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 $G$?


First of all, we use the fact that the order of $G'$ and $G/G'$ are co-prime, hence we can use Schur-Zassenhaus theorem (in particular, statement of theorem 1 here) to show $G$ must be a semidirect product of $G'$ and $G/G'$.

Now we look at how can $G/G'$ act on $G$ via automorphisms. If we manage to show that the only action is the trivial action, we'll have shown there's only one semidirect product of $G'$ and $G/G'$, which is the direct product. Since $G'$ and $G/G'$ are cyclic, the direct product will be abelian.

Let the order of $G'$ be $Q=p_1p_2\ldots p_n$ and order of $G/G'$ be $W=p_{n+1}\ldots p_m$ such that $p_i$s are distint and unrelated. Let $1_N$ the identity of $G'$ and $1_H$ be the identity of $G/G'$. Since both the groups are cyclic, and the action is via automorphisms, the action is completely determined by the value of: $$g_H(g_N)$$ where $g_H$ and $g_N$ are respective generators. If we set it to be identity: $$g_H(g_N) = g_N$$ then the action is the trivial one. Now set it to be something else, and we'll show that can't define a valid action. Let it be: $$g_H(g_N) = (g_N)^r$$ where $0 \leq r < Q$ In that case, \begin{align} (g_H)^2(g_N) &= g_H(g_H(g_N))\\ &= g_H((g_N)^r)\\ &= (g_H(g_N))^r\\ &= (g_N)^{r^2} \end{align} Since we're dealing with cyclic groups, let's just change the notation and write \begin{align} (2 \mod W)(1 \mod Q) \equiv r^2 \mod Q \end{align} Similarly, \begin{align} (W \mod W)(1 \mod Q) \equiv r^W \mod Q \end{align} But $$(0 \mod W)(1 \mod Q) \equiv 1 \mod Q$$ Therefore: $$r^W \equiv 1 \mod Q$$ One solution to this is $r=1$. If $r \neq 1$, then by Lagrange's theorem, $W | T$, where $T$ is size of unit group of modulo $Q$. In particular, $$p_1 | (p_{n+1}-1)\ldots (p_m-1)$$ which means it divides some $p_i-1$, which is false.

Hence the only action is trivial, which shows the only semidirect product is the direct product and the group is abelian.


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