MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
If $G$ is finite, then there are no closed subgroups of infinite index so you can drop your last sentence. – Benjamin Steinberg Jul 8 '14 at 17:44
@Benjamin: the assertion is that the condition holds for finite groups, which, as you mention, is a tautology. – YCor Jul 8 '14 at 18:11
Thanks, I have edited my question. – Pablo Jul 8 '14 at 20:07
up vote 11 down vote accepted

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 $H$.

share|cite|improve this answer
There is nothing special about $SU_2(\mathbb{C})$, $G$ may be any compact connected nonabelian Lie group and $H$ its maximal torus. See – Adam Przeździecki Jul 9 '14 at 8:50

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 $G$.

share|cite|improve this answer
You are right, and I already knew this implication. I'm very interested whether the union of conjugates can contain a neighbourhood of $1$ (equivalently, can it contain an open normal subgroup) – Pablo Jul 8 '14 at 20:03
@Pablo: If $1\in G$ has an open basis consisting of groups then no - if $U$ is an open subgroup of $G$ then $G/U$ is discrete. If $G$ is compact then $G/U$ is finite - thus by intersections we may assume $U$ is normal. By Hausdorff if $H$ is closed then there exists such $U$ that $UH\neq G$ then $UH/U\neq G/U$ and argue as above. If you don't want to cover all of $G$ but an open subgroup $V$ then $V$ is also closed and $1\in V$ has an open basis of subgroups and argue as above. – Adam Przeździecki Jul 9 '14 at 9:27
@AdamPrzezdziecki I disagree. The conjugates of $H$ by $G$ could be bigger than the conjugates of $H$ by $V$. In the finite case, we can certainly have $G \supsetneq V \supsetneq H$ so that $V \subset \bigcup_{g \in G} g H g^{-1}$. For example, $H = \mathbb{Z}/p$, $V = (\mathbb{Z}/p)^2$ and $G = GL_2(\mathbb{Z}/p) \ltimes (\mathbb{Z}/p)^2$. I haven't found an example with $[G:V]<\infty$ and $[V:H]=\infty$ yet, though. – David Speyer Jul 9 '14 at 19:23
I also agree with David and have been thinking about this on and off today. – Benjamin Steinberg Jul 9 '14 at 19:26
Oops - yes the last sentence was added too hastily. – Adam Przeździecki Jul 9 '14 at 20:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.