A nontrivial p-group has nontrivial center I know this is a very common corollary of the class equation. And I know how to do it by using class equation. 
But can you do it bu using group action, maybe find a nice set for G to act on? 
 A: Let $G$ be a nontrivial finite $p$-group.
It is possible to prove without the class equation that in $G$, the number of subgroups of order $p$ is $\equiv 1 \mod{p}$. For example, we could use McKay's proof of Cauchy's theorem. From this it follows that the number of normal subgroups of order $p$ is also $\equiv 1 \mod{p}$, since the number of non-normal subgroups of order $p$ is $\equiv 0 \mod{p}$. This can be seen by having $G$ act on the non-normal subgroups by conjugation, every orbit has size $p^k$ for some $k > 0$.
Thus $G$ contains a normal subgroup $N$ of order $p$. By the normalizer-centralizer lemma, there exists a homomorphism $f: G \rightarrow \operatorname{Aut}(P)$ with $\operatorname{Ker}(f) = C_G(N)$. Now $\operatorname{Aut}(P) \cong \mathbb{Z}_p^*$ has order $p-1$ which is coprime to $p$, so $C_G(N) = G$ follows by $G / \operatorname{Ker}(f) \cong f(G)$ and Lagrange's theorem. Thus $N \leq Z(G)$ which proves that $Z(G)$ is not trivial.
Of course, proving this theorem this way is pretty silly. This is just a guess and my opinion, but I feel like the only easy way to prove this is with the class equation.
A: Consider the action of $G$ on itself by conjugation. Let $\chi_1,\ldots,\chi_n$ be the orbits of $G$ under this action. Since $G$ is a $p$-group, all orbits must have order $1$ or some power of $p$. But
$$|G|=\sum\limits_{i=1}^n |\chi_i|$$
so since the LHS is divisible by $p$, the RHS is divisible by $p$ as well. Since all $|\chi_i|$ which aren't divisible by $p$ are $1$, the number of such $|\chi_i|$ must be divisible by $p$. So there are either no $\chi_i$ of cardinality $1$, or at least $p$. But $\chi_i$ of cardinality $1$ correspond precisely to elements of the center, and since $e$ is in the center there is at least $1$. So there are at least $p$.
