Let $(W,S)$ be a finite irreducible Coxeter-System of rank $n$ and $E$ be a real reflection representation of $W$. Let $x\in E$ and suppose that the isotropy group of $x$ is generated by one element in $S$. Now which are the subgroups of rank $n-1$ that do not stabilize $x$?
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1$\begingroup$ When you say "do not stabilize" do you mean that the stabilizer of $x$ in the subgroup is trivial? $\endgroup$– S. Carnahan ♦Nov 2, 2013 at 0:42
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$\begingroup$ yes, that is what I mean $\endgroup$– TomNov 4, 2013 at 9:01
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$\begingroup$ If $s_1$ to $s_n$ are the roots and $s_1$ is the root which fixes $x$ then the group generated by $s_2$ to $s_n$ would be what I need. However, is this all that can happen? $\endgroup$– TomNov 4, 2013 at 12:38
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$\begingroup$ Tom: There are other possibilities too. Consider do instance what happens in rank 2 cases. $\endgroup$– MishaNov 4, 2013 at 19:39
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$\begingroup$ No I am a bit confused specially by the "answer" of Ben. $\endgroup$– TomNov 5, 2013 at 8:22
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1 Answer
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You have to consider the Coxeter Graph: The subgroups you are looking for are those that you get by removing one edge...
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1$\begingroup$ This is incorrect and does not even attempt to answer the question. $\endgroup$– MishaNov 4, 2013 at 19:41
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$\begingroup$ why? isnt that all clear from the Coxeter Graph?! $\endgroup$– BenNov 6, 2013 at 17:20
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$\begingroup$ Ben: you also have to think about conjugated of standard parabolics. If you think your "solution" is correct, try to write the details. $\endgroup$– MishaNov 6, 2013 at 17:53