(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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
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).)
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2e948f6-fd6f-485f-942e-c931230f8579%2F581e35c6-9fde-4faa-9e0e-df6414eff8d8%2Fcnokq3_processed.jpeg&w=3840&q=75)
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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