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Abelian groups of order 70 are cyclic

  • Thread starter redone632
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1. Homework Statement
Show that every abelian group of order 70 is cyclic.


2. Homework 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:

Dick

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Use Cauchy's theorem with your three primes.
 
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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:

Dick

Science Advisor
Homework Helper
26,258
618
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.
 
14
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Sure. Or you could just take the direct approach and argue that the element abc must have order 70.
Awesome. Thanks!
 
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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