Group isomorphism induces a homomorphism on the quotient group?

I have the following problem:

Let $\phi:G\to H$ be an isomorphism of groups, and let $N$ be a normal subgroup of $G$. Then $\phi(N)$ is a normal subgroup of $H$. Prove that $\phi$ induces an isomorphism such that $G/N\simeq H/\phi(N)$.

How would one go about this problem? Is the first isomorphism theorem handy here? Thanks for any help!

• You can see that $\phi$ induces a homomorphism $G\to H/\phi(N)$. Then you can find its kernel and use the isomorphism theorem – Maxime Ramzi Feb 19 '17 at 8:34
• How exactly would $\phi$ induce a homomorphism on $G\to H/\phi(N)$? – Sir_Math_Cat Feb 19 '17 at 8:35
• It's the composition of $\phi$ with the canonical homomorphism $H \to H/\phi(N)$. – Derek Holt Feb 19 '17 at 8:50

Let us solve the problem in a more general setting:

if $\phi :G \to H$ is a surjective group morphism and $N \unlhd G$, and if $\ker \phi \subseteq N$, then $G/N \simeq H/\phi(N)$.

The hypothesis $\ker \phi \subseteq N$ is essential. Without it, the statement is no longer true in general, as is shown by the example $N = \{e\}$: you would get $G \simeq H$, which is not necessarily true. Another counterexample is given by $\phi(g) = e \ \forall g \in G$ and $N \ne G$ arbitrary: in this case, the conclusion of the problem would be $G/N \simeq H$ which again is not true in general, because the left-hand side depends on the arbitrary $N$ while the right-hand side does not.

Surjectivity is essential in order to have $\phi(N) \unlhd H$ (without surjectivity, you only have $\phi(N) \le H$ in general, even though $N \unlhd G$).

The solution is standard algebraic manipulation, and your intuition is correct, the first isomorphism theorem will do the job. In order to apply it, we'll have to construct a group morphism $f : G \to H/\phi (N)$ having $\ker f = N$. Since we don't have too many ingredients in our problem, we don't really have a choice: if $p : H \to H/\phi (N)$ is the natural projection $h \mapsto p(h) = \hat h$, then define $f = p \circ \phi$. Being a composition of morphisms, it itself will be a morphism.

To find $\ker f$, consider the sequence of equivalences

$$g \in \ker f \iff f(g) = \hat e \iff p(\phi (g)) = \hat e \iff \phi (g) \in \phi (N) \iff \exists n \in N \text{ such that } \\ \phi(g) = \phi(n) \iff \phi(gn^{-1}) = e \iff gn^{-1} \in \ker \phi \subseteq N \iff g \in Nn \subseteq N \iff g \in N$$

which shows that $\ker f = N$, and now the first isomorphism theorem immediately gives you $G/N \simeq H/\phi(N)$.