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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.

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

2 Answers 2

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 $H$.

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There is nothing special about $SU_2(\mathbb{C})$, $G$ may be any compact connected nonabelian Lie group and $H$ its maximal torus. See en.wikipedia.org/wiki/Maximal_torus#Properties –  Adam Przeździecki Jul 9 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 $G$.

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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 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 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 at 19:23
    
I also agree with David and have been thinking about this on and off today. –  Benjamin Steinberg Jul 9 at 19:26
    
Oops - yes the last sentence was added too hastily. –  Adam Przeździecki Jul 9 at 20:03

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