# Hom$(G,A)$ $\cong$Hom $(G/G',A)$

Let $G$ is a group and $\varphi: G\longrightarrow A$ be any group homomorphism wherein $A$ is abelian. Then $$\text{Hom}(G,A)\cong \text{Hom} \left(\frac{G}{G'},A\right)$$

What we have here is, for all homomorphism $\varphi: G\longrightarrow A$:

1. $G'\trianglelefteq \ker(\varphi)\trianglelefteq G$

2. $\frac{G}{G'}/\frac{\ker(\varphi)}{G'}\cong \frac{G}{\ker(\varphi)}\leq A$

I assume $\psi:\frac{G}{G'}\longrightarrow A$ be an arbitrary homomorphism. Just for sure; if $$\Phi: \text{Hom}(G,A)\longrightarrow \text{Hom} (\frac{G}{G'},A)$$ with $\Phi(\varphi)=\psi, \varphi(g)=\psi(gG'), g\in G$ is a proper choice for getting our result? Thanks.

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Yes, that's right. –  Zhen Lin Sep 2 '12 at 11:47
\ker is a command supported by MathJax. In fact \hom is supported as well, but that would be too far of an edit. –  Asaf Karagila Sep 2 '12 at 12:19
@AsafKaragila: Thanks Asaf for noting me that. I'll try to use them as you said. –  Babak S. Sep 2 '12 at 12:25

You know that the commutator subgroup $G'$ is the smallest normal subgroup (with respect to inclusion) in $G$ such that $G/G'$ is abelian. Now I think your choice of $\Phi$ is the correct one: By definition of the commutator subgroup, given any homomorphism $\varphi$ to an abelian group $A$ we have that $G/\ker \varphi \cong$ (some subgroup of an abelian group) that is abelian, from which it follows by the universal property of quotients that we always get a unique group homomorphism

$$\psi : G/G' \longrightarrow A.$$

such that $\psi \circ \pi = \varphi$, where $\pi : G \longrightarrow G/G'$. Now define $\Phi : \hom(G,A) \longrightarrow \hom(G/G',A)$ by $\Phi(\varphi) = \psi$. The uniqueness of $\psi$ guarantees that $\Phi$ is well-defined. To show that $\Phi$ is an isomorphism, you can define an inverse

$$\begin{eqnarray*} \Psi:&\hom(G/G',A)&\to \hom(G,A)\\ &f& \mapsto f \circ \pi.\end{eqnarray*}$$

It is clear that $\Psi$ is well defined. We now check that it is the inverse of $\Phi$. We have given any $f \in \hom(G/G',A)$ that

$$\Phi\circ \Psi(f) = \Phi(f \circ \pi) = f$$

where the last step follows because of the following. $\Phi(f \circ \pi)$ is supposed to be the unique linear map $g : G/G' \to A$ such that $g \circ \pi = f \circ \pi$, clearly setting $f = g$ works and so by uniqueness it follows that $\Phi(f \circ \pi) = f$. Since this holds for all $f \in \hom(G/G',A)$ we conclude that

$$\Phi \circ \Psi = \textrm{id}_{\hom(G/G',A)}.$$

a similar computation shows that $\Psi \circ \Phi$ is the identity on $\hom(G,A)$ from which it follows that $\Psi$ and $\Phi$ are mutual inverses.

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@Thanks for your help. –  Babak S. Sep 2 '12 at 11:56
@BabakSorouh No problem mate! –  user38268 Sep 2 '12 at 11:57