First here is my setup:

Let $W$ be some group, and $C$ a normal subgroup of finite index, and let $W/C=G$. Now let $L$ be a a $G$-module on which $C$ acts trivially, so in particular we get on action of $W$ on $L$.

Now from the Lydon-Hochschild-Serre spectral sequence for group homology we get a sequence

$$ H_{2}(G,L) \overset{d}\longrightarrow H_{1}(C,L)_{G} \overset{cor}\longrightarrow H_1(W,L) \overset{coinf}\longrightarrow H_{1}(G,L) \longrightarrow 0$$

Now also from the definition of Tate-cohomology groups we have the following exact sequence

$$0 \longrightarrow \widehat{H}^{-1}(G,H_{1}(C,L)) \overset{d'}\longrightarrow H_{1}(C,L)_{G} $$

$$\rightarrow H_{1}(C,L)^{G} \longrightarrow \widehat{H}^{0}(G,H_{1}(C,L)) \rightarrow 0 $$

Where we can use that since $C$ acts trivially on $L$, then $H_{1}(C,L) \cong C \otimes L$ and we can use this to give $H_{1}(C,L)$ a $G$-action.

Now using cup products one can use the Tate-Nakayama Lemma to get an isomorphism $H_{2}(G,L) \cong \widehat{H}^{-1}(G,C \otimes L ) $

(Now I cant seem to put a diagram here so ill explain this in words)

I want to show that the composition of maps $$H_{2}(G,L) \cong \widehat{H}^{-1}(G,C \otimes L ) \overset{d'}\rightarrow H_{1}(C,L)_G $$ is the same as

$$H_{2}(C,L) \overset{d}\longrightarrow H_{1}(C,L)_G $$

i.e. that there is a commutative triangle there, but I cant seem to prove this. I can actually join up both the sequences mentioned above using transfer maps and cup products and I can prove that the second and third squares commute, but Its this first square which im having trouble with.

I hope this makes sence, the problem is I cant seem to draw the commutative diagrams I want.

Thank you

First here is my setup:

Let $W$ be some group, and $C$ a normal subgroup of finite index, and let $W/C=G$. Now let $L$ be a a $G$-module on which $C$ acts trivially, so in particular we get on action of $W$ on $L$.

Now from the Lyndon-Hochschild-Serre spectral sequence for group homology we get a sequence

$$ H_{2}(G,L) \overset{d}\longrightarrow H_{1}(C,L)_{G} \overset{cor}\longrightarrow H_1(W,L) \overset{coinf}\longrightarrow H_{1}(G,L) \longrightarrow 0$$

Now also from the definition of Tate-cohomology groups we have the following exact sequence

$$0 \longrightarrow \widehat{H}^{-1}(G,H_{1}(C,L)) \overset{d'}\longrightarrow H_{1}(C,L)_{G} $$

$$\rightarrow H_{1}(C,L)^{G} \longrightarrow \widehat{H}^{0}(G,H_{1}(C,L)) \rightarrow 0 $$

Where we can use that since $C$ acts trivially on $L$, then $H_{1}(C,L) \cong C \otimes L$ and we can use this to give $H_{1}(C,L)$ a $G$-action.

Now using cup products one can use the Tate-Nakayama Lemma to get an isomorphism $H_{2}(G,L) \cong \widehat{H}^{-1}(G,C \otimes L ) $

(Now I cannot seem to put a diagram here so I'll explain this in words.)

I want to show that the composition of maps $$H_{2}(G,L) \cong \widehat{H}^{-1}(G,C \otimes L ) \overset{d'}\rightarrow H_{1}(C,L)_G $$ is the same as

$$H_{2}(C,L) \overset{d}\longrightarrow H_{1}(C,L)_G $$

i.e. that there is a commutative triangle there, but I cannot seem to prove this. I can actually join up both the sequences mentioned above using transfer maps and cup products and I can prove that the second and third squares commute, but it's this first square which I am having trouble with.

I hope this makes sense, the problem is I cannot seem to draw the commutative diagrams I want.

Thank you