Groups of semiprime order

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 Z₄ and Z₂ + Z₂ are not isomorphic. But the groups Z₁₅ and Z₃ + Z₅are 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 Z₄ and Z₂ + Z₂ are different groups, because we assumed p > q. If p = q, then Z_pq is not isomorphic to Z_p + Z_q.

• Number of groups of prime power order
• Orders of finite simple groups
• Homotopy groups of spheres