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Abelian groups of order 70 are cyclic

  1. Nov 14, 2009 #1
    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. jcsd
  3. Nov 14, 2009 #2

    Dick

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    Use Cauchy's theorem with your three primes.
     
  4. Nov 15, 2009 #3
    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: Nov 15, 2009
  5. Nov 15, 2009 #4

    Dick

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    Sure. Or you could just take the direct approach and argue that the element abc must have order 70.
     
  6. Nov 15, 2009 #5
    Awesome. Thanks!
     
  7. Nov 15, 2009 #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.
     
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