I know that if $P$ is a finite $p$-group, say $\mid P \mid = p^a$ for some prime $p$, and that if $N$ is a non-trivial normal subgroup of $P$ then the center of $P$ intersects non-trivially with $N$ (i.e., $Z(P) \cap N \neq \{1\}$).
My question is suppose that I have another $p$-group $Q$, say $\mid Q \mid = q^b$ for some prime $q$, which also shares this property (that for any non-trivial normal subgroup $M$ of $Q$, the center of $Q$ intersects non-trivially with $M$). Is it true that if I create the direct product
$$ P \times Q $$
and give it a group structure component-wise, will this new group $P \times Q$ inherit the same property? If not, is there a simple counter-example? I've tried proving it by assuming that if $S$ is a normal subgroup of $P \times Q$ then looking at $S$ component wise will give normal subgroups of $P$ and $Q$, but I'm not even entirely sure that this fact is true.
Let $N$ be a normal subgroup of $P\times Q$.
Then $N\cap P$ and $N\cap Q$ are normal in $P$ and $Q$ respectively.
If $N\cap P$ is not identity, then this subgroup should intersect center of $P$ non-trivially (same for $Q$).
Q. Can it happen that $N\cap P$ as well as $N\cap Q$ are identity?
No! Because $P$ and $Q$ are Sylow-subgroups of distinct prime, and $|N|$ should be divisible by $p$ or $q$.
Remark: $Z(P\times Q)=Z(P)\times Z(Q)$.