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Thread: Prove H is a subgroup

  1. #1
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    Prove H is a subgroup

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

    Since G is a group under $\displaystyle *$, H is associative under $\displaystyle *$

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

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

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

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

    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 $\displaystyle h,k\in H$ and $\displaystyle k^{-1}\in H$, $\displaystyle h*(k^{-1})^{-1}=h*k=hk\in H$
    since $\displaystyle hk \in H$ (given), then are you trying to show again that $\displaystyle hk \in H$ ? $\displaystyle \smile$
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  5. #5
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    Re: Prove H is a subgroup

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

    Since G is a group under $\displaystyle *$, H is associative under $\displaystyle *$

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

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

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

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

    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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