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

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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 #Related facts for more).

Related facts


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.