A course in abstract algebra by Nicholas Jackson PDF

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20, page 11. 21, page 12. 22, page 12. 23, page 12. 24, page 13. 25, page 13. 26, page 13. 27, page 14. 28, page 14. 30, page 15. 32, page 16. 34, page 17. 36, page 18. 38, page 18. 35, page 17. 37, page 18. 39, page 18. 40, page 18. 42, page 19. 46, page 22. 43, page 21. 45, page 21. 47, page 22. 48, page 22. 50, page 24. 36 Matrices yield an important and rich class of groups, many of which are nonabelian. 44 Matrix groups led naturally to a discussion of geometric operations on vector spaces, and in particular we considered isometries: operations which preserve distance.

Xk ) = ( x1 xk )( x1 xk−1 ) . . ( x1 x3 )( x1 x2 ) of transpositions. 59 The transpositions in Sn generate Sn . In general, there may be multiple ways to represent a given permutation as a product of transpositions. 58 29 30 a course in abstract algebra gives one valid way, but since any cyclic permutation ( x1 x2 . . xk ) can be rewritten as, for example, ( xk x1 x2 . . xk−1 ) (or any other cyclic permutation of that list), the decomposition will not be unique. 60 The symmetric group Sn is generated by any of the following sets of cyclic permutations: (i) (ii) (iii) {(1 2), (1 3), .

3 4)(2 3)(1 2)(2 3)(3 4) . . (k−1 k). (iii) We can write any transposition of the form (k k+1) as the product (1 2 . . n)k−1 (1 2)(1 2 . . n)1−k , and hence by part (ii), the transposition (1 2) together with the n–cycle (1 2 . . n) generate all of Sn . This completes the proof. As remarked above, the decomposition of a permutation σ into transpositions will not, in general, be unique, nor need those transpositions be disjoint. It need not even be the case that two such decompositions have the same number of transpositions; however we can at least say that the parity (odd or even) of the number of transpositions will be the same for any such decompositions of the same permutation.

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A course in abstract algebra by Nicholas Jackson

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