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rt.illustration idea – Is the group of integer factors of a actual linear algebraic group a maximal closed subgroup? Answer

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rt.illustration idea – Is the group of integer factors of a actual linear algebraic group a maximal closed subgroup?

Permit $ G $ breathe a unostentatious linear algebraic group. Permit $ G_ mathbb {R} $ breathe the actual factors of $ G $. Permit $ G_ mathbb {Z} $ breathe the integer factors of $ G $. is $ G_ mathbb {Z} $ a maximally closed subgroup? In different phrases, they’re the one closed subgroups $ H $ from $ G_ mathbb {R} $ in order that
$$ G_ mathbb {Z} subset H subset G_ mathbb {R} $$
solely $ G_ mathbb {R} $ and $ G _ { mathbb {Z}} $?

The respond isn’t any for $ SO_3 $. $ SO_3 ( mathbb {Z}) $ is the 24-element octahedral group (isomorphic to the symmetric group $ S_4 $ ) and this group is a maximally finite subgroup of twists and actually including each different twist and taking the closure seems to breathe all $ SO_3 ( mathbb {R}) $.

The respond too appears to breathe no for $ SL_n $. See Ycor’s second respond right here:

https://math.stackexchange.com/questions/3388661/is-sl-2-mathbb-za-maximal-discrete-subgroup-in-sl-2-mathbb-r

Although that solely says that $ SL_n ( mathbb {Z}) $ is most amongst discrete subgroups I think that on this illustration it’s too most amongst closed subgroups.

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