Proving that a subgroup is conjugate-dense

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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

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 $g \in G$ ,

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