For $N\unlhd G$ , with $C_G(N)\subset N$ we have $G/N$ is abelian Question is that :
let $N\unlhd G$ such that every subgroup of $N$ is Normal in $G$ and  $C_G(N)\subset N$. 
Prove that $G/N$ is abelian.
what could be the possible first thought (though for me it took some time :)) is to use that $C_G(N)$ is Normal subgroup (As in general centralizer is a subgroup). one reason to see this is that $C_G(N)$ is not Normal in General and $C_G(N)$ is not subset of $N$ in general.
As $C_G(N)\subset N$, we have $G/N\leq G/C_G(N)$ 
I some how want to say that $G/C_G(N)$ is abelian and by that conclude that $G/N$ is abelian.
I would like someone to see If my way of approach is correct/simple?? 
I have not yet proved that $G/C_G(N)$ is abelian. I would be thankful if someone can give an idea.
Thank You.

For any $n\in N$, $g\in G$, there is an integer $k$ s.t. $gng^{-1}=n^k$ (as the subgroup generated by $n$ is normal). That implies that $ghn(gh)^{-1}=hgn(hg)^{-1}$ for all $g,h\in G$, $n\in N$, i.e. that $G/$(the kernel of the conjugation action of $G$ on $N$) is Abelian. The kernel is $C_G(N)$,  i.e. $G/C_G(N)$ is Abelian, and thus (as you observed), $G/N$ is Abelian..
