# Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable

## Problem 245

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.

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