Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been reading Class field theory by JS Milne, 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

share|cite|improve this question
You are trying to prove that "the map" is equal to $x \mapsto x \cup \gamma$. But what is "the map"? – m_t_ 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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.