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Math Help - Torsion Subgroup

  1. #1
    Super Member Bernhard's Avatar
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    Torsion Subgroup

    Dummit and Foote Section 2.1 (Definitions and Examples of Subgroups) Exercise 6 reads as follows:

    Let G be an abelian group. Prove that H = {g \in G| |g| is less than infinity} is a subgroup of G (called the torsion subgroup of G). Give an explicit example where this set is not a subgroup when G is not abelian.

    ================================================== ======

    H = {g \in G| |g| is less than infinity}

    Let x,y \in H ie |x| is less than infinity and |y| is less than infinity

    Need to show x y^{-1} \in H

    ie need to show |x y^{-1}| is less than infinity

    ? but how does one do this? Also ? explicit example ?

    Can anyone help?

    Peter
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  2. #2
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    Re: Torsion Subgroup

    suppose |y| = n so that y^n = e. then (y^{-1})^n = (y^n)^{-1} as you can verify by multiplication.

    therefore, (y^{-1})^n = (y^n)^{-1} = e^{-1} = e, so |y^{-1}| is finite.

    so y^{-1} \in H if y \in H.

    now suppose that x,y \in H, so that |x| = m, |y| = n.

    (xy^{-1})^{mn} = x^{mn}(y^{-1})^{mn} (since G is abelian)

     = (x^m)^n((y^{-1})^n)^m = e^ne^m = e, so |xy^{-1}| < \infty, that is xy^{-1} \in H.

    note we didn't show what the order of xy^{-1} actually was, just that it was finite (and less than or equal to mn).

    finding a counter-example is a bit tricky. let's use G = Sym(Z), the group of all bijections of Z.

    define f(n) = -n. it's clear f is a bijection, and f has order 2.
    define g(n) = 1-n. it's also clear g is a bijection, and:

    g(g(n)) = g(1-n) = 1 - (1-n) = 1 - 1 + n = n, so g also has order 2.

    but fg(n) = f(g(n)) = f(1-n) = n-1, and for any positive integer k:

    (fg)^k(n) = n - k \neq n so fg is of infinite order.
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Torsion Subgroup

    Thanks Deveno

    Really helpful - and really neat too!

    Peter
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Re: Torsion Subgroup

    Another counter-example is the infinite dihedral group. This is the group with presentation,

    \langle a, b; a^2=1, aba=b^{-1}\rangle

    so it is basically the regular dihedral group but the rotation element has infinite order.

    So, notice that this group is generated by a and ab, both of which are of order 2. Thus, the torsion elements generated the entire group, and so do not form a subgroup!
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