# the image of a normal subgroup

If $G$ is any group and $N$ is a normal subgroup of $G$ and $\phi\colon:G \to G'$ is a homomorphism of $G$ onto $G'$, prove that the image of $N$, $\phi(N)$, is a normal subgroup of $G'$.

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This was flagged as a duplicate of Showing that if a subgroup is normal, it's homomorphic image is normal, but that is about proof details, whereas this question is asking for a proof. –  robjohn Nov 9 '12 at 16:17

$\phi(a)\phi(N) = \phi(aN) = \phi(Na) = \phi(N)\phi(a)$ for every $a \in G$. Since $\phi$ is surjective, $\phi(N)$ is a normal subgroup of $G'$.
@Lily Since $\phi$ is surjective, every element $x$ of $G'$ can be written as $x = \phi(a)$ for some $a \in G$. Hence, by the above equation, $x\phi(N) = \phi(N)x$ for every $x \in G'$. Therefore $\phi(N)$ is a normal subgroup of $G'$. –  Makoto Kato Oct 14 '13 at 1:10