TITLE: A normal subgroup intersects the center of the $p$-group nontrivially
QUESTION [31 upvotes]: If $G$ is a finite $p$-group with a nontrivial normal subgroup $H$, then the intersection of $H$ and the center of $G$ is not trivial.

REPLY [2 votes]: Can we try induction on $n$ by taking quotients like $G/Z$ WHERE $Z$ IS THE CENTER.
More technically it goes like this.
We take the quotient group $G/Z$ where $Z$ is the center and as $G$ is a $p-group$,$Z$ is non trivial.Hence $o(G/Z)=o(G)/O(Z)$ is less than $o(G)$ to be ideal for application of induction.Now look at the set $S=(Zx|x\in N)$.My claim is that this set is a subgroup and as a matter of fact a normal subgroup.Because $(Zx_1)(Zx_2)=Zx_1x_2$ and hence the closure and the inverse property is trivial.Again for any $x \in G$ and $n \in N$ $(Zx)(Zn)(Zx)^{-1}=Zxnx^{-1}$ and due to $N$ being normal,that belongs to $S$.Hence $S$ is a normal subgroup in $G/Z$.Now due to non-triviality of $Z$ we can apply induction to claim that the intersection of $S$ with the center of  $G/Z$ is non-trivial.That is there is a $a$ not in $Z$ but in $N$ such that $Zax=Zxa$ for all $x \in G$ Hence there is an $a \in N$ such that $axa^{-1}x^{-1}$ belongs to $Z$  for all $x \in G$.Now as $a^{-1}$ belongs to $N$ clearly $xa^{-1}x^{-1}$ be in $N$ and thus $axa^{-1}x^{-1}$ belongs to $N$.But if $N$ and $Z$ has trivial intersection then $ax=xa$ for all $x \in G$ which makes $a \in Z$,a contradiction.