# groups of order pq

We can use Sylow’s theorems to examine a group $G$ of order $pq$, where $p$ and $q$ are primes (http://planetmath.org/Prime) and $p.

Let $n_{p}$ and $n_{q}$ denote, respectively, the number of Sylow $p$-subgroups and Sylow $q$-subgroups of $G$.

Sylow’s theorems tell us that $n_{q}=1+kq$ for some integer $k$ and $n_{q}$ divides $pq$. But $p$ and $q$ are prime and $p, so this implies that $n_{q}=1$. So there is exactly one Sylow $q$-subgroup, which is therefore normal (indeed, fully invariant) in $G$.

Denoting the Sylow $q$-subgroup by $Q$, and letting $P$ be a Sylow $p$-subgroup, then $Q\cap P=\{1\}$ and $QP=G$, so $G$ is a semidirect product of $Q$ and $P$. In particular, if there is only one Sylow $p$-subgroup, then $G$ is a direct product of $Q$ and $P$, and is therefore cyclic.

Given $G=Q\rtimes P$, it remains to determine the action of $P$ on $Q$ by conjugation. There are two cases:

Case 1: If $p$ does not divide $q-1$, then since $n_{p}=1+mp$ cannot equal $q$ we must have $n_{p}=1$, and so $P$ is a normal subgroup of $G$. This gives $G=C_{p}\times C_{q}$ a direct product, which is isomorphic to the cyclic group $C_{pq}$.

Case 2: If $p$ divides $q-1$, then $\operatorname{Aut}(Q)\cong C_{q-1}$ has a unique subgroup (http://planetmath.org/Subgroup) $P^{\prime}$ of order $p$, where $P^{\prime}={\{x\mapsto x^{i}\mid i\in\mathbb{Z}/q\mathbb{Z},i^{p}=1\}}$. Let $a$ and $b$ be generators for $P$ and $Q$ respectively, and suppose the action of $a$ on $Q$ by conjugation is $x\mapsto x^{i_{0}}$, where $i_{0}\neq 1$ in $\mathbb{Z}/q\mathbb{Z}$. Then $G={\langle a,b\mid a^{p}=b^{q}=1,aba^{-1}=b^{i_{0}}\rangle}$. Choosing a different $i_{0}$ amounts to choosing a different generator $a$ for $P$, and hence does not result in a new isomorphism class. So there are exactly two isomorphism classes of groups of order $pq$.

Title groups of order pq GroupsOfOrderPq 2013-03-22 12:51:05 2013-03-22 12:51:05 yark (2760) yark (2760) 22 yark (2760) Example msc 20D20 SylowTheorems SemidirectProductOfGroups