For each prime *p*, there is only one group with *p* elements, the cyclic group with that many elements. It would be plausible to think there is only one group of order *n* if and only if *n* is prime, but this isn’t the case.

If *p* and *q* are primes, then there are ostensibly at least two groups of order *pq*: the cyclic group *Z*_{pq}, and *Z*_{p} + *Z*_{q}, the direct sum of the cyclic groups of orders *p* and *q*. However, there may just be one group of order *pq* after all because the two groups above could be isomorphic.

If *p* = *q* = 2, then *Z*_{4} and *Z*_{2} + *Z*_{2} are not isomorphic. But the groups *Z*_{15} and *Z*_{3} + *Z*_{5} *are* isomorphic. That is, there is only one group of order 15, even though 15 is composite. This is the smallest such example.

Let *p* and *q* be primes with *p* > *q*. If *q* does not divide *p*-1, then there is only one group of order *pq*. That is, all groups of order *pq* are isomorphic to the cyclic group *Z*_{pq}. So when *p* = 5 and *q* = 3, there is only one group of order 15 because 3 does not evenly divide 5-1 = 4. The same reasoning shows, for example, that there must only be one group with 77 elements because 7 does not divide 10.

Now if *q* does divide *p*-1, then there are two distinct groups of order *pq*. One is the cyclic group with *pq* elements. But the other is non-Abelian, and so it cannot be *Z*_{p} + *Z*_{q}. So once again *Z*_{pq} is isomorphic to *Z*_{p} + *Z*_{q}, but there’s a new possibility, a non-Abelian group.

Note that this does not contradict our earlier statement that *Z*_{4} and *Z*_{2} + *Z*_{2} are different groups, because we assumed *p* > q. If *p* = *q*, then *Z*_{pq} is not isomorphic to *Z*_{p} + *Z*_{q}.

## Related posts