Groups of order p squared are abelian

Statement

Let $p$ be a prime number and let $G$ be a group of order $p^2$ . Then $G$ is abelian.

Remarks

By the nontrivial centre theorem and Lagrange's theorem, $|Z(G)|$ is $p$ or $p^2$ . If $|Z(G)| = p$ , then $G/Z(G)$ has order $p$ , so it is cyclic, and this criterion gives a contradiction. Thus $Z(G) = G$ , and $G$ is isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ or to $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ . For $p^3$ the result is false: the dihedral group $D_4$ and the quaternion group have order $8$ and are not abelian.

Tag	`09JZ9P4899`
URI	`m2://09JZ9P4899`
Kind	statement, theorem
Writing	W0 stub
MSC	20D15
Status	This copy of the site has no build data. Run `m2 status 09JZ9P4899`.
Uses	A nontrivial finite p-group has nontrivial centre, If G modulo its centre is cyclic then G is abelian, Lagrange's theorem

Tag

09JZ9P4899

URI

m2://09JZ9P4899

Kind

statement, theorem

Writing

W0 stub

MSC

20D15

Status

This copy of the site has no build data. Run m2 status 09JZ9P4899.

Uses

A nontrivial finite p-group has nontrivial centre, If G modulo its centre is cyclic then G is abelian, Lagrange's theorem

Groups of order p squared are abelian

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