{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# r3 - REVIEW SHEET FOR MATH 411 FINAL EXAM FALL 2011 c 2011...

This preview shows pages 1–3. Sign up to view the full content.

REVIEW SHEET FOR MATH 411 FINAL EXAM, FALL 2011 c 2011 BY SIMAN WONG. ALL RIGHTS RESERVED. General comments about your exams: TWO-HOUR final exam on Wednesday Dec 14, 1:30-3:30pm 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 non-empty 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 2-cycles 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 2-cycles involved is well-defined (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 2-cycles involved is odd or even. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document