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Given a group extension:

$$ 0 \rightarrow N \rightarrow G \rightarrow \frac{G}{N} \rightarrow 0 $$

I need to show that the induced action of $G$ by conjugation is trivial on the Hochschild-Serre spectral sequence.

$\textbf{Attempt 1}$: So far from brown I know that given a $G$-module M, we can have $G$ acting by conjugation, for a $g \in G$ we get: $$ (\phi_g, \alpha_g) : (G,M) \rightarrow (G,M) $$ acting with $\phi_g(x) := gxg^{-1}$ and $\alpha_g(m) := gm$. We know this induces the trivial map on $H_n(G, M)$ but I need to show it induces the trivial map on $$ E_{i,j}^2 = H_i(G/N, H_j(N, M))$$ which I've been really struggling to do. I know that to induce a map on this homology I need a pair of maps $$ (\psi_g, \beta_g) : (G/N, H_j(N, M)) \rightarrow (G/N, H_j(N,M))$$ and I guess that $\psi_g$ will be conjugation as above. But I dont know how to interpret the map $\beta_g$. I know it also needs the property that $\beta_g(ab) = \psi_g(a)\beta_g(b)$ for $a \in G/N$ and $b \in H_j(N,M)$. I also feel like using the usual bar resolution should make this easier to calculate by hand at some later stage but I just cant make progress at the moment.

$\textbf{Attempt 2}$: An alternate route I thought was that $g \in G$ gives an automorphism of the above group extension, which then of course extends to the long exact sequence on homology and then from there this is what we can use to get the exact couple that hochschild-serre comes from but this is only the 1'st page and it seems incredibly messy to track it then through to the derived couple.

Anyone have any ideas how to show this induced action is trivial?

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    $\begingroup$ You should state more clearly what you want to prove. The induced action of what on what? $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '18 at 17:16
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    $\begingroup$ In what way does G act on the spectral sequence? $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '18 at 17:41
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    $\begingroup$ One thing is to have an action of the groups that appear on the second page and another, quite more elaborate, is an action the whole spectral sequence... Do you have a action on H(N,M)? $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '18 at 17:48
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    $\begingroup$ I'd say that making sure of that claim would be a very good step! $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '18 at 18:00
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    $\begingroup$ Well, unless you are able to make precise in what exact way an action on the group extension induces an action on the spectral sequence, that's just hand waving, nothing more than a what-could-possibly-go-wrong argument which is, as you know, entirely useless. $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '18 at 18:23
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We know for any group $G$ and $M$ a $G$-module then $G$ acts by conjugation on iteself and $g\cdot m$ on $M$ induces the trivial action on $H_n(G, M)$. [see Browns cohomology of groups book]

In particular since we can make $H_j(N,M)$ into a $G/N$-module by this same action when $N\trianglelefteq G$ we get by the first point that $G/N$ acts trivially on $H_i(G/N, H_j(N, M))$ by the action described above (conjugation and left multiplication).

Hence for a $g \in G$ we can map $G \rightarrow G/N$ and then act on $H_i(G/N, H_j(N,M))$ in order to get an action of $G$ on $H_i(G/N, H_j(N,M))$ which is trivial since the $G/N$ action was trivial.

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    $\begingroup$ You write «Since we can make $H_j(N,M)$ into a $G/N$-module...». How exactly can you do that? For that you need an action of $G$ on $H_j(N,M)$, and that is precisely what my last commeent wanted you to think about. So: how does $G$ act on $H_j(N,M)$ $\endgroup$ – Mariano Suárez-Álvarez Feb 25 '18 at 16:08
  • $\begingroup$ I know from Brown's book: G acts on the pair (N,M) by conjugation in the first argument and left mult in the second inducing an action on H_j(N,M). It can then be shown that restricting this to N acting on (N,M) the action on homology becomes trivial and hence we can quotient to get a G/N action on H_j(N,M). Is that what you were asking for me to think through? $\endgroup$ – marineabcd Feb 27 '18 at 10:28

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