Please do Exercise 14.3.28 and please show step by step and explain Proposition 14.3.27. Disjoint cycles commute: that is, given two disjointcycles o = (a1, a2,...,aj) and 7 =(b1, b2,..., br) we haveOT = TO = (a1, a2, ..., a;) (b1, b2,..., br)PROOF. We present this proof as a fill-in-the-blanks exercise:Exercise 14.3.28. Fill in the blanks to complete the proof:Recall that permutations are defined as bijections on a set X. In orderto show that the two permutations Or and to are equal, it's enough to showthat they are the same function. In other words, we just need to show thatOT(1) =_<1> for all r e X.We'll define A = {a1,a2, ...,aj} and B = {b1,b2, ..., br}. By hypothesisA and B are disjoint, so A_< 2 >B = _<3> . Given an arbitrary x € X,there are three possibilities: (i) r € A and r ¢ B; (ii) x €<4> andæ ¢ < 6 > ; (ii) æ ¢ _< 7 > _ and r¢ < 8 > .(i) In this case, since r ¢ B it follows that T(x) = x. We then haveOT(x) = o(T(x)) = o(x). Furthermore, since r € A it follows thato(x) E A, so o(x) ¢ B. We then have To(x) = T(0(x)) = o(x). Itfollows that στ(1)-τσ (π). (ii) In this case, since r ¢<9 > _ it follows that < 10 >_(x) = x. Wethen have To(x) = < 11> _ = _< 12 > _(x). Furthermore, since r E< 13 >_it follows that < 14 >(x) E< 15 >, so < 16 > (x)< 17 > . We then have or(x)follows that στ(ο)-τα (x).< 18 >< 19 > _(x). It(iii) In this case, since r ¢ A it follows that < 20 >_(x) = x. Similarlysince r ¢ < 21 >_ it follows that < 22 > _(x) = x.We then haveTO(x)TO(x).= r.< 23 > _ and or(x)= < 24 > . It follows that or(æ) =In all three cases we have or(r) = < 25 > , so therefore or = To.What we've discovered about products of two disjoint cycles is also true forproducts of any number of disjoint cycles. Since disjoint cycles act indepen-dently, they all commute.

All the blanks which are to be filled up are highlighted in Red.

Exercise 14.3.28. Fill in the blanks to complete the proof: Recall that permutations are defined as bijections on a set X. In order to show that the two permutations Or and to are equal, it's enough to show that they are the same function. In other words, we just need to show that OT(1) =_<1> for all r € X. We'll define A = {a1, a2, . .. ,a;} and B = {b1, b2, ..., br}. By hypothesis A and B are disjoint, so A_< 2 > _B = <3> . Given an arbitrary r € X, there are three possibilities: (i) z € A and r ¢ B; (ii) r € <4> _ and z¢ < 6 > ; (iii r ¢ _<7> _ and x¢ _< 8 > . (i) In this case, since r ¢ B it follows that 7(x) = 1. We then have OT(1) = 0(t(x)) = o(x). Furthermore, since r € A it follows that o(x) € A, so o(x) ¢ B. We then have ro(r) = t(0(r)) = o(x). It follows that στ(r ) -τσ (r). %3D %3D

Exercise 14.3.28. Fill in the blanks to complete the proof: Recall that permutations are defined as bijections on a set X. In order to show that the two permutations Or and to are equal, it's enough to show that they are the same function. In other words, we just need to show that OT(1) =_<1> for all r € X. We'll define A = {a1, a2, . .. ,a;} and B = {b1, b2, ..., br}. By hypothesis A and B are disjoint, so A_< 2 > _B = <3> . Given an arbitrary r € X, there are three possibilities: (i) z € A and r ¢ B; (ii) r € <4> _ and z¢ < 6 > ; (iii r ¢ _<7> _ and x¢ _< 8 > . (i) In this case, since r ¢ B it follows that 7(x) = 1. We then have OT(1) = 0(t(x)) = o(x). Furthermore, since r € A it follows that o(x) € A, so o(x) ¢ B. We then have ro(r) = t(0(r)) = o(x). It follows that στ(r ) -τσ (r). %3D %3D

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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