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Math Help - g\dot a=ag^{-1}

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    g\dot a=ag^{-1}

    Let G be a group and let A = G. Show that the maps defined by g\cdot a=ag^{-1}, \ \forall g,a\in G do not satisfy the axioms of a left group action of G on itself.

    I found the opposite. Here is what I did.

    Let g_1,g_2\in G.
    g_1\cdot (g_2\cdot a)=g_1\cdot (ag_2^{-1})=ag_2^{-1}g_1^{-1}=a(g_1g_2)^{-1}=(g_1g_2)\cdot a

    e\cdot a=ae^{-1}=a

    What is wrong?

    Correction: It says do satisfy the axioms.
    Last edited by dwsmith; October 8th 2011 at 10:29 AM.
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