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Can someone double-check this simple binary relation proof?

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data


    *attached



    2. Relevant equations



    3. The attempt at a solution


    Let a,b ∈ H. Then (∀x ∈ S)(a*x = x*a) and (∀x ∈ S)(b*x = x*b). It is easy to see, then, that a*b = b*a. Now let c ∈ S. Then (a*b)*c = c*(a*b) by the associativity of *.

    Q.E.D.
     

    Attached Files:

  2. jcsd
  3. Sep 7, 2011 #2

    micromass

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    The last statement does not follow from associativity. Associativity merely says that

    [tex](a*b)*c=a*(b*c)[/tex]

    I also see no reason to take c in S. You need to prove that a*b in S. That is, you need to show for all c that (a*b)*c=c*(a*b). You don't only need to prove it for the c in S.
     
  4. Sep 7, 2011 #3
    Hey micromass. In order to show that a*b is an element of H, isn't the statement to prove: (for all c in S)((a*b)*c = c*(a*b))? Am i mistaken here?

    *By the way i am taking c in S to be an arbitrary element
     
  5. Sep 7, 2011 #4

    micromass

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    Oops, I had my notation mixed up :frown: Yes, that is correct!
     
  6. Sep 7, 2011 #5
    Heh. So wait... is the proof correct?
     
  7. Sep 7, 2011 #6

    micromass

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    You still need to explain why (a*b)*c = c*(a*b). It doesn't follow from associativity.
     
  8. Sep 7, 2011 #7
    Can i say that since a,b are elements of S (since H is a subset of S), then a*b is in S. But also, since c is in S, then (a*b)*c is in S. and since * is associative on S, a*(b*c) is in S?
     
  9. Sep 7, 2011 #8

    micromass

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    How does thart prove (a*b)*c=c*(a*b)??

    You need to use that a and b are in H.
     
  10. Sep 8, 2011 #9
    ahhh, finally Figured it out micromass! I wasn't using the fact that a and b were in H to commute the result. That is, since (a*b)*c = a*(b*c), for some unknown reason I thought this was the goal. But now we may use that a is in H to write (b*c)*a and continue to use the fact that a,b are in H along with associativity to obtain our result.

    I appreciate your help
     
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