How to prove $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)^{S}\cong\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}(G/S),A)$?

Question

Given that S is a normal subgroup of a finite group G and A is a G-module, I have difficulty in proving $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)^{S}\cong\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}(G/S),A)$. Would anyone please tell me how to deal with this? I appreciate any help.

Answer 1

First, note that $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)^S=\{\phi\in\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)\mid s.\phi=\phi\}$. Now, the action of $S$ on $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)$ is $$(s.\phi)(g)=\phi(gs)$$ so $\phi\in\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)^S$ if, and only if, it is constant on the cosets of $S$ in $G$ (i.e. $\phi(g)=\phi(g')$ whenever $gS=g'S$).

This means that the map $$ \Xi:\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)^S\to\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}(G/S),A) $$ given by $$\Xi(\phi)\left(\sum_{gS\in G/S}a_{gS}\,gS\right)=\sum_{gS\in G/S}a_{gS}\,\phi(g)$$ is well defined.

It is a straightforward computation to see that the map $$ \Xi^{-1}:\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}(G/S),A)\to\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)^S $$ given by $$\Xi^{-1}(\psi)\left(\sum_{g\in G}b_{g}\,g\right)=\sum_{g\in G}b_{g}\,\phi(gS)$$ is indeed the inverse of $\Xi$. In one direction we have \begin{align*} (\Xi^{-1}\circ\Xi)(\phi)\left(\sum_{g\in G}b_{g}\,g\right)&=\sum_{g\in G}b_g \Xi(\phi)(gS)\\&=\sum_{g\in G}b_g\phi(g)\\&=\phi\left(\sum_{g\in G}b_{g}\,g\right). \end{align*} The other direction is similar.