Dummit and Foote 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 $H \leq Z(G)$

Proof: By Corollary 15, $N_G(H)/C_G(H) \lesssim Aut(H)$. The only group of order 17 is $\mathbb{Z_{17}}$, which is cyclic. By proposition 16, $Aut(H) \cong \mathbb{Z_{17}}^* \cong \mathbb{Z_{16}}$. So, $\frac{|N_G(H)|}{|C_G(H)|}$ divides 16 by Lagrange. Since $H \unlhd G$, $N_G(H)=G$, so $\frac{|N_G(H)|}{|C_G(H)|}=\frac{3825}{|C_G(H)|}$. Since $3825=3^2\cdot 5^2 \cdot 17$, the only possibility for the order of $C_G(H)$ is 3825 i.e. $C_G(H)=G$ which means that $H$ commutes with all of $G$ $\therefore H \leq Z(G)$ $\blacksquare$