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

    Since H is an abelian group, we have $\displaystyle {(xy^{-1})}^2=xy^{-1}xy^{-1}=xxy^{-1}y^{-1}=1$. 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 $\displaystyle S_3$.
    Last edited by aliceinwonderland; Nov 3rd 2009 at 11:23 PM.
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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, $\displaystyle (a^{-1})^2=1$. So, $\displaystyle a^{-1}a^{-1}=1$ and thus $\displaystyle a^{-1} \in H$.
    Problem: Let $\displaystyle H$ be any group. Define $\displaystyle S=\{a\in H~|~a^2=1\}$. Prove that $\displaystyle S$ is a subgroup of $\displaystyle H$.

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

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

    $\displaystyle \square$
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