Vipul: New page: {{survey article|conjugate-dense subgroup}} This article discusses general strategies for proving that a subgroup of a group is a conjugate-dense subgroup, i.e., that every element of...

2008-07-15T19:28:10Z

New page: {{survey article|conjugate-dense subgroup}} This article discusses general strategies for proving that a subgroup of a group is a conjugate-dense subgroup, i.e., that every element of...

New page

{{survey article|conjugate-dense subgroup}}

This article discusses general strategies for proving that a subgroup of a group is a [[conjugate-dense subgroup]], i.e., that every element of the whole group is conjugate to some element of the subgroup.

Note that for a finite group, no proper subgroup can be conjugate-dense. More generally, in any group, no proper subgroup of finite index can be conjugate-dense. {{further|[[Union of all conjugates is proper]]}}

Most of the strategies discussed here work not just for subgroups, but for arbitrary subsets. In other words, given a group <math>G</math> and a subset <math>H</math>, these strategies help prove that every element of <math>G</math> is conjugate to some element of <math>H</math>. While the most special case is that where <math>H</math> is a subgroup of <math>G</math>, other cases of interest arise, for example, when <math>H</math> is a union of a few well-chosen subgroups.

==Related techniques==

* [[Proving that a subset generates a group]]
* [[Proving product of subgroups]]
==The general strategy==

Suppose <math>G</math> is a group and <math>H</math> is a subgroup (or more generally, subset) of <math>G</math>. Then, <math>G</math> [[group acts as automorphisms by conjugation|acts on itself by conjugation]]. We want to show that every element of <math>G</math> is in the orbit of some element of <math>H</math>. Equivalently, we want to show that starting with any element <math>g \in G</math>, we can find an element <math>a \in G</math> such that <math>aga^{-1} \in H</math>.

===The step-by-step approach===

In this approach, we think of the elements of <math>H</math> as ''extreme'' elements, and create a gradation in the elements of <math>G</math>. Next, we show that, starting with any arbitrary element <math>g \in G</math>,

Proving that a subgroup is conjugate-dense - Revision history

Vipul: New page: {{survey article|conjugate-dense subgroup}} This article discusses general strategies for proving that a subgroup of a group is a conjugate-dense subgroup, i.e., that every element of...