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Let $N$ be a normal subgroup of a group $G$ such that every subgroup of $N$ is normal in $G$ and $C_G(N)\subseteq N $. Prove that $G/N$ is abelian.

Here, as usual, $C_G\left(N\right)$ means the centralizer of $N$ in $G$ (i.e., those elements of $G$ that commute with everything in $N$).

I think we need to use that every subgroup of $N$ is normal in $G$ but i can't use .Please help me with Hints.

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    $\begingroup$ Answer-in-a-comment by Nicky Hekster from the dupe thread: "Let $n \in N$, and consider $H= \left\langle n \right\rangle$. This subgroup of $N$ is normal in $G$, so by the $N/C$ normalizer/centralizer theorem $G/C_G(H)$ embeds in $\operatorname{Aut}(H)$, which is abelian, since $H$ is cyclic. Hence, $G' \subseteq C_G(H)$, in particluar $G' \subseteq C_G(n)$. Since $C_G(N) = \bigcap_{n \in N} C_G(n)$, this yields $G' \subseteq N$, that is, $G/N$ is abelian." $\endgroup$
    – darij grinberg
    Sep 24, 2019 at 2:33

1 Answer 1

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Let $a \in N$, $b,c \in G$. Since $\langle a \rangle$ is normal in $G$, it is normalized by $b$ and $c$.

The automorphism group of a cyclic group is abelian, so $b^{-1}c^{-1}bc$ centralizes $\langle a \rangle$. Now this is true for all $a \in N$, so $b^{-1}c^{-1}bc \in C_G(N) \le N$.

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