Let $N$ normal in $G$ such that every subgroup of $N$ is normal in $G$ and $C_G(N) ⊆ N$. Prove that $G/N$ is abelian.

Question

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

I tried to show that the commutator subgroup of $G$ is present in $N$ but was unsuccessful.

Answer 1

Hint: pick an $x \in N$ and look at $M=\langle x \rangle$. Since $M \unlhd G$, we have $G/C_G(M)$ embeds homomorphically into Aut$(M)$, which is abelian. So $G' \subseteq C_G(\langle x \rangle)$ for all $x \in N$. Conclude that $G' \subseteq C_G(N)$.