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Math Help - showing identity in proving k ={x in g|x = aha^-1 h in H}

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    Post showing identity in proving k ={x in g|x = aha^-1 h in H}

    Ok the problem is let H subgroup of G, a fixed in G, let K = {x in G|x = aha^-1 h in H} Prove or prove K subgroup of G. Im pretty sure i got closure right and the identity is pretty easy but im stuck on showing inverse in K. We know for some x in G x = aha^-1 and we know x^-1 in G because G group. but would that transfer to x^-1 = (aha^-1)^-1 or what that still doesnt seem to get me where i need to go

    **Sorry about the title i just noticed i put identity when i meant inverse
    Last edited by ChrisBickle; January 22nd 2010 at 08:02 PM.
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    Quote Originally Posted by ChrisBickle View Post
    Ok the problem is let H subgroup of G, a fixed in G, let K = {x in G|x = aha^-1 h in H} Prove or prove K subgroup of G. Im pretty sure i got closure right and the identity is pretty easy but im stuck on showing inverse in K. We know for some x in G x = aha^-1 and we know x^-1 in G because G group. but would that transfer to x^-1 = (aha^-1)^-1 or what that still doesnt seem to get me where i need to go

    **Sorry about the title i just noticed i put identity when i meant inverse

    For any m\in\mathbb{Z}\,,\,\,(aha^{-1})^m=ah^ma^{-1}. You also may want to use (xy)^{-1}=y^{-1}x^{-1} , which is true in any group.

    Tonio
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