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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, but that is about proof details, whereas this question is asking for a proof. – robjohn Nov 9 '12 at 16:17
up vote 4 down vote accepted

$\phi(a)\phi(N) = \phi(aN) = \phi(Na) = \phi(N)\phi(a)$ for every $a \in G$. Since $\phi$ is surjective, $\phi(N)$ is a normal subgroup of $G'$.

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+1 Painfully simple and accurate. – DonAntonio Nov 9 '12 at 16:43
@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. – Lily Oct 14 '13 at 1:00
@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
@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!! – Lily Oct 14 '13 at 1:18
@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. – Makoto Kato Oct 14 '13 at 1:30

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