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

This is a stupid question about group cohomology, but it confuses me a lot. Basically I think that the problem is the fact that I do not really understand Shapiro's lemma.

Say we take a profinite group $G$ and some finite index normal subgroup $H$. Consider the map of $G$-modules $\mathbb{Z} \to \mathbb{Z}[G/H]$ given by $1 \mapsto N_{G/H} = \sum_{\overline{\sigma} \in G/H} \overline{\sigma}$. This map induces a maps in group cohomology $H^i(G,\mathbb{Z}) \to H^i(G,\mathbb{Z}[G/H]) \cong H^i(H,\mathbb{Z})$, where the isomorphism is Shapiro's lemma.

Now my question is: is this just the map obtained when "restricting" from $G$ to the subgroup $H$?

For example if $i = 2$, then the map can be identified with a map $$\text{Hom}_{\text{continuous}}(G,\mathbb{Q}/\mathbb{Z}) \to \text{Hom}_{\text{continuous}}(H,\mathbb{Q}/\mathbb{Z}).$$ Is this just the map obtained by precomposing $G \to \mathbb{Q}/\mathbb{Z}$ with the inclusion $H \to G$?

share|cite|improve this question
The answer to your question is yes, and it is true in wider generality. Have a look at Benson's Representations and Cohomology vol 1, the exercise at the end of ch2 (p.48 in my edition). – m_t_ Dec 12 '11 at 17:16
up vote 2 down vote accepted

Yes, the restriction map and the corestriction map on group cohomology can both be interpreted in terms of the isomorphism provided by Shapiro's lemma. Details can be found on pp 60-61 of 'Cohomology of Number Fields' by Neukirch/Schmidt/Wingberg. You may also find the discussion on pp 67-68 of Milne's notes on Class Field Theory (available at his website) to be helpful.

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.