## Characterization of natural numbers which are only the orders of cyclic groups

Prove the following classification: if $G$ is a finite group of order $p_1p_2 \cdots p_k$ where the $p_i$ are distinct primes such that $p_i$ does not divide $p_j - 1$ for all $i,j$, then $G$ is cyclic.

We proceed by induction on the width of $G$. (Recall that the width of a finite group is the number of prime divisors (including multiplicity) in its prime factorization.)

For the base case, suppose $G$ has width 1. Then $|G| = p$, and clearly $p$ does not divide $p-1$. Moreover, every group of order $p$ (a prime) is cyclic.

For the inductive step, suppose that every finite group of width at most $k$ and whose order’s prime divisors are distinct and satisfy the divisibility requirement is cyclic. Let $G$ be a finite group of width $k+1$, say $|G| = p_1 p_2 \cdots p_{k+1}$, whose order’s prime divisors are distinct and satisfy the divisibility requirement.

Note that every proper subgroup of $G$ has width at most $k$ and satisfies the distinctness and divisibility criteria; by the induction hypothesis, every proper subgroup of $G$ is cyclic. Using this previous exercise, $G$ is not simple (whether or not it is abelian). Let $N \leq G$ be a nontrivial, proper, normal subgroup. Now $N$ is cyclic, and (after a relabeling) we have $|N| = p_1p_2 \cdots p_m$ for some $1 < m < k+1$. By the N/C Theorem, $G/C_G(N) \leq \mathsf{Aut}(N)$. Since $N$ is cyclic, hence abelian, $N \leq C_G(N)$. So $|G/C_G(N)|$ divides $p_{m+1} \cdots p_{k+1}$. By Proposition 16 and properties of the Euler totient function, $|\mathsf{Aut}(N)| = (p_1-1)(p_2-1) \cdots (p_m-1)$. Now if $p_i$ is a prime dividing $|G/C_G(N)|$, then $p_i$ divides some $p_j-1$, a contradiction. Thus $G/C_G(N) = 1$, and we have $N \leq Z(G)$. In particular $Z(G)$ is not trivial. Since $|G/Z(G)|$ properly divides $|G|$, by the induction hypothesis $G/Z(G)$ is cyclic. Hence $G$ is abelian.

By Cauchy's Theorem, for each $p_i$, $G$ has an element $g_i$ of order $p_i$. Since $G$ is abelian, $|g_1g_2 \cdots g_{k+1}| = \mathsf{lcm}(p_1,p_2,\ldots,p_{k+1}) = |G|$. Hence $G$ is cyclic.