A simple question about the center of a finite group

Let $G$ be a finite group. $N$ a normal subgroup of $G$ and $K$ a characteristic subgroup of $N$: $$K \text{ char } N \triangleleft G.$$ If $Z(G/N)=1$ and $Z(G)=1$, does it follow that $Z(G/K)=1$?

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What's the relation between $N$ and $K$? –  Chris Godsil Oct 25 '12 at 2:31
Yes, when G is the trivial group. –  Mustafa Gokhan Benli Oct 25 '12 at 2:42
Sorry,I have corrected it.$K$char$N$. –  Tom Oct 25 '12 at 4:03
As you state the question, no. What it $G$ is any finite group with $Z(G) = 1$ and we take $N = G, K =[G,G]$? Then $Z(G) = Z(G/N) = 1$ and $G/K$ is Abelian (here, $[G,G]$ means the derived group). –  Geoff Robinson Oct 25 '12 at 4:45
This is not a research level question, so is not suitable for MO. With the question as it is worded now, a counterexample is $S_3 \times A_5$, with $N=S_3$, $|K|=3$. –  Derek Holt Oct 25 '12 at 9:12

As it is not entirely clear where the containments are meant to be, I will assume you are asking about the case $N\triangleleft K\triangleleft G$; in this case $Z(G) = Z(G/N) = 1$ is not sufficient in general to guarantee $Z(G/K)$ is also trivial. As an example, take $N$ to be the Klein-4 subgroup of $S_4$ and $K = A_4$. Then $G/N\cong S_3$ so $Z(G) = Z(G/N)$ is trivial, but $G/K$ is abelian (isomorphic to $\mathbb{Z}/2\mathbb{Z}$) and so $Z(G/K)$ is not trivial.
I am sorry.It is not the case.It is $K$char$N\triangleleft G$. Thank you all the same. –  Tom Oct 25 '12 at 4:35