(When) is this homomorphism an isomorphism?

Question

I am interested in the following natural question:

Suppose that we have an onto homomorphism $\phi: G \to \bar{G}$, and that we know that, for some (normal) subgroup $H$ of $G$, $H \cong \phi(H)$ and $G/H \cong \bar{G}/\phi(H)$. Is $G \cong \bar{G}$?

I think a counter-example should exist, but cannot find one. So I am interested in:

1. a counter-example, and
2. if a counter-example does exist, are there any additional restrictions we can impose that make this true?

Answer 1

Following Geoffrey's comment, if the isomorphism is induced by $\phi$ in both cases, we have exact sequences

$$ 1 \to H \to G \to G/H \to 1 $$

and

$$ 1 \to \phi(H) \to \widetilde{G} \to \widetilde{G}/\phi(H) \to 1. $$

The trivial map $1\to 1$, $\phi$, its restriction and induced morphism between the quotients gives a commutative diagram with exact rows.

By hypothesis, all vertical maps but $\phi$ are isomorphisms, so by the five lemma $\phi$ is also an isomorphism. This can also be proved by an elementary diagram chase (which is essentially the proof of the five lemma made easier, since two of the maps are trivial).

Edit: note that, as also observed in the comments, for the quotient by $\phi(H)$ to make sense one should know this subgroup is normal. A sufficient condition would be to assume $\phi$ to be epi.

Answer 2

Here is a counter-example from the world of infinite groups:

Let $F(X)$ denote the free group over the alphabet $X$. Consider the natural homomorphisms $\phi_1:F(a,b,c)\rightarrow F(a,b)$ and $\phi_2:F(a,b)\rightarrow F(a)$. Define $H:=\ker\phi_2\phi_1$, so $\phi_1(H)=\ker(\phi_2)$. Hence, $F(a,b,c)/H\cong F(a,b)/\phi_1(H)$. Now, non-trivial normal subgroups of infinite index in free groups are non-finitely generated free groups, and so are all isomorphic. Hence, $H\cong\phi_1(H)$. This means that the conditions you wish hold. However, $G=F(a,b,c)\not\cong F(a,b)=\overline{G}$.