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(c) Show that Proposition 2 no longer holds without the condition that G is commutative. That is, give an example of a finite non-commutative group G and two elements a, b with order m, n respectively, such that m and n are coprime but the product ab does not have order mn.

(c) Show that Proposition 2 no longer holds without the condition that G is commutative. That is, give an example of a finite non-commutative group G and two elements a, b with order m, n respectively, such that m and n are coprime but the product ab does not have order mn.

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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Problem 4. 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. Let G be a finite commutative group whose order is a product of the form p1 x ...x Pk, with P1,... Pk distinct prime numbers. Then G is a cyclic group. Before giving the proof of this theorem, we introduce the following intermediate results that will be needed: Proposition 2. Let G be a finite commutative group, and let a, b e G be two elements of order m, n respectively. If m and n are coprime, then the product ab has order m x n. Proof of Proposition 2. Let e denote the identity element of G. Since a and b have order m and n respectively, we have (ab)m" = am"n = (a")"(b")" = e"e" = e. %3D Let d denote the order of ab. Since (ab)mn = e, it follows that d < mn. Let us show that m divides d. We have (ab)" = a"b" = a", since b has order n. Thus, the cyclic subgroup (ab) contains a". Moreover, it also contains a" = e since a has order m. Since m and n are coprime by hypothesis, Bézout's Identity gives us two integers k, l e Z such that km + en = 1. Since the cyclic subgroup (ab) is stable under products and inverses, and contains a" and a", it also contains (a")*(a")' = am+n = a' = a. %3D %3D %3D It thus follows that the order m of a divides the order d of ab.
Transcribed Image Text:Problem 4. 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. Let G be a finite commutative group whose order is a product of the form p1 x ...x Pk, with P1,... Pk distinct prime numbers. Then G is a cyclic group. Before giving the proof of this theorem, we introduce the following intermediate results that will be needed: Proposition 2. Let G be a finite commutative group, and let a, b e G be two elements of order m, n respectively. If m and n are coprime, then the product ab has order m x n. Proof of Proposition 2. Let e denote the identity element of G. Since a and b have order m and n respectively, we have (ab)m" = am"n = (a")"(b")" = e"e" = e. %3D Let d denote the order of ab. Since (ab)mn = e, it follows that d < mn. Let us show that m divides d. We have (ab)" = a"b" = a", since b has order n. Thus, the cyclic subgroup (ab) contains a". Moreover, it also contains a" = e since a has order m. Since m and n are coprime by hypothesis, Bézout's Identity gives us two integers k, l e Z such that km + en = 1. Since the cyclic subgroup (ab) is stable under products and inverses, and contains a" and a", it also contains (a")*(a")' = am+n = a' = a. %3D %3D %3D It thus follows that the order m of a divides the order d of ab.
Since a and b play symmetric roles, a similar proof shows that n divides the order d of ab. Thus, both integers n and m divide d. Since n and m are coprime, it follows that their product mn also divides d. Since we proved earlier that d < mn, it follows that d= mn, which concludes the proof of Proposition 2. One can prove by induction the following generalisation of Proposition 2, the proof of which we will omit: Proposition 3. Let G be a commutative group, and let a1,..., ar EG be k> 2 elements of order m1,.. coprime, then the product (a . ak) has order mı x.x mk. ..,m respectively. If the integers m1,..., m, are pairwise ... We are now ready to prove the main theorem: Proof of Theorem 1. The result is true when G| = 1, so we consider the case where |G| > 2. We write the order of G as |G| = P1 x ...x Pk, for some distinct prime numbers P1,.. . Pk. Since the prime numbers p1,..., Pk are divisors of |G|, we can choose elements a1,..., ak of order p1,.., Pk respectively. Since G is a commutative group, and the orders p1,..., Pk are distinct prime numbers, and in particular pairwise coprime, it follows from Proposition 3 that the product az ... az has order pi x x Pk = |G|. Since G is a finite group that contains an element of order |G|, it follows that G is cyclic. Comprehension questions: (a) Explain how Theorem 1 can be used to show that the multiplicative group Z, is cyclic. (b) Show that Theorem 1 no longer holds without the condition on the order of the group G. That is, give an example of a finite commutative group G whose order is not a product of distinct primes and such that G is not cyclic. (c) Show that Proposition 2 no longer holds without the condition that G is commutative. That is, give an example of a finite non-commutative group G and two elements a, b with order m, n respectively, such that m and n are coprime but the product ab does not have order mn. (d) The proof of Proposition 2 is currently incomplete, as it never explicitly uses the hypothesis that G is commutative. Explain what part of the proof uses implicitly the commutativity of G. Give an example to show how that part of the proof would break down without this hypothesis on G.
Transcribed Image Text:Since a and b play symmetric roles, a similar proof shows that n divides the order d of ab. Thus, both integers n and m divide d. Since n and m are coprime, it follows that their product mn also divides d. Since we proved earlier that d < mn, it follows that d= mn, which concludes the proof of Proposition 2. One can prove by induction the following generalisation of Proposition 2, the proof of which we will omit: Proposition 3. Let G be a commutative group, and let a1,..., ar EG be k> 2 elements of order m1,.. coprime, then the product (a . ak) has order mı x.x mk. ..,m respectively. If the integers m1,..., m, are pairwise ... We are now ready to prove the main theorem: Proof of Theorem 1. The result is true when G| = 1, so we consider the case where |G| > 2. We write the order of G as |G| = P1 x ...x Pk, for some distinct prime numbers P1,.. . Pk. Since the prime numbers p1,..., Pk are divisors of |G|, we can choose elements a1,..., ak of order p1,.., Pk respectively. Since G is a commutative group, and the orders p1,..., Pk are distinct prime numbers, and in particular pairwise coprime, it follows from Proposition 3 that the product az ... az has order pi x x Pk = |G|. Since G is a finite group that contains an element of order |G|, it follows that G is cyclic. Comprehension questions: (a) Explain how Theorem 1 can be used to show that the multiplicative group Z, is cyclic. (b) Show that Theorem 1 no longer holds without the condition on the order of the group G. That is, give an example of a finite commutative group G whose order is not a product of distinct primes and such that G is not cyclic. (c) Show that Proposition 2 no longer holds without the condition that G is commutative. That is, give an example of a finite non-commutative group G and two elements a, b with order m, n respectively, such that m and n are coprime but the product ab does not have order mn. (d) The proof of Proposition 2 is currently incomplete, as it never explicitly uses the hypothesis that G is commutative. Explain what part of the proof uses implicitly the commutativity of G. Give an example to show how that part of the proof would break down without this hypothesis on G.
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