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I'm studying the Lyndon-Hochschild-Serre spectral sequence for $H\triangleleft G$: $$ H_p(G/H;H_q(H;A))\Rightarrow H_{p+q}(G;A) $$

where $A$ is a $G$-module.

I was told (w/o giving a proof) that the induced action of $G/H$ over $H_q(H;A)$ is the action induced by the conjugation action of $G$ over $H$. Instead it was told me to check on Weibel for the answer. Now, I know that the action of conjugation of $G$ over $H$ induces an action of $G/H$ over $H_q(H;A)$, BUT I don't see where Weibel proves that conjugation IS the action in this spectral sequence, or where this is implied.

Keep in mind that in the example right after defining this spectral sequence, Weibel implicitly use this fact:

If $H$ is in the center of $G$, $G/H$ acts trivially on $H_*(H;A)$ ....

So the explanation should be somewhere before this statement. But I can't find anything. Is is a consequence of the application of Grothendieck spectral sequence?

I really don't know where to look to find a proof of it on Weibel. Can someone point me where the author proves this fact about the induced action?

UPDATE 1: Pedro Tamaroff suggest me to have a look on Rotman and in fact I found something: This is right after the definition of the LHS-spectral sequence pic1

It refers to Lemma 9.82 which is the following (this is the cohomological version, I'll refer to the homological one pic2

So my question is the following. Is using conjugation the only way to see $H_n(S,A)$ as a $G/S$-module?? Because this is what I think is done. At least, I can't find anywhere where it is explicitly written that the action in LHS is the one by conjugation, and every reference I see seems to use this fact.

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    $\begingroup$ Check Rotman's book on Homological Algebra, the procedure is explained in detail. I don't think Weibel explains it. $\endgroup$ – Pedro Tamaroff Oct 28 '15 at 22:51
  • $\begingroup$ @PedroTamaroff I update my question according to your suggestion. could you please have a look at it? $\endgroup$ – Riccardo Oct 29 '15 at 8:18

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