By Herbert S. Bear
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Additional info for Lectures on Gleason Parts
Let N I be the inverse image of W I in N , and recall that U I = U ∩ U w I . Let L I =
19(v) told us that Bδ ∪ Bδ sδ Bδ is a group, so sδ Bδ sδ ⊆ Bδ ∪ Bδ sδ Bδ . Thus we have our claim. We must show that the BN-pair of L I is strongly split. We have seen that Bw I = X I T , a semi-direct product where X I = U ∩ Bw I = U ∩ U x0 w I . So, given J ⊆ I , we must check X I ∩ (X I )w J X I . We have X I = U ∩ w0 w I wJ w0 w I wJ w0 w I w J U and X I ∩ (X I ) = U ∩ U ∩U ∩U . Knowing that U ∩ U wJ U by the strongly split condition satisfied in G, it suffices to check that U ∩ U w0 w I ∩ U w J ∩ U w0 w I w J = U ∩ U w0 w I ∩ U w J .
We state (and prove) the following results for future reference. 29. Let I ⊆ . Then the following hold. (i) NG (U I ) = PI and U I is the largest normal p-subgroup of PI . (ii) If g ∈ G is such that g U I ⊆ U , then g ∈ PI and g U I = U I . If moreover g U I = U J for some J ⊆ , then I = J . Proof. (i) NG (U I ) contains PI , so NG (U I ) is a parabolic subgroup PJ with J ⊇ I . Assume δ ∈ \ I is such that (U I )sδ = U I . Since X δ ⊆ U I , we have X −δ = (X δ )sδ ⊆ U I ⊆ U , a contradiction. So J = I .