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?
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5$\begingroup$ The definition of "two groups are isomorphic" is precisely "there exists an isomorphism between the two groups." $\endgroup$– angryavianApr 19, 2019 at 5:40
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2$\begingroup$ @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. $\endgroup$– rb612Apr 19, 2019 at 5:43
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1$\begingroup$ 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? $\endgroup$– Four SeasonsApr 19, 2019 at 5:50
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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.