A remark on a theorem of Ritter and Segal

Serge Bouc

A well-known theorem proved independently by Ritter ([4]) and Segal ([5]) states that if P is a p-group, then the natural morphism from the Burnside ring B(P) of P to the Grothendieck ring R_{ℚ}(P) of rational representations of P, mapping a finite P -set X to the permutation module ℚX, is surjective.

The object of this note is to complete this theorem in the following way:

Theorem 1 Let p be a prime number, and P be a finite p-group. If V is a non-trivial simple ℚP -module, then there exist subgroups R ⊃ Q of P, with |R : Q| = p, and an isomorphism of ℚP -modules

V ≃ Ind_{R}^{P}Inf_{R/Q}^{R}Ω_{R/Q}

where Ω_{R/Q} is the augmentation ideal of the group algebra ℚR/Q.

In other words there is an exact sequence of ℚP -modules

0 → V → ℚ(P/Q) → ℚ(P/R) → 0

where the map ℚ(P/Q) → ℚ(P/R) is the natural projection. In particular in R_{ℚ}(P)

V = ℚ(P/Q) − ℚ(P/R)

The following lemma is a special case of this theorem:

Lemma 2 1. Let P be a cyclic p-group. Then the modules

Ind_{R}^{P}Inf_{R/Φ(R)}^{R}Ω_{R/Φ(R)}

for non-trivial subgroups R of G, are the non-trivial irreducible ℚP -modules.

2. Let P be an elementary abelian p-group. Then the modules

Inf_{P/Q}^{P}Ω_{P/Q}

for subgroups Q of index p in P are the non-trivial simple ℚP -modules.

Proof: The number of simple ℚP -modules for a finite group P is equal to the number of conjugacy classes of cyclic subgroups of P (see [6] Chapitre 13 Théorème 29 Corollaire 1). Hence in both cases, it suffices to show that the listed modules are simple and not isomorphic to each other. Indeed the number of non-trivial subgroups of a cyclic group P is equal to the number of (conjugacy classes of) cyclic subgroups of P, minus one, and the number of subgroups of index p in an elementary abelian p-group P is equal to the number of (conjugacy classes of) cyclic subgroups of P, minus one.

If P is cyclic of order p^{n}, and R is a subgroup of P of order p^{d} > 1, then the module Ind_{R}^{P}Inf_{R/Φ(R)}^{R}Ω_{R/Φ(R)} has dimension p^{n−d}(p − 1). Thus as d varies from 1 to n, those modules are not isomorphic to each other. Since moreover

Ind_{R}^{P}Inf_{R/Φ(R)}^{R}Ω_{R/Φ(R)} ≃ Inf_{P/Φ(R)}^{P}Ind_{R/Φ(R)}^{P/Φ(R)}Ω_{R/Φ(R)}

and since inflation takes simple modules to simple modules, it suffices to prove that for any cyclic p-group P, if R is the subgroup of P of order p, the module V = Ind_{R}^{P}Ω_{R} is simple. But looking at scalar products of the character of V with complex characters of P shows that the character of V is the sum of all primitive characters of P. If W is a non-zero rational direct summand of V, with character χ_{W}, then there is a primitive complex character ζ of P such that the scalar product (χ_{W} , ζ) is non-zero, hence equal to 1. It follows that (χ_{W} ,^{s}ζ) = 1 for all automorphism s of the field of p^{n}-th roots of unity. Since the Galois group of this field is transitive on the primitive roots of unity, it follows that (χ_{W} , ζ′) = 1 for all primitive complex character ζ′ of P. This shows that the character of W is equal to the character of V, hence that W = V. Thus V is a simple ℚP -module.

Now if P is elementary abelian, the modules Inf_{P/Q}^{P}Ω_{P/Q}, for |P : Q| = p, are inflated from simple modules for the group of order p, by the previous discussion of the cyclic group case. Hence they are simple. Since the kernel of Inf_{P/Q}^{P}Ω_{P/Q} is Q, it follows that those modules are not isomorphic to each other.

Corollary 3 Let E be an elementary abelian subgroup of P. Then any simple ℚP -module has dimension at most (p − 1)|P : E|.

Proof: Let V be a simple ℚP -module, and E be an elementary abelian subgroup of P. By Frobenius reciprocity, there is a non-zero morphism of ℚP - modules from V to Ind_{E}^{P}Res_{E}^{P}V. Since ℚP is semi-simple, this morphism is split injective, and there is a simple direct summand W of Res_{E}^{P}V such that V is a direct summand of Ind_{E}^{P}W. By lemma 2, the dimension of W is at most p − 1, thus dim_{ℚ} V ≤ |P : E|(p − 1).

Proof of theorem 1: The following proof is inspired by methods of M. Enguehard ([2]), and goes by induction on the order of P. I can suppose that P is neither cyclic nor elementary abelian.

Let V be a non-trivial simple ℚP -module. If V is not faithful, then there exists a non-trivial normal subgroup N of P such that

V ≃ Inf_{P/N}^{P}W

where W is a ℚ(P/N)-module. If V is simple and non-trivial, then so is W. By induction hypothesis, there are subgroups R/N ⊃ Q/N of P/N such that

W ≃ Ind_{R/N}^{P/N}Inf_{R/Q}^{R/N}Ω_{R/Q}

Taking inflation to P gives

V ≃ Ind_{R}^{P}Inf_{R/N}^{R}Ω_{R/Q} ≃ Ind_{R}^{P}Inf_{R/Q}^{R}Ω_{R/Q}

Hence I can suppose that V is faithful. Now there are two cases:

• There exists a normal subgroup E of P which is elementary abelian of order p^{2}. Let L be a simple summand of Res_{E}^{P}V as a ℚE-module, and let I denote the inertial subgroup of L in P, i.e.

I = {x ∈ P | ^{x}L ≃ L}

Then by Clifford theory (see [1] Theorem 11.1) there exists a positive integer e such that

Res_{E}^{P}V ≃ e( ∑_{g∈P/I} ^{g}L)

Let L̃ denote the L-isotypic component of Res_{E}^{P}V (i.e. the submodule generated by all simple summands isomorphic to L). Then

V ≃ Ind_{I}^{P} L̃

Now I contains the centralizer C_{P} (E) of E in P, and the quotient P/C_{P} (E) is a p-subgroup of the automorphism group of E, which has order p(p−1)(p^{2} −1). Hence I has index 1 or p in P.

If |P : I| = p, then V ≃ Ind_{I}^{P} L̃, and L̃ is a simple I-module. It cannot be the trivial module, since in this case V ≃ Ind_{I}^{P} ℚ which is not simple. By induction hypothesis, there exist subgroups R ⊃ Q of P such that

L̃ ≃ Ind_{R}^{I}Inf_{R/Q}^{R}Ω_{R/Q}

Taking induction up to P gives

V ≃ Ind_{R}^{P}Inf_{R/Q}^{R}Ω_{R/Q}

Hence I can suppose I = P, and in this case V = L̃. If V is faithful, then L has to be a faithful simple ℚE-module. By Lemma 2, there are no such simple ℚE-modules, and this is a contradiction.

• The other case is when P has no normal subgroup of order p^{2}. In this case, by Theorem 4.10 of Chapter 5 of [3], the group P is cyclic if p is odd, and if p = 2, the group P is cyclic, or quaternion of order at least 8, dihedral of order at least 16, or semi-dihedral of order at least 16.

If P is cyclic, the result holds by Lemma 2. Hence I can suppose that p = 2, and P is quaternion, dihedral, or semi-dihedral, and P has order 2^{n}, with n ≥ 3.

In this case P has a non-central cyclic subgroup A of index 2. Let C denote the unique subgroup of order 2 of A.

The restriction of V to A is a faithful ℚA-module, hence by lemma 2 it must contain Ind_{C}^{A}Ω_{C} as a direct summand. In particular dim_{ℚ} V ≥ 2^{n−2}.

Now if P contains an elementary abelian subgroup E of order 4, then there is a simple ℚE-module W such that V is a direct summand of Ind_{E}^{P}W. Since W is one dimensional by lemma 2, it follows that dim_{ℚ} V ≤ |P : E| = 2^{n−2}. Thus dim_{ℚ} V = 2^{n−2}, and V is isomorphic to Ind_{E}^{P}W. Moreover W must be non-trivial, since otherwise ℚ is a direct summand of Ind_{E}^{P}W ≃ V. Hence W = Inf_{E/F}^{E} Ω_{E/F}, for some subgroup F of E of index 2, and V has the form required for theorem 1.

If P contains no elementary abelian subgroup E of order 4 (and P is not cyclic), then P is generalized quaternion. In this case C is the only subgroup of P of order 2, and the simple ℚC-modules are one dimensional, isomorphic to ℚ and Ω_{C}. Since V is faithful, the restriction Res_{C}^{P}V must contain Ω_{C} as a direct summand. By Frobenius reciprocity, it follows that V is a direct summand of M = Ind_{C}^{P}Ω_{C}. To complete the proof of theorem 2, it suffices to show that M is simple.

This can be done by the following elementary proof. The quaternion group P of order 2^{n} has the following presentation

P = < x, y | x^{2^{n−1}} = 1, x^{2^{n−2}} = y^{2}, yxy^{−1} = x^{−1} >

The subgroup A is generated by x, and C is generated by x^{2^{n−2}} = y^{2}. The group P is the disjoint union of A and Ay. Thus

M = ℚP ⊗_{ℚC} Ω_{C} = (ℚA ⊗_{ℚC} Ω_{C}) ⊕ (ℚAy ⊗_{ℚC} Ω_{C})

Now the generator x^{2^{n−2}} of C acts by -1 on Ω_{C}. It follows that

ℚA ⊗_{ℚC} Ω_{C} ≃ 𝔽 = ℚ[X]/<X^{2^{n−2}} + 1>

and this is an isomorphism of ℚA-modules if the action of x on 𝔽 is given by multiplication by X. Moreover 𝔽 is a field, since X^{2^{n−2}} + 1 is a cyclotomic polynomial.

Denote by σ the automorphism of 𝔽 sending X to X^{−1} = −X^{2^{n−2}−1}, and by Y the element y ⊗ 1 of ℚAy ⊗_{ℚC} Ω_{C}. Then M ≃ 𝔽 ⊕ 𝔽Y as ℚP -modules, where the action of x is given by left multiplication by X, and the action of y is given by

y(A + BY) = −σ(B) + σ(A)Y, ∀A, B ∈ 𝔽

since moreover y^{2} acts by -1 on Ω_{C}. In other words if M is endowed with the ℚ-algebra structure induced by the following multiplication

(A′ + B′Y)(A + BY) = A′A − B′σ(B) + (A′B + B′σ(A))Y, ∀A, A′, B, B′ ∈ 𝔽

then the generators x and y of P act on M by multiplication by X and Y respectively. To show that M is simple, it suffices to show that M is a skew field. But for A, B ∈ 𝔽

(A + BY)(σ(A) − BY ) = Aσ(A) + Bσ(B) ∈ 𝔽

If A = a_{0}+a_{1}X+…+a_{2^{n−2}−1}X^{2^{n−2}−1} and B = b_{0}+b_{1}X+…+b_{2^{n−2}−1}X^{2^{n−2}−1}, with coefficients a_{i} and b_{i} in ℚ, for 0 ≤ i < 2^{n−2}, then the coefficient of 1 in Aσ(A) + Bσ(B) is equal to a_{0}^{2} + a_{1}^{2} + … + a_{2^{n−2}−1}^{2} + b_{0}^{2} + b_{1}^{2} + … + b_{2^{n−2}−1}^{2}. This is positive if A + BY is non-zero, hence Aσ(A) + Bσ(B) ≠ 0 is invertible in 𝔽, hence in M. Thus M is a skew-field, as was to be shown.

In view of theorem 1, it is natural to ask when conversely, being given two subgroups R ⊃ Q with |R : Q| = p, the module Ind_{R}^{P}Inf_{R/Q}^{R}Ω_{R/Q} is an irreducible ℚP -module. A possible answer is the following:

Proposition 4 Let p be a prime number, and P be a p-group. Let R ⊃ Q be subgroups of P, with |R : Q| = p. Then the following conditions are equivalent:

1. The module Ind_{R}^{P}Inf_{R/Q}^{R}Ω_{R/Q} is an irreducible ℚP -module.

2. If S is any subgroup of P such that R ∩ S ⊆ Q, then |S| < |R|.

3. The group N_{P} (Q)/Q is cyclic or generalized quaternion, the group R/Q is its unique subgroup of order p, and if S is any subgroup of P such that |S| > |Q|, then S ∩ N_{P} (Q) ⊈ Q.

Proof: Suppose first that the ℚP -module Ind_{R}^{P}Inf_{R/Q}^{R}Ω_{R/Q} is irreducible. Then its character is orthogonal (for the scalar product) to any irreducible module of the form S_{U,V} = Ind_{U}^{P}Inf_{U/V}^{U} Ω_{U/V}, where U ⊃ V are subgroups of P with |U : V | = p and |U| > |R|.

Those irreducible modules, together with the trivial one, generate the submodule M of R_{ℚ}(P) generated by the permutation modules ℚ(P/S) = Ind_{S}^{P} ℚ, for |S| ≥ |R|: clearly S_{U,V} ∈ M if |U| > |R|, since S_{U,V} = Ind_{V}^{P} ℚ − Ind_{U}^{P}ℚ in R_{ℚ}(P). Conversely, if S is a subgroup of P with |S| ≥ |R|, then Ind_{S}^{P} ℚ is the sum of the trivial module and of some irreducible modules S_{U,V}. And if S_{U,V} is a direct summand of Ind_{S}^{P} ℚ, then in particular

dim_{ℚ} S_{U,V} = |P : U|(p − 1) ≤ dim_{ℚ} Ind_{S}^{P} ℚ − 1 = |P : S| − 1 < |P : S|

Thus

|S| ≤ (p − 1)|S| < |U|

and it follows that |U| > |S| ≥ |R|.

Thus if W_{R,Q} = Ind_{R}^{P}Inf_{R/Q}^{R}Ω_{R/Q} is irreducible, then its character is orthogonal to the characters Ind_{S}^{P} 1, for |S| ≥ |R|. Since the character of W_{R,Q} is equal to

Ind_{Q}^{P}1 − Ind_{R}^{P}1

and since the scalar product of Ind_{Q}^{P}1 with Ind_{S}^{P} 1 is equal to the number of double cosets QxS, for x ∈ P, it follows that

|Q\P/S| = |R\P/S|

for any subgroup S of P with |S| ≥ |R|. This is equivalent to requiring that QxS = RxS for any x ∈ P, or equivalently that R ⊆ Q.^{x}S. Now this is equivalent to R = Q(R∩^{x}S), or R∩^{x}S ⊈ Q since Q has index p in R. It follows that if S is a subgroup of P with R ∩ S ⊆ Q, then |S| < |R|. In other words, condition 1 implies condition 2.

Conversely, if condition 2 holds, then the previous discussion shows that the character of the module W_{R,Q} is orthogonal to the characters of all the simple modules S_{U,V}, for |U| > |R|. Now for dimension reasons, the module W_{R,Q} cannot have a simple direct summand S_{U,V}, for |U| < |R|. It follows that W_{R,Q} has a direct summand S_{U,V}, with |U| = |R|. Then S_{U,V} and W_{R,Q} have the same dimension, hence they are isomorphic. Hence W_{R,Q} is irreducible, and this completes the proof of the equivalence of conditions 1 and 2.

If condition 2 holds, and if S is a subgroup of P containing Q, and not containing R, the intersection R ∩ S is equal to Q, since Q is maximal in R. It follows that |S| < |R|, or equivalently that |S| ≤ |Q|. Hence S = Q, and R is the smallest element of ]Q, P]. Hence it is the smallest element of ]Q, N_{P} (Q)], and the group N_{P} (Q)/Q has a single subgroup R/Q of order p. It follows that N_{P} (Q)/Q is cyclic or generalized quaternion. If S is any subgroup of P with |S| > |Q|, or equivalently |S| ≥ |R|, then R ∩ S ⊈ Q. Since R ⊆ N_{P} (Q), it follows that S ∩ N_{P} (Q) ⊈ Q and 2 implies 3.

Conversely if 3 holds, and if S is any subgroup of P with |S| ≥ |R|, then S ∩N_{P} (Q) ⊈ Q and Q.(S ∩N_{P} (Q)) ≠ Q. Since R/Q is the unique minimal nontrivial subgroup of N_{P} (Q)/Q, it follows that R ⊆ Q.(S∩N_{P} (Q)), or equivalently R = Q.(R ∩ S ∩ N_{P} (Q)) = Q.(R ∩ S). Thus R ∩ S ⊈ Q, and 3 implies 2.

Theorem 1 has the following easy consequence:

Corollary 5 Let P be a p-group, and V be a finite dimensional ℚP -module. Then

dim_{ℚ} V ≡ dim_{ℚ} V^{P} mod. (p − 1)

More generally, for k ∈ ℕ denote by N_{k} the intersection of all subgroups S of P with |P : S| = p^{k}. Then

dim_{ℚ} V ≡ dim_{ℚ} V^{N_{k}} mod. p^{k}(p − 1)

Proof: The first congruence is a special case of the second one, when k = 0. To prove this second congruence, decompose V as a direct sum of simple modules

V = n_{1}ℚ ⊕ ⨁_{S_{R,Q}∈Σ} n_{R,Q}S_{R,Q}

for integers n_{1} and n_{R,Q}, where Σ is a set of mutually non-isomorphic non-trivial simple modules S_{R,Q}. Now since N_{k} is a normal subgroup of P, the set of fixed points S_{R,Q}^{N_{k}} is a ℚP -submodule of S_{R,Q}. Hence it is zero or the whole of S_{R,Q}. This last case occurs exactly when N_{k} acts trivially on S_{R,Q}, i.e. when N_{k} ⊆ Q. It follows that

V^{N_{k}} = n_{1}ℚ ⊕ ⨁_{S_{R,Q}} n_{R,Q}S_{R,Q}

where the sum runs over those simple modules S_{R,Q} ∈ Σ for which Q ⊇ N_{k}. The previous two equations show that

dim_{ℚ} V = dim_{ℚ} V^{N_{k}} + ∑_{S_{R,Q}} n_{R,Q} dim_{ℚ} S_{R,Q}

where the sum runs over those simple modules S_{R,Q} ∈ Σ for which Q ⊉ N_{k}. This implies |P : Q| > p^{k}, i.e. |P : R| ≥ p^{k}. Now the dimension of S_{R,Q} is equal to |P : R|(p − 1), and the congruence follows.

Acknowledgments: I wish to thank the referee for suggesting substantial simplifications of the proof of theorem 1.

References

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