Prove that the only homomorphism between a simple non-abelian group G and abelian group A is trivial

Question

Prove that the only homomorphism between a simple non-abelian group $G$ and abelian group $A$ is trivial.

OK. So G is a perfect group (G' = G) and A is abelian (A' = {1})

Answer 1

If we name the homomorphism $\varphi: G \to A$ then the kernel of $\varphi$ must be a normal subgroup of $G$. So, since $G$ is simple and $\ker{\varphi}$ is normal in $G$ two cases can happen: $\ker{\varphi}=\{e\}$ or $\ker{\varphi}=G$.

If $\ker{\varphi}=\{e\}$ then $\varphi$ is one-to-one and the image of $\varphi$ would be a subgroup of $A$ which is isomorphic to $G$. But since $A$ is Abelian all of its subgroups must be Abelian too, while $G$ is non-Abelian. Therefore it can't be isomorphic to any subgroup of $A$. This forces $\ker{\varphi}=G$. Which means that $\forall g \in G: \varphi(g)=0$ and $\varphi$ is the trivial homomorphism.

Answer 2

Let $g \in G$ since $G'=G$ then exist $g_1, g_2 \in G$ such that $g_1g_2g_1^{-1}g_2{^-1} = g$.

Then we get : $f(g) = f(g_1g_2g_1^{-1}g_2^{-1}) = f(g_1)f(g_2)f(g_1^{-1})f(g_2^{-1}) ∈ A' = \{e\}$