Common Contracts
1 similar null contracts
On the Zero Slice of the Sphere SpectrumMarch 30th, 2015
is thick. For n = 2, any subgroup of Sn is e or Sn, and our statement is trivial. Hence, we may assume inductively that n ≥ 3 and the theorem is proved for n−1. Since the action of N (H) on V ≥H −V >H is free and the space F = QuotN(H),H (Th(V /V ≥H)) is solid, it is sufficient by Lemma 3.5 to see that F is thick as an N (H)/H-space. Since H ƒ= e, Sn, there exists n > i ≥ 2 such that ji ƒ= 0.
