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Math Help - Algebra, group action

  1. #1
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    Algebra, group action

    Studying for a final and had the following practice problem, which I can't answer. Could anyone help?

    Let G be a finite group acting on a finite set. Prove that <a>=<b> for two elements a,b in G then I(a)=I(b), where for any element g in G, I(g)={s in S: gs=s}.

    Thank you.
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    Studying for a final and had the following practice problem, which I can't answer. Could anyone help?

    Let G be a finite group acting on a finite set. Prove that <a>=<b> for two elements a,b in G then I(a)=I(b), where for any element g in G, I(g)={s in S: gs=s}.

    Thank you.
    I presume you are talking about G-sets.
    The notation I will use is, G for the group.
    And X for the non-empty set.
    The binary operation * will simply be juxtaposition of two elements,
    *:G\times X\to X
    The standard way to show,
    X_a=X_b
    It to show,
    X_a\subseteq X_b \mbox { and }X_b\subseteq X_a
    Now,
    X_a=\{x \in X| ax=x\}
    X_b=\{x \in X|bx=x\}

    Let,
    x\in X_a
    Then,
    x\in X_b
    Because,
    b\in <a>, b=a^c
    Thus,
    bx=a^cx=a^{c-1}x=...=ax=x implies x\in X_b
    Similarly,
    a\in <b>,a=b^d
    Thus,
    ax=b^dx=b^{d-1}x=...=bx=x implies x\in X_a.
    Thus,
    X_a\subseteq X_b\mbox{ and }X_b\subseteq X_a\to X_a= X_b

    (The facts that the sets are finites and groups are finite was not necessary).
    Last edited by ThePerfectHacker; December 2nd 2006 at 07:06 PM.
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