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Math Help - abelian groups

  1. #1
    Junior Member
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    abelian groups

    Let G be an abelian group (possibly infinite and let the set T={a in G|a^m=e, m>1 depending on a}. Prove that:

    a) T is a subgroup of G.
    b) G/T has no element - other than its identity element - of finite order.

    Please show steps. This section confuses me so much.

    Thank you.
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  2. #2
    Super Member Gamma's Avatar
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    Subgroup

    All you gotta do to show something is a subgroup is show it is closed under the group operation and closed under inverses (the rest is inherited from the mother group).

    This is the torsion subgroup. Let a,b\in T. Then \exists m, n \in \mathbb{N} such that a^m=1=b^n. We wish to show ab\in T.
    This is particularly easy because G is assumed to be abelian. (ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=1^n 1^m=1
    Thus ab\in T as desired.

    Let a\in T \subset G \Rightarrow a^{-1}\in G, suppose a^n=1.

    (a^{-1})^n=a^{-n}=(a^n)^{-1}=1^{-1}=1

    Thus a^{-1}\in T as desired
    T is a subgroup of G.

    You just modded out by everything of finite order by definition of T. G/T cannot have anything other than the coset 1T itself of finite order, think about it. G is abelian so every subgroup is normal so this is a well defined group.

    Just a note, if G were finite everything is torsion by LaGrange, so G/T is trivial. The only interesting things are for infinite abelian groups.
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