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:
- $\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e$
Proof
Suppose $G$ is abelian.
By definition of abelian group:
- $\map Z G = G$
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:
- $\ds \order {\map Z G} = \order G - \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
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
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Exercise $5.17$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 52$ Lemma $1$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 48.4$ Conjugacy
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.20$