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, $Z \left({G}\right) = G$.

This will always be the case for $r = 1$, as a group of prime order is cyclic, and a cyclic group is abelian.


 * 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 Conjugacy Class Equation:


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

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

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.