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 $G'$.
the image of a normal subgroup
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$\begingroup$
abstract-algebra
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1This was flagged as a duplicate of [Showing that if a subgroup is normal, it's homomorphic image is normal](http://math.stackexchange.com/q/219$0$69), but that is about proof details, whereas this question is asking for a proof. – 2012-11-09
1 Answers
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$\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 $G'$.
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0@Makoto, thanks so much for your patience. :) – 2013-10-14