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Math Help - Abstract Algebra (left cosets)

  1. #1
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    Abstract Algebra (left cosets)

    Suppose that H is a Normal Subgroup of G . a, b in G. Prove that (aH)(bH)=abH
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  2. #2
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    Re: Abstract Algebra (left cosets)

    This follows directly from the definition of a normal subgroup.
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    Re: Abstract Algebra (left cosets)

    Can you show me a more detailed solution?
    I know if H is normal sub group in G then hah^-1 is in H for all h in H , and all a in G
    And same apply for another element in G, b in G then hbh^-1 in H for all h in H and b in G
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    Re: Abstract Algebra (left cosets)

    Quote Originally Posted by mohammedlabeeb View Post
    I know if H is normal sub group in G then hah^-1 is in H for all h in H , and all a in G
    This is a wrong definition. The correct one is that aha-1 ∈ H for all h ∈ H and a ∈ G, which is equivalent to saying that aH = Ha for all a ∈ G.
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    Re: Abstract Algebra (left cosets)

    What is my next step?
    If I start with the LHS: (aH)(bH)
    Is it right if I say ah1 bh2 ?
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    Re: Abstract Algebra (left cosets)

    Hi,
    I think maybe you want a detailed "set theoretic" proof. Here is one:

    Abstract Algebra (left cosets)-mhfgroups10.png
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    Re: Abstract Algebra (left cosets)

    '
    Last edited by mohammedlabeeb; April 19th 2013 at 10:22 PM.
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    Re: Abstract Algebra (left cosets)

    Quote Originally Posted by johng View Post
    Hi,
    I think maybe you want a detailed "set theoretic" proof. Here is one:

    Click image for larger version. 

Name:	MHFgroups10.png 
Views:	7 
Size:	11.0 KB 
ID:	28040
    you are the best
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    Re: Abstract Algebra (left cosets)

    I have attached my solution please let me know if it is right or how can I fix it? CCI04202013_00001.bmp
    Quote Originally Posted by mohammedlabeeb View Post
    you are the best
    Quote Originally Posted by johng View Post
    Hi,
    I think maybe you want a detailed "set theoretic" proof. Here is one:

    Click image for larger version. 

Name:	MHFgroups10.png 
Views:	7 
Size:	11.0 KB 
ID:	28040
    Follow Math Help Forum on Facebook and Google+

  10. #10
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    Re: Abstract Algebra (left cosets)

    Quote Originally Posted by mohammedlabeeb View Post
    I have attached my solution please let me know if it is right or how can I fix it?
    The solution is OK, but you proved the two key facts several times:

    (1) ∀a ∈ G ∀h1 ∈ H ∃h2 ∈ H, ah1 = h2a
    (2) ∀a ∈ G ∀h1 ∈ H ∃h2 ∈ H, h1a = ah2.

    Note that (1) is equivalent to aH ⊆ Ha and to aHa-1 ⊆ H for all a ∈ G, and (2) is equivalent to Ha ⊆ aH and to H ⊆ aHa-1 for all a ∈ G. Thus, (1) and (2) together are equivalent to aH = Ha for all a and to aHa-1 = H for all a.

    I recommend internalizing these three equivalent ways of saying that H ⊲ G.

    If we use the fact that aH = Ha as the definition of H ⊲ G, then the original claim is trivial. Indeed, (aH)(bH) = a(Hb)H (by normality) = a(bH)H = (ab)HH = (ab)H. This is why I said it follows directly from the definition of a normal subgroup.
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