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Let $H\to G$ be a subgroup of a finite group, let $M$ be an $H$-module, $N$ be the induced $G$-module, by Shapiro lemma we have $H^i(G,N)\cong H^i(H,M)$. What is the map on the level of cochains? Given a cochain $G^i\to N$, what is the corresponding $H^i\to M$? (If we use inclusion $H^i\to G^i$, what is the map $N\to M$?)

Carlo Beenakker
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    See Cohomology of Number Fields, (1.6.4). –  Jul 15 '18 at 13:49
  • @TKe Thanks for the reference, I’ll try to work out the details! One thing a bit confusing, if we are constructing the reverse map $H^i(H,M)\to H^i(G,N)$, then a cocycle $H^i\to M$ can associate a cocycle $H^i\to N$ by inclusion, but then how does it extends to a cocycle on $G^i$? –  Jul 15 '18 at 14:44
  • See also the exercise in Chapter III.8 of Ken Brown's "Cohomology of Groups". – Chris Gerig Jul 15 '18 at 20:24
  • @Qixiao : you have to take a transversal for $H$ in $G$... again, the details are in NSW, Cohomology of Number Fields. By the way, the isomorphism the other way around is easy: Shapiro's isomorphism is the restriction from $G$ to $H$, followed by the map $N\to M$ which evaluates at 1 -- thinking of the induced module as $Hom_{\mathbb{Z}H}( \mathbb{Z} G, M)$. – Pierre Jul 17 '18 at 07:41

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