Center of a finite group 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 $G/N$?
 A: 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 $|G/N|=2$.
A: 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 $M$.
