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Let $G$ be a group, $H \leq G$ a subgroup, and $x \in G$. If $xHx^{-1} \subset H$, then we may not have $xHx^{-1} = H$. Here is a simple counterexample.

But this don't prevent the hope to finding appropriate conditions to guarantee an equality. Important special cases such as when $H$ is finite or $H$ has finite index in $G$, or when $G, H$ are connected compact Lie groups are easily shown to be true. I want to know if this is true when $G$ is a Lie group and $H$ is a closed subgroup (not necessarily connected), and more generally for topological groups; we may also consider algebraic groups or profinite groups. You may impose some mild condition when appropriate. In any case, I want the most general statement.

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  • Any group is a Lie group with the discrete topology, so your hope of getting this for disconnected closed subgroups is vain. – Marc van Leeuwen Dec 03 '15 at 12:38
  • @Marc van Leeuwen But this is not interesting, a Lie group is a manifold, so we usually require it to have a countable basis for its topology. – Lao-tzu Dec 03 '15 at 12:44
  • @Marc van Leeuwen Sorry for the typo, fixed now. – Lao-tzu Dec 03 '15 at 12:47
  • I think there are countable groups that give counterexamples, so I don't see how assuming a countable basis for the topology would be of help in proving such a property. – Marc van Leeuwen Dec 03 '15 at 12:50
  • @Marc van Leeuwen In fact the link I gave is countable group. But if you regard any group as a Lie group with the discrete topology, a genuine Lie group have positive dimension would seems weird. – Lao-tzu Dec 03 '15 at 12:52
  • Descending chain condition. – Myself Dec 03 '15 at 13:42
  • @Myself Could you say some more details? – Lao-tzu Dec 03 '15 at 13:44
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    Unfortunately I don't know much about topological groups, I just wanted to remark that if $H^x < H$, then $H>H^x > H^{x^2} > \dots $ is an infinite descending chain. So if you are in a situation where such chains do not exist (for instance by a dimension argument), then this situation cannot occur. – Myself Dec 03 '15 at 14:03
  • For instance, for the group $G(k)$ of rational points of a linear algebraic group $G$ and a subgroup $H(k)<G(k)$, this will not happen because of dimension reasons coupled with the fact that $H/H^\circ$ is finite. For topological groups, to the best of my knowledge (which is very limited) $H<G$ could be closed, with $H/H^\circ$ "wild" (infinite), even when $G$ is connected. So there you would need to impose some DCC on closed subgroups. – Myself Dec 03 '15 at 14:09
  • @Myself Thank you very much for the idea of DCC. – Lao-tzu Dec 03 '15 at 14:44

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