Abelian simple groups have prime order

Statement

Let $G$ be an abelian group. Then $G$ is simple if and only if $G$ is finite of prime order. In that case, $G$ is cyclic, and $G \cong \mathbb{Z}/p\mathbb{Z}$ for a prime number $p$ .

Remarks

In an abelian group, each subgroup is normal, so a simple abelian group has no subgroup except $\{e\}$ and itself. Then $G = \langle g \rangle$ for each $g \neq e$ , so $G$ is cyclic; $\mathbb{Z}$ is not simple, because $2\mathbb{Z}$ is a proper nontrivial subgroup. A finite cyclic group of order $n$ has a subgroup of order $d$ for each divisor $d$ of $n$ , so $n$ must be prime.

Tag	`0JJV80RXHD`
URI	`m2://0JJV80RXHD`
Kind	statement, proposition
Writing	W0 stub
MSC	20E32, 20K01
Status	This copy of the site has no build data. Run `m2 status 0JJV80RXHD`.
Uses	Simple group, Cyclic group, Cauchy's theorem, Subgroups of cyclic groups are cyclic

Tag

0JJV80RXHD

URI

m2://0JJV80RXHD

Kind

statement, proposition

Writing

W0 stub

MSC

20E32, 20K01

Status

This copy of the site has no build data. Run m2 status 0JJV80RXHD.

Uses

Simple group, Cyclic group, Cauchy's theorem, Subgroups of cyclic groups are cyclic

Abelian simple groups have prime order

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