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

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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 $1 \le i,j \le r$.

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 $p_1p_2 \dots p_r$ 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 $i,j$.