Odd-order cyclic group equals derived subgroup of holomorph

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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

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].