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Math Help - Group of Unitary Roots

  1. #1
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    Group of Unitary Roots

    I came up with a new group.
    Let \mathbb{G}= \bigcup_{n=1}^{\infty}U_n where U_n = \{z\in \mathbb{C} | z^n = 1\}.

    Then \mathbb{G} under complex multiplication will form a group.

    I hope this is group is not isomorphic to any of the known ones. And hopefully has some interesting properties.

    We know the following:
    1) \mathbb{G} is an abelian group.
    2) \mathbb{G} is a countable group.
    3*) \mathbb{G} has a cyclic subgroup for every finite order.

    What are the answers to the following questions:
    1)Is \mathbb{G} is free abelian group?
    2)Is \mathbb{G} a finitely generated abelian group?
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  2. #2
    Super Member Rebesques's Avatar
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    Hm, it certainly cannot be both, as then it would be isomorphic to some \mathbb{Z}^k.


    ps. where do u come up with these?
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  3. #3
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    Quote Originally Posted by Rebesques View Post
    Hm, it certainly cannot be both, as then it would be isomorphic to some \mathbb{Z}^k.
    Yes. This group is neither free abelian nor finitely generated. The easiest way to see this is to consider the torsion subgroup of \mathbb{G}. It quite remarkable, I think, that T[\mathbb{G}] = \mathbb{G} because this is an infinite group and its torsion subgroup is itself! But anyways, if  \mathbb{G} \simeq \mathbb{Z}^k then that would be a contradiction because T[\mathbb{Z}^k] = \{ 0 \} and if \mathbb{G} was finitely generated abelian group, that is \mathbb{G} \simeq \mathbb{Z}^k \times \mathbb{Z}_{p_1^{a_1}} \times ... \times \mathbb{Z}_{p_m^{a_m}} then T[\mathbb{G}] \simeq \mathbb{Z}_{p_1^{a_1}} \times ... \times \mathbb{Z}_{p_m^{a_m}} which is clearly impossible because one is finite and other oter infinite.

    ps. where do u come up with these?
    What do you mean?
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