Groups of order pq

In this post, we will classify groups of order pq, where p and q are primes with p<q. It turns out there are only two isomorphism classes of such groups, one being a cyclic group the other being a semidirect product.

Let $G$ be the group of order $pq$ .

Case 1: $p$ does not divide $q-1$ .

By Sylow’s Third Theorem, we have $n_p\equiv 1\pmod p$ , $n_p\mid q$ , $n_q\equiv 1\pmod q$ , $n_q\mid p$ .

Since $n_q\mid p$ , $n_q=1$ or $p$ . Since $p<q$ and $n_q\equiv 1\pmod q$ , we conclude $n_q=1$ . Similarly, since $n_p\mid q$ , $n_p=1$ or $q$ . Since $p\nmid q-1$ , $n_p\equiv 1\pmod p$ implies $n_p=1$ .

Let $P$ , $Q$ be the Sylow $p$ -subgroup and Sylow $q$ -subgroup respectively. By Lagrange’s Theorem, $P\cap Q=\{1_G\}$ . Thus $|P\cup Q|=p+q-1$ . Since $\displaystyle pq\geq 2q>p+q>p+q-1,$ there is a non-identity element in $G$ which is not in $P\cup Q$ . Its order has to be $pq$ , thus $G$ is cyclic. Therefore $G\cong\mathbb{Z}_{pq}$ .

Case 2: $p$ divides $q-1$ .

From previous arguments, $n_q=1$ hence $Q$ is normal. Thus $QP=PQ$ so $PQ$ is a subgroup of $G$ . $\displaystyle |PQ|=\frac{|P||Q|}{|P\cap Q|}=pq,$ thus $G=PQ$ . $\text{Aut}(Q)\cong(\mathbb{Z}/q\mathbb{Z})^*\cong\mathbb{Z}_{q-1}$ is cyclic, thus it has a unique subgroup $P'$ of order $p$ , where $P'=\{x\mapsto x^i\mid i\in\mathbb{Z}_q, i^p=1\}$ .

Let $a$ and $b$ be generators for $P$ and $Q$ respectively. Suppose the action of $a$ on $Q$ by conjugation is $x\mapsto x^{i_0}$ , where $i_0^p=1$ . (We may conclude this since the action of $a$ on $Q$ by conjugation is an automorphism which has order 1 or $P$ , thus it lies in $P'$ .)

If $i_0=1$ , then $G=P\times Q\cong\mathbb{Z}_{pq}$ .

If $i_0\neq 1$ , then $\displaystyle G=PQ=\langle P,Q\rangle=\langle a,b\mid a^p=b^q=1, aba^{-1}=b^{i_0}\rangle.$ Choosing a different $i_0$ amounts to choosing a different generator $a$ for $P$ , and hence does not result in a new isomorphism class.