In particular, we will use Sylow’s theorem (3) and (4), and its corollary in the proof below.
For (b), recall that a group $G$ is solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such that the factor groups $G_i/G_{i-1}$ are all abelian groups for $i=1,2,\dots, n$.
Proof.
(a) The group $G$ has a normal Sylow $p$-subgroup
By Sylow’s theorem, the number $n_p$ of Sylow $p$-subgroups of $G$ satisfies $n_p\equiv 1 \pmod{p}$ and $n_p$ divides $q$.
The only such number is $n_p=1$.
Thus $G$ has the unique Sylow $p$-subgroup $P$ of order $p$.
Since $P$ is the unique Sylow $p$-subgroup, it is a normal subgroup of $G$.
(b) The group $G$ is solvable
Let $P$ be the normal Sylow subgroup of $G$ obtained in (a).
Then we have the following subnormal series
\[\{e\} \triangleleft P \triangleleft G,\]
where $e$ is the identity element of $G$.
The factor groups are $G/P$ and $P/\{e\}\cong P$.
The order of the group $P$ is the prime $p$, and hence $P$ is an abelian group.
The order $|G/P|=|G|/|P|=pq/q=q$ is also a prime, and thus $G/P$ is an abelian group.
Thus the factor groups are abelian. Thus $G$ is a solvable group.
A Group of Order $20$ is Solvable
Prove that a group of order $20$ is solvable.
Hint.
Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.
See the post summary of Sylow’s Theorem to review Sylow's theorem.
Proof.
Let $G$ be a group of order $20$. The […]
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 […]
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 […]
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 $P$ […]
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 […]
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 […]
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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