Dear Ján, Dear Sunil,

I want to say two things about the niceness of the representation ring ofthe kind of groups we are looking at. Essentially, when trying to compute thefiltration, there are two types of representations which we like: 1-dimensionalrepresentations, and representations which factor via a group which we havealready studied.

About 1-dimensional representations first. Let G be any finite group. Letme call two (real) characters ρ1 and ρ2 equivalent (for lack of a better name)if there is a 1-dimensional representation χ such that ρ1 = χρ2. (I will neveruse “equivalent” to mean isomorphic). The number of equivalence classes tellsyou something about the number of generators for R(G) as a module over thesubring generated by 1-dimensional characters.

Lemma. Suppose that G fits into an exact sequence

1 −→ V −→ G −→W −→ 1

where V and W are F2-vector spaces, and V is central. Then the number ofequivalence classes of irreducible, real characters of G is

2dimV+dimW−dimH1(G,F2) .

So for example if G = V ×W this number is 20 = 1, which makes sense. If Gis extraspecial (V = Z/2) this number is 21 = 2, which was proved by Quillenand is the crucial point in the argument.

Proof. Let ρ be any irreducible representation. Its restriction to V , bySchur’s lemma, defines a 1-dimensional character χρ ∈ H1(V ). Consider itsclass in the cokernel of H1(G) −→ H1(V ): it depends only on the equivalenceclass of ρ. We will see that ρ 7→ χρ is a bijection between the equivalenceclasses and the elements of this cokernel. This will give the result, as the kernelof H1(G) −→ H1(V ) is H1(W ).

If ρ1 and ρ2 map to the same class, we may assume that they define preciselythe same χρ. If this is the trivial character of V , then ρ1 and ρ2 factor via W ,so they are 1-dimensional and there is nothing to prove. Otherwise, let K bethe kernel of χρ; it is a hyperplane in V and a central subgroup of G. Now ρ1

and ρ2 both factor through the extraspecial group G = G/K, and the action ofthe central Z/2 = V/K is non-trivial in both ρ1 and ρ2. By Quillen, ρ1 and ρ2

must be equivalent. So ρ 7→ χρ is injective.

For the surjectivity, one may use Quillen’s result again which asserts theexistence of the appropriate representation of G. Or one may pick an irreduciblesummand in χρ ↑G. QED

The second lemma I have to offer exploits the fact that we are dealing withgroups which embed in products of easy groups.

Lemma. Let G embed in a finite product H =∏Gi where each Gi is

either Z/2, Z/4 or D4 (aka D8). Let ρ be an irreducible representation of G.Then there exist 1-dimensional characters χ1, χ2, . . . , χk such that, if ∆i is theunique irreducible representation of Gi which is non-trivial on the central Z/2,one has

(χ1 + χ2 + · · ·+ χk)ρ = ∆i1∆i2 · · ·∆ir .

1

(Thinking of course of each ∆i as a representation of G.)

So if R(G) were a field it would be generated by 1-dimensional representa-tions and the ∆i’s . . . I wish I could say something better.

Proof. Each Gi has a distinguished Z/2 in its center, and together theyform an F2-vector space VH which is a central subgroup of H. Moreover thequotient WH = H/VH is also an F2-vector space. Call V = VH ∩ G and W =WH ∩G (by which I mean G/V ), so there is an exact sequence

1→ V → G→W → 1 .

Each map pi : V → Gi lands in fact in the central Z/2, so it can be seen as anelement of the dual V ∗. Assume for simplicity that the numbering is arrangedso that p1, . . . , ps form a basis of V ∗.

Given ρ, we form χρ ∈ V ∗ as above, and write

χρ = pδ11 · · · pδss

with each δi = 0 or 1. Putτ = ∆δ1

1 · · ·∆δss .

The restriction of τ to V defines, by Schur’s lemma, a character which isnone other than χρ, by construction. So by the previous lemma (and its proof),each irreducible component of τ is of the form χiρ for a 1-dimensional χi. QED

Best wishes,

Pierre

2