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

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Revision as of 12:26, 12 November 2008 by Vipul (talk | contribs) (New page: ==Statement== Suppose <math>G</math> is a finite group and <math>\operatorname{Aut}(G)</math> is its automorphism group. Suppose, further, that the orders of ...)

## Statement

Suppose is a finite group and is its automorphism group. Suppose, further, that the orders of and are relatively prime. Then, there are two possibilities:

- is the trivial group.
- is isomorphic to the cyclic group of order equal to , where the are pairwise distinct primes, and does not divide for any .

## Proof

### Proof outline

- 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.
- Next, we show that for every prime divisor of the order of the group, the -Sylow subgroup must be cyclic of order . Thus, the whole group is cyclic of order where the are distinct primes.
- Finally, we observe that the automorphism group of a group of this form has order . For this to be relatively prime to we need to impose the additional condition that does not divide for any .