Abelian groups of order 70 are cyclic

In summary, the conversation discusses the task of proving that every abelian group of order 70 is cyclic. The first attempt was to prove the contrapositive, but it was not successful. The idea of using the primes 2, 5, and 7 was mentioned, and it was suggested to use Cauchy's theorem. The final proof involves using Cauchy's theorem to show that there must be elements of order 2, 5, and 7, which form normal subgroups and lead to the group being isomorphic to the direct product of cyclic groups of those orders. As these orders are relatively prime, the direct product is cyclic, and therefore the group of order 70 is cyclic.
  • #1
redone632
14
0

Homework Statement


Show that every abelian group of order 70 is cyclic.

Homework Equations


Cannot use the Fundamental Theorem of Finite Abelian Groups.

The Attempt at a Solution


I've tried to prove the contrapositive and suppose that it is not cyclic then it cannot be abelian. But that has lead no where quickly.

Something tells me that I need to use the fact that 2*5*7 = 70 and 2 5 7 are all primes. But nothing is clicking. We haven't done the Fundamental Theorem of Finite Abelian Groups so there must be a way to prove this without it. If someone can point me in the right direction that would help a lot!
 
Last edited:
Physics news on Phys.org
  • #2
Use Cauchy's theorem with your three primes.
 
  • #3
Dick said:
Use Cauchy's theorem with your three primes.

Hmmm I think I got it. I just want to make sure it's right.

Since 2, 5, 7 are primes that divide 70. Then by Cauchy's Theorem there must be an elements of order 2, 5, 7 say, a, b, c respectively. Since [itex]G[/itex] is abelian, then every subgroup must be normal. Therefore, the subgroups generated a, b, c are distinct and normal. [itex]G[/itex] is then the internal direct product [itex]<a> \times <b> \times <c>[/itex]. Then by a theorem, [itex]G[/itex] is isomorphic to [itex]<a> \oplus <b> \oplus <c>[/itex]. As [itex]|a|[/itex], [itex]|b|[/itex], [itex]|c|[/itex] are relatively prime, [itex]<a> \oplus <b> \oplus <c>[/itex] is cyclic. Therefore [itex]G[/itex] is cyclic.
 
Last edited:
  • #4
redone632 said:
Hmmm I think I got it. I just want to make sure it's right.

Since 2, 5, 7 are primes that divide 70. Then by Cauchy's Theorem there must be an elements of order 2, 5, 7 say, a, b, c respectively. Since [itex]G[/itex] is abelian, then every subgroup must be normal. Therefore, the subgroups generated a, b, c are distinct and normal. [itex]G[/itex] is then the internal direct product [itex]<a> \times <b> \times <c>[/itex]. Then by a theorem, [itex]G[/itex] is isomorphic to [itex]<a> \oplus <b> \oplus <c>[/itex]. As [itex]|a|[/itex], [itex]|b|[/itex], [itex]|c|[/itex] are relatively prime, [itex]<a> \oplus <b> \oplus <c>[/itex] is cyclic. Therefore [itex]G[/itex] is cyclic.

Sure. Or you could just take the direct approach and argue that the element abc must have order 70.
 
  • #5
Dick said:
Sure. Or you could just take the direct approach and argue that the element abc must have order 70.

Awesome. Thanks!
 
  • #6
There are two groups of order 21, even though it's isomorphic to the direct product of the Cyclic Group of Order 3 and the Cyclic Group of order 7. 3 and 7 are co-prime.

EDIT: Nevermind, didn't read the "abelian" in the problem. Your proof is good.
 

What is an Abelian group of order 70?

An Abelian group of order 70 is a mathematical structure consisting of a set of 70 elements and a binary operation (usually denoted as + or *) that follows the commutative property, meaning that the order of the elements in the operation does not affect the result.

What does it mean for an Abelian group of order 70 to be cyclic?

A cyclic group is a group in which all elements can be generated by repeatedly applying a single element. In the case of an Abelian group of order 70, this means that there exists one element, called the generator, that can be combined with itself a certain number of times to produce all 70 elements in the group.

Why is it significant that an Abelian group of order 70 is cyclic?

It is significant because it is a rare occurrence. In general, not all groups are cyclic, and for an Abelian group of order 70 to be cyclic means that all 70 elements can be generated by a single element, making it a simpler and more structured group.

How can you prove that an Abelian group of order 70 is cyclic?

To prove that an Abelian group of order 70 is cyclic, one can use a theorem called the Fundamental Theorem of Finite Abelian Groups, which states that any finite Abelian group can be written as a direct product of cyclic groups. By factoring 70 into its prime factors (2, 5, and 7), we can show that the group is isomorphic to a direct product of cyclic groups, proving its cyclic nature.

What are some real-world applications of Abelian groups of order 70 being cyclic?

Cyclic groups have many applications in various fields of mathematics and science, but one specific example is in cryptography. Cyclic groups are used in the Diffie-Hellman key exchange algorithm, which is used to securely exchange information over an insecure channel. This algorithm relies on the fact that certain cyclic groups are difficult to break, making them useful in encrypting data.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
938
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
622
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
7K
  • Linear and Abstract Algebra
Replies
1
Views
730
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top