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 $G$.

Every Group of Order 72 is Not a Simple Group
Prove that every finite group of order $72$ is not a simple group.
Definition.
A group $G$ is said to be simple if the only normal subgroups of $G$ are the trivial group $\{e\}$ or $G$ itself.
Hint.
Let $G$ be a group of order $72$.
Use the Sylow's theorem and determine […]

Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4
Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.
Hint.
Use Sylow's theorem.
(See Sylow’s Theorem (Summary) for a review of Sylow's theorem.)
Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]

Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]

Sylow Subgroups of a Group of Order 33 is Normal Subgroups
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]

A Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is Normal
Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$.
Then prove that $H$ is a normal subgroup of $G$.
(Michigan State University, Abstract Algebra Qualifying […]

Are Groups of Order 100, 200 Simple?
Determine whether a group $G$ of the following order is simple or not.
(a) $|G|=100$.
(b) $|G|=200$.
Hint.
Use Sylow's theorem and determine the number of $5$-Sylow subgroup of the group $G$.
Check out the post Sylow’s Theorem (summary) for a review of Sylow's […]

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