2
$\begingroup$

I need some help in proving this result from Weibel or Brown (page $48$ corollary $6.3$).

If $G$ is a group and $N$ a normal subgroup of it, then the conjugation action of $G$ on $N$ induces an action of $G/N$ on $H_{\ast}N$.

My Attempt Now abstractly I think I know how to get an action of $G/N$ on $H_{\ast}N$. Take a $G$-proj. resolution of $\mathbb{Z}$ and see it as an $N$-proj. resolution. Clearly, $G$ acts on it, and by definition of $H_*(N)$ we take the $N$ coinvariants, so clearly the action factor through $G/N$.

BUT, I really never used the fact that $G$ acts on $N$, nor the preceding result (this result should be an immediate corollary)

Fix $g_0 \in G$, and let $\alpha \colon G \to G$ be given by $\alpha(g)=g_0gg_0^{-1}$. then $\alpha_* \colon H_*G \to H_*G$ is the identity.

Therefore I feel that I'm overlooking something, in particular I'm not so convinced that taking the $N$-coinvariants with respect to the $G$ action would lead to the canonical action of $G/N$ on the homology.

Therefore could someone show me what is this induced action, and where this conjugation and preceding result comes into play in this corollary?

N.B This question is clearly related to this one. The reason why I decided to ask a separate question is that I have specific doubts about its proof , I want some comments on my approach and the linked question is pretty old.

$\endgroup$
1
$\begingroup$

The question was not to construct some action, but that specifically the conjugation action gives rise to the action. Choosing a $G$-projective resolution of $\mathbb{Z}$ and viewing it as $N$-projective, you can use the homomorphism $\alpha_{g_0}:N\rightarrow N$ given by $\alpha_{g_0}(n)=g_0ng_0^{-1}$ and the chain map $\tau_{g_0}(x)=g_0x$ as before to obtain a homomorphism $G\rightarrow \text{End}(H_*N)$ and that conjugation by $N$ is the identity on $H_*N$ to get the $G/N$-action.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.