Why is $\operatorname{Hom}_{\mathrm{Groups}}(G,A)$ isomorphic to $\operatorname{Hom}_{\mathrm{Ab}}(G/[G,G],A)$?

Question

This question is inspired by an exercise from the Weibel's book on Homological Algebra (beginning of chapter 6 on Group Cohomology).

Let $G$ be a group and $A$ be a $G$-module. My question simply is:

Why are the groups $\operatorname{Hom}_{\mathrm{Groups}}(G,A)$ and $\operatorname{Hom}_{\mathrm{Ab}}(G/[G,G],A)$ isomorphic?

Any hint will be particularly appreciated. Thank you

Answer 1

Any group homomorphism from a group $G$ to an abelian group $A$ must factor through the abelianization of $G$, that is, the quotient $G/[G,G]$.

In other words, if $G$ is a group and $A$ is an abelian group, then for any group homomorphism $f:G\to A$, there is a unique group homomorphism $\varphi$ from the abelian group $G/[G,G]$ to the abelian group $A$ such that $f=\varphi\circ q$, where $q:G\to G/[G,G]$ is the quotient map.

(In your situation, $A$ is an abelian group with extra structure, namely, a $G$-action. However this does not affect facts about group homomorphisms to $A$.)

The isomorphism $\operatorname{Hom}_{\mathrm{Groups}}(G,A)\longrightarrow\operatorname{Hom}_{\mathrm{Ab}}(G/[G,G],A)$ sends $f$ to its corresponding $\varphi$.

(Relevant Wikipedia entry)