Center of Group of Prime Power Order is Non-Trivial

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Theorem

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


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

$\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e$


Proof

Suppose $G$ is abelian.

From Group equals Center iff Abelian:

$\map Z G = G$

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 $\map Z G \ne G$ and therefore $G \setminus \map Z G \ne \O$.

Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $G \setminus \map Z G$ is partitioned.

From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element.


From the Conjugacy Class Equation:

$\displaystyle \order {\map Z G} = \order G - \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$


From Number of Conjugates is Number of Cosets of Centralizer:

$\order {\conjclass {x_j} } \divides \order G$


Let $\map {N_G} x$ be the normalizer of $x$ in $G$.

Then:

$\forall j: 1 \le j \le m: \index G {\map {N_G} {x_j} } > 1 \implies p \divides \index G {\map {N_G} {x_j} }$


Since $p \divides \order G$, it follows that:

$p \divides \order {\map Z G}$

and the result follows.

$\blacksquare$


Sources