The behavior of quotient groups under homomorphisms

We're learning normal subgroups, kernels, homomorphisms and isomorphisms in abstract algebra right now. I'm trying to tie the ends together:

I know that if $G$ is a group, $N$ a normal subgroup of $G$, and $\phi: G\to G′$ is a homomorphism then $\phi(N)$ is a normal subgroup of $G′$.

But can I say that quotient group $G/N$ is isomorphic to $G'/ \phi(N)$?

Let's assume that $\phi$ is a surjective homomorphism.

• Are you assuming that $\phi$ is a surjective homomorphism? – cws Oct 22 '14 at 4:20
• Yes, come to think of it -- that probably makes sense since a trivial or non-surjective homomorphisms would clearly make the statement false. – user138798 Oct 22 '14 at 4:27

Consider the homomorphism $\phi:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ defined by $(1,0),(0,1) \mapsto 1$ and note that $\langle (1,0) \rangle$ is a proper normal subgroup of $\mathbb{Z} \times \mathbb{Z}$. It is clear that $\phi \langle (1,0) \rangle=\mathbb Z$ since $(x,0) \mapsto x$. But $$\frac{\mathbb Z \times \mathbb Z}{\langle (1,0) \rangle} \cong \mathbb Z \not \cong 1 \cong \mathbb Z /\mathbb Z.$$
You can actually just consider $N=G' \times \{1_{G'}\}$ and $G=G' \times G'$ then do essentially the same coordinate mapping I did in the integer example (and have $G'$ be nontrivial) for a class of examples.
Note that if $\phi: G \to G'$ is an isomorphism then it is true. See this question.