Order of Element Divides Order of Group
Theorem
Given a finite group \(G\) with an element \(g\) of finite order, \(\mathrm{ord}(g) \mid G\).
Proof
For any \(g \in G\), we have that \(\langle g \rangle\) is a subgroup of \(G\) with size \(\mathrm{ord}(g)\).
Then, from Lagrange's theorem we have that
\[ |G| = [G : \langle g \rangle]|\langle g \rangle|\]
and hence
\[ \mathrm{ord}(g) = \langle g \rangle \mid |G|.\]