Let \(G\) be a finite group and let \(g \in G\). Prove that the order of \(g\), denoted \(\operatorname{ord}(g)\), divides the order of the group \(|G|\).
Solution
The cyclic subgroup generated by \(g\) has order \[ |\langle g\rangle|=\operatorname{ord}(g). \] Since \(\langle g\rangle\le G\), Lagrange’s theorem gives \[ |G|=[G:\langle g\rangle]\,|\langle g\rangle|. \] Therefore \[ \boxed{\operatorname{ord}(g)\mid |G|}. \] In particular, writing \(|G|=m\operatorname{ord}(g)\) gives \[ g^{|G|}=(g^{\operatorname{ord}(g)})^m=e. \]