Why is $\hom(G \times H, A) \cong \hom(G,A) \times \hom(H,A)$?

Question

Let $A$ be abelian group and $G$ and $H$ be groups. Why does the following hold?

$$\hom(G \times H, A) \cong \hom(G,A) \times \hom(H,A)$$

I tried my best and couldn't come up with an isomorphism. Any help?

This claim has been made here.

Answer 1

Let $\phi$ be a homomorphism from $G \times H$ to $A$. Then you get a homomorphism from $G$ to $A$ by sending $g$ to $\phi(g,1_H)$. You also get another homomorphism from $H$ to $A$ by sending $h$ to $\phi (1_G,h)$. Thus given the homomorphism $\phi$, you get a pair of homomorphisms, one from $G$ to $A$, and another from $H$ to $A$. This is your function $$\textrm{hom}(G\times H, A) \rightarrow \textrm{hom}(G,A) \times \textrm{hom}(H,A)$$ which you need to show is a bijection.

Answer 2

Consider$$\begin{array}{rccc}\Phi\colon&\hom(G\times H,A)&\longrightarrow&\hom(G,A)\times\hom(H,A)\\&\varphi&\mapsto&\left(\left(\begin{array}{ccc}G&\longrightarrow&A\\g&\mapsto&\varphi(g,e)\end{array}\right),\left(\begin{array}{ccc}H&\longrightarrow&A\\h&\mapsto&\varphi(e,h)\end{array}\right)\right).\end{array}$$