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(a) S6 (b) S7 (c) Ss

(a) S6 (b) S7 (c) Ss

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
Problem 3CEXP
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Please do Exercise 14.3.38 part A and B. Please show step by step and explain

(a) S6 (b) S7 (c) Ss
Transcribed Image Text:(a) S6 (b) S7 (c) Ss
Example 14.3.37. We know that every permutation in S, is the product of disjoint cycles. Let us list all possible cycle lengths and number of cycles for the permutations of Sz. • First of all, S; contains the identity, which has no cycles. • Second, some permutations in S; consist of a single cycle. The single cycle could have length 2, 3, 4, or 5 (remember, we don't count cycles of length 1). • Third, some permutations in S3 consist of the product of two disjoint cycles. To enumerate these, suppose first that one of the cycles is a cycle of length 2. Then the other cycle could be a cycle of length 2 (for instance in the case (12)(34)) or a cycle of length 3 (as in the case (14)(235)). There are no other possibilities, because we only have 5 elements to permute, and a larger disjoint cycle would require more elements. • It's not possible to have three or more disjoint cycles, because that would require at least six elements. To summarize then, the possible cycle structures for permutations in S3 are: • The identity single cycles of lengths 5, 4, 3, or 2 two disjoint cycles of lengths 2 and 3; and two disjoint cycles of lengths 2 and 2 Exercise 14.3.38. Following Example 14.3.37, list all possible cycle struc- tures of permutations in the following:
Transcribed Image Text:Example 14.3.37. We know that every permutation in S, is the product of disjoint cycles. Let us list all possible cycle lengths and number of cycles for the permutations of Sz. • First of all, S; contains the identity, which has no cycles. • Second, some permutations in S; consist of a single cycle. The single cycle could have length 2, 3, 4, or 5 (remember, we don't count cycles of length 1). • Third, some permutations in S3 consist of the product of two disjoint cycles. To enumerate these, suppose first that one of the cycles is a cycle of length 2. Then the other cycle could be a cycle of length 2 (for instance in the case (12)(34)) or a cycle of length 3 (as in the case (14)(235)). There are no other possibilities, because we only have 5 elements to permute, and a larger disjoint cycle would require more elements. • It's not possible to have three or more disjoint cycles, because that would require at least six elements. To summarize then, the possible cycle structures for permutations in S3 are: • The identity single cycles of lengths 5, 4, 3, or 2 two disjoint cycles of lengths 2 and 3; and two disjoint cycles of lengths 2 and 2 Exercise 14.3.38. Following Example 14.3.37, list all possible cycle struc- tures of permutations in the following:
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