---
tag: XN73MXR9RE
kind: proof
name: Proof that abelian simple groups have prime order
msc: ["20E32", 20K01]
proves: 0JJV80RXHD
uses: [071DDFVG4J, 0C093D1T4M, 0JPDTEDCFD, 0YT586TKXF, 031M8HX7DA, 3C53TD, 05MM4EK4SA, 0736AXH2ME, 0B85Q41RP6]
sources:
  - url: https://en.wikipedia.org/wiki/Simple_group
    locator: Examples
---

## Proof

We prove the two directions, then the last claim.

1. Let $G$ be [[abelian|071DDFVG4J]] and [[simple|0C093D1T4M]]. By definition $G \neq \{e\}$, so we can pick $g \in G$ with $g \neq e$.
2. Each subgroup of $G$ is normal, because [[subgroups of abelian groups are normal|0JPDTEDCFD]]. So the only subgroups of $G$ are $\{e\}$ and $G$.
3. The [[cyclic subgroup|0YT586TKXF]] $\langle g \rangle$ contains $g \neq e$, so by step 2 it equals $G$.
4. Suppose that $g$ has infinite order. Then the powers $g^k$, $k \in \Z$, are distinct, by [[this result|031M8HX7DA]]. So $g^2 \neq e$.
5. In that case $g \notin \langle g^2 \rangle$. Otherwise $g = g^{2k}$ for some $k \in \Z$, and the powers $g^1$ and $g^{2k}$ with $1 \neq 2k$ would be equal.
6. So $\langle g^2 \rangle$ is a subgroup other than $\{e\}$ and $G$, which contradicts step 2. Hence $g$ has finite order $n$.
7. By [[this result|031M8HX7DA]] and step 3, $|G| = |\langle g \rangle| = n$, and $n > 1$ because $g \neq e$.
8. Suppose that $n$ is not prime. Then $n = ab$ with integers $1 < a < n$ and $1 < b < n$.
9. Let $x = g^a$. Then $x^b = g^n = e$. For $0 < j < b$ we have $0 < aj < n$, so $x^j = g^{aj} \neq e$. So $x$ has order $b$.
10. By [[this result|031M8HX7DA]], $|\langle x \rangle| = b$ with $1 < b < n$. So $\langle x \rangle$ is neither $\{e\}$ nor $G$, which contradicts step 2.
11. Hence $G$ is finite and its order $n$ is a prime number. This proves the first direction.
12. Conversely, let $G$ be abelian of prime order $p$. Then $G \neq \{e\}$.
13. Let $N$ be a normal subgroup of $G$. By [[Lagrange's theorem|3C53TD]], $|N|$ divides $p$, so $|N| = 1$ or $|N| = p$. That is, $N = \{e\}$ or $N = G$. So $G$ is simple.
14. Finally, let $G$ be abelian and simple, so $|G| = p$ is prime by step 11. By [[groups of prime order are cyclic|05MM4EK4SA]], $G$ is [[cyclic|0B85Q41RP6]].
15. The group $\Z/p\Z$ is cyclic of order $p$, generated by the class of $1$. By [[this result|0736AXH2ME]], $G \cong \Z/p\Z$.

## Idea

In an abelian group, every subgroup is normal, so simplicity means there are no subgroups at all besides the trivial ones. Any element $g \neq e$ then generates the whole group. A proper power of $g$ would generate a smaller nontrivial subgroup unless the order of $g$ is prime.