$G\times H$ is not a simple group for $|G|, |H|\geq 2$

Question

I would appreciate if you could please express your opinion about my proof and maybe give me a hint where you deem suitable.

Proof:

Define a homomorphism as follows $\phi: (G\times H)\to G$ by $\phi[(g,h)]=g$. We then show that this homomorphism is well-defined (omitted).

Now, for any $(e, h) \in G\times H$, $\phi[(e,h)]=e$. Therefore, $\ker\phi$ is not trivial. But $\ker\phi$ is a normal subgroup of $G\times H$. Therefore, $G\times H$ is not simple.

Answer 1

Note that both $H,G$ are normal in $H\times G$: $$(h,g)(h',1)(h^{-1},g^{-1}) = (hh'h^{-1},gg^{-1})=({}^hh',1)\in H\times 1$$The same works for $1\times G$.