I have this question for the proof of a lemma I was reading.
Let $G$ be an abelian group and $M$ a $G$-module with that I have the question about this part:
"Since e $G$ is an abelian group it's well known that for any $g\in G$, the map $g:M\rightarrow M$ induces identity homomorphism on all homology groups."
I assume that the map from above it is the multiplication by $g$. Now I was trying the next thing (The idea is from the chapter 3, section VIII of Kenneth Brown Group Cohomology).
Since $G$ is abelian the map $\alpha=id:G\rightarrow G$ satisfy that the map $g:M\rightarrow M$ is compatible with $\alpha$; i.e. $g(hm)=\alpha(h)g(m)$. Now if $F$ is a the bar resolution of $\mathbb{Z}$ over $\mathbb{Z}G$, the map $\tau:F\rightarrow F$ given by $\tau(x)=g^{-1}x$ is a chain map compatible with $\alpha$.
Therefore I have the chain map $\tau\otimes g:F\otimes_{G} M\rightarrow F\otimes_{G} M$ given by $x\otimes m\mapsto g^{-1}x\otimes gm$.
We have that $F$ is also a right $G$-module but since $G$ is abelian, we have that $gx=xg$ for all $x\in F$ and $g\in G$. Thus $g^{-1}x\otimes gm=xg^{-1}\otimes gm=x(g^{-1}g)\otimes m=x\otimes m$
Thus this chain map is the identity and therefore induces identity homomorphism on all homology groups.
That's the only idea I could think, Is that correct, specially the last part where I assume that $F$ is also a right module.
Any suggestions are welcome.
Tensoring the inhomogeneous bar resolution with M we obtain a sequence of abelian groups:$$\cdots \stackrel{d}\to M^{G\times G\times G} \stackrel{d}\to M^{G\times G} \stackrel{d}\to M^{G} \stackrel{d}\to M\to 0. $$
For $m\in M$ and $g_1,g_2,\cdots,g_k\in G$, let $[g_k,g_{k-1},\cdots,g_1,m]\in M^{G^k}$ be the element with co-ordinate $m$ on the summand $M$ corresponding to $(g_1,\cdots,g_k)$ and all other co-ordinates $0$. Then the above differentials are given by: $$d\colon[g_k,g_{k-1},\cdots,g_1,m]\mapsto $$ $$ [g_{k-1},\cdots,g_1,m]-[g_kg_{k-1},\cdots,g_1,m]+[g_k,g_{k-1}g_{k-2},\cdots,g_1,m]-\cdots +(-1)^k[g_k,g_{k-1},\cdots,g_1m] $$ The maps induced by multiplication by $h\in G$ are given by: $$h_*\colon[g_k,g_{k-1},\cdots,g_1,m]\mapsto[g_k,g_{k-1},\cdots,g_1,hm].$$ These are isomorphisms of abelian groups, regardless of $G$ being abelian. However, if $h$ is central in $G$ then $h_*$ is a chain map. That is, if $h$ is central in $G$ then $h_*d=dh_*$ as $hg_1=g_1h$.
We define $I\colon M^{G^k}\to M^{G^{k+1}}$ for each $k\geq 0$ by: $$ I\colon [g_k,g_{k-1},\cdots,g_1,m]\mapsto $$ $$ [h,g_k,g_{k-1},\cdots,g_1,m]-[g_k,h,g_{k-1},\cdots,g_1,m]+\cdots+(-1)^k[g_k,g_{k-1},\cdots,g_1,h,m]. $$
Again assume that $h$ is central in $G$. The key result is: $$dI+Id=1-h_*.\qquad\qquad[1]$$
To see this, consider $(dI+Id)[g_k,g_{k-1},\cdots,g_1,m]$ and note that all terms of $Id [g_k,g_{k-1},\cdots,g_1,m]$ are cancelled out by the corresponding term in $dI[g_k,g_{k-1},\cdots,g_1,m]$.
The remaining terms in $dI[g_k,g_{k-1},\cdots,g_1,m]$ of the form $\pm [g_k,g_{k-1},\cdots,hg_r,\cdots,g_1,m]$ and $\mp [g_k,g_{k-1},\cdots,g_rh,\cdots,g_1,m]$ cancel out, as $h$ is central in $G$. This leaves the two desired terms:$$(dI+Id)[g_k,g_{k-1},\cdots,g_1,m]=[g_k,g_{k-1},\cdots,g_1,m]-[g_k,g_{k-1},\cdots,g_1,hm],$$ and we have proven $[1]$.
Finally note that if $z$ is a cycle, then $dz=0$ so by $[1]$: $$z-h_*z=dIz+Idz=dIz,$$ which is a boundary. Thus $z$ and $h_*z$ represent the same element in homology.