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Math Help - <a>=<a^-1>

  1. #1
    Junior Member
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    <a>=<a^-1>

    Let H be subgroup of G, show that
    i) a\notin H \Rightarrow H\cap Ha=\emptyset
    ii) b\notin Ha \Rightarrow Ha\cap Hb = \emptyset
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  2. #2
    Junior Member
    Joined
    Oct 2009
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    We will prove the contrapositive statements.


    (i)

    Suppose x\in H\cap Ha. Then x=h_1=h_2a for some h_1,h_2\in H, \therefore\ a=h_2^{-1}h_1\in H.


    (ii)

    Suppose x\in Ha\cap Hb. Then x=h_1a=h_2b for some h_1,h_2\in H, \therefore\ b=h_2^{-1}h_1a\in Ha.
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