Odd-order cyclic group equals derived subgroup of holomorph

Statement
Suppose $$G$$ is an fact about::odd-order cyclic group: a fact about::cyclic group of odd order. Then, $$G$$ equals the fact about::commutator subgroup of the holomorph $$G \rtimes \operatorname{Aut}(G)$$.

Breakdown at the prime two
The analogous statement is not true for all groups of even order. In fact, the commutator subgroup 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) uses::Cyclic implies Aut-Abelian: The automorphism group of a cyclic group is Abelian.
 * 2) uses::Inverse map is automorphism iff Abelian: For an Abelian group, the map sending every element to its inverse is an automorphism.
 * 3) uses::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 commutator subgroup of the holomorph of $$G$$: the semidirect product $$K = G \rtimes \operatorname{Aut}(G)$$.

Proof:


 * 1) $$G$$ contains the commutator 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 commutator 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 $$[K,K]$$.