Exercise 4.4.13 (If $|G| = 203$, $H \unlhd G$ and $|H| = 7$, then $G$ is abelian.)

Question

Let $G$ be a group of order $203$. Prove that if $H$ is a normal subgroup of order $7$ in $G$ then $H \le Z(G)$. Deduce that $G$ is abelian in this case.

Richard Ganaye · Accepted Answer

\begin{proof} 
 By hypothesis,
$$|G| = 203 = 7 \cdot 29,$$
and $|H| = 7$, $H\unlhd G$.

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, $\mathrm{Aut}(H) \simeq (\mathbb{Z}/7\mathbb{Z})^	imes$, where $7$ is a prime number, thus
$$|\mathrm{Aut}(H)| = 6.$$
By Lagrange's Theorem,
$$|G/C_G(H)| = |K|	ext{ divides } 6,$$
therefore 
$$|G/C_G(H)| \in \{1,2,3,6\}.$$
But $|G/C_G(H)|  = |G| / |C_G(H)|$ is a divisor of $|G| = 203$, and $2, 3, 6$ are not divisors of $203$, 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
$$H \leq Z(G).$$

\bigskip
Since $H \leq Z(G) \leq G$, by Lagrange's Theorem, $7$ divides $|Z(G)|$ and $|Z(G)|$ divides $7 \cdot 29$, thus $|Z(G)| = 7$ or $|Z(G)| = 7 \cdot 29 =  |G|$, so
$$Z(G) = H \qquad 	ext{or} \qquad Z(G) = G.$$
Assume for the sake of contradiction that $Z(G) =H$. Then $|G/Z(G)| = 29$ is prime, thus $G/Z(G)$ is cyclic. By Exercise 3.1.36, $G$ is abelian, but in this case $Z(G) = G 
e H$. This is a contradiction, which shows that $Z(G) 
e H$. 

In conclusion $Z(G) = G$, so $G$ is abelian.

\end{proof}

Exercise 4.4.13 (If $|G| = 203$, $H \unlhd G$ and $|H| = 7$, then $G$ is abelian.)

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Exercise 4.4.13 (If $|G| = 203$, $H \unlhd G$ and $|H| = 7$, then $G$ is abelian.)

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