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I'm trying to understand Chapter 12, Section 11 in Cartan + Eilenberg's Homological Algebra, which concerns finite groups with periodic cohomology. Unfortunately I am jumping right to this section in the book (I've been working in Serre's Local Fields, and I'm doing the exercise at the end of Chapter 8, Section 4, which refers to the above section in Cartan + Eilenberg), so I'm a little disconcerted with the notation change (and things like using $(\Pi:1)$ for the order of the group $\Pi$ - what is up with that?)

I suppose I have two main questions.

  1. How do we know that, given a finite group $G$ with periodic cohomology, say with $${\widehat{H}}{}^n(G,A)\cong\!\!\!{\widehat{H}}{}^{n+q}(G,A)$$ for all $n\in\mathbb{Z}$ for some $q\in\mathbb{N}$, these isomorphisms must be given by cup-producting with a fixed element $g\in\widehat{H}{}^q(G,\mathbb{Z})$? This seems to be an implicit assumption in their investigation, and while I can very well believe that it's true (the cup-product satisfies some universal property, if I understand correctly), I don't see what's barring the isomorphisms from being "accidental".

  2. The fact that the period $q$ is necessarily even (unless $G$ is trivial in which case $q=1$) seems very mysterious to me. Of course, this is the key property for the exercise I'm doing (defining a generalization of the Herbrand quotient), so I would like to have a firm grasp of why it's true. I can more or less follow the reasoning in Cartan + Eilenberg for why this is true, but it's just a proof by contradiction by making a computation using the cup product, and using the fact that $\mathbb{Z}/2\mathbb{Z}$ has periodic cohomology with even period. Furthermore, I again am not seeing why cup-producting with an element of $\widehat{H}{}^q(G,\mathbb{Z})$ is necessarily involved. So, are there any more intuitive explanations of why group cohomology, if it is periodic, has even period? Are there any references other than Cartan + Eilenberg I can look at for this fact?

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up vote 4 down vote accepted

As stated, neither 1 nor 2 is true. For example, (for $p>2$) $H^{\bullet}(\mathbb F_p;\mathbb F_p)=\mathbb F_p[\varepsilon,t]/\varepsilon^2$ ($\deg\varepsilon=1$, $\deg t=2$) — so (corresponding Tate) cohomology are 1-periodic, but this periodicity is not given by cup-product (of course, these groups are also 2-periodic, and 2-periodicity is given by cup-product).

P.S. One actually needn't compute ring structure in $H^{\bullet}(\mathbb F_p;\mathbb F_p)$ to show that the periodicity is not given by cup-product: indeed, it can't be given by an element of $H(G;\mathbb Z)$ because $H^1(\mathbb F_p;\mathbb Z)=0$; it can't be even given by cup-product with an element of $H^1(\mathbb F_p;\mathbb F_p)$ — because (by supercommutativity) square of such element must have order 2 and there are no such elements in $H^2$ (which is not very surprising for a vector space over a field of $char\ne2$).

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That's true for $p>2$, but for $p=2$ you get a polynomial ring in one variable of degree one. So the period is one, but this time the periodicity is a cup product. – m_t_ Jun 10 '11 at 7:51
@mt_ You're right, $p$ should be $>2$, of course. But for $p=2$ 1-periodicity is still not given by a cup-product with an element of $H(G;\mathbb Z)$. – Grigory M Jun 10 '11 at 9:22
@Grigory: Many thanks for your answer, I guess I am misunderstanding what's going on in Cartan + Eilenberg. It seemed to me that the only sensical generalization of the Herbrand quotient to any finite group $G$ with periodic cohomology was $$\prod_{i=0}^{q-1}h^i(A)^{(-1)^i}$$ (where $h_i(A)=|\widehat{H}{}^i(G,A)|$), which would only be "balanced" if the period $q$ was even. Perhaps your answer indicates that if the period is odd, we should simply "go around twice" (which would always produce an answer of 1). – Zev Chonoles Jun 10 '11 at 17:14
@Zev One don't even need to know the ring structure — see update (well, it also more or less explains how to find the ring structure, actually: $\varepsilon^2$ has to be zero and multiplication by $t$ comes from the action of $H(G;\mathbb Z)$ — but feel free to ignore this). As for Herbrand quotient etc, I can't say anything, I'm afraid. – Grigory M Jun 10 '11 at 17:53
@Grigory: I've had a look at Brown's Cohomology of Groups, Chapter 6, Section 9, which is also on periodic cohomology, and if I understand correctly, his Theorem 9.1 shows that there is cup-product-induced periodicity if and only if there is some periodicity, but the proof does not guarantee that the periods are the same. He leaves as an exercise at the end of the section that if $G$ is non-trivial then the period of any cup-product-induced periodicity must be even. (This distinction was not as clear in Cartan + Eilenberg). So I think that's how your example jibes with the theory. – Zev Chonoles Jun 11 '11 at 6:31

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