Proving that a subgroup is conjugate-dense

This is a survey article related to:conjugate-dense subgroup
View other survey articles about 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 information: 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 $G$ 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 $g \in G$,