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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.

Mikasa
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  • @all who upvoted my question: I thank every one who appreciated my work, but it would be so helpful for me if you can kindly spare some time for me and give some suggestion/hint. Thank You. –  Sep 02 '13 at 06:49
  • I guess that the upvotes to your question (at least mine) came from people who found this exercise fun. I had not seen it before, and had to think about it. Eventually I came up with essentially the same solution as user8268. Anyway, the point is that there are many reasons for upvoting a question (as well as many reasons for upvoting an answer). – Jyrki Lahtonen Sep 02 '13 at 07:13
  • @JyrkiLahtonen : Does that mean what i have done on the way to arrive at a conclusion does not matter. :( –  Sep 02 '13 at 07:16
  • User8268's answer does prove that $G/C_G(N)$ is abelian, so the step in your idea of a solution, $G/C_G(N)$ abelian $\implies$ $G/N$ abelian, does come into play. I don't feel qualified to compare the merits of different ideas to this question, as so far I am only aware of one solution :-) – Jyrki Lahtonen Sep 02 '13 at 07:21
  • @SteveD : Yes, I see that It is a possible duplicate of the other Question. But I feel that there is a difference between the way it was asked there and the way it is asked here. As it is a rule of the forum Not to have same question in more than one way, please feel free to do whatever is necessary. –  Sep 02 '13 at 07:25

2 Answers2

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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..

user8268
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  • i do not fully understand your idea. It is all getting mixed up for me. could you please edit this if possible. –  Sep 02 '13 at 07:11
  • @PraphullaKoushik please let me know which part is not clear – user8268 Sep 02 '13 at 07:19
  • I am unable to relate your term "factors through abelian group" to what ever i know in your Answer. –  Sep 02 '13 at 07:22
  • @Praphulla: A key component in the answer is that the automorphism group of a cyclic group is abelian (you know that, right?). So the conjugation actions by $g$ and $h$ on the cyclic subgroup $\langle n \rangle$ must commute. Therefore conjugation by $ghg^{-1}gh^{-1}$ is the identity. Therefore $ghg^{-1}gh^{-1}\in C_G(N)$. +1, of course. – Jyrki Lahtonen Sep 02 '13 at 07:25
  • @user8268 : though you have not changed much, It looks to me more clear now. Thank you. :) –  Sep 02 '13 at 07:31
  • @JyrkiLahtonen : Yes, yes. Now i understand. I some how feel (atleast for me) that the way you say something is much more clear than the way i see and you make me to think about it. Thank You. –  Sep 02 '13 at 07:34
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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 $$ xN \cdot yN = yN\cdot xN \text{ in } G/N $$

Prahlad Vaidyanathan
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