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

Statement

Suppose $G$ is an Odd-order cyclic group (?): a Cyclic group (?) of odd order. Then, $G$ equals the Commutator subgroup (?) of the holomorph $G \rtimes \operatorname{Aut}(G)$.

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

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

Proof

Given: A cyclic group $G$ of odd order.

To prove: $G$ equals the derived subgroup of the holomorph of $G$: the semidirect product $K = G \rtimes \operatorname{Aut}(G)$.

Proof:

1. $G$ contains the derived subgroup of $K$: By fact (1), $\operatorname{Aut}(G)$ is Abelian, so $K/G \cong \operatorname{Aut}(G)$ is Abelian. Thus, $G$ contains the commutator subgroup $[K,K]$.
2. The derived subgroup of $K$ contains $G$: For this, note (fact (2)) that the inverse map is an automorphism of $G$, say, denoted by an element $\sigma \in \operatorname{Aut}(G)$. The commutator between any $g \in G$ and $\sigma$ is $g^2$, so the set of squares of elements of $G$ is in $[K,K]$. By fact (3), every element of $G$ is the square of an element of $G$, so $G$ is contained in $[K,K]$.