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If $f:G\to H$ is a homomorphism of groups, $H$ is abelian and $N$ is a subgroup of $G$ containing the kernel of $f$, then $N$ is normal in $G$.

Help please.

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    Please supply some additional information, so that we can help you better. What have you done? Where are you stuck?2012-12-16

3 Answers 3

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$f:G\to H$ is a homomorphism so according to first theorem of homomorphism we know that $\frac{G}{\ker f}\leq H$. $H$ is abelian so $\frac{G}{\ker f}$ is abelian quotient group. This means as @Don noted $G'\subset \ker f$. But you pointed $N$ has $\ker f$ as an subgroup, so $G'\subset N$ and again @Don's neat answer help you to get the answer.

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    Nicely put, dear friend $\ddot\smile$2013-04-13
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Hints:

1) For any group $\,G\,\,,\,H\triangleleft G\,$ , we have that $\,G/H\,$ abelian $\,\Longleftrightarrow G':=[G:G]\leq H\,$

2) If $\,H\leq G\,$ , then $\,G'\leq H\Longrightarrow H\triangleleft G\,$

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Look at $f(g^{-1}ng)$ for $g\in G$ and $n\in N$:

Since $H$ is abelian, we obtain $f(g^{-1})\cdot f(n)\cdot f(g)=f(g^{-1})\cdot f(g)\cdot f(n)=f(n)$.

So $f(g^{-1}ng) \cdot f(n^{-1})= e_H \Rightarrow g^{-1}ng\cdot n^{-1} \in \ker f \Rightarrow g^{-1}ng \in \ker f \cdot n \subset N$.

Now $\forall n \in N \, \forall g \in G \quad g^{-1}ng \in N$ means that $N$ is normal in $G$.