How to find the normalizer of a finite subroup in a Lie group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:20:20Z http://mathoverflow.net/feeds/question/101145 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101145/how-to-find-the-normalizer-of-a-finite-subroup-in-a-lie-group How to find the normalizer of a finite subroup in a Lie group? Gang Han 2012-07-02T14:06:08Z 2012-07-02T14:06:08Z <p>If a group \$G\$ is generated by finitely many subgroups \$G_i\$ and \$H\$ a subgroup of \$G\$, under which conditions can \$N_G(K)\$, the normalizer of \$K\$ in \$G\$, be generated by all the normailizers \$N_{G_i}(K)\$?</p> <p>More concretely, let \$G\$ be a compact connected Lie group and \$K\$ a finite abelian subgroup of \$G\$ whose centralizer in \$G\$ is \$K\$, i.e., \$K\$ is a maximal finite abelian subgroup of \$G\$. Assume that \$G\$ is generated by closed connected abelian subgroups \$G_i\$, \$i=1\cdots n\$. (Each \$G_i\$ is a torus.) It is known that \$K\$ normailizes each \$G_i\$. Is \$N_G(K)\$, the normalizer of \$K\$ in \$G\$, generated by all the normailizers \$N_{G_i}(K)\$,\$i=1\cdots n\$ ?</p> <p>This is true for all the examples I know. But I cannot find a proof. Your help is very appreciated!</p>