Show that the order of any element in a finite group divides the order of the group.
Solution
Let \(G\) be finite and let \(g\in G\).
1. The cyclic subgroup \(\langle g\rangle\) has order equal to the order of \(g\).
By definition, the order \(|g|\) is the least positive integer \(n\) such that \(g^n=e\). Then \[ \langle g\rangle=\{e,g,g^2,\ldots,g^{n-1}\} \] has exactly \(n=|g|\) elements.
2. Lagrange’s theorem gives\[ |G|=[G:\langle g\rangle]\,|\langle g\rangle|. \]
The left cosets of the subgroup \(\langle g\rangle\) partition the finite group \(G\), and each coset has \(|\langle g\rangle|\) elements. Their number is the index \([G:\langle g\rangle]\).
3. Therefore \(|g|\) divides \(|G|\).
By 1 and 2, \[ |G|=[G:\langle g\rangle]\,|g|, \] and the index is a positive integer.