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I am interested in finding a method for determining all the matrix subgroups of a matrix group that have a specific connected component. This is what I thought would work from what I have read so far is:

Suppose we have a matrix group $G$ (for example $SO(n)$, $USp(2n)$, $U(n)$ etc.). Further we have a connected subgroup of $G$, $G_1$. I want to find a way to determine all the subgroups $H \subset G$ that have connected component $G_1$.

From what I have read it seems to me that $H \subset N_G(G_1)$, the normaliser of $G_1$ in $G$. Further, there would exists a subgroup $H_1 \subset N_G(G_1)/G_1$ such that all the components of $H$ would be $hG_1$ for $h \in H_1$. Thus to determine all possible $H$ it is sufficient to determine all possible subgroups of $N_G(G_1)/G_1$.

Does this work? Is it true that $H \subset N_G(G_1)$? If not, is there a method known for doing something like this? Any solution or reference would be greatly appreciated.

Mastrel
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  • Interesting question, never thought about it. This shows the limit of Cartan's theorem: you only get connected sbgp out of lie sbalgs. I have a strong feeling that there should be an (algebraic) topology way to do it, since your question is also related to the question that how many components are there in the sbgp. – Troy Woo Aug 29 '14 at 18:54
  • If your connected subgroup is trivial, the larger group is $SU(2)$, and you require your subgroup to be closed, then this is the Mckay correspondence. – Nate Aug 29 '14 at 19:13

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