Abelian quotient group I'm stuck on the following practice problem.   Any hints would be appreciated.
Suppose $N$ is a normal subgroup of  $G$ such that every subgroup of $N$ is normal in $G$ and
$C_{G}(N) \subset N$. Prove that $G/N$ is abelian.
I'm not sure how to use the fact that $C_{G}(N) \subset N$.
Thanks
 A: Let $n\in N$, and consider the action of $G$ on $\langle n\rangle$. This embeds $G/C_G(\langle n\rangle)$ into $Aut(\langle n\rangle)$, an abelian group. Doing this for all cyclic subgroups of $N$ gives an embedding of $G/C_G(N)$ into a direct product of abelian groups. We are done then, because that means $G/C_G(N)$ is abelian, and $G/N$ is a quotient of that group.
A: First of all, don't get stuck on what is given. This is the wrong place to look when you start on a proof. Rather, you should look at what you need to prove. In this case, we want to show that $G/N$ is abelian. What does it mean for a group to be abelian?
Well, the definition states that a group $G$ is abelian if for all $g, h \in G$ we have $gh = hg$. So this means we need to pick any two elements from $G/N$ and show that they commute under the group's operation.
I'll let you think about it from there. Let me just emphasize that whenever you write a proof, you need to start with the definition of what you are trying to prove. This almost always gives you a guide as to how to start your proof.
