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 Zpq, and Zp + Zq, 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 + Z5 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 Zpq. 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 Zp + Zq. So once again Zpq is isomorphic to Zp + Zq, 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 Zpq is not isomorphic to Zp + Zq.