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 prime factorization of $20$ is $20=2^2\cdot 5$.
Let $n_5$ be the number of $5$-Sylow subgroups of $G$.

By Sylow’s theorem, we have
\[n_5\equiv 1 \pmod{5} \text{ and } n_5|4.\] It follows from these constraints that we have $n_5=1$.

Let $P$ be the unique $5$-Sylow subgroup of $G$.
The subgroup $P$ is normal in $G$ as it is the unique $5$-Sylow subgroup.

Then consider the subnormal series
\[G\triangleright P \triangleright \{e\},\] where $e$ is the identity element of $G$.
Then the factor groups $G/P$, $P/\{e\}$ have order $4$ and $5$ respectively, and hence these are cyclic groups and in particular abelian.

Therefore the group $G$ of order $20$ has a subnormal series whose factor groups are abelian groups, and thus $G$ is a solvable group.

The post A Group of Order $ is Solvable first appeared on Problems in Mathematics.

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, read the post Sylow’s Theorem (summary).

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.

Related Question.

The similar problems are

• Group of order 18 is solvable
• A group of order $pqr$ contains a normal subgroup of order either $p, q$, or $r$

The post Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable first appeared on Problems in Mathematics.

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 that the factor groups $G_i/G_{i-1}$ are all abelian groups for $i=1,2,\dots, n$.

Proof.

Since $18=2\cdot 3^2$, the number $n_3$ of Sylow $3$-subgroups is $1$ by the Sylow theorem.
(Sylow’s theorem implies that $n_3 \equiv 1 \pmod{3}$ and $n_3$ divides $2$.)
Hence the unique Sylow $3$-subgroup $P$ is a normal subgroup of $G$.

The order of $P$ is $9$, a square of a prime number, thus $P$ is abelian.
(See A group of order the square of a prime is abelian.)
Also, the order of the quotient group $G/P$ is $2$, thus $G/P$ is an abelian (cyclic) group.

Thus we have a filtration
\[G \triangleright P \triangleright \{e\}\] whose factors $G/P, P/\{e\}$ are abelian groups, hence $G$ is solvable.

Related Question.

Check the following similar questions.

• Group of order pq has a normal Sylow subgroup and solvable
• A group of order $pqr$ contains a normal subgroup of order either $p, q$, or $r$

The post Group of Order 18 is Solvable first appeared on Problems in Mathematics.