proof that a nontrivial normal subgroup of a finite $p$-group $G$ and the center of $G$ have nontrivial intersection

Define $G$ to act on $H$ by conjugation; that is, for $g\in G$, $h\in H$, define

$$g\cdot h=gh{g}^{-1}$$

Note that $g\cdot h\in H$ since $H◁G$. This is easily seen to be a well-defined group action.

Now, the set of invariants of $H$ under this action are

$$G_H=\{h\in H|g\cdot h=h\forall g\in G\}=\{h\in H|gh{g}^{-1}=h\forall g\in G\}=H\cap Z(G)$$

The class equation theorem states that

$$|H|=|G_H|+\sum _{i=1}^r[G:G_{x_i}]$$

where the $G_{x_i}$ are proper subgroups of $G$, and thus that

$$|G_H|=|H|-\sum _{i=1}^r[G:G_{x_i}]$$

We now use elementary group theory to show that $p$ divides each term on the right, and conclude as a result that $p$ divides $|G_H|$, so that $G_H=H\cap Z(G)$ cannot be trivial.

As $G$ is a nontrivial finite $p$-group, it is obvious from Cauchy’s theorem that $|G|=p^n$ for $n>0$. Since $H$ and the $G_{x_i}$ are subgroups of $G$, each either is trivial or has order a power of $p$, by Lagrange’s theorem. Since $H$ is nontrivial, its order is a nonzero power of $p$. Since each $G_{x_i}$ is a proper subgroup of $G$ and has order a power of $p$, it follows that $[G:G_{x_i}]$ also has order a nonzero power of $p$.