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.
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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 $p$. |
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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 $G$. |
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