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Free actions if there exists a homeomorphism h: T n + l -) T n + l such that CP2(g) h = h CPI(g) for every g E G. Now let EPz<7Tp3, G) be the set of epimorphisms a of 7Tp3 onto G such that [G: N a ] = 2. for a E EP2(7TP3, G), {3 E Aut(7TP3). The collection EPZ<7T IU 3,G)* of Aut(7T IU 3)-orbits is the set of equivalence classes of EP2( 7T IU 3, G) under this action. This means that epimorphisms a l and a2 are equivalent if there exists an element {3 in AUt(7TP3) such that a 2 = a l o {3-I. A proof of the following theorem can be found in [6, p.

If the votes are counted in a random order, with all sequences equally likely, we may wonder what the probability is that the first candidate never trails. Using O's to designate votes for the first candidate and l's to designate votes for the second, we are asking for the fraction of sequences with k l's and k O's such that every prefix (initial segment) has at least as many O's as l's. These sequences are the ballot sequences of length 2k. To compute the desired probability, we count the ballot sequences and divide bye:).

Consider i such that p ~ i ~ qk + p; we have p - 1 ~ i - 1 ~ qk + p - 1 and qk + 2(p - 1) - (i - 1) = qk + 2p - i - 1. By the induction hypothesis, a' has at least qk + 2p - i - 1 O-linearizations in which at least i - 1 O-intervals are q-good. By' the augmentation property, the replacement of the missing 0 converts these to O-linearizations of a in which at least i O-intervals are q-good. Since every O-interval in b is q-good, b provides the additional needed O-linearization. • The extended Strong Cycle Lemma is best possible in the sense that all its lower bounds may hold with equality simultaneously.

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