# The cohomology of finite $G$-modules

This is to some extent a continuation of an earlier question of mine. Now that I'm all cleared up on what it means for a finite group to have periodic cohomology, I have another question; first I will put up some background info. Sorry about the long read.

I am working on the exercise at the end of Chapter 8, Section 4 in Serre's Local Fields (p.134), which asks the reader to generalize the Herbrand quotient $h(A)$ of a $G$-module $A$ to the case when $G$ is any finite group with periodic cohomology, not necessarily a finite cyclic group. If $q\in\mathbb{N}$ is the period of the cohomology of $G$ - i.e., $q$ is the smallest natural number such that there exists a $g\in \widehat{H}{}^q(G,\mathbb{Z})$ for which $$\widehat{H}{}^n(G,A)\,\,\,\,\xrightarrow{\,\,\,\,\,\,g\,\,\,{\small \smile}\,\,\,\cdot\,\,\,\,\,\,}\,\,\,\,\widehat{H}{}^{n+q}(G,A)$$ is an isomorphism for all $n\in\mathbb{Z}$ (see my previous question for what's going on here) - then I've defined the generalized Herbrand quotient of a $G$-module $A$ to be $$h(A)=\prod_{n=0}^{q-1}|\widehat{H}{}^n(G,A)|^{(-1)^n}$$ As it turns out, the period $q$ must be even, so this definition is "balanaced", i.e. a generalized Herbrand quotient always looks like $$h(A)=\frac{|\widehat{H}{}^0(G,A)|\cdot|\widehat{H}{}^2(G,A)|\cdots|\widehat{H}{}^{q-2}(G,A)|}{|\widehat{H}{}^1(G,A)|\cdot|\widehat{H}{}^3(G,A)|\cdots|\widehat{H}{}^{q-1}(G,A)|}.$$ I've already answered the first part of the exercise, which is to prove that if $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence of $G$-modules ($G$ being any finite group with periodic cohomology), and at least two of $h(A)$, $h(B)$, and $h(C)$ are defined, then the third is as well and $h(B)=h(A)\cdot h(C)$. So I feel pretty confident that this definition is correct. My current question concerns the other part of the exercise, which is to show that if $A$ is a finite $G$-module, then $h(A)=1$.

Hypothesis: if $G$ is a finite group with periodic cohomology, and $A$ is a finite $G$-module, then $$|\widehat{H}{}^{-1}(G,A)|=|\widehat{H}{}^{0}(G,A)|.$$ Because $|{}_NA|\cdot|NA|=|A|$, this is eqiuvalent to $$|I_GA|=[A:A^G].$$

I know that this is true when $G$ is a finite cyclic group (which always have periodic cohomology); in fact this is exactly how they proved the statement $h(A)=1$ for the standard Herbrand quotient, because for a finite cyclic group the period of the cohomology is always $q=2$, so $$h(A)=\frac{|\widehat{H}{}^0(G,A)|}{|\widehat{H}{}^1(G,A)|}=\frac{|\widehat{H}{}^0(G,A)|}{|\widehat{H}{}^{-1}(G,A)|}=1.$$ My motivation for this hypothesis is that it is in fact equivalent to the statement that $$|\widehat{H}{}^{n}(G,A)|=|\widehat{H}{}^{m}(G,A)|$$ for any $n,m\in\mathbb{Z}$, which would clearly imply the desired $h(A)=1$ for the generalized Herbrand quotient. It's equivalent to this stronger statement by the use of dimension shifting; if it's true that $|\widehat{H}{}^{-1}(G,A)|=|\widehat{H}{}^{0}(G,A)|$ for any finite $G$-module $A$, then given a short exact sequence $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ where $B$ is finite and co-induced (for example, $B=\text{Hom}_{\mathbb{Z}}(\mathbb{Z}[G],A)\cong$ $\bigoplus_{G}A$), then $C$ is a finite $G$-module for which $$\widehat{H}{}^{n}(G,A)\cong\widehat{H}{}^{n-1}(G,C)\text{ for all }n\in\mathbb{Z},$$ hence
$$|\widehat{H}{}^{0}(G,A)|=|\widehat{H}{}^{-1}(G,C)|=|\widehat{H}{}^{0}(G,C)|=|\widehat{H}{}^{1}(G,A)|.$$ There is a dual process that works in the other direction with induced modules (in fact induced = co-induced since $G$ is finite). Repeating as necessary we can get that $|\widehat{H}{}^{n}(G,A)|=|\widehat{H}{}^{m}(G,A)|$ for any $n,m\in\mathbb{Z}$.

My question is whether we actually need the assumption that $G$ have periodic cohomology for this to be true. I strongly suspect that we do, but I'm not fluent enough in group cohomology yet to be very good at constructing examples or counter-examples to anything. So,

Question: Does there exist a finite group $G$ and finite $G$-module $A$ such that $$|\widehat{H}{}^{-1}(G,A)|\neq|\widehat{H}{}^0(G,A)|?$$

It may be useful to know that TFAE for a finite group $G$:

• $G$ has periodic cohomology.

• Every abelian subgroup of $G$ is cyclic.

• Every $p$-subgroup of $G$ is either cyclic or is a generalized quaternion group.

• Every Sylow subgroup of $G$ is either cyclic or is a generalized quaternion group.

So if you can construct a counterexample using a group with periodic cohomology, thereby showing my hypothesis to be incorrect, that would be especially useful!

Postscript: Of course, I would really appreciate any help on this question, regardless of whether it's along the lines of my current approach above. If you know the answer, please only post the smallest hints you can conceivably make.

My initial idea was

Hey, $A$ is a finite abelian group! Write it as a direct sum of cyclic $p$-groups $\bigoplus_i\mathbb{Z}/p_i^{k_i}\mathbb{Z}$, use the fact that cohomology factors over finite direct sums to get $$\widehat{H}{}^{n}(G,A)\cong\bigoplus_i \widehat{H}{}^{n}(G,\mathbb{Z}/p_i^{k_i}\mathbb{Z})\text{ for all }n\in\mathbb{Z}$$ and hence $$h(A)=\prod_i h(\mathbb{Z}/p_i^{k_i}\mathbb{Z}),$$ then use the previous part of the problem (the generalized Herbrand quotient is multiplicative on exact sequences) on the exact sequence $$0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/p_i^{k_i}\mathbb{Z}\rightarrow0$$ to get that $h(\mathbb{Z})=h(\mathbb{Z})\cdot h(\mathbb{Z}/p_i^{k_i}\mathbb{Z})$, hence $h(\mathbb{Z}/p_i^{k_i}\mathbb{Z})=1$ for all $i$, so we're done!

What I then realized was that this really only works if the action of $G$ on $A$ is trivial; just because $A$ can be decomposed nicely as an abelian group doesn't mean that it decomposes as a $G$-module. Similarly, there may very well not be an exact sequence $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/p_i^{k_i}\mathbb{Z}\rightarrow0$ as $G$-modules, because $\mathbb{Z}$ can only be acted on trivially or by negation, while $\mathbb{Z}/p_i^{k_i}\mathbb{Z}$ will have lots of automorphisms.

Currently, I think this might be salvageable by breaking up the cohomology groups $\widehat{H}{}^n(G,A)$ into their $p$-primary components, or in other words proving that $h(A)=1$ by proving for each $p$ dividing $|G|$ that the number of $p$'s in the numerator and denominator are the same, but I don't know how I might go about doing that, which is why I'm trying another approach for now.

I know. I'm just jibing you. I'm not done reading this yet but it might be helpful to you to read the chapter in Lang's "Algebra" on the Euler Characteristic of which the Hereband quotient (including your generalizations) is a special case. It's very straight forward. But if you don't feel like that, the main fact to take away from it is the Euler characteristic (Hereband quotient) commutes with taking homology. Using this fact, showing $h(A) = 1$ becomes trivial. –  jspecter Jun 14 '11 at 20:28
Dear Zev, Here is one comment on your proof strategy, in case you don't know this: if $A$ is a finite abelian group, then the decomposition of $A$ as a direct sum of Sylow subgroups is functorial. Namely, if we let $p$ be a prime, and let $A[p^{\infty}]$ denote the $p$-power torsion subgroup of $A$ (which is a subgroup, because $A$ is abelian), then the natural map $\oplus_p A[p^{\infty}] \to A$ is an isomorphism. Furthermore, since the formation of $A[p^{\infty}]$ is functorial, if $A$ is a $G$-module, then $A[p^{\infty}]$ will be a $G$-submodule. –  Matt E Jun 15 '11 at 1:40
If you look at the $p$-adic filtration on each $A[p^{\infty}]$, you furthermore obtain a filtration by $G$-submodules each quotient of which is killed by $p$. Thus many questions about group cohomology with finite coefficients can be reduced to the case when the module $A$ is $p$-torsion for some prime $p$. This is not quite as simple as the case of $\mathbb Z/p$ that you wanted to consider, but is not too far from it, and is the basis for many arguments in the cohomological proofs of local and global class field theory (which I guess is what you're working towards). Regards, –  Matt E Jun 15 '11 at 1:42