Every Group of Prime Order Cyclic

Theorem

For any finite group \(G\) where \(|G|\) is prime, then \(G = \langle g \rangle\) for some \(g \in G\), that is, \(G\) is cyclic.

Proof

Let \(G\) be a finite group of prime order. Then, for any element \(g \in G\), \(\langle g \rangle \leq G\) and hence by Lagrange's theorem

\[ |G| = [G : \langle g \rangle] |\langle g \rangle|.\]

Given that \(G\) is prime, the only possible values for \(|\langle g \rangle|\) are \(1\) or \(|G|\). If \(|\langle g \rangle| = 1\), then \(g = \mathrm{id}\). In this case we select a different element \(g'\), which must exist since \(G\) is of prime order and so has at least two elements, and the identity is unique. This element therefore must satisfy \(|\langle g' \rangle| = |G|\). As a subgroup of equivalent order, \(\langle g' \rangle = G\).

A direct consequence of this is the fact that every group of prime order is abelian.