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

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?

• The definition of "two groups are isomorphic" is precisely "there exists an isomorphism between the two groups." Apr 19, 2019 at 5:40
• @angryavian yes. It's possible OP is thinking that two groups are isomorphic only if every homomorphism is an isomorphism, but that's not the case. Apr 19, 2019 at 5:43
• I see what is going on. This has led me to think about another question originally posted here: math.stackexchange.com/questions/1028605/…. How does providing a counter-example (as both the answers have done) prove that the two quotient groups are not isomorphic, in general? Couldn't it be possible that there exists an isomorphism? Apr 19, 2019 at 5:50

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.