Let \(G\) be a finite group of order \(n\). Prove that \[ g^n=e \] for every \(g\in G\).
Solution
For any \(g\in G\), the cyclic subgroup \(\langle g\rangle\) has order \(|g|\). By Lagrange’s theorem, \[ |g|\mid |G|=n. \] Write \(n=k|g|\). Then \[ g^n=g^{k|g|}=(g^{|g|})^k=e. \] Thus every element satisfies \(g^{|G|}=e\).