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Math Help - Prove H is a subgroup

  1. #1
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    Prove H is a subgroup

    H\neq\emptyset, H\subseteq (G,*), and G is close under inverses h,k\in H and hk,h^{-1}\in H. Prove H\leq G

    Since G is a group under *, H is associative under *

    h^{-1}*(hk)=(h^{-1}h)*k=e*k=k\in H and e\in H

    e=hh^{-1}=h^{-1}h\Rightarrow h^{-1}\in H

    Since h,k\in H and k^{-1}\in H, h*(k^{-1})^{-1}=h*k=hk\in H

    Therefore, since H is closed, associative, has the identity, and has inverses, H\leq G

    Is this correct?
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  2. #2
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    Re: Prove H is a subgroup

    can you write the original problem ? its little confusing since its not clear which are the 'givens'
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    Re: Prove H is a subgroup

    Quote Originally Posted by issacnewton View Post
    can you write the original problem ? its little confusing since its not clear which are the 'givens'
    Everything to the left of the word "prove" is given.
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    Re: Prove H is a subgroup

    Quote Originally Posted by dwsmith View Post
    [tex]
    Since h,k\in H and k^{-1}\in H, h*(k^{-1})^{-1}=h*k=hk\in H
    since hk \in  H (given), then are you trying to show again that hk \in  H ? \smile
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  5. #5
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    Re: Prove H is a subgroup

    Quote Originally Posted by dwsmith View Post
    H\neq\emptyset, H\subseteq (G,*), and G is close under inverses h,k\in H and hk,h^{-1}\in H. Prove H\leq G

    Since G is a group under *, H is associative under *

    h^{-1}*(hk)=(h^{-1}h)*k=e*k=k\in H and e\in H

    e=hh^{-1}=h^{-1}h\Rightarrow h^{-1}\in H

    Since h,k\in H and k^{-1}\in H, h*(k^{-1})^{-1}=h*k=hk\in H

    Therefore, since H is closed, associative, has the identity, and has inverses, H\leq G

    Is this correct?
    What you did is either irrelevant or wrong, except for the associative part.

    You already know that H is closed under * and that any h in H has inverse in H. So all you have to do is to prove that e is in H. just multiply h with h^-1.
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