# 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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