We can use Sylow's theorems to examine a group of order , where and are primes and .

Let and denote, respectively, the number of Sylow -subgroups and Sylow -subgroups of .

Sylow's theorems tell us that for some integer and divides . But and are prime and , so this implies that . So there is exactly one Sylow -subgroup, which is therefore normal (indeed, fully invariant) in .

Denoting the Sylow -subgroup by , and letting be a Sylow -subgroup, then and , so is a semidirect product of and . In particular, if there is only one Sylow -subgroup, then is a direct product of and , and is therefore cyclic.

Given , it remains to determine the action of on by conjugation. There are two cases:

Case 1: If does not divide , then since cannot equal we must have , and so is a normal subgroup of . This gives a direct product, which is isomorphic to the cyclic group .

Case 2: If divides , then has a unique subgroup of order , where . Let and be generators for and respectively, and suppose the action of on by conjugation is , where in . Then . Choosing a different amounts to choosing a different generator for , and hence does not result in a new isomorphism class. So there are exactly two isomorphism classes of groups of order .