Given $H \leq G$, prove isomorphism of $H$ and $gHg^{-1}$

Question

All I am given is that $H \leq G$ and I have to prove that $H \cong gHg^{-1}$. I first verified that $gHg^{-1} \leq G$. Then, I tried the map $f(h) = ghg^{-1}$ and it indeed turned out to be an isomorphism. However, I was wondering if there is a possibility that someone could come up with a map between those two groups which is not an isomorphism. Putting it in other words, how come I can conclude that the two groups are isomorphic in general, just because I happened to find "a map" that turned out to be an isomorphism?

Answer 1

Just because two groups are isomorphic doesn't mean that all homomorphisms between them are isomorphisms. Just consider the trivial homomorphism, $h:G\to G'$ given by $h(g)=e\,\forall g\in G$, for nontrivial $G, G'$.

Furthermore, in the linked answer it is shown that we do not always have $G_1/H_1\cong G_2/H_2$ when certain conditions are met. To show something is not always true you just need one counterexample.