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Prove a group is abelian

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=57698&stc=1&d=1365589730.jpg


    2. Relevant equations



    3. The attempt at a solution

    For the first question, since [f(a)][g(a)] is in C, can I just say that since C is a ring, it is an abelian group, then the four axioms are proven? Then just show closure? Probably not I'm guessing. Associativity of multiplying functions seems so fundamental to me that I really don't know what to do...

    For the second question, shall I start by using composition of functions and then the properties of composition of a function and its inverse, and then go on about the identity function?
     

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    Last edited by a moderator: Apr 10, 2013
  2. jcsd
  3. Apr 10, 2013 #2
    I'd say question one is asking you to flesh out the group structure.

    1. What's the identity element? (make sure it's a member of C^A).

    2. For each element f, what's the inverse element? (make sure it's well defined
    - namely, why can you find f^-1 for each f? )

    Closure is trivial but should be mentioned, namely why is (fg) a well
    defined member of C^A when f and g are?

    Associativity (and Abelian-ness) of course follow from the properties of C,
    but you should have a proof showing f(gh) =(fg)h.

    For question 2. I don't know what A-hat is. What do you know
    about group homomorphisms at this point?
     
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