A normal subgroup intersects the center of the $p$-group nontrivially 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.
 A: Perhaps a slightly less computationally-intensive argument is to simply note that since $H$ is normal, it must be a union of conjugacy classes. Each conjugacy class of $G$ has $p^i$ elements for some $i$; since $H$ contains at least one conjugacy class with $p^0 = 1$  elements (the class of the identity), and $|H|\equiv 0 \pmod{p}$, it must contain other classes with just one element, which must be classes of central elements of $G$.
A: $H$ is normal, consider $G$ acting on $H$ by conjugation.
The class equation yields
$$ \left| H \right| = \left| H^G \right| + \sum_i [G:Stab_{h_i}] \, , $$
where $ H^G = \{ h \in H  \mid \ ghg^{-1} = h , \ \forall g \in G \} $ are the fixed points and $ Stab_{h_i} = \{ g \in G \ \mid \ gh_ig^{-1} = h_{i} \}\leqslant G $ is the stabilizer of a $h_i \in H$.
Observe that in this case $ H \cap Z(G) = H^G $.
$p$ divides $\left| H \right|$ and $[G:Stab_{h_i}]$ for every non trivial orbit, so it divides $\left| H^G \right|$.
In particular, $H^G$ is not empty, so there is an element of $H$ that is also in the center of $G$.
A: 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.
A: Let $a_{1}, . . . , a_{k}$ be representatives of the conjugacy classes of $G$, ordered such that $a_{m} \in H$ and $a_{m+1}, \cdots , a_{k} \notin H$. The conjugacy class $C(a_{i})$ have either $C(a_{i}) \subset H$ or $C(a_{i}) \cap H = \{e\}$. First arrange the $\{a_{1}, . . . , a_{m}\}$ so that the first $r$  represent conjugacy classes of size 1, (i.e. elements in $H \cap Z$) and the latter $m − r$ represent classes of size larger than 1. Then we can write the
class equation for $H \cap Z = H$ as: $$|H| = \sum\limits_{i=1}^{m} |C(a_{i}) \cap H| = |H \cap Z| + \sum\limits_{i=r}^{m} |C(a_{i})| =  |H \cap Z| + \sum\limits_{i=r}^{m}\frac{|G|}{|N(a_{i})|}$$
As $|H| < p^{n}$ every term in the sum is divisible by $p$ so $|H \cap Z|$ from above is divisible by $p$.
