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

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It would be way more useful if you'd posted your insights, ideas, effort, background and/or things you already know about this problem. -1 –  DonAntonio Sep 1 '12 at 23:52

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.

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"The action"...you mean, I presume, the action by conjugation , right? I guess the OP could know this, but it is not immediate from his post, which gives no background, ideas, etc. at all, and not everybody knows about the injection $$N_G(H)/C_G(H) \to Aut(H)$$ –  DonAntonio Sep 1 '12 at 23:45
+1 for very nice approach. –  Babak S. Sep 2 '13 at 7:11

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.

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I downvoted, as this is not even close to an answer. –  user641 Sep 1 '12 at 19:11
@SteveD Sure it isn't a complete answer. It's too lengthy for a comment, though, and it points the OP in the right direction to solve this problem on their own. A "practice problem" sounds like homework, so giving a complete answer would be doing the OP a disservice, IMO. –  Code-Guru Sep 1 '12 at 19:18
It's not an answer in any way, it just rewords some definitions. –  user641 Sep 1 '12 at 19:23
I agree with anon: the OP's post history shows he understand what a commutator is, and the info in the OP's profile says they are a grad student, preparing for comps. –  user641 Sep 1 '12 at 19:45
To rely on reading the OP's post history is too stretching the work one could do to guess the OP's background: I agree with Guru in this, the OP should have posted his question with way more ideas, background, things already known, etc. Yet, I think Steve's point is the main one here: Guru's writing doesn't come even close to be anything ressembling a hint of a possible answer (he didn't even mentioned that $\,G/N\,$ abelian $\,\Longleftrightarrow G'\leq N\,$...!) , and I'd advice him to delete his post as it is useless and will probably bring upon him lots of downvotes. –  DonAntonio Sep 1 '12 at 23:50