If N and every subgroup of N is normal in G then G/N is abelian .

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.

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 $b^{-1}c^{-1}bc \in C_G(N) \le N$.