# 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 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. ## Author: mathtuition88

https://mathtuition88.com/

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