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Thread: <a>=<a^-1>

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

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


    (i)

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


    (ii)

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