# Difference between revisions of "Odd-order cyclic group equals derived subgroup of holomorph"

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==Statement== | ==Statement== | ||

− | Suppose <math>G</math> is a [[fact about::cyclic group]] of odd order. Then, <math>G</math> equals the [[fact about::commutator subgroup]] of the [[fact about::holomorph | + | Suppose <math>G</math> is a [[fact about::cyclic group]] of odd order. Then, <math>G</math> equals the [[fact about::commutator subgroup]] of the [[fact about::holomorph of a group|holomorph]] <math>G \rtimes \operatorname{Aut}(G)</math>. |

==Related facts== | ==Related facts== |

## Revision as of 23:07, 1 December 2008

## Statement

Suppose is a Cyclic group (?) of odd order. Then, equals the Commutator subgroup (?) of the holomorph .

## Related facts

### Corollaries

- Odd-order cyclic group is fully characteristic in holomorph
- Odd-order cyclic group is characteristic in holomorph

## Facts used

- Cyclic implies Aut-Abelian: The automorphism group of a cyclic group is Abelian.
- Inverse map is automorphism iff Abelian: For an Abelian group, the map sending every element to its inverse is an automorphism.
- kth power map is bijective iff k is relatively prime to the order

## Proof

**Given**: A cyclic group of odd order.

**To prove**: equals the commutator subgroup of the holomorph of : the semidirect product .

**Proof**:

- contains the commutator subgroup of : By fact (1), is Abelian, so is Abelian. Thus, contains the commutator subgroup .
- The commutator subgroup of contains : For this, note (fact (2)) that the inverse map is an automorphism of , say, denoted by an element . The commutator between any and is , so the set of squares of elements of is in . By fact (3), every element of is the square of an element of , so is contained in .