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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.

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

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.

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    $\begingroup$ Thanks for the example! I worked through it and it makes sense! $\endgroup$ – user138798 Oct 22 '14 at 16:05
  • $\begingroup$ Awsome! Glad to help. $\endgroup$ – Paul Plummer Oct 23 '14 at 5:15

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