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Math Help - Subgroup proof is this right?

  1. #1
    Senior Member sfspitfire23's Avatar
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    Subgroup proof is this right?

    Show that a^2=1 is a subgroup of H Is my inverse right?

    For the inverse, (a^{-1})^2=1. So, a^{-1}a^{-1}=1 and thus a^{-1} \in H.
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  2. #2
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    Quote Originally Posted by sfspitfire23 View Post
    show that a^2=1 is a subgroup of h is my inverse right?

    For the inverse, (a^{-1})^2=1. So, a^{-1}a^{-1}=1 and thus a^{-1} \in h.
    Assume H is an abelian group. Let K be a subset of H such that K=\{a \in H | a^2=1\}. We show that K is a subgroup of H. It suffices to show that whenever x and y are in K, then xy^{-1} is also in K (link).

    Since H is an abelian group, we have {(xy^{-1})}^2=xy^{-1}xy^{-1}=xxy^{-1}y^{-1}=1. Thus K is a subgroup of H.

    Anyhow I don't think K is necessarily a subgroup of H if H is a non-abelian group. Take an example of S_3.
    Last edited by aliceinwonderland; November 4th 2009 at 12:23 AM.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Show that a^2=1 is a subgroup of H Is my inverse right?

    For the inverse, (a^{-1})^2=1. So, a^{-1}a^{-1}=1 and thus a^{-1} \in H.
    Problem: Let H be any group. Define S=\{a\in H~|~a^2=1\}. Prove that S is a subgroup of H.

    Proof (of inverse): Given a^{-1}\in H, we have to show that a^{-1}\in S if a\in S.

    (a^{-1})^2=a^2(a^{-1})^2=1 so a^{-1}\in S

    \square
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