Let d denote the order of ab. Since (ab)m= 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" since a has order m. Since m and n are coprime by hypothesis, Bézout's Identity gives us two integers k, te Z such that km+ tn 1. Since the cyclic subgroup (ab) is stable under products and inverses, and contains a" and a", it also contains %3D e (a")*(a")' = am+n = a' = a. It thus follows that the order m of a divides the order d of ab. %3D
Let d denote the order of ab. Since (ab)m= 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" since a has order m. Since m and n are coprime by hypothesis, Bézout's Identity gives us two integers k, te Z such that km+ tn 1. Since the cyclic subgroup (ab) is stable under products and inverses, and contains a" and a", it also contains %3D e (a")*(a")' = am+n = a' = a. It thus follows that the order m of a divides the order d of ab. %3D
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
please send handwritten solution for part e
![Problem 4. Read the following mathematical text and answer the associated comprchension
1/3
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 pi 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, beG 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)mn = am"nn = (a")"(b")" = e"e" = e.
Let d denote the order of ab. Since (ab)m =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")' = a*m+en = a'
It thus follows that the order m of a divides the order d of ab.
=a.
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 a,,...,a, € G be k> 2
elements of order m
coprime, then the product (a,a) has order m¡ × .…x m.
m, respectively. If the integers m 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| = P xx P, for some distinct
prime numbers P Pk. Since the prime numbers p..., Pa are divisors of |G],
we can choose elements a,.. a of order p... P respectively. Since G is a
commutative group, and the orders p....pa are distinct prime numbers, and
in particular pairwise coprime, it follows from Proposition 3 that the product
a.a, has order p x.x P. - |G|. Since
element of order |G|, it follows that G is cyclic.
is a finite group that contains an
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c7a5306-15ff-4c28-af31-3b9d67d52d5e%2F8a1789dc-9054-4df3-ae01-a3fb9377795e%2Flhryy88_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 4. Read the following mathematical text and answer the associated comprchension
1/3
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 pi 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, beG 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)mn = am"nn = (a")"(b")" = e"e" = e.
Let d denote the order of ab. Since (ab)m =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")' = a*m+en = a'
It thus follows that the order m of a divides the order d of ab.
=a.
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 a,,...,a, € G be k> 2
elements of order m
coprime, then the product (a,a) has order m¡ × .…x m.
m, respectively. If the integers m 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| = P xx P, for some distinct
prime numbers P Pk. Since the prime numbers p..., Pa are divisors of |G],
we can choose elements a,.. a of order p... P respectively. Since G is a
commutative group, and the orders p....pa are distinct prime numbers, and
in particular pairwise coprime, it follows from Proposition 3 that the product
a.a, has order p x.x P. - |G|. Since
element of order |G|, it follows that G is cyclic.
is a finite group that contains an
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.
![(e) In the proof of Proposition 2 (last line of page 2), the following statement is
not properly justified:
“ It thus follows that the order m of a divides the order d of ab. "
Give a complete justification of this statement, by explaining carefully how
this statement is a consequence of what comes before it.
(f) A student proposed the following argument to generalise the proof of Theo-
rem 1 to the case where G is a commutative group whose order is no longer
assumed to be a product of distinct primes:
Student's proposed generalisation: Let G be a finite commutative group.
Then G is cyclic.
Student's proposed proof. The result is true when G| = 1, so we consider
the case where |G| > 2. We write a prime decomposition for the order of
G as |G| = pi' x ...x p, for some distinct prime numbers p1, ..., Pk and
, p are divisors of G|, we can
choose elements a1,.., ak of order p', ..., p respectively. Since p1,.. , Pk
are distinct prime numbers, it follows that pi",...,P are pairwise coprime.
Since G is commutative group, and the orders of a1,..., ak are pairwise
coprime, it follows from Proposition 3 that the product a1ak has order
x p = |G]. Since G is a finite group that contains an element of
%3D
positive integers a1,...,ak. Since pi',
....
...
...
order |G|, it follows that G is cyclic.
We know from part (b) that Theorem 1 does not hold without the hypothesis
on the order of G. Where exactly is the error in the student's proposed
proof? Give an example to illustrate how that part of the argument breaks
down without additional assumptions on G, and explain why the problem
you identified in the student's proposed proof is not an issue in the proof of
Theorem 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c7a5306-15ff-4c28-af31-3b9d67d52d5e%2F8a1789dc-9054-4df3-ae01-a3fb9377795e%2F7myp59_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(e) In the proof of Proposition 2 (last line of page 2), the following statement is
not properly justified:
“ It thus follows that the order m of a divides the order d of ab. "
Give a complete justification of this statement, by explaining carefully how
this statement is a consequence of what comes before it.
(f) A student proposed the following argument to generalise the proof of Theo-
rem 1 to the case where G is a commutative group whose order is no longer
assumed to be a product of distinct primes:
Student's proposed generalisation: Let G be a finite commutative group.
Then G is cyclic.
Student's proposed proof. The result is true when G| = 1, so we consider
the case where |G| > 2. We write a prime decomposition for the order of
G as |G| = pi' x ...x p, for some distinct prime numbers p1, ..., Pk and
, p are divisors of G|, we can
choose elements a1,.., ak of order p', ..., p respectively. Since p1,.. , Pk
are distinct prime numbers, it follows that pi",...,P are pairwise coprime.
Since G is commutative group, and the orders of a1,..., ak are pairwise
coprime, it follows from Proposition 3 that the product a1ak has order
x p = |G]. Since G is a finite group that contains an element of
%3D
positive integers a1,...,ak. Since pi',
....
...
...
order |G|, it follows that G is cyclic.
We know from part (b) that Theorem 1 does not hold without the hypothesis
on the order of G. Where exactly is the error in the student's proposed
proof? Give an example to illustrate how that part of the argument breaks
down without additional assumptions on G, and explain why the problem
you identified in the student's proposed proof is not an issue in the proof of
Theorem 1.
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