Center of Group of Prime Power Order is Non-Trivial

Theorem
The center of a group whose order is the power of a prime is non-trivial:


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

Proof

 * If $$G$$ is abelian, then the result is certainly true, because then from Center of Abelian Group is Whole Group, $$Z \left({G}\right) = G$$.


 * So, suppose $$G$$ is non-abelian. Thus $$Z \left({G}\right) \ne G$$ and therefore $$G - Z \left({G}\right) \ne \varnothing$$.

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

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

From Conjugacy Class Equation:


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

Now from Number of Conjugates is Number of Cosets of Centralizer:
 * $$\left|{\mathrm{C}_{x_j}}\right| \backslash \left|{G}\right|$$.

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

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

Since $$p \backslash \left|{G}\right|$$, it follows $$p \backslash \left|{Z \left({G}\right)}\right|$$ and the result follows.