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.
 A: 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..
A: Edit: Please ignore the following "proof". It is incorrect (as pointed out in the comments below). However, I will leave it up here because it might have some ideas in there that are worth knowing in the future.
For any $x,y \in G$, define
$$
\alpha = xyx^{-1}y^{-1}
$$
We want to show that $\alpha \in N$. Consider the subgroup
$$
H = \langle \alpha\rangle \cap N < N
$$
By hypothesis, $H\vartriangleleft G$. Hence for any $z\in N$,
$$
z\alpha z^{-1} \in H \subset \langle \alpha \rangle
$$
Hence, $\exists k\in \mathbb{N}$ such that
$$
z\alpha z^{-1} = \alpha^k
$$
Since the map $w \mapsto zwz^{-1}$ is an isomorphism, it follows that $o(\alpha^k) = o(\alpha)$ and hence
$$
(k,o(\alpha)) = 1
$$
Hence, there exist $a,b \in \mathbb{Z}$ such that $ak + bo(\alpha) = 1$, and so
$$
\alpha = (\alpha^k)^a \alpha^{bo(\alpha)} = (\alpha^k)^a = (z\alpha z^{-1})^a \in H \subset N
$$
Hence, $xyx^{-1}y^{-1} \in N$, so
$$
xN \cdot yN = yN\cdot xN \text{ in } G/N
$$
