Page 1 : WHC Call US CAPIVOSNy SO OE i, , Theorem 3.14. , 5 t of ., Every permutation on a finite set A can be expressed as a product of disjoint, cycles., , Proof : Let o be any permutation on a finite set A., , A = {q), a, 43,°° "> On-1s Gn} ., , Now select any element a; and find (ay) if a(a1) = 41 then (a;) is the first, cycle of the factorization., , If o(a,) # a, but a(a)) = ap, then find (a2)., , If (a2) = a, then (a; a) is a cycle of factorization., , If oa) # a, but A(a2) = a3, then find (a5)., , If (a3) = a, then (a; az a3) is the cycle. Continuing this way, find, Q(ay_;}) = ax and since a is a one-one and onto map hence a, is different from, aj, G2," °° , Qx—j, If oax) = a, then (a) a2 a3 - - * ax_; ax) is a cycle of factorization., , If aay) # a, then let O(ax) = as) and continue., , Since A is finite, this process must terminate after a finite number of steps., , After finding the first cycle factor, we select any other element in the above, cycle which is not there in the first cycle and continue in the, , exhausting all the elements of A we get the required factorization, Example 3.25. :, , rGasaorer de 2 es 967, ed armor oa 248s aG ag, = (1248) (359) (67),, , same way. Thus after, ‘ Oo, , which is product of 3 cycles., , eT re, , a

Page 3 : Therefore, P(A) => P(A + 1), , Thus, P(n) is true Vn € N. + transpositions., t of transp', , Hence, every cycle can be expressed as a produc, , «Now consider (a, a) (a a1) = (1) number of pairs of these type, , Thus if we multiply the original product by ie number of transpositions jn, , which will not change the cycle, but it will increase, , the cycle ; . oe, , ct of transpositions in infinitely, , Thus every cycle can be expressed as a produ y, , many ways fF, 3.7 Even and Odd Permutations, , Definition 3.16., , A permutation in S,, is called even if it can be expressed as a product of even, number of transpositions and it is called odd if it can be expressed as a product of odd, number of transpositions (where S,, is a set of all permutations on 7” objects)., , Note 3.6., 1) A product of two even permutations is even., 2) A product of two odd permutations is even., , 3) A product of one even_and one odd permutation is odd., Theorem 3.16., , In a symmetric group S,, (” > 2) the number of odd ‘ «5, the number of even permutations. : Permutations is equal t0, , Proof: Let {f), fo, -- - fp} be the set of ‘ i, {81, 82, °° * Sg} be the set of all even Permutations, ley Permutations and, , Now multiplying {f;, fo, - - - Fp}, , by fi we 2, get 2, are distinct and they are all even Permutations a Fifa, + - ff, which, , e, , umber and total numbe:, , are q in number. T of even permutations, , s PS, , __ Now multiplying {81, g2, Bi) by f, we (1), which are all distinct and they are all odd Permutat get fig, fig>, figs, « - > fide, and an even permutation is odd, but these are Pr ‘ons, since the st ten of, , : ; i, permutations are p in number. N Number and total number of odd

Page 4 : - gsp (2), , From (1) and (2) p = q and p+ gq =(|n = total number of permutations on 7, objects., , oO, , nis, , eqs, , Theorem 3.17, , A subset A, consisting of all the even permutations of S,, (m 2 2) is a subgroup, of the symmetric group S,,., , Proof :1) Let fg ¢€ A, Then f and g are both even., Thus f- g is also an even permutation and fg € An., Therefore 4, is closed w. r. t. composition of permutations., 2) Forf, g,he€ A, where A, Cc S, fg, A € Sy and (Sp, -) is a group., Therefore Associative law holds, (f- 9). h=f- (g A). . 3) Identity permutation in an even permutation., Since I = (? e = & ®) (é ®) product of two permutations., , Te An., , 4) If f-! is the inverse of an even permutation then f—! cannot be odd otherwise, f - $7! will be odd or I will be odd., , Hence, Vf € An, fo € An, , The inverse of every even permutation exists in A,. Hence (A,, -) is a group, and AnC Sp., , Therefore, (A, ) is a subgroup of (5, °). o, Note 3.7. :, , The subgroup A,, is called alternating subgroup of S,,., Example 3.27. :, , Express the following as a product of cycles and find whether it is odd or, even.

Page 5 : . 1234567, Solution : 1) sa eee pou Be A ae 4 OO, , =(1 5) (2 4) (2 3) (2 6), which is an even permutation., , eeeres, » (oro er eg spree 2204 6) 578), , =(1 3) (1 2) (4 6) (5 7) (5 8), which is an odd permutation.