If a group $G \cong H$, and if $H$ is a simple group, is $G$ also a simple group?

Question

I know that is the case, but I asked this question because I cannot find a formal way to prove it. Something is missing in my mind, I don't know what, that's why I asked the question. I don't expect a comprehensive proof, just a hint would be enough.

Answer 1

I am probably overexplaining things, but here 's a detailed argument.

Let $G$ and $K$ isomorphic groups and let $H$ normal subgroup of $G$ and suppose $\phi:G\to K$ is the isomorphism. You want to prove that $\phi(H)$ is normal subgroup of $K$. So let $x\in K$. We need to prove that $x\phi(H)x^{-1} \subset \phi(H)$. But $\phi$ surjective, so there exists $g\in G$ such that $\phi(g)=x$. Therefore $$x\phi(H)x^{-1}=\phi(g)\phi(H)\phi(g)^{-1}=\phi(gHg^{-1})=\phi(H)$$ which is what we wanted.

I leave it up to you to prove that $\phi(H)$ is proper, non trivial (note that in the above we used only the fact that $\phi$ is surjective, you will need to use injectivity to prove it's proper, non trivial). If you take the contrapositive of what we just proved, you got what you asked for.

Answer 2

Of course. If $H$ and $G$ are isomorphic, they are for all intents and purposes the same group.