Please do Exercise 14.3.28 and please show step by step and explain
Extracted text: Proposition 14.3.27. Disjoint cycles commute: that is, given two disjoint cycles o = (a1, a2,...,aj) and 7 = (b1, b2,..., br) we have OT = 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 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 e X. We'll define A = {a1,a2, ...,aj} and B = {b1,b2, ..., br}. By hypothesis A 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> and æ ¢ < 6=""> ; (ii) æ ¢ _< 7=""> _ and r¢ < 8=""> . (i) In this case, since r ¢ B it follows that T(x) = x. We then have OT(x) = o(T(x)) = o(x). Furthermore, since r € A it follows that o(x) E A, so o(x) ¢ B. We then have To(x) = T(0(x)) = o(x). It follows that στ(1)-τσ (π).
Extracted text: (ii) In this case, since r ¢<9> _ it follows that < 10="">_(x) = x. We then 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. Similarly since r ¢ < 21="">_ it follows that < 22=""> _(x) = x.We then have TO(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 for products of any number of disjoint cycles. Since disjoint cycles act indepen- dently, they all commute.