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Direct Product of two Groups

  1. May 19, 2009 #1
    Suppose you had the following:

    (A,*) and (B,[tex]\nabla[/tex])

    So to prove associativity, since I know that both A and B are groups, their direct product will be a group. Could I do the following

    ai , bi[tex] \in A,B


    Since A and B are groups, I know they have distributing everything via the proper binary operations (I got kind of lazy at this point). Can I just multiply both sides by an a-11b-11 and so on until i get something like b3=b3

    I just want to make sure this is sort of hte process one uses to prove the direct product is a group. By the way this question comes from Dummit and Foote, it says that the proof of this is left as a straightforward excercise.
  2. jcsd
  3. May 20, 2009 #2


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    Hi chaotixmonjuish! :smile:
    Yup, it is straightforward …

    i] what is [(a1,b1)(a2,b2)]?

    ii] so what is [(a1,b1)(a2,b2)](a3,b3)?

  4. May 20, 2009 #3
    Your hint kind of confused me.
  5. May 20, 2009 #4
    After multiplying everything

    I'm changing the triangle to *'


    Then once more:


    Since we know (A,*) and (B,*') is a group, we know there exists identity elements ai^1 and bi-1. We can use those identities to get all the a's equal to each other or the b's equal. That's shows associativity.
  6. May 20, 2009 #5


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    hmm … a bit complicated …

    try starting with [(a1,b1)(a2,b2)](a3,b3) = ([a1a2]a3,[b1b2]b3) :wink:
  7. May 20, 2009 #6
    Outside of that little thing, am I on the right track?
  8. May 21, 2009 #7


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    hmm … I got very confused by your …
    … which I still don't see the need for. :confused:

    on looking back, your method seems to be the same as mine (with the first line left out)

    This really is a "straightforward exercise" (as your book calls it), but you seem to have a knack for making it complicated.

    Just start with my line, use the associative law on each side of the comma separately, then separate out (a1,b1), and you're done :smile:
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