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 .

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 , 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 $G=Q⋊P$, 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 $\mathrm{Aut}(Q)\cong C_{q-1}$ 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 $i_0\ne 1$ in $\mathbb{Z}/q\mathbb{Z}$.Then $G=⟨a,b\mid a^p=b^q=1,ab{a}^{-1}=b^{i_0}⟩$.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 $pq$.