Finite non-abelian $p$-group cannot split over its center 
Show that a finite non-abelian $p$-group cannot split over its center.

I'd be happy for some clues.
 A: Suppose $P=Z(P)H$ with $1<H<P$ nontrivial and $H\cap Z(P)=1$. Two questions:


*

*Can $Z(H)\subseteq H$ and $Z(P)$ intersect nontrivially?

*Can $Z(H)=1$? Observe that $H$ is a $p$-group.


Show that every element of $Z(P)Z(H)$ commutes with every element of $Z(P)H=P$, and therefore derive the containment $Z(P)Z(H)\subseteq Z(P)\implies Z(H)\subseteq Z(P)$, contra trivial intersection.
A: This definition is not very common, so it may be worth  mentioning here:
Definition. Let $G$ be a group. A subgroup $K$ of $G$ is said to be co-central in $G$ if $G=Z(G)K$. 
Theorem. Let $G$ be a group. If $K$ is co-central in $G$, then $Z(K)\subseteq Z(G)$.
Proof. Let $z\in Z(K)$, and $x\in G$. We need to show that $zx=xz$. Since $G=Z(G)K$, there exists $a\in Z(G)$ and $k\in K$ such that $x=ak$. Then
$$\begin{align*}
zx &= zak\\
&= azk &&\text{(since }a\in Z(G)\text{)}\\
&= akz &&\text{(since }z\in Z(K)\text{ and }k\in K\text{)}\\
&= xz.
\end{align*}$$
Thus, $z\in Z(G)$, as claimed. $\Box$
Now assume that we can write a $p$-group $P$ as $P=Z(P)H$ with $H$ a subgroup. Then by the lemma, $Z(H)\subseteq Z(P)$. If $P$ were split over $Z(P)$, then we would have $Z(P)\cap H = \{e\}$, hence $Z(H)=\{e\}$. 
Why is that a very big problem for $H$, which can only be solved if $H=\{e\}$?
