Proving that a subgroup is conjugate-dense

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.

Most of the strategies discussed here work not just for subgroups, but for arbitrary subsets. In other words, given a group $$G$$ and a subset $$H$$, these strategies help prove that every element of $$G$$ is conjugate to some element of $$H$$. While the most special case is that where $$H$$ is a subgroup of $$G$$, other cases of interest arise, for example, when $$H$$ 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 $$G$$ is a group and $$H$$ is a subgroup (or more generally, subset) of $$G$$. Then, $$G$$ acts on itself by conjugation. We want to show that every element of $$G$$ is in the orbit of some element of $$H$$. Equivalently, we want to show that starting with any element $$g \in G$$, we can find an element $$a \in G$$ such that $$aga^{-1} \in H$$.

The step-by-step approach
In this approach, we think of the elements of $$H$$ as extreme elements, and create a gradation in the elements of $$G$$. Next, we show that, starting with any arbitrary element $$g \in G$$,