Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ is a normal subgroup of $G$.
Then use Sylow’s theorem show that the number of $p$-Sylow subgroups is $1$ for $p=3, 11$.
Proof.
Since the order of $G$ is $33=3\cdot 11$, we consider $p$-Sylow subgroups of $G$ for $p=3, 11$.
Let $n_p$ be the number of $p$-Sylow subgroups of $G$. It suffices to show that $n_p=1$ for $p=3, 11$. In fact, by Sylow’s theorem any two $p$-Sylow subgroups are conjugate. If there is only one $p$-Sylow subgroup $P$, then $g^{-1}Pg=P$ for any $g\in G$, hence $P$ is normal.
By Sylow’s theorem, the number of $p$-Sylow subgroups of $G$ satisfies
\[n_p \equiv 1 \pmod p\]
and $n_3|11$ and $n_{11}|3$.
From these two constraints, we see that we must have $n_3=n_{11}=1$.
Therefore, for each $p$, there is a unique $p$-Sylow subgroup of $G$.
By the consideration above, we conclude that any $p$-Sylow subgroup of $G$ is a normal subgroup of $G$.
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 […]
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 […]
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 […]
A Group of Order $pqr$ Contains a Normal Subgroup of Order Either $p, q$, or $r$
Let $G$ be a group of order $|G|=pqr$, where $p,q,r$ are prime numbers such that $p<q<r$.
Show that $G$ has a normal subgroup of order either $p,q$ or $r$.
Hint.
Show that using Sylow's theorem that $G$ has a normal Sylow subgroup of order either $p,q$, or $r$.
Review […]
Group of Order 18 is Solvable
Let $G$ be a finite group of order $18$.
Show that the group $G$ is solvable.
Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such […]
Subgroup Containing All $p$-Sylow Subgroups of a Group
Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.
Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.
Then show that $N$ contains all $p$-Sylow subgroups of […]
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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