---
tag: 09JZ9P4899
kind: statement
role: theorem
name: Groups of order p squared are abelian
msc: [20D15]
uses: [0RZCTT30H5, 0AH9Q6YNMT, 3C53TD]
sources:
  - url: https://en.wikipedia.org/wiki/P-group
    locator: Examples
  - url: https://en.wikipedia.org/wiki/P-group
    locator: Up to p3
---

## Statement

Let $p$ be a prime number and let $G$ be a [[group|G5S1M9]] of order $p^2$.
Then $G$ is [[abelian|082GHNPJ53]].

## Remarks

By the [[nontrivial centre theorem|0RZCTT30H5]] and [[Lagrange's theorem|3C53TD]], $|Z(G)|$ is $p$ or $p^2$.
If $|Z(G)| = p$, then $G/Z(G)$ has order $p$, so it is cyclic, and [[this criterion|0AH9Q6YNMT]] 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|0JZ27DMJ5D]] $D_4$ and the [[quaternion group|0NRJGW864D]] have order $8$ and are not abelian.

## Explanations

### L1 human

A group of order $p^2$ has a centre with more than one element.
If the centre were not the whole group, the quotient by the centre would be cyclic, and that is impossible.
So all elements commute.