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I have been reading Class field theory by JS Milne http://www.jmilne.org/math/CourseNotes/CFT.pdf, and im stuck on the chapter about group cohomology and would like some hints, specifically about cup products, on page 79 he has remark 3.5 which goes as follows:

Let G be a finite cyclic group and let $m$ by its size, now let $\gamma \in H^2(G,\mathbb{Z})$, correspond under the isomorphism $ H^2(G,\mathbb{Z}) \cong \operatorname{Hom}(G,\mathbb{Q}/\mathbb{Z})$ to the map sending the generator $\sigma \in G$ to $1/m$. Then the map $H^r(G,M) \rightarrow H^{r+2}(G,M)$ is $x \mapsto x \cup \gamma$. (note these are tate groups not just cohomology groups).

I wanted to prove this, but im not quite sure how, I think I need some clarification on how cup products work.

Thank you

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You are trying to prove that "the map" is equal to $x \mapsto x \cup \gamma$. But what is "the map"? –  mt_ Oct 10 '12 at 14:04
    
He means that the periodicity isomorphism is given by the cup-product (because any finite cyclic group is periodic of order 2). To the OP: Check out Ken Brown's classic textbook. –  Chris Gerig Oct 10 '12 at 17:01
    
Is this isomorphism in some sense related to the duality theorems? Or is it just a dellusion? –  awllower Dec 8 '12 at 15:04

1 Answer 1

Serre Local Fields, VIII, Section 4 gives the following hint: 'the period isomorphisms are given by cup product with theta: this follows, for example, from the formulas for the cup product given in Cartan-Eilenberg (Homological Algebra) p. 252." Another place to look is Weiss, Cohomology of Groups.

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