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Math Help - Proving a subgroup relation

  1. #1
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    Post Proving a subgroup relation

    Heres the question:
    Let G be a group a element in G and H is a subgroup of G. Define K={aha(inverse) such that h element in H. Prove K is a subgroup of G.

    Heres what i got.
    Let x,y elements in K then there exists c,d elements in H such that x=aca(inverse) and y = ada(inverse). then xy equals (aca(inverse))(ada(inverse)) group the a and a inverse in the middle to get a(cd)a(inverse) c,d elements in H so thats in K.

    This is where i get stuck. We know the identity is in H because H subgroup of G how do we show e is in K. Also how do we find the inverse of this and prove its in K.
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  2. #2
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    Hi,

    e \in H so aea^{-1} \in K (according to the definition of K) but aea^{-1}=e.

    b \in K means that there exists c\in H such that b=aca^{-1}. We have c^{-1} \in H so ac^{-1}a^{-1}\in K. But ac^{-1}a^{-1} = (aca^{-1})^{-1}= b^{-1}.
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  3. #3
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    Quote Originally Posted by ChrisBickle View Post
    Heres the question:
    Let G be a group a element in G and H is a subgroup of G. Define K={aha(inverse) such that h element in H. Prove K is a subgroup of G.

    Heres what i got.
    Let x,y elements in K then there exists c,d elements in H such that x=aca(inverse) and y = ada(inverse). then xy equals (aca(inverse))(ada(inverse)) group the a and a inverse in the middle to get a(cd)a(inverse) c,d elements in H so thats in K.

    This is where i get stuck. We know the identity is in H because H subgroup of G how do we show e is in K. Also how do we find the inverse of this and prove its in K.
    Identity: aea^{-1} = aa^{-1} = e
    Inverse: Consider b =ah^{-1}a^{-1}. Can you see why b\in K?
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  4. #4
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    It is still alittle fuzzy to me...lookin at the post above i see what he is doing and i understand the conclusion but i dont get why (ac)inverse(a)inverse = (aca(inverse))inverse...is it a property of something im missing?
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  5. #5
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    Quote Originally Posted by ChrisBickle View Post
    It is still alittle fuzzy to me...lookin at the post above i see what he is doing and i understand the conclusion but i dont get why (ac)inverse(a)inverse = (aca(inverse))inverse...is it a property of something im missing?
    Either by using the fact that (ab)^{-1} = b^{-1}a^{-1} \Rightarrow (aca^{-1})^{-1} = ((ac)(a^{-1}))^{-1} = (a^{-1})^{-1}(ac)^{-1} = ac^{-1}a^{-1}

    Or simply going by the fact that (aca^{-1})(ac^{-1}a^{-1}) = (ac^{-1}a^{-1})(aca^{-1}) = e
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