and n − 1-cohooks. The only such symbols are those parametrizing the trivial and the Steinberg character (see [12, Th. 2.5]). Thus no examples arise.

For the non-generic case S = O_{8}^{−}(2) the candidates can be checked with the computer.

3G. The even-dimensional split orthogonal groups.

Finally, let G = Spin_{2n}^{+}(q), n ≥ 4. If (n, q) ≠ (4, 2) we have the generic situation. The characters ψ_{i} corresponding to the primes ℓ_{i} as in Table 3.5 are unipotent with labels

(3.16)  \binom{n − 1}{1}, and \binom{1 n}{0 1} (n odd) resp. \binom{2 n − 1}{0 1} (n even) ,

[7, (5A), (6A)] with degrees

q \frac{(q^{n} − 1)(q^{n−2} + 1)}{q^{2} − 1}, q^{2} \frac{q^{2n−2} − 1}{q^{2} − 1}, \frac{q^{3}}{2} \frac{(q^{n} − 1)(q^{n−1} + 1)(q^{n−2} + 1)(q^{n−3} − 1)}{(q^{2} − 1)^{2}}

[2, 13.8]. The standard estimate rules out all three possibilities.

If χ is neither ℓ_{1}- nor ℓ_{2}-projective, then arguing as in the previous cases we find that χ is unipotent. The unipotent characters are parametrized by symbols Λ of rank n and defect congruent to 0 mod 4. If χ_{Λ}(1) is not divisible by ℓ_{1} then Λ has to have a n − 1-cohook. If χ_{Λ}(1) is not divisible by ℓ_{2} then it has an n-hook if n is odd respectively an n − 1-hook if n is even. One finds that for odd n only the trivial and the Steinberg character can occur, while for even n there are the two further symbols

\binom{n − 1}{1}  and  \binom{0 … n − 3 n − 1}{1 … n − 2 n − 1}.

The two corresponding characters χ_{i} have degree divisible by l(2n − 4). If n ≥ 6 then the ℓ_{3}-Sylow subgroup of G is cyclic, for ℓ_{3} = l(2n − 4). The character ψ_{3} connected to the trivial character on the ℓ_{3}-Brauer tree with symbol

\binom{n − 2}{2}

has degree congruent to 0 mod q^{n} − 1, as have χ_{i} for i = 1, 2, so we cannot have χ_{i}(1)^{2} − 1 = ψ_{3}(1). Finally, if n = 4 then

χ_{1}(1)^{2} − 1 = q^{2}(q^{2} + 1)^{4} − 1 = (q^{2} − q + 1)(q^{2} + q + 1)(q^{6} + 3q^{4} + 2q^{2} − 1).

Since the order of G equals q^{12}(q^{2} − 1)(q^{4} − 1)^{2}(q^{6} − 1) the gcd of χ_{1}(1)^{2} − 1 with |G| is a divisor of 5^{4}(q^{2} − q + 1)(q^{2} + q + 1). In particular, q^{6} + 3q^{4} + 2q^{2} − 1 ≤ 5^{4}, which is a contradiction for q ≥ 3. On the other hand, the degree χ_{2}(1) = q^{7}(q^{2} + 1)^{2} is already too large. This completes the study of the generic case.

For S = O_{8}^{+}(2) the result is verified from the Atlas tables.

The proof of Proposition 3.4 is complete.

4. The exceptional groups of Lie type

Proposition 4.1. Table 7.1 is correct if S = F^{∗}(G/Z(G)) is an exceptional group of Lie type.

For the proof we let G be a connected reductive simple simply-connected algebraic group and F : G → G a Frobenius map such that the finite group G := G^{F} of fixed points is a non-solvable exceptional group of Lie type. We let T_{1}, T_{2} (and T_{3}) be F -stable maximal tori of G such that T_{i} := T_{i}^{F} have orders as given in Table 4.2.

4.2. Tori T_{1}, T_{2} and T_{3}.
G  T_{1}  T_{2}  T_{3}
2B_{2}(q^{2})  q^{2} + \sqrt{2}q + 1  q^{2} − \sqrt{2}q + 1
2G_{2}(q^{2})  q^{2} + \sqrt{3}q + 1  q^{2} − \sqrt{3}q + 1
2F_{4}(q^{2})  q^{4} + \sqrt{2}q^{3} + q^{2} + \sqrt{2}q + 1  q^{4} − \sqrt{2}q^{3} + q^{2} − \sqrt{2}q + 1  Φ_{12}
G_{2}(q)  Φ_{6}  Φ_{3}
3D_{4}(q)  Φ_{12}
F_{4}(q)  Φ_{12}  Φ_{8}
E_{6}(q)  Φ_{12}Φ_{3}  Φ_{9}  Φ_{8}Φ_{2}Φ_{1}
2E_{6}(q)  Φ_{18}  Φ_{12}Φ_{6}  Φ_{8}Φ_{2}Φ_{1}
E_{7}(q)  Φ_{18}Φ_{2}  Φ_{14}Φ_{2}  Φ_{12}Φ_{3}Φ_{1}
E_{8}(q)  Φ_{30}  Φ_{24}  Φ_{20}

We first consider the Suzuki and Ree groups.

Let G = ^{2}B_{2}(q^{2}), q^{2} ≥ 8. Then all odd Sylow subgroups of G are cyclic. The orders of the tori T_{1}, T_{2} are pairwise coprime and larger than 1. Let ℓ_{i} be a prime dividing |T_{i}|, for i = 1, 2. It is straightforward to work out the Brauer trees for the ℓ_{i}. In both cases the trivial character is connected to the Steinberg character, hence this does not lead to examples. Only the unipotent characters of degrees

χ(1) ∈ \{1, \frac{1}{\sqrt{2}}q(q^{2} − 1) (twice), q^{4}\}

are neither ℓ_{1}- nor ℓ_{2}-projective. The degree of the Steinberg character is too large. If χ(1) = \sqrt{2}q(q^{2} −1)/2, then ψ has to have degree ψ(1) = (q^{2} −2)(q^{4} +1)/2. This is the degree of an irreducible character of an almost simple group with socle G if and only if (q^{2} −2)/2 divides the order of a field automorphism of G. The outer automorphism group of G has order f where q^{2} = 2^{f}. Clearly this implies that q^{2} = 8, leading to the example in Table 7.1.

Next let G = ^{2}G_{2}(q^{2}), q^{2} ≥ 27 (the group ^{2}G_{2}(3) is isomorphic to L_{2}(8).3 so was treated before). All Sylow ℓ-subgroups for ℓ ≠ 2, 3 are cyclic. Let ℓ_{i} be a prime dividing |T_{i}|, for i = 1, 2. The Brauer trees for G were found by Hiss [9, Th. 4.1]. Again, in both cases the trivial character is connected to the Steinberg character, so this case can not occur. The only faithful irreducible characters χ not of ℓ_{1}- or ℓ_{2}-defect 0 are unipotent, of degrees

χ(1) ∈ \{1, \frac{1}{\sqrt{3}}q(q^{4} − 1) (twice), q^{6}\}

(see [2, 13.9]). The largest degree of an irreducible character of the automorphism group of G is at most f(q^{4} − 1)(q^{2} + \sqrt{3}q + 1) where q^{2} = 3^{f}. Hence the degree of all such χ ≠ 1 is too large for χ ⊗ χ^{∗} − 1 to be irreducible.

Now let G = ^{2}F_{4}(q^{2}), q^{2} ≥ 8. Let ℓ_{i} ≠ 3 be primes dividing the orders of T_{i}, i = 1, 2, 3. Then the Sylow ℓ_{i}-subgroups of G are cyclic. The characters connected to the trivial character on the ℓ_{i}-Brauer tree of G have degrees

\begin{matrix}\frac{q^{4}}{4} (q^{2} + 1)^{2}(q^{2} ∓ \sqrt{2}q + 1)^{2}(q^{4} − q^{2} + 1)(q^{4} ± \sqrt{2}q^{3} + q^{2} ± \sqrt{2}q + 1), \\ \frac{q^{4}}{2} (q^{4} + 1)^{2}(q^{8} − q^{4} + 1) \end{matrix}

respectively ([9, Th. 4.5]). The minimal degree of a faithful character of G is at least \frac{1}{\sqrt{2}}q^{9}(q^{2} − 1) [11, 5.3.9]. The standard estimate shows that this does not lead to solutions for q^{2} ≥ 8. The irreducible character degrees χ(1) > 1 not divisible by ℓ_{i}, i = 1, 2, 3, are

\{\frac{q^{4}}{3} (q^{8} − 1)^{2} (twice), q^{24}\} .

But these satisfy χ(1)^{2} − 1 ≥ \sqrt{|G|}, so do not lead to examples.

The group S = ^{2}F_{4}(2)′ is contained in the Atlas and gives rise to a new example.

For the remaining seven families of exceptional groups we will again say that we are in the generic situation if

(1) Zsigmondy primes ℓ_{i} exist for the tori T_{i} listed in Table 4.2, and

(2) the group G is the full covering group of its simple quotient group. It follows for example from [1, Cor. 3.13] that in this case the ℓ_{i}-Sylow subgroups of G are cyclic for all relevant i.

Let first G = G_{2}(q). The generic situation occurs for q ≥ 5. The real stems of the Brauer trees for ℓ_{1} and ℓ_{2} are easily determined. The characters ψ_{i} connected to the trivial character have degrees

\frac{q}{2}(q + 1)^{2}(q^{2} − q + 1), \frac{q}{6}(q + 1)^{2}(q^{2} + q + 1) .

The minimal degree of a faithful representation of G is at least q(q^{2} − 1) [11, 5.3.9]. By the standard estimate this yields no examples.

If χ is neither ℓ_{1}- nor ℓ_{2}-projective then since the tori T_{1}, T_{2} have coprime orders χ is unipotent. By [2, 13.9] the only possibilities are

χ(1) ∈ \{1, \frac{q}{3}(q^{2} − 1)^{2}, q^{6}\} .

But for the latter two we have that χ(1)^{2} − 1 > \sqrt{|G|}, so that no example arises.

For the non generic groups G_{2}(3), G_{2}(4) we can use the Atlas tables, and G_{2}(2) is isomorphic to U_{3}(3).2.

Next, let G = ^{3}D_{4}(q), which is always generic. The character connected to 1 on the ℓ_{1}-Brauer tree is unipotent of degree

\frac{q^{3}}{2} (q^{3} + 1)^{2}

as can be seen by Harish-Chandra induction. The minimal degree of a faithful representation of G is at least q^{3}(q^{2} − 1) [11, 5.3.9]. By the standard estimate this yields no examples. By the known character table of G, the only character degrees χ(1) > 1 not divisible by Φ_{12} are

\{\frac{q^{3}}{2} (q^{3} + 1)^{2}, \frac{q^{3}}{2} (q^{3} − 1)^{2}, q^{12}, (q^{6} − 1)^{2}\}

[4]. But in all cases we find that χ(1)^{2} − 1 > \sqrt{|G|}, so this gives no examples.

Now let G = F_{4}(q) and q ≠ 2. The real stems of the Brauer trees are easily determined by Harish-Chandra induction. It turns out that the characters connected to 1 on the ℓ_{i}-Brauer tree have degrees

\frac{q}{2}Φ_{2}^{2}Φ_{6}^{2}Φ_{8} and q^{2}Φ_{3}^{2}Φ_{6}^{2}Φ_{12} .

The minimal degree of a faithful representation of G is at least q^{6}(q^{2} − 1) [11, 5.3.9]. The standard estimate shows that this does not lead to solutions for any q. Since |T_{1}|, |T_{2}| are coprime, irreducible characters χ of G with degree not divisible by ℓ_{1} and ℓ_{2} are unipotent. From the list in [2, 13.9] their degrees are

χ(1) ∈ \{1, \frac{q^{4}}{4} Φ_{1}^{4}Φ_{2}^{4}Φ_{3}^{2}Φ_{6}^{2}, q^{24}\} .

But the latter two degrees are such that χ(1)^{2} − 1 > \sqrt{|G|}, so this gives no examples. The tables for F_{4}(2) and its exceptional covering group are found in the Atlas [3]. For G = E_{6}(q) the real stems of the Brauer trees were determined in [8, Th. 12.5].

The degrees of the ψ_{i}, i = 1, 2, 3, are given by

qΦ_{8}Φ_{9}, q^{2}Φ_{4}Φ_{5}Φ_{8}Φ_{12}, \frac{q^{3}}{2} Φ_{4}^{2}Φ_{5}Φ_{9}Φ_{12} .

The minimal degree of a faithful representation of G is at least q^{9}(q^{2} − 1) [11, 5.3.9]. The standard estimate shows that this gives a contradiction unless q = 2 and i = 3. But in this case the integer ψ_{3}(1) + 1 is not a square. The only irreducible characters with degree not divisible by ℓ_{i}, i = 1, 2, 3, are the trivial and the Steinberg character. So no possibility occurs.

For G = ^{2}E_{6}(q) the real stems of the Brauer trees can easily be found using Harish-Chandra induction. The degrees of the ψ_{i}, i = 1, 2, 3, are given by

q^{2}Φ_{4}Φ_{8}Φ_{10}Φ_{12}, \frac{q^{3}}{2} Φ_{3}^{2}Φ_{8}Φ_{10}Φ_{18}, \frac{q^{3}}{2} Φ_{2}^{4}Φ_{6}^{2}Φ_{10}Φ_{18} .

The minimal degree of a faithful representation of G is at least q^{9}(q^{2} − 1) [11, 5.3.9]. The standard estimate shows that this does not lead to solutions for any q. On the other hand, the only irreducible characters with degree not divisible by ℓ_{i}, i = 1, 2, 3, are the trivial and the Steinberg character. Thus no possibility occurs.

The character table of the six-fold exceptional cover of ^{2}E_{6}(2) is not contained in the Atlas. Nevertheless, at least the table of the simple group^{2}E_{6}(2) is given. But note that for any irreducible character χ ∈ Irr(6.^{2}E_{6}(2)) the representation corresponding to χ ⊗ χ^{∗} is trivial on the center. So if χ ⊗ χ^{∗} − 1 is irreducible, it must be a character of ^{2}E_{6}(2) already. To handle the full covering group it is therefore sufficient to verify that no degree for ^{2}E_{6}(2) has the form n^{2} − 1 for some integer n. This can efficiently be done with the computer. The group ^{2}E_{6}(2).2 is also contained in [3]. To treat the extensions by outer automorphisms of order 3 we just check that no degree of ^{2}E_{6}(2) or ^{2}E_{6}(2).2 is of one of the forms n^{2} − 1 or (n^{2} − 1)/3 for some integer n.

For G = E_{7}(q) the real stems of the Brauer trees were determined in [8, Th. 12.6]. The degrees of the ψ_{i}, i = 1, 2, 3, are given by

qΦ_{7}Φ_{12}Φ_{14}, q^{2}Φ_{3}^{2}Φ_{6}^{2}Φ_{9}Φ_{12}Φ_{18}, \frac{q^{3}}{2} Φ_{2}^{4}Φ_{6}^{2}Φ_{7}Φ_{10}Φ_{14}Φ_{18} .

The minimal degree of a faithful representation of G is at least q^{15}(q^{2} − 1) [11, 5.3.9]. The standard estimate shows that this gives a contradiction unless q = 2 and i = 3. But for q = 2 the integer ψ_{3}(1) + 1 is not a square. Thus we may assume that χ(1) is prime to ℓ_{i} for i = 1, 2, 3. Then χ must be unipotent, and from the list in [2, 13.9] it follows that either χ = 1 or χ is the Steinberg character. This completes the case of E_{7}.

For G = E_{8}(q) the real stems of the Brauer trees were determined in [8, Th. 12.7]. The degrees of the ψ_{i}, i = 1, 2, 3, are given by

qΦ_{4}^{2}Φ_{8}Φ_{12}Φ_{20}Φ_{24}, q^{2}Φ_{5}Φ_{7}Φ_{10}Φ_{14}Φ_{15}Φ_{20}Φ_{30}, \frac{q^{3}}{2} Φ_{2}^{4}Φ_{6}^{2}Φ_{7}Φ_{8}Φ_{10}^{2}Φ_{14}Φ_{18}Φ_{24}Φ_{30} .

The minimal degree of a faithful representation of G is at least q^{27}(q^{2} − 1) [11, 5.3.9]. The standard estimate shows that this does not lead to solutions for any q. If on the other hand χ(1) is not divisible by any ℓ_{i}, i = 1, 2, 3, then by the comprimeness of |T_{1}| and |T_{2}| it must be unipotent, and by the list in [2, 13.9] it is the trivial or the Steinberg character. Thus no case arises.

This completes the proof of Proposition 4.1.

5. The alternating groups

In order to ensure the existence of enough cyclic Sylow subgroups of the symmetric group 𝔖_{n} we will need the following:

Proposition 5.1 (Tschebyschef). Let n ≥ 17. Then there always exist at least three different primes ℓ with n/2 < ℓ ≤ n.

For a short proof see [13]. To treat the covering group of 𝔄_{n} we need a result on character values.

Lemma 5.2. Let n ≥ 8. Then for any irreducible complex character χ of 𝔄_{n} or 𝔖_{n} there exists an odd k ≤ n − 4 such that χ does not take value −1 on the class (k)(2)^{2}.

Proof. By [3] we may assume that n ≥ 14. Let k ≤ n − 4 be odd. Let χ_{λ} be the irreducible character of 𝔖_{n} indexed by the partition λ ⊢ n. If λ has no k-hook then by the Murnaghan-Nakayama formula [10, 2.4.7] we have that χ(π) = 0 for elements π of cycle shape (k)(2)^{2}. Clearly this also holds for the irreducible constituents of χ|_{𝔄_{n}} since the class (k)(2)^{2} does not split under this restriction.

Hence it remains to consider those λ which have k-hooks for all odd k ≤ n − 4. First assume that n is odd. Then the partitions λ ⊢ n having (n − 4)- (n − 6)- and (n − 8)-hooks are easily classified. Up to conjugates they are given in the subsequent table. From the Murnaghan-Nakayama formula it is immediate to compute the values of the corresponding 𝔖_{n}-characters on the given classes.

λ  (1)  (n − 4)(2)^{2} (n − 6)(2)^{2}
(n)  1  1
(n − 1)(1)  −1  1
(n − 2)(2)  2  1
(n − 2)  −1  −2
(n − 3)(3)  −2  1
(n − 3)  1  −2
(n − 4)(4)  1  −1
(n − 4)(2)  −1  1
(n − 5)(5)  n(n − 1)(n − 2)(n − 3)(n − 5)/120  −1  −1
(n − 5)(3)(2) n(n − 1)(n − 2)(n − 5)(n − 7)/24  −1  −1
(n − 6)(5)  1  2
(n − 7)(4)(2)  1  −2

In all cases, either the degree of χ_{λ} is even and λ is not self-dual, so χ_{λ}((2)^{2}) must also be even. Or, the value on (n − 4)(2)^{2} or (n − 6)(2)^{2} is different from −1.

Now assume that n ≥ 14 is even. As above one arrives at the table giving the partitions λ ⊢ n having (n − 5)- (n − 7)- and (n − 9)-hooks, where we have omitted the λ with (n − 5)-core of shape (4)(1) or (2)(1)^{3} since the corresponding characters vanish on (n − 5)(2)^{2}, and the values of the corresponding characters χ_{λ} of 𝔖_{n}.

\begin{matrix}λ & (n − 5)(2)^{2} & (n − 7)(2)^{2} \\ (n) & 1 & 1 \\ (n − 2)(2) & 1 & 2 \\ (n − 2) & −2 & −1 \\ (n − 3)(2) & 1 & −1 \\ (n − 4)(3) & −1 & 1 \\ (n − 4) & 1 & −1 \\ (n − 5)(4) & 2 & −1 \\ (n − 5)(2) & −1 & 2 \\ (n − 6)(6) & −1 & −2 \\ (n − 7)(6) & 1 & 1 \\ (n − 7)(4)(3) & 1 & 1 \\ (n − 9)(3)^{2}(2) & 1 & −2 \end{matrix}

The result follows. □

Proposition 5.3. Table 7.1 is correct if S = F^{∗}(G/Z(G)) is an alternating group.

Proof. The case of alternating groups of degree n ≤ 10 can be handled with GAP. So assume that G = 2.𝔄_{n} with n ≥ 11. The irreducible characters of 𝔖_{n} are parametrized by partitions λ ⊢ n. They are all real, so a non-real irreducible character of 𝔄_{n} can not extend to 𝔖_{n}. Thus the non-real irreducibles of 𝔄_{n} are parametrized by self-dual partitions λ ⊢ n (such that moreover the length of the main diagonal is odd), in such a way that any such λ stands for the two components of the restriction of the corresponding 𝔖_{n}-character. The degree of χ_{λ} is given by the hook length formula: χ(1) is equal to n!/2 divided by all the hook lengths of λ.

Now let ℓ be a prime such that n/2 < ℓ ≤ n. Then the ℓ-Sylow subgroup of G is cyclic (of order ℓ). If χ_{λ} is not ℓ-projective then λ has to have an ℓ-hook, so the Young diagram for λ has to contain a box with hook length ℓ. Since ℓ > n/2 this box must lie either in the first row or the first column. If λ is self-dual, the conjugate box also has hook length ℓ. As ℓ divides |G| only once, this means that the box lies on the diagonal, and hence is the first box in the first row.

In particular, ℓ is uniquely determined by λ and χ_{λ} must be ℓ′-projective for all other primes ℓ′ with n/2 < ℓ′ ≤ n.

The character connected to the trivial character on the ℓ-Brauer tree of 𝔖_{n} has label

(5.4)  \{ \begin{matrix}(ℓ − 1)(n − ℓ + 1) & if 2ℓ > n + 1 \\ (n − ℓ)^{2}(1) & if 2ℓ = n + 1. \end{matrix}

These partitions are different for two different ℓ. Now let λ be a self-dual partition of n. By the above χ_{λ} is ℓ-projective for all but one prime between (n + 1)/2 and n. By Proposition 5.1 for n ≥ 17 there exist at least three different primes between (n+1)/2 and n. So then χ ⊗ χ^{∗} − 1 contains at least two constituents of the form (5.4) for different ℓ and can not be irreducible. The argument also applies to n = 13 since there are again three primes between 7 and 13. For the remaining cases n = 11, 12, 14, 15, 16 we give explicitly the list of degrees of the ψ_{i} for the two primes ℓ_{1} < ℓ_{2} between (n + 1)/2 and n.

\begin{matrix}n & 11 & 12 & 14 & 15 & 16 \\ ψ_{1}(1) & 132 & 132 & 637 & 1638 & 3640 \\ ψ_{2}(1) & 10 & 54 & 77 & 350 & 1260 \end{matrix}

Visibly none of these has the form m^{2} − 1 for an integer m. Thus the characters of 𝔄_{n} (or 𝔖_{n}) do not lead to solutions.

Now let χ be a faithful irreducible character of a double cover 2.𝔄_{n} or 2.𝔖_{n}. The preimages of elements of cycle shape (k)(2)^{2}, where k is odd, lie in a unique class of 2.𝔄_{n} (see [15, §7]), so χ vanishes on this preimage class. Thus χ ⊗ χ^{∗} − 1, which is a character of 𝔄_{n}, 𝔖_{n} respectively, has value −1 on elements of cycle shape (k)(2)^{2}. But then it can not be irreducible by Lemma 5.2. □

6. The sporadic groups

Proposition 6.1. Table 7.1 is correct if S = F^{∗}(G/Z(G)) is a sporadic simple group.

Proof. This follows from the character tables in the Atlas [3]. The necessary calculations can be done using the computer algebra system GAP [14]. □

7. The result

The correctness of Table 7.1 and hence the Theorem now follows by the classification of finite simple groups.

In the second column of Table 7.1 we have indicated to which group of outer automorphisms (in Atlas notation) the characters χ and ψ extend, thus yielding examples for all intermediate extensions. The last column gives the number of pairs of complex conjugate characters with the given degree. Note that in all examples apart from G = ^{2}B_{2}(8).3 the representation affording the character ψ is an irreducible projective representation for the simple group F^{∗}(G/Z(G)).

7.1. Faithful characters with almost irreducible tensor square.

\begin{matrix}G & χ(1) & ψ(1) & # \\ SU_{n}(2) & (2^{n} − (−1)^{n})/3 & 8(2^{n−2} − (−1)^{n})(2^{n−1} − (−1)^{n−1})/9 & 1 \\ GU_{n}(2), 3|n & (2^{n} − (−1)^{n})/3 & 8(2^{n−2} − (−1)^{n})(2^{n−1} − (−1)^{n−1})/9 & 3 \\ S_{2n}(3) & (3^{n} + (−1)^{n})/2 & 3(3^{n} − (−1)^{n})(3^{n−1} − (−1)^{n−1})/4 & 1 \\ Sp_{2n}(3) & (3^{n} − (−1)^{n})/2 & 3(3^{n} + (−1)^{n})(3^{n−1} + (−1)^{n−1})/4 & 1 \end{matrix}

G  χ(1) ψ(1) #_{2.𝔄_{5}  2  3 2}_{3.𝔄_{6}  3  8 2}_{2.𝔄_{7}  4 15 1}_{6.𝔄_{7}  6 35 2}_{L_{2}(7)  3  8 1}_{4_{1}.L_{3}(4) 2_{3}  8 63 2}_{6.L_{3}(4) 2_{1}  6 35 1}_{6_{1}.U_{4}(3) 2_{2}  6 35 1}_{^{2}B_{2}(8).3  14 195 3}_{^{2}F_{4}(2)′  26 675 1} G  χ(1)  ψ(1) #
2.M_{12} 2  10  99 1
2.M_{22} 2  10  99 1
M_{23}  45  2024 1
3.J_{3}  18  323 2
M_{24}  45  2024 1
2.Ru  28  783 1
6.Suz  12  143 1
3.ON  342 116963 2
J_{4}  1333 1776888 1

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(1996), 2093–2167. I.W.R., Im Neuenheimer Feld 368, D–69120 Heidelberg, Germany

E-mail address: malle@urania.iwr.uni-heidelberg.de