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Math Help - Prove that an Abelian group of order 6 must be cyclic (without Lagrange's thm)

  1. #1
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    Prove that an Abelian group of order 6 must be cyclic (without Lagrange's thm)

    I am trying from two different approached. The first is to try and prove that if |a|=2,3,4,5 then it cannot be in the group for some element a in G.

    My other approach was going to be to find an arbitrary order 6 group and then work on the proof from there. I dont seem to know what elements would be in there? G={e, a, b, ab, b^2, ab^2} ?

    Thanks in advance.
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    Member Black's Avatar
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    There are only two groups of order 6 (up to isomorphism): D_3 and \mathbb{Z}_6. Since the group in question is abelian, it can't be isomorphic to D_3.
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    Quote Originally Posted by Black View Post
    There are only two groups of order 6 (up to isomorphism): D_3 and \mathbb{Z}_6. Since the group in question is abelian, it can't be isomorphic to D_3.

    If he can't use Lagrange then I can't see how can he use that there are only two groups, up to isomorhpism, of order 6...

    I suppose that building a Cayley table, using abelian and a try-and-error process will eventually work, but I can't see a straightfroward way to do it without Lagrange.

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Fundamental theorem of finitely generated abelian groups?
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