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Suppose that $G$ is finite group with two normal subgroups $N$ and $K$ such that $K<N$. Is true that the center of $G/K$ is subset of the center of $G/N$?

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$G/K$ is not necessarily a subset of $G/N$, so what do you mean precisely? If $Z(G/K)=L/K$ and $Z(G/N)=M/N$, do you want to know if $L \subset M$? –  Nicky Hekster Oct 23 '12 at 14:12
    
If $hK\in Z(G/K)$ then are you wanting to prove that $hK\in Z(G/K)$? Well, if $hK\in Z(G/K)$ then $[g, h]\in K\leq N$ for all $g\in G$. So, we're done. –  user1729 Oct 23 '12 at 14:26
    
@user1729 I think you mean "wanting to prove that $hN \in Z(G/N)$" right? –  Nicky Hekster Oct 23 '12 at 14:30
    
@NickyHekster Yup, thanks. –  user1729 Oct 23 '12 at 14:34
    
Thanks for all comments. Yes Nicky I want to know if $L$ is subset of $M$? –  Morad Oct 23 '12 at 14:47
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2 Answers 2

For showing that what @Nicky pointed, consider the Quaternion group $$Q_8=\{+1,-1,+i,-i,+j,-j,+k,-k\}$$ of order $8$. Let $K=\{+1,-1\}, N=\{+1,-1,+i,-1\}$ which are two normal proper subgroups of $Q_8$ such that $K<N$. Now see that $|G/K|=4$ and $|G/N|=2$.

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@Morad: Read the final comment. –  B. S. Oct 23 '12 at 15:07
    
"Groupie" Babak! +1 –  amWhy Feb 12 '13 at 0:06
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My comments answer your question. I shall expand on them here.

If $K\leq N$ are both normal subgroups of a group $G$ with $Z(G/N)=M/N$ and $Z(G/K)=L/K$ then you are wanting to show that $L\subset M$. However, this doesn't really work. Suppose $G$ is the cyclic group of order $4$, with elements $\{0, 1, 2, 3\}$, and take $K=\{0, 2\}$ and $N=G$. Then $Z(G/K)=\{0K, 1K\}$ so we can take $M=\{0, 1\}$. On the other hand, $G/N$ is trivial, so we shall write $G/N=\{3N\}$, and so $L=\{3\}$. This contradicts your assertation.

This is, of course, silly. But entirely valid!

Basically, you need to take $L=\{h: hK\in Z(G/K)\}$ and $M=\{h: hN\in Z(G/N)\}$. Then your theorem works, as if $h\in L$ then

$\begin{align*} &hKgK=gKhK \:\forall\: g\in G\\ \Rightarrow &hgK=ghK\:\forall\: g\in G\\ \Rightarrow &hgh^{-1}g^{-1}\in K\:\forall\: g\in G\\ \Rightarrow &hgh^{-1}g^{-1}\in N\: \forall\: g\in G \: (as\: K\leq N)\\ \Rightarrow &h\in M\\ \Rightarrow &L\subset M \end{align*}$

as required.

On the other hand, I am not sure if you were but you could have been using the correspondence theorem to take $L$ and $M$ as subgroups of $G$ which contain $K$ and $N$ respectively. The above proof still works in this case, by the uniqueness of $L$ and $M$.

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Generally the notation $L/K$ is used when $L$ is a subgroup, not a transversal. –  Jack Schmidt Oct 24 '12 at 11:03
    
True. I tried to write down my reasons for using $L$ and $M$, but it got long so I have decided to delete them and say: basically I just copied what NickyHekster wrote. Also, $L$ and $M$ are, at different points in my answer, transversals, common-or-garden subsets, and subgroups of $G$. Switching notation would be messy. –  user1729 Oct 24 '12 at 11:32
    
@user1729: Thank you very much. Ok there are many examples that this doesn't really work, but is there any example of non-abelian group such that $Z(G/N)=1$? –  Morad Oct 24 '12 at 13:00
    
@Morad: The problem wasn't with the group being abelian or anything like that - it was just because I took a transversal for $Z(G/N)$ which had empty intersection with $Z(G/K)$. Heck - you can take two transversals for $G/K$ with empty intersection! That is to say, my counter-example always works, even if $N=K$. You need to re-formulate your question, probably into one of the two ways I suggested. –  user1729 Oct 24 '12 at 13:08
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@Morad: Can you explain what you think happens when $Z(N)=1$? My point is, if $N$ and $K$ are both proper subgroups of $G$ then we can (often? always?) take transversals for $G/K$ and $G/N$ which intersect trivially. $Z(G/N)$ is just a subset of the transversal of $G/N$, and so intersects the transversal for $G/K$ trivially. –  user1729 Oct 29 '12 at 12:05
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