Exercise 4.4.12 (If $|G| = 3825$, $H \unlhd G$ and $|H| = 17$, then $H \leq Z(G)$)

Question

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

Richard Ganaye · Accepted Answer

\begin{proof} By hypothesis,
$$|G| = 3825 = 3^2\cdot 5^2\cdot17,$$
and $|H| = 17$, $H\unlhd G$.

Since $H \unlhd G$, we know that
$$N_G(H) = G.$$
By Corollary 15, $N_G(H)/C_G(H) = G/C_G(H)$ is isomorphic to a subgroup $K$ of $\mathrm{Aut}(H)$. Moreover, by Proposition 16, $\mathrm{Aut}(H) \simeq (\mathbb{Z}/17\mathbb{Z})^	imes$, where $17$ is a prime number,  thus
$$|\mathrm{Aut}(H)| = 16.$$
By Lagrange's Theorem,
$$|G/C_G(H)| = |K|	ext{ divides } 16 = 2^4,$$
therefore 
$$|G/C_G(H)| = 2^k,\qquad 	ext{where }  0 \leq k \leq 4.$$
But $|G/C_G(H)|  = |G| / |C_G(H)|$ is a divisor of $|G| = 3825$, which is odd, thus $|G/C_G(H)|$ is odd, and is a power of $2$. Hence 
$$|G/C_G(H)| = 1.$$
Therefore
$$C_G(H) = G.$$
In other words, every element of $H$ commutes with every element of $G$, so
$$H \leq Z(G).$$
\end{proof}

Exercise 4.4.12 (If $|G| = 3825$, $H \unlhd G$ and $|H| = 17$, then $H \leq Z(G)$)

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Exercise 4.4.12 (If $|G| = 3825$, $H \unlhd G$ and $|H| = 17$, then $H \leq Z(G)$)

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