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Math Help - Conjugation question

  1. #1
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    Conjugation question

    If G is a group, K \lhd G is proper and \forall x \in G \backslash K, x^2 = e, show that \forall t \in G \backslash K, \forall a \in K, t^{-1} at = a^{-1}.

    I'm completely confused on how to solve this. I just seem to be going round in circles.

    Can anyone help?
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  2. #2
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    Quote Originally Posted by Giraffro View Post
    If G is a group, K \lhd G is proper and \forall x \in G \backslash K, x^2 = e, show that \forall t \in G \backslash K, \forall a \in K, t^{-1} at = a^{-1}.
    If a\in K and  t \in G \setminus K then  at \in G \setminus K. Thus (at)^{-1} = at, and also t^{-1} = t.
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  3. #3
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    Thanks! I thought it was something blindingly obvious.
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