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I wish to prove whether this is true or false.

If $\phi: G \rightarrow H$ is a group homomorphism, $N \vartriangleleft G$, then $G/N \cong \phi(G)/\phi(N)$.

I'm not even sure if $N$ being normal in $G$ implies that $\phi(N)$ is normal in $H$, but I can't think of an immediately obvious counter example.

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It is true that $\phi (N)$ will be normal in $\phi(G)$ (though not in $H$ itself!), and thus $\phi(G)/\phi(N)$ will be a group (to see this, use the fact that $\phi$ is a group homomorphism).

However, the claim itself is false; take $N \leq G$ and $\phi$ to be the trivial map, so $\phi(G)/\phi(N)$ is trivial but $G / N$ is not.

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  • $\begingroup$ What do you mean by $\phi(G)/\phi(N)$ being trivial? $\endgroup$ – Irregular User Jun 7 '16 at 22:58
  • $\begingroup$ I mean that $\phi(G)/\phi(N)$ is the trivial group, i.e., the group consisting of just the identity element. $\endgroup$ – Devlin Mallory Jun 7 '16 at 23:00
  • $\begingroup$ Ah right, and what is the trivial map? The map that sends every $g \in G$ to $1_H$? $\endgroup$ – Irregular User Jun 7 '16 at 23:03
  • $\begingroup$ Yep, that's correct! $\endgroup$ – Devlin Mallory Jun 7 '16 at 23:07
  • $\begingroup$ Thanks for the confirmation! $\endgroup$ – Irregular User Jun 7 '16 at 23:09
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It is true when $N$ contains $\ker \phi$. That is part of the isomorphism theorems.

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