If $p,q$ are primes, then there are only two groups of size $pq$ up to an isomorphism?

Question

Let $p,q$ be primes. Show there are exactly two groups of order $pq$, up to an isomorphism

$\mathbb{Z}_p\times\mathbb{Z}_q\simeq\mathbb{Z}_{pq}$

I am not too sure what this question is asking or if it even makes sense. I am rather new to group theory, and would appreciate it if someone could guide me through this problem, or direct me to the actual question incase the above makes no sense.

I did some research and it has to do with something regarding the Sylow Theorem's. The above "$\times$" indicates a direct product. I am however, not familiar with semi-direct products.

I apologize regarding the ambiguity of the question and the lack of work. Any help would be much appreciated. Thanks in advance!

Answer 1

If $p=q$, there are two groups of order $p^2$, namely $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_{p} \oplus \mathbb{Z}_p$. If $p,q$ are distinct primes, suppose $p<q$. Then there are two cases.

If $p$ does not divide $q-1$, then the only group of order $pq$ is the cyclic one. The assumption $p$ does not divide $q-1$ in conjunction with Sylow's theorems allows one to conclude that there are unique subgroups of order $p$ and $q$ respectively. The union of these two subgroups has cardinality $p+q-1$, and since $p+q-1<pq$ you can obtain an element of order $pq$ in the ambient group.

If $p$ does divide $q-1$, then we have two groups: in addition to the cyclic group of order $pq$, there is also going to be (up to isomorphism) one unique non-abelian (indeed, centreless) group of order $pq$, which can be obtained using semi-direct products. An example of this is $S_3$, which is not isomorphic to $\mathbb{Z}_6$.