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Unformatted text preview: REVIEW SHEET FOR MATH 411 FINAL EXAM, FALL 2011 c 2011 BY SIMAN WONG. ALL RIGHTS RESERVED. General comments about your exams: • TWOHOUR final exam on Wednesday Dec 14, 1:303:30pm in LGRC A201 • no book/notes/calculator (not that calculator will help!) • Show your work , and use complete English sentences! • be prepared to recite definitions as well as precise statements of theorems • the final exam is cumulative – you need to know materials covered on both Midterms and all of your problem sets! Permutations • for any nonempty set S , a permutation on S is a bijective set function from S to itself • composition of bijections is a bijection, and the inverse of a bijection is also a bijection, so Perm( S ), the set of permutations on the set, forms a group under composition • if S is a finite set of size n then #Perm( S ) = n ! • in the special case where S = { 1 ,...,n } , we write S n for Perm( S ); it is called the permu tation group on n letters • if n ≤ 3 then S n is isomorphic to D n , but that is false for n > 3 Properties of permutations • Permutations in S n have two different decompositions: (1) as a product of disjoint cycles – this decomposition is unique up to reordering – this decomposition allows us to inverse of a permutation, compose two permuta tions, and compute the order of an element in S n ; e.g. if σ 1 ,...,σ m are pairwise disjoint then (1) ord( σ 1 ··· σ m ) = LCM (ord( σ 1 ) ,..., ord( σ m )) . the proof of this useful fact make uses of, among other things, the crucial prop erty that disjoint permutations in S n commute . Here is a concrete example: consider the permutation (13)(514)(23)(51): 1 → 5 → 5 → 1 → 3 3 → 3 → 2 → 2 → 2 2 → 2 → 3 → 3 → 1 (132) 4 → 4 → 4 → 5 → 5 5 → 1 → 1 → 4 → 4 o (45) Thus (13)(514)(23)(51) = (132)(45), so by (1), we get ord ( (13)(514)(23)(51) ) = 6. (2) as a product of 2cycles – thanks to the disjoint cycle decomposition, it suffices to do this for a cycle: ( a 1 ··· a r ) = ( a r a r 1 ) ··· ( a r a 1 ) – this decomposition is not unique , but it is a crucial fact that the parity of the number of 2cycles involved is welldefined (in other words: the number of 2 cycles involved modulo 2). This allows us to declare that an element in S n is odd or even depends on whether the number of 2cycles involved is odd or even....
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 Winter '11
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