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

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.

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