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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 1 ,..., m are pairwise disjoint then (1) ord( 1 m ) = LCM (ord( 1 ) ,..., ord( 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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