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Basic group theory question

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data

    I need to determine [tex]dad^1[/tex] for each element d in the left-coset formed by acting on the elements in [tex]C_G(a)[/tex] with the element c such that c is not an element of the subgroup [tex]C_G(a)[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I don't really understand what the question is asking me to do? I'm fairly new to group theory and understand that d are the elements in the left-coset of the subgroup and that a is the fixed element of the group G but I'm not really familiar with [tex]d^1[/tex], why is there a 1 in the exponent of the element?

    Can anyone start me off here? Or at least give me an outline of the goal?
     

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  3. May 7, 2014 #2

    Fredrik

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    I would guess that it's a prime, not a 1. d and d' are just the notations they use for two different elements of that set. (Or is it perhaps ##d^{-1}##, the inverse of ##d##?)

    Is that the full problem statement? What is the definition of ##C_G(a)##?
     
  4. May 7, 2014 #3

    pasmith

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    Should that not be [itex]dad^{-1}[/itex]?

    How does [itex]C_G(a)[/itex] relate to the element [itex]a[/itex]?

     
  5. May 7, 2014 #4
    I have updated the opening post with an attachment of the question, hopefully its more clear
     
  6. May 7, 2014 #5

    ChrisVer

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    It obviously meant to be the inverse (it's easier to lose a minus than misclick ' for 1)...
    In any case, what you are asked to do is to find [itex]dad^{-1}[/itex] with [itex]a\in G[/itex] and [itex]d \in cG[/itex], c does not belong to a's centralizer, that means that it doesn't commute with a...
    you should try to write d and d^-1 and check what you can do...
     
  7. May 7, 2014 #6
    Thanks for the hints and clarification
     
  8. May 8, 2014 #7
    I'm really confused still, does the left coset commute with any of the elements? also does it obey any group axioms, or is this not relevant

    Am I supposed to start with d or d^-1 = something and keep doing valid operations on it until i get to dad^-1 = something?
     
  9. May 8, 2014 #8

    Fredrik

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    I'm also very confused by the problem statement. It appears to have at least one more typo.

    The first thing you need to do is to write down the definitions of the relevant sets. The only difficulty is that the problem statement is so confusing. I would guess that "some element cG" actually means "some element ##c\in G##". I don't think it can mean "some element of cG", because ##cG=\{cg|g\in G\}=G##. (For all ##g\in G##, we have ##g=c(c^{-1}g)\in cG##).
     
  10. May 8, 2014 #9
    Is there a different definition for a coset of a centralizer vs a coset of a standard subgroup? And I think you are correct about the 2nd typo, I have emailed my professor to check though.
     
  11. May 8, 2014 #10

    micromass

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    I think it's this. Take an element ##a\in G## and define

    [tex]C_G(a) = \{b\in G~\vert~ba=ab\}[/tex]

    Take an element ##c\in G## such that ##c\notin C_G(a)##, then we consider the coset ##cC_G(a)##. The question is to calculate ##dad^{-1}## for each ##d\in cC_G(a)##. And with that, I think you need to show that ##dad^{-1}## does not depend on ##d\in C_G(a)##, but only of ##a## and ##c##.
     
  12. May 8, 2014 #11
    Is this what I am finding? The conjugates of elements in the centralizer?

    http://www.proofwiki.org/wiki/Conjugates_of_Elements_in_Centralizer
     
  13. May 8, 2014 #12

    Fredrik

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    That's a very similar result. But I think we have told you enough to give the problem a shot now. I would recommend that you do it without looking at solutions of similar problems. Just write down the definitions of the sets mentioned in the problem statement, and then you're almost done.
     
  14. May 8, 2014 #13
    Of course, sorry, this is just the 1st ever problem i've attempted in group theory and its a struggle to get my head around.

    From the definition of the left coset is it correct to say ##cS_G(a)=\{d~\epsilon~ G:\exists~ g ~\epsilon~ S_G(a):d=cg\}##

    so ##dad^{-1}=cgad^{-1}## ?
     
  15. May 8, 2014 #14

    Fredrik

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    Yes, but you can simplify the notation to ##cS_G(a)=\{cg:g\in S_G(a)\}##.

    Good start, but you need to handle ##d^{-1}## as well.
     
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