$G$ contains a normal $p$-Sylow subgroup and $p$ divides the order of the center I am looking at the following: 
Let $G$ be a non-abelian finite group with center $|Z|>1$. 
I want to show that if $G/Z$ is a $p$-group, for some prime $p$, then $G$ contains a normal $p$-Sylow subgroup and $p\mid |Z|$. 
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I have done the following: 
Since $G/Z$ is a $p$-group, it contains $p$-Sylow subgroups, say $P$. 
From the correspondence theorem we have that there is a bijective mapping between the subgroups of $G$ that contain $Z$ and the subgroups of $G/Z$, 
$$\phi (A)\mapsto A/Z, \ A\in G$$ 
So, the corresponding $p$-Sylow of $G$ exists and it is the $PZ$. 
Is this correct? 
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We have that $|G/Z|=p^n$, for some $n$. 
So, $\frac{|G|}{|Z|}=p^n \Rightarrow |Z|=\frac{|G|}{p^n}$. 
Since $G$ contains a $p$-Sylow sybgroup we have that $|G|=p^ma$, where $p\not\mid a$. 
So, we have to show that $m\geq n$, or not? 
How could we show that? 
 A: For the first question, why is $PZ$  a $p$-group? It might not be so you cannot answer like this.
Hint : You should try to consider the action of conjugation of $G$ on its $p$-Sylows.

 Consider the action of conjugation of $G$ on its $p$-Sylows. You know (from the second theorem of Sylow) that this action is transitive. But this action of conjugation factors naturally (since $Z$ acts trivially by conjugation) through $G/Z$. Let $n_p$ be the number of $p$-Sylows, from the second theorem of Sylow $n_p=1$ mod $p$. Since $G/Z$ is a $p$-group the number of fixed elements of $G/Z$ acting on the $p$-Sylows is congruent to $n_p$ mod $p$. In particular, there is a fixed $p$-Sylow for the action of $G/Z$ whence of $G$. By transitivity there cannot but be only one $p$-Sylow.

For the second question, you should use what you have already done. 
Hint : Let $P$ be the unique $p$-Sylow. Assume that $|Z|$ is prime to $p$ then $P\cap Z$ is trivial. Using the fact that $G/Z$ is a $p$-group, show that $G=Z\times P$ and find an element in $G$ which is central but not in $Z$ (which is a contradiction).

 If $|Z|$ is prime to $p$, since $G/Z$ is a $p$-group we have an isomorphism of group between $G/Z$ and $P$. Since $Z$ and $P$  are both normal subgroups of $G$, with trivial intersection and generating $G$, we have $G=Z\times P$. Finally the center of $P$ is not trivial (since it is a $p$-group) so $Z(G)=Z\times Z(P)\neq Z=Z(G)$ which is a contradiction.

A: $PZ$ may not be Sylow-$p$ subgroup of $G$ (take $G=\mathbb{Z}_3\times Q_8$.)
The hypothesis $G/Z$ is a $p$-group implies that $G$ is nilpotent. In nilpotent groups, all Sylow subgroups are normal, and so $G$ is direct product of them. Since center of a $p$-group is non-trivial, this almost completes your expectation (fill up the details).
