# Odd-order cyclic group equals derived subgroup of holomorph

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

## 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 .