$H^n(S,A)$ as a $G/S$ module given $S\lhd G$: non-unique $G/S$ actions?

56 Views Asked by At

My question is about a lemma in Joseph J. Rotman's textbook, An Introduction to Homologyical Algebra (2nd Edition).

Let $G$ be a group and let $A$ be a $G$-module. In his Lemma 9.82(ii) on P567, he states that

If $S$ is a normal subgroup of $G$, then $H^n(S,A)$ is a $G/S$-module. Moreover, if $A$ is a trivial $G$-module, then $H^n(S,A)$ is a trivial $G/S$-module.

I understand Rotman's construction (see P564-568) that makes $H^n(S,A)$ into a $G/S$-module, and indeed, according to his construction, if $A$ is a trivial $G$-module then $H^n(S,A)^{G/S} = H^n(S,A)$, i.e. $H^n(S,A)$ is a trivial $G/S$-module. However, if we alternatively define the action of $g \in G$ on $H^n(S,A)$ as

$$(g(f))(g_1,g_2,...,g_n) = g.f(g^{-1}g_1g,g^{-1}g_2g,...,g^{-1}g_ng)$$

for any cocycle $f\colon G\times G\times\cdots \times G \rightarrow A$ (really, $f$ is a representative cochain), where $g.a$ denotes the action of $g\in G$ on $a\in A$, as is defined in Ehud Meir's answer in https://mathoverflow.net/questions/212636/the-term-h1n-ag-n-in-the-inflation-restriction-exact-sequence, this seem to imply that $H^n(S,A)$ can be a nontrivial $G/S$-module even in the case of $A$ being a trivial $G$-module.

To me, this seems to imply that the $G/S$-action on $H^n(S,A)$ is not uniquely defined. If this is true, then these two $G/S$-actions would lead to to different transgression maps and hence different Lyndon-Hochschild-Serre spectral sequences; these of course should not alter the cohomology of the total group $H^n(G,A)$.

To summarize my question: given a group $G$, a $G$-module $A$, and a normal subgroup $S\lhd G$. Is it true that the $G/S$-action on $H^n(S,A)$ is not uniquely defined?