Groups of order 3825 (D&F 4.4.12)

Let \(G\) be a group of order 3825. Prove that if \(H\) is a normal subgroup of order 17 in \(G\) then \(H \le Z(G)\).

Since \(H\) has prime order, \(\text{Aut}(H)\) is isomorphic to \((\mathbb{Z}/17\mathbb{Z})^\times\) which has order 16. Since \(H \triangleleft G\), \(G/C_G(H)\) is isomorphic to a subgroup of \(\text{Aut}(G)\), meaning its order is either 1, 2, 4, 8 or 16. But only 1 divides \(|G| = 3825\), so it follows that \(C_G(H) = G\) and \(H \le Z(G)\).