THE GROUP OF ENDOTRIVIAL MODULES FOR THE SYMMETRIC AND ALTERNATING GROUPS

JON F. CARLSON, DAVID J. HEMMER, AND NADIA MAZZA

Abstract. We complete a classification of the groups of endotrivial modules for the modular group algebras of symmetric groups and alternating groups. We show that for n ≥ p^{2} the torsion subgroup of the group of endotrivial modules for the symmetric groups is generated by the sign representation. The torsion subgroup is trivial for the alternating groups. The torsion free part of the group is free abelian of rank one if n ≥ p^{2} + p and has rank 2 if p^{2} ≤ n < p^{2} + p. This completes the work begun earlier by Carlson, Mazza and Nakano.

1. Introduction

Endotrivial modules were first defined for p-groups by Dade [8], though they had earlier appeared in a celebrated paper of Hall and Higman. Early work saw them as the building blocks for the endopermutation modules which are the sources, in the sense of J. A. Green’s theory of vertices and sources, of the irreducible modules of p-solvable groups. These modules occur in several other situations in modular representation theory. For p-groups a classification of the endotrivial modules was completed by the first author and Thévenaz [7], building on the work of several others. Subsequently, the endopermutation modules were classified by Bouc [3].

In this paper, we consider endotrivial modules for symmetric and alternating groups. Our motivation comes from the fact that taking the tensor product with an endotrivial module is a self equivalence (functor) on the stable module category, that is, the localized category of modules modulo projectives. Thus the endotrivial modules define a distinguished subgroup of the Picard group of all self equivalences of the stable module category. With that in mind, the first and third authors together with Nakano determined the group of endotrivial modules for finite groups of Lie type [5] and made some progress on the symmetric and alternating groups. Specifically, the paper [6] found the group of endotrivial modules for all symmetric and alternating groups in characteristic 2 and for all S_{n} and A_{n} for n < p^{2} in the case that p is odd.

Our main result is that in almost all cases for n ≥ p^{2}, the group of endotrivial modules for the symmetric group S_{n} is isomorphic to ℤ⊕ℤ/2 and for the alternating group A_{n} is isomorphic to ℤ. The only exception is for p^{2} ≤ n < p^{2} + p, where the torsion free parts of both groups is the sum of two copies of ℤ rather than only one. The class of the sign representation generates the copy of ℤ/2 in the case of the symmetric groups. The class of the Heller shift Ω(k) of the trivial module k is a generator for the torsion free part of both groups. In the case that p^{2} ≤ n < p^{2} + p, there is another generator for the torsion free part of the group of endotrivial modules which remains somewhat elusive. We have some information on the structure of this generator, but it is not precise. A general discussion is given in the last section.

Date: October 22, 2008.

2000 Mathematics Subject Classification. Primary 20C20, 20C30.

Research of the first author was supported in part by NSF grant DMS-0654173.

Research of the second author was supported in part by NSF grant DMS-0556260.

Research of the third author was supported in part by Swiss National Foundation Research Fellowship PA002-113164/1.

This paper was written while all three authors were visiting the Mathematical Sciences Research Institute in Berkeley in the spring of 2008. The authors wish to thank MSRI for support and for its very stimulating atmosphere.

2. Notation and definitions

Let k be a field of characteristic p which is a splitting field for the symmetric group S_{n} and all of its subgroups. When defining subgroups of the symmetric group we assume the natural ordering on the letters unless otherwise indicated. For example, S_{a} is the collection of all permutations on {1, …, a}.

For two subgroups H and K of a finite group G, we let [G/H] denote a complete set of representatives for the left H-cosets in G and we let [H\G/K] be a complete set of representatives for the H-K double cosets in G. For elements g, h of a group G and for a subgroup H of G, we write ^{g}h instead of ghg^{−1} and ^{g}H for gHg^{−1}.

We consider finitely generated left modules over group algebras. We denote by mod(kG) the category of finitely generated kG-modules and by stmod(kG) the corresponding stable module category. Given a group inclusion H ↪ G we denote the induction and restriction functors between mod(kG) and mod(kH) by Ind_{H}^{G} and Res_{H}^{G}, respectively. If M and N are kG-modules, we write Hom_{k}(M, N) for the kG-module of all k-linear maps from M to N. If N = M, we write End_{k} M instead of Hom_{k}(M, M), and if N = k is the trivial kG-module, we write M^{∗} = Hom_{k}(M, k) for the k-linear dual of M. Let M ⊗ N be the tensor product of two modules M and N over the base field k with diagonal action of the group G. We write M | N to mean that the module M is isomorphic to a direct summand of N.

For a kG-module M, let Ω^{n}(M) be the kernel of a projective cover P ⟶ M of M and let Ω^{−1}(M) be the cokernel of the injective hull M ↪ Q. Iterating we define Ω^{n}(M) = Ω(Ω^{n−1}(M)) and Ω^{−n}(M) = Ω^{−1}(Ω^{n−1}(M)). We remind the reader that kG is a self injective ring, and hence injective modules are also projective.

Definition 2.1. A kG-module M is endotrivial provided End_{k} M ≅ k ⊕ (proj), or equivalently, End_{k} M ≅ k in stmod(kG).

Recall that Hom_{k}(M, N) ≅ M^{∗} ⊗ N as kG-modules. Consequently, the tensor product of two endotrivial modules is endotrivial. This allows us to define the group of endotrivial modules, whose elements are equivalence classes of endotrivial modules.

Definition 2.2. Two endotrivial kG-modules are equivalent if they are isomorphic in stmod(kG). That is, [M] = [N] if M ⊕ P ≅ N ⊕ Q for projective modules P and Q. The group of endotrivial kG-modules is the set T (G) of equivalence classes [M] of endotrivial kG-modules M, with the operation given by the rule [M] + [N] = [M ⊗ N].

Clearly, T (G) is abelian, and we have that 0 = [k] and −[M] = [M^{∗}]. Furthermore, if p does not divide the order of G, then every module is projective. In this case, the definition of an endotrivial module does not have much meaning, as every object in the stable category is equivalent to the zero object, and also every module is an endotrivial module, by a strict interpretation of the definition. In that case, we set T (G) = {0}.

3. Properties of endotrivial modules

In this section we recall some basic properties of the group T (G) that will be of use to us.

Theorem 3.1. Let G be a finite group. The group T (G) is finitely generated. Thus, the torsion subgroup T T (G) of T (G) is finite and there is a torsion free subgroup T F (G) of T (G) of finite rank, such that T (G) ≅ T T (G) ⊕ T F (G).

(a) [6, Theorem 2.3(a)] The modules Ω^{n}(k) are endotrivial and their classes form a cyclic direct summand of T (G).

(b) [6, Theorem 2.3(b)] Let n denote the number of conjugacy classes of maximal elementary abelian p-subgroups of p-rank 2 in G. Then the rank of T F (G) is n if G has p-rank at most 2 and is n + 1 if the p-rank of G is greater than 2.

(c) [6, Theorem 2.3(d)] Let P be a Sylow p-subgroup of G.

(i) The torsion subgroup T T (P) is trivial except in the case that P is cyclic, quaternion or semidihedral.

(ii) If T T (P) is trivial, then T T (G) is generated by the classes [M] of indecomposable endotrivial kG-modules M such that Res_{P}^{G} M ≅ k ⊕ (proj), for a projective kP -module (proj).

Note that in general, a module with vertex P and trivial source is not endotrivial. Nevertheless, for a subgroup H which contains a Sylow p-subgroup, the groups T (G) and T (H) are related to each other by the following.

Proposition 3.2. [5, Proposition 2.6] Let H be a subgroup of G that contains a Sylow p-subgroup P of G, and let M be an indecomposable endotrivial kG-module. The following hold.

(a) If N_{G}(P) ≤ H then the restriction map \xymatrix{Res_{H}^{G} : T (G) \ar[r] & T (H)} is injective.

The kH-module Res_{H}^{G} M is endotrivial and has a unique indecomposable nonprojective direct summand. This summand has vertex P and is isomorphic to the kH-Green correspondent of M.

(b) Suppose that H is a normal subgroup of G. Then Res_{H}^{G} M is endotrivial and indecomposable. Thus, if P is noncyclic and is a normal Sylow p-subgroup of G, then T T (G) is isomorphic to the group of one-dimensional kG-modules, i.e. T T (G) ≅ G/G′P.

More generally for any indecomposable endotrivial kG-module M and for any subgroup H of G, we have that Res_{H}^{G} M ≅ M_{0} ⊕(proj), where M_{0} is an indecomposable endotrivial kH-module and (proj) is a projective kH-module. In particular, for any endotrivial kG-module M, there is a unique indecomposable endotrivial direct summand M_{0} of M such that M ≅ M_{0} in stmod(kG). Notice also that Dim M ≡ ±1 mod |P |, if p is odd, whereas Dim M ≡ ±1 mod \frac{|P |}{2}, if p = 2.

4. Subgroup structure

In this section we collect some information concerning the p-local structure of the symmetric and alternating groups for p an odd prime. We write G for the symmetric group S_{n} of degree n, for an integer n greater or equal than p^{2}, and we write A for the alternating subgroup of the same degree n as G. Hence the Sylow p-subgroups of G are not abelian and are all contained in A.

For H and W two finite groups, with W a transitive subgroup of some symmetric group S_{n}, the wreath product of H and W is the group G = H ≀ W, isomorphic to a semidirect product

(H^{(1)} × ⋯ × H^{(n)}) ⋊ W , with H^{(i)} ≅ H for all i and

with W acting on the set {H^{(i)} | 1 ≤ i ≤ n} by permutation of the superscripts {1, …, n}. The normal subgroup H^{(1)} × ⋯ × H^{(n)} of G is called the base subgroup. More generally, we define inductively iterated wreath products H^{≀i} = (H^{≀(i−1)}) ≀ H, for all i ≥ 2, and for all transitive subgroups H of some symmetric group.

Some detail of the structure of the Sylow p-subgroup of G and its normalizer can be found in [1]. Let N be the normalizer of a Sylow p-subgroup P of G. The normalizer N_{A} of P in A has index 2 in N. For i ≥ 0, write N_{i} = N_{S_{p^{i}}}(P_{i}) for the normalizer of a Sylow p-subgroup P_{i} ≅ C_{p}^{≀i} in S_{p^{i}}. Then, N_{i} ≅ P ⋊ (C_{p−1})^{i}. In the general case, we write

n = ∑_{0≤i≤s} a_{i}p^{i} with s ≥ 2, and a_{s} ≠ 0

for the p-adic expansion of n. There are isomorphisms

P ≅ ∏_{1≤i≤s} (P_{i}^{a_{i}}) and N ≅ ∏_{0≤i≤s} (N_{i} ≀ S_{a_{i}})

For a group H, let us denote by H′ its derived subgroup. We thank Jørn Olsson for providing the proof of the next Lemma.

Lemma 4.1. Assume n = p^{t} for some integer t ≥ 1 and set N = N_{t} and P = P_{t}. Then, N′ = P and N_{A}′ = P.

Proof. Since the factor group N/P is abelian, we have that P contains N′.

Conversely, we need to show that N/N′ has order prime to p. We proceed by induction on t, for t ≥ 1. If t = 1, then N_{1} ≅ P_{1} ⋊ C_{p−1} and we can easily verify that the p-cycle generating P_{1} is a commutator in N_{1}. So P ⊆ N′. Assume now t > 1 and set Q = Q_{t} for the base subgroup of P_{t}. That is Q_{t} ≅ P_{t−1}^{p}. By [12, Lemma 4.2], we have that Q_{t} ◁ N_{t}, which implies that N_{t} = N_{H_{t}}(P_{t−1} ≀ P_{1}), where H_{t} is the subgroup of G containing P_{t} and isomorphic to S_{p^{t−1}} ≀ S_{p}. Now, by [12, Proposition 1.5], the factor group N_{t}/Q′_{t} is isomorphic to N_{S_{p^{t−1}}}(P_{t−1})/P′_{t−1} × N_{1}. Since Q′_{t} ≤ N′_{t}, we obtain, by induction, that N_{t}/N′_{t} ≅ (C_{p−1})^{t−1} × C_{p−1}.

The statement for the alternating group is immediate.

□

We end with an important observation that will be useful in the next section.

Proposition 4.2. Consider the above notation and assume that n = p^{s}, for s ≥ 1. Let H be a subgroup of G isomorphic to S_{n−1}. There exist elements σ_{1}, σ_{2}, …, σ_{s} ∈ H, each of order p − 1, such that

N = ⟨P, σ_{1}, σ_{2}, …, σ_{s}⟩ ≅ P_{s} ⋊ (C_{p−1})^{s}

Furthermore if σ_{i,j} = σ_{i}σ_{j} for all 1 ≤ i, j ≤ s, then N_{A} = ⟨P, σ_{i,j} , 1 ≤ i, j ≤ s⟩, and the σ_{i,j} can be chosen in a given subgroup of A isomorphic to A_{n−1}.

Proof. We proceed by induction on s. Clearly, the statement holds in case s = 1, where σ_{1} is just a p − 1 cycle. Assume that s ≥ 1 and that the statement holds for n = p^{s−1}. By [1, (1.3)],

N_{S_{p^{s}}}(P_{s})/P_{s} = N_{S_{p^{s−1}·p}}(P_{s−1} ≀ P_{1})/(P_{s−1} ≀ P_{1}) = \frac{N_{S_{p^{s−1}}}(P_{s−1})}{P_{s−1}} × \frac{N_{S_{p}}(P_{1})}{P_{1}}.

With the above inductive statement we can construct specific instances of the normalizer of the Sylow subgroup. That is, the normalizer of P_{1} is generated by the cycles

σ_{1} = (1, 2, …, p) and μ_{1,1} = (σ_{1} ↦ σ_{1}^{ℓ}),

where ℓ generates the group of units in 𝔽_{p}. For example, if p = 5, we can take ℓ = 2 and hence μ_{1,1} = (1, 2, 4, 3), which fixes the letter 5. In general, μ_{1,1} can be taken to be the (p − 1)-cycle, (1, ℓ, ℓ^{2}, …, ℓ^{p−1}), where ℓ^{i} should be read as the residue of ℓ^{i} modulo p, or as the element in 𝔽_{p}. The cycle fixes the letter p (which is zero in 𝔽_{p}).

Then N_{S_{p^{2}}}(P_{2}) is generated by

σ_{1}, σ_{2} = ∏_{i=1}^{p}(i, i+p, …, i+(p−1)p), μ_{2,1} = ∏_{i=0}^{p−1} σ_{2}^{i}μ_{1,1}σ_{2}^{−i} and μ_{2,2} = (σ_{2} ↦ σ_{2}^{ℓ}).

Here σ_{1} and μ_{1,1} are the cycles given exactly as above, but now considered to be elements of S_{p^{2}}. The elements σ_{1} and σ_{2} generate P_{2}. The element μ_{2,1} is the product of the conjugates of μ_{1,1} by powers of σ_{2}. These conjugates are disjoint cycles and hence μ_{2,1} commutes with σ_{2} and normalizes the normal subgroup of P_{2} generated by σ_{1} and its conjugates by powers of σ_{2}. The elements σ_{2} and μ_{2,2} generate a subgroup isomorphic to N_{S_{p}}(P_{1}).

The general case is similar. The group P_{s} is generated by σ_{1}, …, σ_{s} where σ_{s} is the product of p^{s−1} disjoint p-cycles

σ_{s} = ∏_{i=1}^{p^{s−1}}(i, i + p^{s−1}, …, i + (p − 1)p^{s−1}).

Then the normalizer N_{S_{p^{s}}}(P_{s}) is generated by P_{s}, by

μ_{s,j} = ∏_{i=0}^{p−1} σ_{s}^{i}μ_{s−1,j}σ_{s}^{−i}, for j = 1, …, s − 1, and by μ_{s,s} = (σ_{s} ↦ σ_{s}^{ℓ}).

The element μ_{s,s} is a product of p^{s−1} (p−1)-cycles, conjugation by each one of which takes the corresponding p-cyclic factor of σ_{s} to its ℓ^{th} power and fixes the other factors. The elements σ_{1}, …, σ_{s} in the statement of the proposition can be taken to be the elements μ_{s,1}, …, μ_{s,s} in the above construction. It is a straightforward exercise to show that these element commute with one another. It is also clear that every one of these elements stabilizes the letter n = p^{s}, and hence they are contained in S_{p^{s}−1} as asserted. The normalizer of any other Sylow p-subgroup is conjugate to this one, and hence the conjugate elements stabilize some letter.

The last statement of the proposition is straightforward from this analysis and Equation 2.1 of [1].

□

5. The torsion subgroup

Let H be a subgroup of G = S_{n} isomorphic to S_{n−1}. Let N be the normalizer of a Sylow p-subgroup P of G. We let A, N_{A}, H_{A} = H ∩ A be as in the previous section.

By Proposition 3.2, the indecomposable torsion endotrivial kG-modules are among the Green correspondents of the one-dimensional kN-modules. So suppose that χ is a one-dimensional kN-module which has an endotrivial kG-Green correspondent M. In this section we show that χ, and thus M, is necessarily either k or ϵ, where ϵ denotes the one-dimensional sign representation.

Throughout this section let χ and M be as above. Assume that χ_{A} is a onedimensional kN_{A}-module with kA-Green correspondent M_{A} which we assume to be endotrivial.

We recall the results from [5, Thms. A, B]

Proposition 5.1. T T (S_{n}) = ⟨[ε]⟩ and T T (A_{n}) = {[k]}, for all 3p ≤ n < p^{2}.

Our objective is to prove the following.

Theorem 5.2. Let p ≠ 2 and assume that n ≥ p^{2}. Then T T (S_{n}) = ⟨[ε]⟩ and T T (A_{n}) = {[k]}.

We proceed by an inductive argument which varies depending on the p-adic expansion of n. The most difficult cases are the ones in which the the expansion has only one term. These situations are treated first. The base case of the induction is given by Proposition 5.1.

Suppose that n = p^{s}. The next result is needed for this case.

Lemma 5.3.

(a) There are exactly (p − 1)^{s} one-dimensional kN-modules, corresponding to a choice of a (p − 1)^{st} root of unity for each σ_{i}.

(b) The restriction to H∩N of the one-dimensional kN-modules remain pairwise nonisomorphic.

(c) There are exactly \frac{(p−1)^{s}}{2} one-dimensional kN_{A}-modules all of which are pairwise nonisomorphic upon restriction to H_{A} ∩ N_{A}.

Proof. The first statement is straightforward from Proposition 3.2 and from Lemma 4.1. The second claim follows from the choice of the elements σ_{i} that determine the action of N on any one-dimensional kN-module. Namely, by Proposition 4.2, σ_{i} ∈ H ∩ N, for all i. Part (c) follows from parts (a) and (b). □

The case p = 3 and n = 9 is treated separately, because Proposition 5.1 does not apply and S_{8} has torsion endotrivial modules which are not one-dimensional. In this case, N has 4 one-dimensional modules. Explicitly, we checked using the algebra software MAGMA (cf. [2]), that the Green correspondents for these modules are k, ϵ, M and M ⊗ ϵ where M has dimension 118, which is not congruent to ±1 mod 81. Hence M is not endotrivial. A similar statement holds for M_{A}.

Thus we have proved the smallest case (p^{s} = 9) of the following.

Proposition 5.4. Let s ≥ 2. If p^{s} > 9, then assume also that T T (S_{p^{s}−1}) = ⟨[ε]⟩, and T T (A_{p^{s}−1}) = {[k]}. We have that T T (S_{p^{s}}) = ⟨[ε]⟩ and T T (A_{p^{s}}) = {[k]}.

Proof. Assume that p^{s} > 9. We know that M | Ind_{N}^{G} χ. Tensoring by ϵ if necessary, we can assume without loss of generality that Res_{H}^{G} M ≅ k⊕(proj), and so there is a nontrivial map in Hom_{kH}(k, Res_{H}^{G} M). Likewise, Hom_{kH_{A}}(k, Res_{H_{A}}^{A} M_{A}) is nonzero. Using the Mackey Formula and the Eckmann-Shapiro Lemma, we get that

0 ≠ Hom_{kH}(k, Res_{H}^{G} Ind_{N}^{G} χ) ≅ Hom_{kH}(k, ⨁_{x∈[H\G/N]} Ind_{^{x}N∩H}^{H} Res_{^{x}N∩H}^{^{x}N} ^{x}χ)
≅ ∏_{x∈[H\G/N]} Hom_{k(^{x}N∩H)}(k, Res_{^{x}N∩H}^{^{x}N} ^{x}χ).  (1)

By Lemma 5.3(b), this latter is nonzero only if χ = k, in which case M ≅ k. Thus T T (G) = ⟨[ε]⟩ as desired. Similarly, Hom_{kH_{A}}(k, Res_{H_{A}}^{A} Ind_{N_{A}}^{A} χ_{A}) is nonzero if and only if χ_{A} = k. □

Suppose that n = 2p^{s} and s ≥ 2. The normalizer of the Sylow p-subgroup P has the form N ≅ N_{s} ≀ S_{2}, where N_{s} is the normalizer of the Sylow p-subgroup of S_{p^{s}}. Let J be the subgroup of G containing N and isomorphic to a wreath product S_{p^{s}} ≀ S_{2}. We proceed as in [6, § 6 and 8]. Let S = S_{p^{s}} × S_{p^{s}} be the Young subgroup of G for the partition (p^{s}, p^{s}). Note that J = SN and S is a normal subgroup of index 2 in J. Write N_{S} = N ∩ S. Likewise, let us set A = A_{2p^{s}} and J_{A} = J ∩ A.

Proposition 5.5. Assume that T T (S_{p^{s}}) = ⟨[ε]⟩, T T (A_{p^{s}}) = {[k]}. Assume also that T T (S_{2p^{s}−2}) = ⟨[ε]⟩. Then we have that T T (S_{2p^{s}}) = ⟨[ε]⟩ and T T (A_{2p^{s}}) = {[k]}.

Proof. Let χ be a 1-dimensional kN-module with an endotrivial kJ-Green correspondent L. By the Green correspondence and the Mackey Formula, we have that

Res_{S}^{J} L | Res_{S}^{J} Ind_{N}^{J} χ ≅ Ind_{N_{S}}^{S} χ_{N_{S}}

since J = SN, and where χ_{N_{S}} = Res_{N_{S}}^{N} χ. Therefore, Res_{S}^{J} L is a direct summand of Ind_{N_{S}}^{S} χ_{N_{S}}.

Now, the conditions that S is normal in J and that L is an indecomposable endotrivial module imply that Res_{S}^{J} L is an indecomposable endotrivial module, by Proposition 3.2. Thus, L_{S} = Res_{S}^{J} L is the kS-Green correspondent of χ_{N_{S}}. Note that the Green correspondence is well-defined in this case, since N_{S} = N_{S}(P).

Let K be a subgroup of S containing N_{S} and isomorphic to a direct product S_{p^{s}} × N_{s}, where N_{s} is the normalizer of the Sylow p-subgroup of S_{p^{s}}. By our assumption, L_{S} is an indecomposable endotrivial module. So Res_{K}^{S} L_{S} ≅ U ⊕(proj), for some indecomposable endotrivial kK-module U, which also satisfies the condition that Res_{N_{S}}^{K} U ≅ χ_{S} ⊕ (proj). Now, because K has a nontrivial normal p-subgroup, Res_{N_{S}}^{K} U has no nonzero projective summand. That is, U must be a direct summand of χ_{S} induced to K, and the restriction back to N_{S} consists entirely of modules whose vertices contain that normal subgroup. Therefore, Res_{N_{S}}^{K} U ≅ χ_{S} and U has dimension 1. A similar argument, using the fact that a set of coset representatives of K in S can be taken to normalize a p-subgroup of K shows that Res_{K}^{S} L_{S} ≅ U, and so L_{S} and L also have dimension 1. Likewise, the indecomposable torsion endotrivial kJ_{A}-modules have dimension one.

We first handle the case of the symmetric groups. There are exactly four 1dimensional kJ-modules, which form a Klein four group. That is, T T (J) is generated by the sign representation and χ = k ≀ ε, which is the one-dimensional module on which S acts by the trivial representation and elements not in S act by multiplication by −1. Relative projectivity shows that the kG-Green correspondent M of χ is a Young module. Namely, M is isomorphic to a direct summand of the permutation module Ind_{S}^{G} k = M^{(p^{s},p^{s})}. It is well known that the indecomposable summands of a permutation module M^{λ} are Young modules labeled by partitions greater or equal than λ in the dominance order. In addition, the Young module Y^{λ} occurs exactly once. Now, S is the only proper Young subgroup of G of index prime to p, and therefore, Y^{(p^{s},p^{s})} is the only indecomposable direct summand of M^{(p^{s},p^{s})} with vertex P. By the Krull-Schmidt Theorem, we conclude that M is isomorphic to Y^{(p^{s},p^{s})}.

Therefore, the question is reduced to determining whether Y^{(p^{s},p^{s})} is an endotrivial module. From [11, Theorem 5.1] we prove that this is not the case. Explicitly, if L is a subgroup of G isomorphic to S_{2p^{s}−1} then Res_{L}^{G} Y^{(p^{s},p^{s})} has a direct summand V of the form

V = ⨁_{0≤i≤s} Y^{(p^{s}+p^{i}−1,p^{s}−p^{i})}.

Recall that s ≥ 2 and that a Young module Y^{λ} is projective if and only if λ is p-restricted (that is, the difference of any two consecutive parts of λ is less than p, cf. [10, Theorem 2]). In particular, V has at least two direct summands which are not projective, and so Res_{L}^{G} Y^{(p^{s},p^{s})} is not endotrivial. A fortiori, neither is Y^{(p^{s},p^{s})}. This shows that T T (G) = ⟨[ε]⟩ ≅ ℤ/2, whenever G = S_{2p^{s}}, with s ≥ 2.

We now turn to the alternating groups. As in [6], the above argument does not apply to the non trivial 1-dimensional kJ_{A}-modules. These form a Klein four group, generated by the restriction of a 2-dimensional kJ-module. Namely, let V be the subgroup of J of index 8 and isomorphic to a direct product A_{p^{s}} × A_{p^{s}}. Then, V is normal in J. Because the factor group J/V is dihedral of order 8, there is a simple kJ-module U of dimension 2. The same argument of [6, § 8] says that Res_{J_{A}}^{J} U splits as the direct sum of two 1-dimensional conjugate modules χ ⊕ ^{x}χ, and that Res_{S}^{J} U ≅ λ⊕(ελ), where λ ≅ k_{S_{p}^{s}} ⊗ε_{p^{s}} affords a signed permutation module, usually denoted as M^{(p^{s}|p^{s})}, and with Young vertex S, in the sense of Grabmeier (cf. [9, § 1.3]). Moreover, M^{(p^{s}|p^{s})} contains the kG-Green correspondent Y of U (in fact, Y = M^{(p^{s}|p^{s})}, as this latter is indecomposable). We have that Res_{A}^{G} Y = Y_{A} ⊕ Y_{A}^{x}, were Y_{A} is isomorphic to the kA-Green correspondent of χ. Now, if Y_{A} is endotrivial, then so is Res_{B}^{A} Y_{A}, for any subgroup B of A. Hence, take B ≅ S_{2p^{s}−2} and assume that Res_{B}^{A} Y_{A} is endotrivial. By hypothesis, we have that Res_{B}^{A} Y_{A} ≅ μ ⊕ (proj), where μ ≅ ε, or μ ≅ k. In particular, it follows that 0 ≠ Hom_{B}(μ, Res_{B}^{A} Y_{A}). By the Mackey formula and the Eckmann-Shapiro lemma and using the fact that A = J_{A}B, we obtain

0 ≠ Hom_{B}(μ, Res_{B}^{A} Ind_{J_{A}}^{A} χ) ≅ Hom_{B}(μ, Ind_{J_{A}∩B}^{B} Res_{J_{A}∩B}^{J_{A}} χ) ≅
≅ Hom_{J_{A}∩B}(k, Res_{J_{A}∩B}^{J_{A}} χ) ,

since Res_{J_{A}∩B}^{B} μ = k. However, a direct computation shows that Res_{J_{A}∩B}^{J_{A}} χ is not trivial. So we have a contradiction with the assumption that Y_{A} is endotrivial. □

Suppose that n = ap^{s}, for 3 ≤ a < p and s ≥ 2. Again, we assume without loss of generality and by induction that Res_{H}^{G} M ≅ k ⊕ (proj) and that Res_{H_{A}}^{A} M_{A} ≅ k⊕(proj), where H ≅ S_{ap^{s}−1} as before. Moreover, Res_{N}^{G} M ≅ χ⊕(proj) and Res_{N_{A}}^{A} M_{A} ≅ χ_{A}⊕(proj), by the Green correspondence. In this case, N ≅ N_{s}≀S_{a}, which has 2(p − 1)^{s} modules of dimension one as in the previous case. However, these modules are distinguished by their restrictions to the two subgroups N_{s} and S_{a−1}. Both of these subgroups are contained in H and hence, the restrictions of M to both of these subgroups must be a trivial module plus a projective module. We conclude that χ = k is the trivial kN-module and its Green correspondent M is the trivial kG-module.

Suppose the p-adic expansion of n has two or more terms. We write

n = a_{0} + a_{1}p + ⋯ + a_{s}p^{s} for s ≥ 2, a_{s} ≠ 0 and n ≠ a_{s}p^{s}

Recall that

N ≅ ∏_{i≥0} (N_{i} ≀ S_{a_{i}}).

We know that the kN-Green correspondent of M has dimension one. Without loss of generality, we assume by induction that

Res_{S_{asp^{s}}}^{G} M ≅ k ⊕ (proj) and that Res_{A′}^{A} M_{A} ≅ k ⊕ (proj)

where A′ = S_{a_{s}p^{s}} ∩ A. Now, each of the subgroups N_{i} ≀ S_{a_{i}} for i < s is conjugate to a subgroup of S_{a_{s}p^{s}}, and the corresponding claim holds for the corresponding subgroups of N_{A}. Thus Res_{N}^{G} M ≅ k ⊕ (proj) and Res_{N_{A}}^{A} M_{A} ≅ k ⊕ (proj). Hence M and M_{A} are are the Green correspondents of k and are trivial modules.

With all of this we can complete the proof of the main theorem.

Proof of Theorem 5.2. We perform induction on s where s is the largest degree in the p-adic expansion of n, beginning with s = 2. For n = p^{2}, we invoke Propositions 5.1 and 5.4. We use the proof of the case that the p-adic expansion of n has more than one term to prove the Theorem for n < 2p^{s}. Then we apply Proposition 5.5. Now we use the proof in the case that n = ap^{s} for a ≥ 3, and the proof in the case of more than one term in the p-adic expansion of n. Applied in the proper order, these results prove the theorem for all n such that n < p^{s+1}. We can now use Proposition 5.4 to prove the theorem for n = p^{s+1}. This last is the induction step. □

6. Torsion free complements

Recall our assumption that p > 2. It is an easy calculation to see that for n ≥ p^{2} + p, the groups G = A_{n} and G = S_{n} have no maximal elementary abelian p-subgroups of p-rank 2. As a consequence, by Theorem 3.1, the torsion free part T F (G) of the group T (G) is isomorphic to ℤ and is generated by [Ω(k)]. This is the major part of our investigation of T F (G).

In the case that p^{2} ≤ n < p^{2} + p, a Sylow p-subgroup P of G has the form C_{p} ≀ C_{p}, and P has two conjugacy classes of maximal elementary abelian subgroups. These are the base subgroup E_{1} ≅ C_{p}^{p}, which is normal in P, and E_{2} = ⟨x, y⟩ ≅ C_{p}^{2}, where ⟨x⟩ = E_{1} ∩ E_{2} is the center of P, and y is a noncentral p-element not in E_{1}. We can take y to be a generator of the second C_{p} in the wreath product expression for P.

We have proved the first part of the following.

Theorem 6.1. Let G denote either the symmetric group S_{n} or the alternating A_{n}.

(1) If n ≥ p^{2} + p, then T F (G) = ⟨[Ω(k)]⟩ ≅ ℤ.

(2) Suppose that p^{2} ≤ n < p^{2}+p. Then T F (G) ≅ ℤ^{2}. The class [Ω(k)] generates one direct summand of T F (G). The other summand is generated by the class [M] of an indecomposable endotrivial kG-module M, having the property that

Res_{E_{1}}^{G} M ≅ k ⊕ (proj) and Res_{E_{2}}^{G} M ≅ Ω^{2pr}(k) ⊕ (proj)

for some integer r with 1 ≤ r ≤ p − 1.

Proof. Theorem 3.1 shows that the rank of T F (G) in case (2) is 2, once we verify that E_{1} and E_{2} are maximal elementary abelian p-subgroups of G. That is, the only question is whether E_{2} is conjugate to a subgroup of E_{1} in S_{n}. This is not possible because of the cycle structure of the elements of E_{2}. Specifically, every nonidentity element of E_{2} is the product of p p-cycles. On the other hand, there is a subgroup F of index p in E_{1} which has no element that is a product of p p-cycles. Explicitly, consider H = E_{1} ∩ S_{p^{2}−p}. Then, any subgroup of order p^{2} in E_{1} has a nontrivial intersection with H and hence contains a nonidentity element which does not have the cycle structure of the nonidentity elements of E_{2}.

In case (2) we know that the class of Ω(k) generates a summand of T F (G). The sum of the restriction maps

\xymatrix{T F (G) \ar[r] & T F (E_{1}) ⊕ T F (E_{2}) ≅ ℤ ⊕ ℤ}

is an injection. The other generating class [M] can be taken to have its restriction to E_{1} be zero in T F (E_{1}), by making a suitable translation. Hence we can assume that the restriction of M to E_{1} is the direct sum of k plus a projective module. The restriction to E_{2} is isomorphic to the direct sum of Ω^{t}(k) plus a projective module, with t ≠ 0. By taking a dual if necessary we can assume that t > 0.

The restriction map Res_{E_{2}}^{G} factors through the restriction map Res_{P}^{G} to the Sylow p-subgroup P of G that contains E_{1} and E_{2}. Consequently, t is bounded below by the same value that is obtained for T F (P). This value is 2p by [7]. It also follows that t = 2pr is a multiple of 2p. On the other hand, t is bounded above by the minimal degree of a cohomology element of H^{∗}(G, k) whose restriction to the center Z of P is not nilpotent (see the proof of Theorem 3.1 of [5]). Now by the analysis of [4] (see Proposition 5.1) there is an element γ in degree 2p(p − 1) in integral cohomology whose restriction to Z is not zero. Because Z is cyclic and γ has even degree this element is not nilpotent and thus γ restricts to a nonzero element in the mod-p cohomology. Therefore, 2p ≤ t ≤ 2p(p − 1) and t is divisible by 2p as asserted. □

We conclude with some partial information on the missing generator. From [5] and [7], we can find explicit generators for T (N_{G}(P)). In the case of the symmetric group S_{n}, with p^{2} ≤ n < p^{2} + p, this latter result can be improved in the following way. Consider P, E_{1} and E_{2} = ⟨x, y⟩ as above. Set H = N_{G}(E_{1}). Then, H contains N_{G}(P) and has the form H ≅ (N_{1} ≀ S_{p}) × S_{a}, with a = n − p^{2}, in the notation of Section 4. Notice that the inflation of an endotrivial module from N_{1} ≀ S_{p} to H is endotrivial, and that any endotrivial kH module can be obtained up to equivalence in this way. Thus our task is reduced to finding generators for T (H) in case n = p^{2}, which we assume henceforth. Now, [7, Theorem 3.1] gives us that

T (P) = ⟨[Ω(k)], [M]⟩ with M = Ω^{−2}(Ω_{P/⟨y⟩}^{2}(k))

Here, Ω_{P/⟨y⟩}^{2}(k) is the unique indecomposable direct summand of the tensor product Ω_{P/⟨y⟩}(k) ⊗ Ω_{P/⟨y⟩}(k), where Ω_{P/⟨y⟩}(k) is the kernel of the map k[P/⟨y⟩] → k, that sends a coset u⟨y⟩ ∈ P/⟨y⟩ to 1. We observe that Ω_{P/⟨y⟩}(k) extends to H. Indeed, let C be a complement of E_{1} in the base subgroup of H. Thus, C ≅ C_{p−1}^{p}. In fact, H = E_{1} ⋊ CS_{p}, with ⟨y⟩ ≤ H. Consider the permutation module L = k[H/CS_{p}]. Then

Res_{P}^{H} L ≅ Res_{P}^{H} Ind_{CS_{p}}^{H} k ≅ ⨁_{x∈[P \H/CS_{p}]} Ind_{(CS_{p})^{x}∩P}^{P} k ≅ k[P/⟨y⟩]

by Mackey formula and since H = P CS_{p}. Thus, the relative syzygy Ω_{H/CS_{p}}(k), that is the kernel of the map L → k, restricts to P to an indecomposable module, isomorphic to Ω_{P/⟨y⟩}(k). Consequently, Ω_{P/⟨y⟩}^{2}(k) extends to Ω_{H/CS_{p}}^{2}(k), proving simultaneously that this latter is endotrivial. The same holds for the translate M = Ω^{−2}(Ω_{H/CS_{p}}^{2}(k)). Obviously, the restriction of M to A∩H is an indecomposable endotrivial module. This proves the following.

Proposition 6.2. Consider the symmetric group G = S_{n} and the alternating group A = A_{n}, with p^{2} ≤ n < p^{2} + p. In the same notation as above, with H = N_{G}(E_{1}) and H_{A} = N_{A}(E_{1}), we have that

T (H) = T T (H) ⊕ ⟨[Ω(k)], [M]⟩ and T (H_{A}) = T T (H_{A}) ⊕ ⟨[Ω(k)], [Res_{H_{A}}^{H} M]⟩

where T T (H) and T T (H_{A}) are generated by the one-dimensional modules, and where the module M satisfies

Res_{E_{1}}^{H} M ≅ k ⊕ (proj) and Res_{E_{2}}^{H} M ≅ Ω^{2p}(k) ⊕ (proj) .

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Department of Mathematics, University of Georgia, Athens, GA 30602, USA

Department of Mathematics, University at Buffalo, SUNY, 244 Mathematics Building, Buffalo, NY 14260, USA

E-mail address: dhemmer@math.buffalo.edu

Department of Mathematical Sciences, University of Aberdeen, Aberdeen, Scotland, UK