After seeing that groups of order $pq$ are never simple, we now study their structure more precisely. Let $p$ and $q$ be primes with $p>q$. The Sylow $p$-subgroup is forced to be normal, and the remaining question is how a subgroup of order $q$ acts on it by conjugation. In this lesson, the focus keyword is groups of order pq. The main result says that such a group is either cyclic or a nonabelian semidirect product, and the nonabelian case can occur only when $q$ divides $p-1$. :::definition[Group of Order pq] Let $p$ and $q$ be prime integers with $p>q$. A group of order pq is a finite group $G$ such that $|G|=pq$. ::: :::theorem Let $G$ be a group of order $pq$, where $p$ and $q$ are prime integers with $p>q$. Then the Sylow $p$-subgroup of $G$ is normal. ::: :::proof Given that $G$ is a group of order $pq$, where $p$ and $q$ are prime integers with $p>q$. To prove that the Sylow $p$-subgroup of $G$ is normal. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. By Sylow's third theorem, $$\begin{align*} n_p &\equiv 1 \pmod p, \\ n_p &\mid q. \end{align*}$$ Thus $n_p=1$ or $n_p=q$. Since $q<p$, the congruence $q\equiv 1 \pmod p$ is impossible. Therefore $$\begin{equation*} n_p=1. \end{equation*}$$ The unique Sylow $p$-subgroup is normal. Hence the Sylow $p$-subgroup of $G$ is normal. $\square$ ::: :::theorem Let $G$ be a group of order $pq$, where $p$ and $q$ are prime integers with $p>q$. If $q\nmid (p-1)$, then $G$ is cyclic. ::: :::proof Given that $G$ is a group of order $pq$, $p>q$, and $q\nmid (p-1)$. To prove that $G$ is cyclic. By the preceding theorem, the Sylow $p$-subgroup $P$ is normal in $G$. By Cauchy's theorem, choose $a\in G$ of order $q$, and let $Q=\langle a\rangle$. Also choose $b\in P$ with $P=\langle b\rangle$. Since $P$ is normal, conjugation by $a$ induces an automorphism of $P$. Therefore there exists an integer $r$ such that $$\begin{equation*} a^{-1}ba=b^r. \end{equation*}$$ Because $a^q=e$, applying this conjugation $q$ times gives $$\begin{equation*} b=b^{r^q}. \end{equation*}$$ Thus $$\begin{equation*} r^q\equiv 1 \pmod p. \end{equation*}$$ If the action is nontrivial, then $r$ has order $q$ in $(\mathbb{Z}/p\mathbb{Z})^\times$, whose order is $p-1$. Hence $q\mid (p-1)$, contradicting the hypothesis. Therefore the action is trivial, so $a$ commutes with $b$. Since $G=PQ$, $P$ and $Q$ are cyclic of relatively prime orders and commute elementwise. Thus $$\begin{equation*} G\cong P\times Q\cong C_p\times C_q\cong C_{pq}. \end{equation*}$$ Hence $G$ is cyclic. $\square$ ::: This preview shows how the arithmetic condition $q\mid(p-1)$ controls groups of order $pq$ when $p>q$. Choose two primes or use a preset from the examples and exercises. The Sylow $p$-subgroup is always normal, so the remaining question is whether a subgroup of order $q$ can act nontrivially on $C_p$. When $q\nmid(p-1)$, no nontrivial action exists and every group of order $pq$ is cyclic. :::scientific-preview[Order pq Structure Explorer]