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Let $G$ be a group, $H$ be an abelian group, $\varphi:G \to H$ be a homomorphism onto $H$. Let $N$ be a subgroup of $G$ that contains $\operatorname{ker}(\varphi)$. Show that $N$ is a normal subgroup of $G$.

I don't understand the word "contains", if it says $N=\operatorname{ker}(\varphi)$ then I write:

Let $n \in N$, and $g \in G$ be arbitrary, then $\varphi(gng^{-1})=\varphi(g)\varphi(n)\varphi(g)^-1=\varphi(g)\varphi(g)^-1=e$, so $N$ is normal.

But in this question, there can be a $y \in N$ and also $y$ is not an element of $\operatorname{ker}(\varphi)$. Or should I prove first that $N=\operatorname{ker}(\varphi)$? Thanks for any help in advance.

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    $N$ need not be equal to the kernel for this. The important thing to remember is that $H$ is abelian, so any subgroup of $H$ is normal. Then use the correspondence theorem. – 2017-01-02

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