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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](http://math.stackexchange.com/q/219069), but that is about proof details, whereas this question is asking for a proof.2012-11-09

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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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    @Makoto, I am wondering what is the role that "phi is surjective" in this proof? Will the statement that "the image of a normal subgroup of G under homomorphism is still a normal subgroup of G' still be valid when phi is not surjective? I cannot recall if every homomorphism is surjective, and don't follow that why you say "since phi is surjective". Thank you.2013-10-14
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    @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 $G'$.2013-10-14
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    @Makoto, thank you so much for your clarification. Just want to be more clear, is surjective a built-in property for homomorphism? I remember there's a property that "a homomorphism is 1-1 iff Kernel={identity element}". But I cannot recall if there's a similar condition for a homomorphism to be surjetive(onto)? Thank you!!2013-10-14
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    @Lily You're welcome. I don't know what you mean by "a built-in property". In general, there's no such condition as Kernel = 1 for a homomorphism to be surjective. However, if we only consider abelian groups. Such a condition exists. Namely Cokernel = 0.2013-10-14
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    @Makoto, by the question itself, it only says phi is a homomorphism, didn't mention phi is surjective, I'm confused that why we can say phi is surjective in the proof? Thanks for your kindly help. ps. Sorry, my question might be very stupid...>"<2013-10-14
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    @Lily No problem. The question says it is "onto". It means surjective.2013-10-14
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    @Makoto,OMG... I didn't see the original post carefully @_@ I'm so sorry..... so if it's without phi being onto, we cannot draw the same conclusion right?2013-10-14
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    @Lily You are right.2013-10-14
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    @Makoto, thanks so much for your patience. :)2013-10-14