Coprime automorphism group implies cyclic with order a cyclicity-forcing number

Statement

Suppose $G$ is a finite group and $\operatorname {Aut} (G)$ is its automorphism group. Suppose, further, that the orders of $G$ and $\operatorname {Aut} (G)$ are relatively prime. Then, there are two possibilities:

• $G$ is the trivial group.
• $G$ is isomorphic to the cyclic group of order equal to $p_{1}p_{2}\dots p_{r}$, where the $p_{i}$ are pairwise distinct primes, and $p_{i}$ does not divide $p_{j}-1$ for any $1\leq i,j\leq r$.

Note that thinking of $1$ as an empty product of primes, we see that the first case can be merged into the second.

(A natural number is termed a cyclicity-forcing number if it satisfies this second condition. See #Related facts for more).

Related facts

• Classification of cyclicity-forcing numbers: Incidentally, every group of order $n$ is cyclic for a natural number $n$ if and only if $n$ is a product of distinct primes with no primes dividing one less than any other prime. In other words, the $n$ for which the only group of order $n$ is cyclic are the same as the $n$ for which the order of the automorphism group is coprime to the order of the group.
• Trivial automorphism group implies trivial or order two

Proof

Proof outline

1. We first show that the group must be Abelian, otherwise the order of the inner automorphism group would divide the order of the group as well as of its automorphism group.
2. Next, we show that for every prime divisor $p$ of the order of the group, the $p$-Sylow subgroup must be cyclic of order $p$. Thus, the whole group is cyclic of order $p_{1}p_{2}\dots p_{r}$ where the $p_{i}$ are distinct primes.
3. Finally, we observe that the automorphism group of a group of this form has order $(p_{1}-1)(p_{2}-1)\dots (p_{r}-1)$. For this to be relatively prime to $p_{1}p_{2}\dots p_{r}$ we need to impose the additional condition that $p_{i}$ does not divide $p_{j}-1$ for any $i,j$.