# Abelian groups of order 70 are cyclic

1. Nov 14, 2009

### redone632

1. The problem statement, all variables and given/known data
Show that every abelian group of order 70 is cyclic.

2. Relevant equations
Cannot use the Fundamental Theorem of Finite Abelian Groups.

3. 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: Nov 15, 2009
2. Nov 14, 2009

### Dick

Use Cauchy's theorem with your three primes.

3. Nov 15, 2009

### redone632

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.

Last edited: Nov 15, 2009
4. Nov 15, 2009

### Dick

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

5. Nov 15, 2009

### redone632

Awesome. Thanks!

6. Nov 15, 2009

### 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.

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