Difference between revisions of "Odd-order cyclic group equals derived subgroup of holomorph"

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==Statement==
 
==Statement==
  
Suppose <math>G</math> is a [[fact about::cyclic group]] of odd order. Then, <math>G</math> equals the [[fact about::commutator subgroup]] of the [[fact about::holomorph of a group|holomorph]] <math>G \rtimes \operatorname{Aut}(G)</math>.
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Suppose <math>G</math> is an [[fact about::odd-order cyclic group]]: a [[fact about::cyclic group]] of odd order. Then, <math>G</math> equals the [[fact about::commutator subgroup]] of the [[fact about::holomorph of a group|holomorph]] <math>G \rtimes \operatorname{Aut}(G)</math>.
  
 
==Related facts==
 
==Related facts==
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===Breakdown at the prime two===
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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===
 
===Corollaries===

Revision as of 23:10, 1 December 2008

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 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

Facts used

  1. Cyclic implies Aut-Abelian: 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 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].