Abelian groups of order 70 are cyclic

[redone632]

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!

 

[Dick]

Use Cauchy's theorem with your three primes.

 

[redone632]

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 G is abelian, then every subgroup must be normal. Therefore, the subgroups generated a, b, c are distinct and normal. G is then the internal direct product <a> \times <b> \times <c>. Then by a theorem, G is isomorphic to <a> \oplus <b> \oplus <c>. As |a|, |b|, |c| are relatively prime, <a> \oplus <b> \oplus <c> is cyclic. Therefore G is cyclic.

 

[Dick]

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 G is abelian, then every subgroup must be normal. Therefore, the subgroups generated a, b, c are distinct and normal. G is then the internal direct product <a> \times <b> \times <c>. Then by a theorem, G is isomorphic to <a> \oplus <b> \oplus <c>. As |a|, |b|, |c| are relatively prime, <a> \oplus <b> \oplus <c> is cyclic. Therefore G is cyclic.

Sure. Or you could just take the direct approach and argue that the element abc must have order 70.

 

[redone632]

Sure. Or you could just take the direct approach and argue that the element abc must have order 70.

Awesome. Thanks!

 

[Quantumpencil]

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.