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(b). Let N = {1,2,3,...,}. Let G = Sym(N) be the group of all bijec- tive functions from N→ N. Then it is clear that G is an infinite group (under composition) as there are infinite bijections from N to N. Con- sider a = (1, 2) and b = (2,3). That is a is the function such that a(1) = 2,a(2) = 1 and a (n) = n for all ne N. It is clear a = a-¹ since a² = I. Let b = (2, 3) denote the function b(2) = 3, b(3) = 2 and b(n) = n for all ne N. Then we have

(b). Let N = {1,2,3,...,}. Let G = Sym(N) be the group of all bijec- tive functions from N→ N. Then it is clear that G is an infinite group (under composition) as there are infinite bijections from N to N. Con- sider a = (1, 2) and b = (2,3). That is a is the function such that a(1) = 2,a(2) = 1 and a (n) = n for all ne N. It is clear a = a-¹ since a² = I. Let b = (2, 3) denote the function b(2) = 3, b(3) = 2 and b(n) = n for all ne N. Then we have

Advanced Engineering Mathematics
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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ABSTRACT ALGEBRA UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS LEGIBLY. NO LONG EXPLANATION NEEDED.
(b). Let N = {1,2,3,...,}. Let G = Sym(N) be the group of all bijec- tive functions from N→ N. Then it is clear that G is an infinite group (under composition) as there are infinite bijections from N to N. Con- sider a = (1,2) and b = (2,3). That is a is the function such that a(1) = 2,a(2) = 1 and a (n) = n for all ne N. It is clear a = a-¹ since a² = I. Let b = (2, 3) denote the function b(2) = 3, b(3) = 2 and b(n) = n for all ne N. Then we have a=¹ba = a oboa = (1, 2) (2, 3) o (1,2) = (1,3) It is clear that a-¹ba = (1,3) b = (2,3). (1)
Transcribed Image Text:(b). Let N = {1,2,3,...,}. Let G = Sym(N) be the group of all bijec- tive functions from N→ N. Then it is clear that G is an infinite group (under composition) as there are infinite bijections from N to N. Con- sider a = (1,2) and b = (2,3). That is a is the function such that a(1) = 2,a(2) = 1 and a (n) = n for all ne N. It is clear a = a-¹ since a² = I. Let b = (2, 3) denote the function b(2) = 3, b(3) = 2 and b(n) = n for all ne N. Then we have a=¹ba = a oboa = (1, 2) (2, 3) o (1,2) = (1,3) It is clear that a-¹ba = (1,3) b = (2,3). (1)
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