Proof that a nontrivial finite p-group has nontrivial centre by the class equation

Proof

We reduce the class equation modulo $p$. Let $e$ be the identity of $G$.

1. First, $\abs{G} = p^r$ with $r \geq 1$. Let $q$ be a prime that divides $\abs{G}$; one exists since $\abs{G} > 1$. By Cauchy's theorem, some element of $G$ has order $q$.
2. Since $G$ is a $p$-group, that order is a power of $p$. As $q$ is prime, $q = p$.
3. So $p$ is the only prime factor of $\abs{G}$, and $\abs{G} = p^r$ by the fundamental theorem of arithmetic. Since $\abs{G} > 1$, $r \geq 1$.
4. Each positive divisor $d$ of $p^r$ is a power of $p$. Indeed, a prime factor of $d$ divides $p^r$. So it divides $p$ by Euclid's lemma, and it equals $p$.
5. By the class equation, $\abs{G} = \abs{Z(G)} + \sum_{i=1}^{k} [G : C_G(a_i)]$, where $\operatorname{Cl}(a_1), \dots, \operatorname{Cl}(a_k)$ are the conjugacy classes with more than one element.
6. By the class size formula, $[G : C_G(a_i)] = \abs{\operatorname{Cl}(a_i)}$, which divides $\abs{G} = p^r$ and is greater than $1$.
7. By step 4, each $[G : C_G(a_i)]$ is $p^j$ with $j \geq 1$, so $p$ divides it. Also $p$ divides $\abs{G}$, by step 3.
8. Therefore $p$ divides $\abs{Z(G)} = \abs{G} - \sum_{i} [G : C_G(a_i)]$.
9. The centre contains $e$, so $\abs{Z(G)} \geq 1$. A positive multiple of $p$ is at least $p \geq 2$. So $Z(G) \neq \set{e}$.