Odd-order cyclic group equals derived subgroup of holomorph
From Groupprops
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
- Odd-order cyclic group is fully characteristic in holomorph
- Odd-order cyclic group is characteristic in holomorph
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
.