Classification of groups of order a product of two distinct primes

Statement

Suppose $p$ and $q$ are distinct prime numbers with $p < q$ . Then, there are two possibilities for the number of isomorphism classes of groups of order $p q$ :

If $p$ does not divide $q - 1$ , then there is only one isomorphism class of groups of order $p q$ , namely, the cyclic group.
If $p$ divides $q - 1$ , then there are two possibilities: the cyclic group of order $p q$ and the semidirect product $Z_{q} ⋊ Z_{p}$ where $Z_{q}$ is thought of as the additive group of integers mod $q$ and $Z_{p}$ is identified with the subgroup of order $p$ in $Z_{q}^{*}$ , which is cyclic of order $q - 1$ .

Related facts

Facts used

Every Sylow subgroup is cyclic implies metacyclic

Proof

Specific case: The group is of order $2 p$

See Classification of groups of an order two times a prime

Direct proof using Sylow results

Suppose $G$ is a group and $| G | = p q$ for $p < q$ primes. Let $n_{p}$ , $n_{q}$ be the number of Sylow $p$ -subgroups and Sylow $q$ -subgroups respectively.

Then by Sylow's theorem, $n_{p} \equiv 1 mod p$ , $n_{p} ∣ q$ , and $n_{q} \equiv 1 mod q$ , $n_{q} ∣ p$ .

Since $p < q$ we require $n_{q} = 1$ , as otherwise, if $n_{q} = p$ , $p \equiv 1 mod q$ , contradicting that $p < q$ .

Thus $n_{q} = 1$ . So, the Sylow $q$ -subgroups are normal subgroups.

We have $n_{p} = 1$ or $q$ , so we split the problem into two cases:

Case 1: $n_{p} = 1$

If $n_{p} = 1$ , the Sylow $p$ -subgroups are also normal. Letting $P$ be a particular subgroup of order $p$ and $Q$ be a particular subgroup of order $q$ , we see that for $g \in P$ , $h \in Q$ , $g h = h g$ .

We must also have $P \cap Q = {e}$ - by considering the orders of the elements in $P$ and $Q$ we see only the identity can be in both.

We also have $⟨ g h ⟩ \leq P Q \leq G$ . Taking $g$ and $h$ to be non-trivial elements of $P$ and $Q$ , by the fact $g$ and $h$ commute we see that the order of $⟨ g h ⟩$ is $p q$ . Hence $⟨ g h ⟩ = P Q = G$ , which tells us that $G = P \times Q$ is isomorphic to the cyclic group of order $p q$ .

Case 2: $n_{p} = q$

By the condition on $n_{p}$ , this can only happen when $q \equiv 1 mod p$ . Take particular Sylow subgroups $P$ and $Q$ as before.

We again have $P \cap Q = {e}$ by the same reasoning as in the previous case.

We see that the order of $Q P$ is $p q$ (pick $h$ and $g$ in $Q$ and $P$ non-trivial and consider the elements $h^{j} g^{i}$ for $j = 1, . . ., q$ , $i = 1, . . ., p$ . It follows these are distinct as $P \cap Q = {e}$ .)

This is the condition for $G$ to be a semidirect product $G ≅ Q ⋊_{φ} P$ , where $φ$ is the conjugation action.

Hence we have found a second group of order $p q$ , in the case that $q \equiv 1 mod p$ .

Proof using fact (1) (strong thing to use in proof)

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