Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer
1  
"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.

share|improve this answer
    
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
1  
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
3  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.