Let \(p\) be a prime integer, and let \(G\) be a group of order \(|G| = p\). Prove that: (1) \(G\) is cyclic (\(G \cong \mathbb{Z}_p\)). (2) \(G\) is simple (has no non-trivial proper normal subgroups).
Solution
Choose \(g\in G\setminus\{e\}\). By Lagrange, \[ |\langle g\rangle|\mid |G|=p. \] Since \(g\neq e\), the subgroup \(\langle g\rangle\) is nontrivial, so its order is \(p\). Hence \(\langle g\rangle=G\), and \(G\cong C_p\).
Again by Lagrange, every subgroup \(H\le G\) has order dividing \(p\), so \(H=1\) or \(H=G\). Thus \(G\) has no nontrivial proper normal subgroup and is simple.