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  1. #1
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    show that...

    Let H be subgroup of abelian group G, then show that:
    i) $\displaystyle (aH)^-1=a^-1H$
    ii) $\displaystyle (Ha)^2=Ha^2$
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    Quote Originally Posted by GTK X Hunter View Post
    Let H be subgroup of abelian group G, then show that:
    i) $\displaystyle (aH)^-1=a^-1H$
    ii) $\displaystyle (Ha)^2=Ha^2$
    Can you define $\displaystyle (aH)^-1$ and $\displaystyle (Ha)^2$ for me plz?
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  3. #3
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    Quote Originally Posted by GTK X Hunter View Post
    Let H be subgroup of abelian group G, then show that:
    i) $\displaystyle (aH)^-1=a^-1H$
    ii) $\displaystyle (Ha)^2=Ha^2$

    Do you mean in the quotient group $\displaystyle G/H$?
    This follows at once from the definitions of product in quotient groups. Read and understand well these definitions.

    Tonio
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  4. #4
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    take any $\displaystyle h\in H$
    i) $\displaystyle (ah)^{-1}=h^{-1}a^{-1}=a^{-1}h^{-1}=a^{-1}h_o$
    ii) $\displaystyle haha=hhaa=h_1a^2$
    am i right? ^_^
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  5. #5
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    Quote Originally Posted by GTK X Hunter View Post
    take any $\displaystyle h\in H$
    i) $\displaystyle (ah)^{-1}=h^{-1}a^{-1}=a^{-1}h^{-1}=a^{-1}h_o$
    ii) $\displaystyle haha=hhaa=h_1a^2$
    am i right? ^_^
    It's simpler then that.

    $\displaystyle \left(aH\right)\left(a^{-1}H\right)=(aa^{-1})H=H=e\implies a^{1}H=\left(aH\right)^{-1}$. Similarly for the second.
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