Consider a group of size $n$, where $n$ is the product of distinct unrelated primes (two primes $p$ and $q$ are unrelated if $q \nmid (p-1)$ and $p \nmid (q-1)$). The claim is that there is only one group (up to isomorphism) of this order (the cyclic one). I'm reading a proof of this claim from this article, where the proof is the latter part of proposition 1. The gist of the proof is, if $G$ is abelian, you're already done, because you can pick elements with order equal to each of the prime divisors of $n$, and the product of those elements has order equal to the order of the group (that's where the commutativity comes in).
Next step in the proof is showing that $G$ cannot not be abelian, by assuming it is. Then the proof claims groups of order $n$ are metacyclic, i.e. they have a normal subgroup $G'$ such that $G'$ and $G/G'$ are both cyclic. Since $G$ is not even abelian, let alone cyclic, $G'$ is non-trivial. Now here comes the part where I'm stuck at:
[T]here has to exist a relation between a prime divisor of $|G:G'|$ and a prime divisor of $|G'|$. Otherwise $G'$ would be contained in the center of $G$, and thus be a direct factor of $G$. This is clearly not possible.
My question is, if $|G'|$ and $|G:G'|$ have no related prime divisors, then why should $G'$ be in the center of $G$?