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

From Groupprops

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==Facts used== | ==Facts used== | ||

− | # [[uses::Cyclic implies | + | # [[uses::Cyclic implies abelian automorphism group]]: The automorphism group of a cyclic group is Abelian. |

− | # [[uses::Inverse map is automorphism iff | + | # [[uses::Inverse map is automorphism iff abelian]]: For an Abelian group, the map sending every element to its inverse is an automorphism. |

# [[uses::kth power map is bijective iff k is relatively prime to the order]] | # [[uses::kth power map is bijective iff k is relatively prime to the order]] | ||

+ | |||

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

## Latest revision as of 22:58, 1 July 2017

This article states and (possibly) proves a fact that is true for certain kinds of odd-order groups. The analogous statement is not in general true for groups of even order.

View other such facts for finite groups

## Contents

## Statement

Suppose is an Odd-order cyclic group (?): a Cyclic group (?) of odd order. Then, equals the Commutator subgroup (?) of the holomorph .

## Related facts

### Breakdown at the prime two

The analogous statement is not true for all groups of even order. In fact, the derived subgroup of the holomorph of a cyclic group of even order is the subgroup comprising the squares in that group, which has index two in the group.

### Corollaries

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

## Facts used

- Cyclic implies abelian automorphism group: 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 derived subgroup of the holomorph of : the semidirect product .

**Proof**:

- contains the derived subgroup of : By fact (1), is Abelian, so is Abelian. Thus, contains the commutator subgroup .
- The derived 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 .