Every Sylow 11-Subgroup of a Group of Order 231 is Contained in the Center $Z(G)$
Problem 464
Let $G$ be a finite group of order $231=3\cdot 7 \cdot 11$.
Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$.
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Hint.
Prove that there is a unique Sylow $11$-subgroup of $G$, and consider the action of $G$ on the Sylow $11$-subgroup by conjugation.
Check out the post “Sylow’s Theorem (summary)” for a review of Sylow’s theorem.
Proof.
We first claim that there is a unique Sylow $11$-subgroup of $G$.
Let $n_{11}$ be the number of Sylow $11$-subgroups in $G$.
By Sylow’s theorem, we know that
\begin{align*}
&n_{11}\equiv 1 \pmod{11}\\
&n_{11}|21.
\end{align*}
By the first condition, $n_{11}=1, 12, 23 \cdots$ and only $n_{11}=1$ divides $21$.
Thus, we have $n_{11}=1$ and there is only one Sylow $11$-subgroup $P_{11}$ in $G$, and hence it is normal in $G$.
Now we consider the action of $G$ on the normal subgroup $P_{11}$ given by conjugation.
The action induces the permutation representation homomorphism
\[\psi:G\to \Aut(P_{11}),\]
where $\Aut(P_{11})$ is the automorphism group of $P_{11}$.
Note that $P_{11}$ is a group of order $11$, hence it is isomorphic to the cyclic group $\Zmod{11}$.
Recall that
\[\Aut(\Zmod{11})\cong (\Zmod{11})^{\times}\cong \Zmod{10}.\]
The first isomorphism theorem gives
\begin{align*}
G/\ker(\psi) \cong \im(\psi) < \Aut(P_{11})\cong \Zmod{10}.
\end{align*}
Hence the order of $G/\ker(\psi)$ must be a divisor of $10$.
Since $|G|=231=3\cdot 7 \cdot 11$, the only possible way for this is $|G/\ker(\psi)|=1$ and thus $\ker(\psi)=G$.
This implies that for any $g\in G$, the automorphism $\psi(g): P_{11}\to P_{11}$ given by $h\mapsto ghg^{-1}$ is the identity map.
Thus, we have $ghg^{-1}=h$ for all $g\in G$ and $h\in H$.
It yields that $P_{11}$ is in the center $Z(G)$ of $G$.
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