# Difference between revisions of "Coprime automorphism group implies cyclic with order a cyclicity-forcing number"

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(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 ...) |
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* <math>G</math> is the trivial group. | * <math>G</math> is the trivial group. | ||

* <math>G</math> is isomorphic to the [[cyclic group]] of order equal to <math>p_1p_2 \dots p_r</math>, where the <math>p_i</math> are pairwise distinct primes, and <math>p_i</math> does not divide <math>p_j - 1</math> for any <math>1 \le i,j \le r</math>. | * <math>G</math> is isomorphic to the [[cyclic group]] of order equal to <math>p_1p_2 \dots p_r</math>, where the <math>p_i</math> are pairwise distinct primes, and <math>p_i</math> does not divide <math>p_j - 1</math> for any <math>1 \le i,j \le r</math>. | ||

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+ | ==Related facts== | ||

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+ | * [[Classification of cyclicity-forcing numbers]]: Incidentally, every group of order <math>n</math> is cyclic for a natural number <math>n</math> if and only if <math>n</math> is a product of distinct primes with no primes dividing one less than any other prime. In other words, the <math>n</math> for which the only group of order <math>n</math> is cyclic are the same as the <math>n</math> for which the order of the automorphism group is coprime to the order of the group. | ||

==Proof== | ==Proof== |

## Revision as of 20:28, 15 December 2008

## 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 .

## Related facts

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

## 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 .