Center of Group of Prime Power Order is Non-Trivial

Theorem
Let $G$ be a group whose order is the power of a prime.

Then the center of $G$ is non-trivial:


 * $\forall G: \left|{G}\right| = p^r: p \in \mathbb P, r \in \N_{>0}: Z \left({G}\right) \ne \left\{{e}\right\}$

Proof
Suppose $G$ is abelian.

From Group equals Center iff Abelian:
 * $Z \left({G}\right) = G$

and the result is seen to be true as $G$ is itself non-trivial.

From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$.

So, suppose $G$ is non-abelian.

Thus $Z \left({G}\right) \ne G$ and therefore $G \setminus Z \left({G}\right) \ne \varnothing$.

Let $\operatorname{C}_{x_1}, \operatorname{C}_{x_2}, \ldots, \operatorname{C}_{x_m}$ be the conjugacy classes into which $G \setminus Z \left({G}\right)$ is partitioned.

From Conjugacy Classes of Center Elements are Singletons, all of these will have more than one element.

From the Conjugacy Class Equation:


 * $\displaystyle \left|{Z \left({G}\right)}\right| = \left|{G}\right| - \sum_{j \mathop = 1}^m \left|{\operatorname{C}_{x_j}}\right|$

From Number of Conjugates is Number of Cosets of Centralizer:
 * $\left|{\operatorname{C}_{x_j}}\right| \mathrel \backslash \left|{G}\right|$

Let $N_G \left({x}\right)$ be the normalizer of $x$ in $G$.

Then:
 * $\forall j: 1 \le j \le m: \left[{G : N_G \left({x_j}\right)}\right] > 1 \implies p \mathrel \backslash \left[{G : N_G \left({x_j}\right)}\right]$

Since $p \mathrel \backslash \left|{G}\right|$, it follows that:
 * $p \mathrel \backslash \left|{Z \left({G}\right)}\right|$

and the result follows.