FOLLOW THE EXERCISE 2. Let a, b E R and let fab: R R be the

FOLLOW THE EXERCISE

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2. Let a, b E R and let fab: R → R be the function defined as fab(x) = ax + b for all x E R. Use this for the problems below. a) Prove that G = {fab | a, b = R, and a # 0} is a group, where the operation is composition. Note: you only have to show properties (G2) and (G3) of a group since the associative property (G1) is always true for composition of functions. Exercise: Prove that $:( 1₁ +) → (52, +) given by 6 (a) = 5a Hae I is an Isomorphism where 522 = {... -10₁ - 5₁0, 5, 10, 15.. 3 = { all integer multiples of 53. Proof: show is 1-1 Assume da = $b where a,b & Z. That is 5a = 5b where a₁b²Z. WTS: a = b. By assumption 5a = 56 multiply both sides by 2/56, ½ (5a) = 2/5 (56) So₁ a = b Therefore, is 1-1. (2) Show & is on to Assume: 64 522 WTS &Z such that 5a = b 5a=b Solving for a multiply both sides by 15. 금 (59) 금 6 So, a = 1/² b be I because be 57 by assumption since b is a multiple of 5, b multiplied 1/5 is an integer. Also we know $(a)=b since Therefore is onto. $(a) = 5a = 5 ( ¼/ b) = b Show $(a+b) = $(a) + 6(b) ta,be Z. starting with the left side, $(a+b)= 5 (a+b) = 5a+ 5b = $(a) + (b) Therefore, & is an ismorphism that is (Z₁ +) ~ (572, +) 2. Let a, b E R and let fab: R → R be the function defined as fab(x) = ax + b for all x E R. Use this for the problems below. a) Prove that G = {fab | a, b = R, and a # 0} is a group, where the operation is composition. Note: you only have to show properties (G2) and (G3) of a group since the associative property (G1) is always true for composition of functions. Exercise: Prove that $:( 1₁ +) → (52, +) given by 6 (a) = 5a Hae I is an Isomorphism where 522 = {... -10₁ - 5₁0, 5, 10, 15.. 3 = { all integer multiples of 53. Proof: show is 1-1 Assume da = $b where a,b & Z. That is 5a = 5b where a₁b²Z. WTS: a = b. By assumption 5a = 56 multiply both sides by 2/56, ½ (5a) = 2/5 (56) So₁ a = b Therefore, is 1-1. (2) Show & is on to Assume: 64 522 WTS &Z such that 5a = b 5a=b Solving for a multiply both sides by 15. 금 (59) 금 6 So, a = 1/² b be I because be 57 by assumption since b is a multiple of 5, b multiplied 1/5 is an integer. Also we know $(a)=b since Therefore is onto. $(a) = 5a = 5 ( ¼/ b) = b Show $(a+b) = $(a) + 6(b) ta,be Z. starting with the left side, $(a+b)= 5 (a+b) = 5a+ 5b = $(a) + (b) Therefore, & is an ismorphism that is (Z₁ +) ~ (572, +)

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a To show G fab a b R and a 0 is a group under composition G2 Closure T... View the full answer