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Math Help - Torsion subgroup

  1. #1
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    Torsion subgroup

    Let G be an abelian group. Prove that \{g\in G:|g|< \infty\} is a subgroup of G. Give an explicit example where this set is not a subgroup when G is non-abelian.

    First if G is infinite, then the torsion group would be empty. So suppose G is finite. I am supposed to use the subgroup criterion to show this but not entirely sure with what to do.

    Subgroup criterion is a subset H of a group is a subgroup iff H is nonempty and for all x,y in H xy^{-1}\in H.

    Let y=x, |x|<\infty, and x\in H.
    By property 2, xy^{-1}\in H. Therefore, xx^{-1}=e\in H.
    Since e and x are in H, we have ex^{-1}=x^{-1}\in H

    Correct?

    Not sure about the example when G is non-abelian.
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  2. #2
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    Re: Torsion subgroup

    Quote Originally Posted by dwsmith View Post
    Let G be an abelian group. Prove that \{g\in G:|g|< \infty\} is a subgroup of G. Give an explicit example where this set is not a subgroup when G is non-abelian.

    First if G is infinite, then the torsion group would be empty.
    no. (Q-{0},*) is infinite, but its torsion group is not empty (-1 has order 2).

    So suppose G is finite.
    if G is finite, every element has to have finite order, so for a finite group, Tor(G) = G.

    this is really quite simple. suppose a,b are of finite order m and n, respectively.

    then (ab^{-1})^{mn} = a^{mn}(b^{-1})^{mn} because G is abelian

     = (a^m)^n(b^{-1})^{nm}= (a^m)^n(b^{-n})^m = (a^m)^n((b^n)^{-1})^m = e^n(e^{-1})^n = ee = e

    so ab must be of finite order.

    for the non-abelian case, the only example that comes to mind is the free group on two generators a,b subject to the relations a^2 = b^2 = e.

    clearly ab is of infinite order.
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