$G \cong H$ and $G$ is simple. Then $H$ is simple as well.

Question

I think it must be true. Yet I have no rigorous proof for that. So, what I need to prove is that "group being simple" is invariant under isomorphism.

That if $G \cong H$, then either both are simple groups, or both are not simple.

Answer 1

Hint: Let $f:H\to G$ be an isomorphism and let $K\subseteq H$ be a normal subgroup. What can you say about $f(K)\subseteq G$?

Answer 2

Recall a group is simple if it doesn't have a normal subgroup other than itself and the identity group, or put another way "no non trivial normal subgroup".

So a natural proof by contradiction arises.

Since $G \cong H$ we can consider an isomorphism $\pi: H \rightarrow G$.

Suppose now that $H$ has a non trivial normal subgroup $K$. we will show that $\pi(K)$ must be a non trivial normal subgroup of $G$.

1. $\pi(K)$ has the same number of elements as $K$ since $\pi$ is an bijection (since isomorphisms are bijections), so if it is a subgroup of G, then it must be a non trivial subgroup.

2. This question: Generalized Isomorphism Theorem for Groups, shows that the images of normal groups of $H$ must be normal subgroups of $G$ under the isomorphism $\pi$

3. We combine both, to conclude that $\pi(K)$ is a normal subgroup of G, AND is a non trivial subgroup (so it isn't the whole group or the identity), thus it MUST be a non trivial normal subgroup, a contradiction if we started by assuming $G$ is simple.

4. Thus we conclude that if $G$ is simple and $G \cong H$ then it must be the case that $H$ is simple.