TITLE: Union of conjugates of a closed subgroup of a compact group
QUESTION [6 upvotes]: Let $G$ be a compact Hausdorff group, $H \leq G$ a closed subgroup of infinite index in $G$. 
Is it possible that the conjugates of $H$ cover some open neighbourhood of $1$ in $G$ (or the whole of $G$)?
If this is possible, I would like to know whether there are conditions on $G$ rendering this impossible.

REPLY [11 votes]: The usual example is $G = {\rm SU}_2({\bf C})$ and $H$ the diagonal subgroup.
Every unitary matrix is diagonalizable, and thus contained in a
conjugate of $H$.

REPLY [4 votes]: If your group $G$ is profinite and if $H$ is a proper closed subgroup, then $G$ cannot be a union of conjugates of $H$.  This is because there must be a finite image $G_0$ of $G$ in which the image $H_0$ of $H$ is proper.  Since the conjugates of $H_0$ cannot cover $G_0$ by the well-known result for finite groups, the conjugates of $H$ cannot cover $G$.