Basic Results on Constrained Sets of Involutions

Let us assume that Si K) K) (L). Among the elements in (L) not contained in Si K) K) we pick s such that (s) is as small as possible. 

For the remainder of this section, we assume L to be constrained. Instead of we shall write . 

By way of contradiction, we assume that Q is not empty. We pick an element q in Q, and we do this in such a way that q) is as small as possible. 

Since q tl and q) = t) + 1, g = tl. Thus, the right hand side of the above equation is equal to UpqrUtiq. 

Let us first prove that I e R. If 1 = ni, this follows from our hypothesis that 0 L) C i . If 2 ni, we obtain from Lemma 2.3.8(h) that / G //. Thus, 

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