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(c) Consider the subgroups H = (−1) and K = (5) of G. Show that GHxKZ₂ x Z2m-2. (d) As an application to number theory, find the number of positive integers x less than 128 such that x¹000 - 1 is divisible by 128. (Hint: Show that this is the same as the number of elements (a, b) € Z₂ × Z25 such that (1000 ā, 1000 6) = (0,0).)

(c) Consider the subgroups H = (−1) and K = (5) of G. Show that GHxKZ₂ x Z2m-2. (d) As an application to number theory, find the number of positive integers x less than 128 such that x¹000 - 1 is divisible by 128. (Hint: Show that this is the same as the number of elements (a, b) € Z₂ × Z25 such that (1000 ā, 1000 6) = (0,0).)

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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2(c)(d)

2. It is relevant in number theory to understand the group structure of Z. In this problem we will work out the structure of Z when n = 2m. The cases m = 1 and m = 2 are easy to compute directly: we have Z {1} is the trivial group, and Z₁ = {1,3} ≈ Z₂. So we suppose m ≥ 3, and let G = Zm. (a) Use mathematical induction to show that for every integer k ≥ 0, we have 52%: = 1+ 2+2q for some odd integer q. (b) Show that the order of 5 in G is 2m-2 (c) Consider the subgroups H = = 1) and K = (5) of G. Show that G≈H×K≈ Z₂ × Z2m-2 1000 (d) As an application to number theory, find the number of positive integers x less than 128 such that x 1 is divisible by 128. (Hint: Show that this is the same as the number of elements (a, b) € Z2 × Z25 such that (1000 ā, 1000 6) (0,0).) =
Transcribed Image Text:2. It is relevant in number theory to understand the group structure of Z. In this problem we will work out the structure of Z when n = 2m. The cases m = 1 and m = 2 are easy to compute directly: we have Z {1} is the trivial group, and Z₁ = {1,3} ≈ Z₂. So we suppose m ≥ 3, and let G = Zm. (a) Use mathematical induction to show that for every integer k ≥ 0, we have 52%: = 1+ 2+2q for some odd integer q. (b) Show that the order of 5 in G is 2m-2 (c) Consider the subgroups H = = 1) and K = (5) of G. Show that G≈H×K≈ Z₂ × Z2m-2 1000 (d) As an application to number theory, find the number of positive integers x less than 128 such that x 1 is divisible by 128. (Hint: Show that this is the same as the number of elements (a, b) € Z2 × Z25 such that (1000 ā, 1000 6) (0,0).) =
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