(a) S6 (b) S7 (c) Ss
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please do Exercise 14.3.38 part A and B. Please show step by step and explain

Transcribed Image Text:(a) S6
(b) S7
(c) Ss

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:
Expert Solution
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