$G/Z(G)$ is cyclic then is abelian? my question is if $G/Z(G)$ is order of p then is commutative then is abelian group but if G is abelian then $G=Z(G)$ therefore $G/Z(G)$ is not order of p?Is it contradiction?
 A: It shows $|G/Z(G)|=$ prime $p$ can't hold. The only way $G/Z(G)$ can be cyclic is in the most obvious of ways: when it is trivial. It is a useful fact that shows up a lot in exercises. E.g., reasoning by contradiction at some step you end up concluding that the index of $Z(G)$ in $G$ is prime, and, voilà, you get that $G$ must be abelian.
A: What you would have shown (if you had formulated more carefully) is that $G/Z(G)$ can never be of prime order. This seems to remove any interest from the statement, since its hypothesis is impossible to satisfy.
However this is because the statement is just too limited. If you look at the proof, it actually shows something more general:

If $G$ is a group, $H$ is a central subgroup ($[G,H]=\{e\})$, and $G/H$ is cyclic, then $G$ is Abelian.

In this form there can be useful applications.
Proof: Let $H$ be as stated and $g\in G$ such that $gH$ generates the quotient $G/H$. Then $G$ is generated by the fully commuting set of elements $\{g\}\cup H$, and is therefore and Abelian group. $\Box$
A: Note that $G/H$ abelian does not always imply $G$ abelian.
Your argument seems to assume that this holds.
It is not even true that $G/H$ cyclic always implies $G$ abelian. (Take $G=S_3$ and $H=A_3$.)
However, it does when $H=Z(G)$. That is the point here.
