Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
(b) Explain how Proposition 3 can be used to show that the multiplicative group Z is not cyclic.

(b) Explain how Proposition 3 can be used to show that the multiplicative group Z is not cyclic.

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
See similar textbooks
icon
Related questions
Question
please send solution for part b only handwritten solution accepted
Problem 2. Read the following mathematical text and answer the associated comprehension questions at the end. The goal of this text is to prove the following theorem: Theorem 1. Forn> 3, the multiplicative group Z. is not cyclic. We start by proving several intermediate results: Proposition 2. For m 1, the additive group Z contains at most one element of order 2. Proof. The proposition is true if m is odd. Indeed, a group of odd order does not contain any element of order 2 by Lagrange's Theorem. Thus, it is enough to consider the case where the order of Zm is of the form m = 20 for some integer e>1. Let ā e Zu be an element of order 2 with a an integer such that 0 < a < 2l – 1. We thus have a +ā = 0 in Z2e, i.e. 2a is divisible by 20. This implies that a is divisible by l. Since 0 < a < 2l-1, it follows that we either have a 0 or a = l. Thus, we either have a = 0, which is the identity element and has order 1, or a = 1, which is an element of order 2. Thus, the only element of order 2 of Zu is 7. Proposition 3. Let G be a finite cyclic group. Then G contains at most one element of order 2. Proof. Let m be the order of G. We know from the course that a cyclic group of order m is isomorphic to the additive group Zm, so let us consider a group isomorphism f : G → Zm. If G does not contain any element of order 2, there is nothing further to do, so let us assume that G contains an element r of order 2. We have a2 = e in G. We thus have f(a2) = f(e) in Zm, and by properties of group homomorphisms, this implies that f(r) + f(x) = 0. %3D Thus, f(x) is an element of Z, of order at most 2. However, since r is non-trivial, f(r) cannot have order 1. Thus, it has order exactly 2. By Proposition 2, it thus follows that f(x) is the unique element of Zm of order 2. Let y E G be another element of order 2 of G. The same reasoning shows that f(y) is also the unique element of order 2 of Zm. Thus, f(r) = f(y), and by injectivity of f, it follows that r = y. Thus, G contains exactly one element of order 2.
Transcribed Image Text:Problem 2. Read the following mathematical text and answer the associated comprehension questions at the end. The goal of this text is to prove the following theorem: Theorem 1. Forn> 3, the multiplicative group Z. is not cyclic. We start by proving several intermediate results: Proposition 2. For m 1, the additive group Z contains at most one element of order 2. Proof. The proposition is true if m is odd. Indeed, a group of odd order does not contain any element of order 2 by Lagrange's Theorem. Thus, it is enough to consider the case where the order of Zm is of the form m = 20 for some integer e>1. Let ā e Zu be an element of order 2 with a an integer such that 0 < a < 2l – 1. We thus have a +ā = 0 in Z2e, i.e. 2a is divisible by 20. This implies that a is divisible by l. Since 0 < a < 2l-1, it follows that we either have a 0 or a = l. Thus, we either have a = 0, which is the identity element and has order 1, or a = 1, which is an element of order 2. Thus, the only element of order 2 of Zu is 7. Proposition 3. Let G be a finite cyclic group. Then G contains at most one element of order 2. Proof. Let m be the order of G. We know from the course that a cyclic group of order m is isomorphic to the additive group Zm, so let us consider a group isomorphism f : G → Zm. If G does not contain any element of order 2, there is nothing further to do, so let us assume that G contains an element r of order 2. We have a2 = e in G. We thus have f(a2) = f(e) in Zm, and by properties of group homomorphisms, this implies that f(r) + f(x) = 0. %3D Thus, f(x) is an element of Z, of order at most 2. However, since r is non-trivial, f(r) cannot have order 1. Thus, it has order exactly 2. By Proposition 2, it thus follows that f(x) is the unique element of Zm of order 2. Let y E G be another element of order 2 of G. The same reasoning shows that f(y) is also the unique element of order 2 of Zm. Thus, f(r) = f(y), and by injectivity of f, it follows that r = y. Thus, G contains exactly one element of order 2.
We are now ready to prove Theorem 1: Proof of Theorem 1. We know that an element k E Zyn belongs to Z, if and only if k and 2" are coprime, that is, if and only ifk is odd. In particular, the elements a := 2"-1 -1 and b:= 2n-1 +1 belong to Z, and we have a² = (2"-I - T)² : 2n x2n-2 + T = T, %3D 2n and (2"-1+1)2: 2" x2n-2 + ī =T. 2n Thus, Z contains at least two distinct elements of order 2. Since cyclic groups contain at most one element of order 2 by Proposition 2, this implies that Z is not cyclic. Comprehension questions: (a) Following the same approach as in the proof of Proposition 2, show that for an integer m > 1, the additive group Zm either contains no element of order 3 or contains exactly two elements of order 3. (b) Explain how Proposition 3 can be used to show that the multiplicative group Z is not cyclic. (c) In the proof of Proposition 3, the following statement is not justified: "However, since a is non-trivial, f(x) cannot have order 1." Properly justify this statement. (d) Show that Theorem 1 does not hold for = 1 and n 2. That is, show that the multiplicative groups Z and Z are cyclic. (e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitly use the hypothesis that n > 3. In what parts of the proof is that assumption implicitly used?
Transcribed Image Text:We are now ready to prove Theorem 1: Proof of Theorem 1. We know that an element k E Zyn belongs to Z, if and only if k and 2" are coprime, that is, if and only ifk is odd. In particular, the elements a := 2"-1 -1 and b:= 2n-1 +1 belong to Z, and we have a² = (2"-I - T)² : 2n x2n-2 + T = T, %3D 2n and (2"-1+1)2: 2" x2n-2 + ī =T. 2n Thus, Z contains at least two distinct elements of order 2. Since cyclic groups contain at most one element of order 2 by Proposition 2, this implies that Z is not cyclic. Comprehension questions: (a) Following the same approach as in the proof of Proposition 2, show that for an integer m > 1, the additive group Zm either contains no element of order 3 or contains exactly two elements of order 3. (b) Explain how Proposition 3 can be used to show that the multiplicative group Z is not cyclic. (c) In the proof of Proposition 3, the following statement is not justified: "However, since a is non-trivial, f(x) cannot have order 1." Properly justify this statement. (d) Show that Theorem 1 does not hold for = 1 and n 2. That is, show that the multiplicative groups Z and Z are cyclic. (e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitly use the hypothesis that n > 3. In what parts of the proof is that assumption implicitly used?
Expert Solution
steps

Step by step

Solved in 5 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS