Exercise 4.5.24 (If $|G| = 231$, then $Z(G)$ contains a Sylow $11$-subgroup of $G$)

Question

Prove that if $G$ is a group of order $231$ then $Z(G)$ contains a Sylow $11$-subgroup of $G$ and a Sylow $7$-subgroup is normal in $G$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $|G| = 231 = 3 \cdot 7 \cdot 11$. By Sylow's Theorem,
\begin{align*}
&n_{11} \mid 3 \cdot 7 = 21,\qquad n_{11} \equiv 1 \pmod {11},\
&n_7 \mid 3\cdot 11 = 33,\qquad n_{7} \equiv 1 \pmod 7,\
\end{align*}
Therefore
$$n_{11} = 1,\qquad n_{7} = 1,$$
So $G$ contains a normal Sylow $11$-subgroup $P$ and a normal $7$-subgroup $Q$.

\bigskip
Since $P$ is normal in $G$, $G$ acts by conjugation on $P$. This action affords a homomorphism
$$
\varphi 
\left\{
\begin{array}{ccl}
 G &	o &\mathrm{Aut}(P)\
 g & \mapsto &\varphi_g
      \left\{
      \begin{array}{ccl}
      P & 	o &P\
      x & \mapsto & gxg^{-1}
      \end{array}
      ight.
 \end{array}
 ight.
 $$
Since $|P| = 11$ is prime, $P$ is cyclic of order $11$, thus $|\mathrm{Aut}(P)| = 10$, and $|G| = 3\cdot 7 \cdot 11$, thus $\mathrm{g.c.d}(|\mathrm{Aut}(P)|, |G|) = 1$. This shows that the homomorphim $\varphi$ is trivial ($|\varphi(G)| = |G|/ |\ker(\varphi)|$, thus $|\varphi(G)|$ is a divisor of $|\mathrm{Aut}(P)|$ and $|G|$, so  $|\varphi(G)| = 1$ and $\varphi(G) = \{\mathrm{id}_P\}$).

Therefore, for all $g\in G$, $\varphi_g = \mathrm{id}_P$, so for all $x \in P$, $gxg^{-1} =x$. This proves that
$$P \leq Z(G).$$

\end{proof}

Exercise 4.5.24 (If $|G| = 231$, then $Z(G)$ contains a Sylow $11$-subgroup of $G$)

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Exercise 4.5.24 (If $|G| = 231$, then $Z(G)$ contains a Sylow $11$-subgroup of $G$)

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