Question about normal subgroup I am stuck on this question:

Let $f: G \rightarrow H$ be a group homomorphism.  Suppose $H$ is abelian.  Let $N$ be a subgroup of $G$ that contains $\ker f$.  Prove that $N \trianglelefteq G$.

This is my attempt:
Since $H$ abelian, $f(G)$ is abelian. So, by $1^{st}$ isomorphism theorem $G/\ker f \cong f(G)$.
Now, let $a \in G, n\in N,$ I claim that $an=na$ for all $a \in G$ and $n \in N$.
 To see that, assume for the contrary there exists $a \in G$, $n \in N$ such that $an \neq na$. Clearly, $n \notin N$ because otherwise will contradicts theorem 5.5 that says $\ker f \unlhd G$. So we must have $n \in N$ such that $n \notin \ker f$.
I know that $f(an)=f(na)$ because of the property of an abelian group, but I could not see here any contradiction since we don't know anything about $f$ except it is a group homomorphism.
I would appreciate any help with that.
Thanks in advance.
 A: You know that $G/\operatorname{Ker} f$ is abelian. Then $N/\operatorname{Ker} f$ is a subgroup of an abelian group and is therefore a normal subgroup of $G/\operatorname{Ker} f$. Therefore $N$ is normal in $G$.
A: If $f\colon G\to H$ is a surjective group homomorphism, then the map $\Phi\colon K\to f(K)$ is a bijection between the set of all subgroups of $G$ containing $\ker f$ and the subgroups of $H$. Under this map normality of subgroups is preserved, in the sense that a subgroup $K$ of $G$ containing $\ker f$ is normal in $G$ if and only if $f(K)$ is normal in $H$.
Some textbooks name this result “correspondence theorem” or “$n$-th homomorphism theorem” (for some value of $n$).
In your case, the homomorphism is not necessarily surjective, but if you consider the surjective $f_0\colon G\to f(G)$ (the same as $f$, but with a different codomain), you have that $f(G)$ is abelian, hence every subgroup thereof is normal.
You can also prove the statement directly. Let $N$ be a normal subgroup of $G$ containing $\ker f$; let $g\in G$ and $x\in N$. Then
$$
f(gxg^{-1})=f(g)f(x)f(g)^{-1}=f(x)
$$
due to $H$ being abelian. Therefore $x^{-1}(gxg^{-1})\in\ker f$, hence
$$
x^{-1}(gxg^{-1})\in N
$$
and therefore $gxg^{-1}\in N$.
