Let \(p\) be any prime number.
Let \(G\) be a group with order \(p\).
Let \(H\) be any subgroup of \(G\). Then, by Lagrange’s Theorem, \(|H|\) must divide \(p\) and thus, \(|H|=1\) or \(|H|=p\). In either case, we get that \(H\) is a trivial subgroup of \(G\).
Thus, \(G\) is simple and there is no nontrivial normal subgroup of \(G\).
Thus, any group with prime order is simple.