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


*

*$\pi(K)$ 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.

*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 $\pi$

*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.

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