[a repost from SE due to the lack of response]

Given a group $G$, let $A$ be a $G$-module and let $N\trianglelefteq G$. If I understand it correctly, the superscript "G/N" in the third term of the standard inflation-restriction exact sequence $$ 0\to H^1(G/N,A^N)\to H^1(G,A)\to H^1(N,A)^{G/N}\to H^2(G/N,A^N)\to H^2(G,A) $$ means the fixed points under the action of $G/N$ on the first cohomology group $H^1(N,A)$. But how is this action defined? Is there an elementary definition (in terms of cocycles) that does not refer to the Lyndon–Hochschild–Serre spectral sequence?

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    $\begingroup$ When in doubt, consult Ken Brown's Cohomology of Groups! $\endgroup$ Jul 30, 2015 at 16:17

1 Answer 1


You do not need LHS spectral sequence for this action. The functors $H^*(N,-)$ are the derived functors of $(-)^N:G\text{-mod}\rightarrow G/N\text{-mod}$, so they will carry a structure of $G/N$-modules. Explicitly, on the level of cocycles, it can be described as follows: if you have a one cocycle $f:N\rightarrow A$ and $g\in G$, then the action on $f$ will be given by $g\cdot(f)(n) = g\cdot f(g^{-1}ng)$

  • $\begingroup$ For this to be an action of $G/N$, $N$ must act trivially modulo $B^1(N,A)$. I do not see why this holds in general. Suppose $g\in N$. Then $(gf-f)(n)=g\cdot f(g^{-1}ng)-f(n)$. Why is this always of the form $ga-a$ for some $a\in A$? (Note that it is not assumed that $N$ is abelian. Otherwise $a=f(n)$, of course.) $\endgroup$
    – Anvita
    Jul 30, 2015 at 13:22
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    $\begingroup$ Assume that $g\in N$. Then by using the cocycle condition we have $gf(g^{-1}ng) - f(n) = gf(g^{-1}) + f(n) + nf(g) - f(n) = gf(g^{-1}) + nf(g)$. Now, since $f(gg^{-1}) = 0$ we have $f(g) + gf(g^{-1})) = 0$ so $gf(g^{-1}) = -f(g)$. This implies that the difference is $nf(g) - f(g)$ so we can take $a=f(g)$. $\endgroup$
    – Ehud Meir
    Jul 30, 2015 at 13:47

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