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.

  • 1
    $\begingroup$ Are you assuming that $\phi$ is a surjective homomorphism? $\endgroup$ – cws Oct 22 '14 at 4:20
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.