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Assume that $G$ is a connected compact Lie group (or a connected complex reductive group), and $K$ is a diagonalizable subgroup of $G$. It is known that the Weyl group $W_G(K)$ of $K$, defined as $N_G(K)/Z_G(K)$, is always finite.

If $K$ is a maximal torus of $G$, then $W_G(K)$ is generated by root transvections. My question is, if $K$ is a disconnected diagonalizable subgroup, how to determine $W_G(K)$? For example, if $K$ is a maximal finite diagonalizable subgroup, how to determine $W_G(K)$? Any reference or anwsers are very appreciated!

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Your terminology "Weyl group" is nonstandard in this generality and unrelated to the question being asked. Originally Weyl groups are only defined relative to a maximal torus, in the setting of compact Lie groups or reductive algebraic groups over an arbitrary algebraically closed field. [As Borel pointed out, those finite groups do play an essential role in Weyl's study of finite dimensional representations but actually appeared earlier in Cartan's work.] Also, "root transvection" is unclear: do you mean "reflection with respect to a root"? –  Jim Humphreys Jul 8 '12 at 13:23
Thanks! Sorry I made a mistake, if $K$ is a maximal torus of $G$ , then $W_G(K)$ is generated by reflections with respect to a root. –  Gang Han Jul 25 '12 at 12:35
P.S. Instead of writing an "answer", you should edit your original question and take into account the first line of my comment. If you edit the terminology and notation to avoid reference to Weyl, there is still a question there, though it probably doesn't have a reasonable answer. –  Jim Humphreys Jul 27 '12 at 16:09
For a finite $K$, $N_G(K)/Z_G(K)$ is contained in the group of automorphisms of $K$, which is finite, so it is in theory feasible to check each automorphism and see if there is an element of $G$ realizing it. If $K$ is contained in a unique maximal torus, which for $SU(n)/SL_n$ happens if its eigenspaces are one-dimensional, then it's a subgroup of the Weyl group of that torus that preserves a certain lattice extension of the root lattice, which it's easy to check. –  Will Sawin Sep 5 '12 at 15:26
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