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(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.

(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.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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. For n> 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 = 2l for some integer e>1. Let ā e Ze be an element of order 2 with a an integer such that 0 < a < 2l – 1. We thus have a +ā = 0 in Z, 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 = e. Thus, we either have a = 0, which is the identity element and has order 1, or a = l, which is an element of order 2. Thus, the only element of order 2 of Ze 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 a? = 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. Thus, f(r) is an element of Z, of order at most 2. However, since r is non-trivial, f(ax) cannot have order 1. Thus, it has order exactly 2. By Proposition 2, it thus follows that f(r) 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(x) 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. For n> 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 = 2l for some integer e>1. Let ā e Ze be an element of order 2 with a an integer such that 0 < a < 2l – 1. We thus have a +ā = 0 in Z, 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 = e. Thus, we either have a = 0, which is the identity element and has order 1, or a = l, which is an element of order 2. Thus, the only element of order 2 of Ze 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 a? = 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. Thus, f(r) is an element of Z, of order at most 2. However, since r is non-trivial, f(ax) cannot have order 1. Thus, it has order exactly 2. By Proposition 2, it thus follows that f(r) 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(x) 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 E Zn belongs to Z, if and only if k and 2" are coprime, that is, if and only if k is odd. In particular, the elements a = 2n-1 I and b:= 2"-1+ 1 belong to Z, and we have (2"-I - T) = 2x2n-2 a? = 2 + T= I, and 62 = (2"-1 + T)? = 2" x2"-2 + 2n + T T. 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 n 1 and n 2. That is, show that the multiplicative groups Z and Z} are cyclic. (c) 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? %3D
Transcribed Image Text:We are now ready to prove Theorem 1: Proof of Theorem 1. We know that an element E Zn belongs to Z, if and only if k and 2" are coprime, that is, if and only if k is odd. In particular, the elements a = 2n-1 I and b:= 2"-1+ 1 belong to Z, and we have (2"-I - T) = 2x2n-2 a? = 2 + T= I, and 62 = (2"-1 + T)? = 2" x2"-2 + 2n + T T. 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 n 1 and n 2. That is, show that the multiplicative groups Z and Z} are cyclic. (c) 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? %3D
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