Non-Abelian Group of Order $pq$ and its Sylow Subgroups

Group Theory Problems and Solutions in Mathematics

Problem 293

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 $q$.

 
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Hint.

Use Sylow’s theorem. To review Sylow’s theorem, check out the post Sylow’s Theorem (summary). Read the corollary there as well to understand the proof below.

Proof.

Let $n_p, n_q$ be the number of $p$-Sylow subgroups and $q$-Sylow subgroups, respectively.
Then by Sylow’s theorem, we have
\[n_p\equiv 1 \pmod p, \qquad n_p|q \tag{*}\] and
\[n_q \equiv 1 \pmod q, \qquad n_q|p. \tag{**}\] Since $q \equiv 1 \pmod p$, we have $q>p$. Thus $n_q$ must be $1$ from (**).
Hence, $G$ has a unique $q$-Sylow subgroup, and it is normal.

From (*), the possibilities for $n_p$ are either $1$ or $q$.
We eliminate the possibility of $n_p=1$ as follows.
If $n_p=1$, then $G$ has a unique $p$-Sylow subgroup, and hence it is normal.
Let $P, Q$ be the unique normal $p$-Sylow subgroup and $q$-Sylow subgroup of $G$, respectively. Then $P, Q$ are normal subgroup of order $p$ and $q$, and hence $P \cap Q=\{e\}$, where $e$ is the identity element of $G$.
Since the order $|PQ|=pq=|G|$, these conditions imply that we have
\[G\cong P \times Q.\] Since $P$ and $Q$ are groups of prime order, hence it is cyclic, in particular abelian.
The direct product of abelian group is abelian, hence $G$ is abelian.
This is a contradiction.

Thus, we must have $n_p=q$.


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  1. 06/18/2017

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