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  1. #1
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    show that...

    Let H be subgroup of abelian group G, then show that:
    i) (aH)^-1=a^-1H
    ii) (Ha)^2=Ha^2
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  2. #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) (aH)^-1=a^-1H
    ii) (Ha)^2=Ha^2
    Can you define (aH)^-1 and (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) (aH)^-1=a^-1H
    ii) (Ha)^2=Ha^2

    Do you mean in the quotient group 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 h\in H
    i) (ah)^{-1}=h^{-1}a^{-1}=a^{-1}h^{-1}=a^{-1}h_o
    ii) haha=hhaa=h_1a^2
    am i right? ^_^
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by GTK X Hunter View Post
    take any h\in H
    i) (ah)^{-1}=h^{-1}a^{-1}=a^{-1}h^{-1}=a^{-1}h_o
    ii) haha=hhaa=h_1a^2
    am i right? ^_^
    It's simpler then that.

    \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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