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 $$p_1p_2 \dots p_r$$, where the $$p_i$$ are pairwise distinct primes, and $$p_i$$ does not divide $$p_j - 1$$ for any $$1 \le i,j \le r$$.

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 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 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 $$p_1p_2 \dots p_r$$ 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 $$p_1p_2\dots p_r$$ we need to impose the additional condition that $$p_i$$ does not divide $$p_j - 1$$ for any $$i,j$$.