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Exact(12)
Let S < G be a split torus.
Let H be a reductive X -group and S ⊂ H a split torus.
If S ⊂ H is a split torus then Z H ( S ) is a closed reductive subgroup (see [41, XIX, 2.2]).
This is a closed subgroup of H which is a split torus and which is not properly included in any other split torus of H.
This is a simple simply connected R -group of absolute type S L 4, R. It contains a split torus S whose R –points are matrices of the form x 0 0 x - 1 where x ∈ R ×.
But then m : S × G m, R → H yields a split torus of H that properly contains S (since the multiplication map has finite kernel), which contradicts the maximality of S. □.
Similar(48)
In view of the last Lemma and Proposition 8.1 it will suffice to show that if S is a maximal split torus of G, then H Z a r 1 ( R, Z G ( S ) ) = 1.
Let G be a reductive group scheme over R. We say that a maximal split torus S of G is generically maximal split if S K is a maximal split torus of G K. Let S be a generically maximal split torus of G.
By Theorem 7.1 m is a MAD subalgebra if and only if T d is a maximal split torus of G, in which case m = m ( T d ).
Let S be a maximal split torus of our simply connected R GgR Gp G.
As before, S ′ is also a maximal split torus of G.
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