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
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.
- Odd-order cyclic group is fully characteristic in holomorph
- Odd-order cyclic group is characteristic in holomorph
- 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
Given: A cyclic group of odd order.
To prove: equals the derived subgroup of the holomorph of : the semidirect product .
- 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 .