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Free product of non-trivial groups is non-abelian

  1. Dec 15, 2011 #1
    Hello

    I have to show that the free product of a collection of more than one non-trivial group is non-abelian.

    But doesn't this just follow from the definition of the free product?
    Or how would you tackle this question?
     
  2. jcsd
  3. Dec 15, 2011 #2

    Office_Shredder

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    Well... what definition of free product are you using, and how does non-abelian follow from it?
     
  4. Dec 15, 2011 #3
    You mean the definition of the free product?

    As a set, the free product  G consists of all
    words g_1g_2.. .g_m of arbitrary finite length m > 0, where each letter g_i belongs to
    a group G_ i and is not the identity element of G_ i , and adjacent letters g_i and g_(i+1‚)
    belong to different groups G .

    For a counterexample to being abelian i thought about:
    [itex]\pi_1(S^1 \vee S^1)[/itex] which corresponds to the free product of the fundamental groups of S^1 and which is not abbelian
     
  5. Dec 15, 2011 #4

    Office_Shredder

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    So if you have two groups G and H, and you look at G*H, can you identify two elements in G*H which do not commute?
     
  6. Dec 15, 2011 #5
    Hi Office_Shredder

    Thanks for your answer.
    >So if you have two groups G and H, and you look at G*H, can you identify two elements in G*H which do not commute?

    Thats what I meant when I wrote in the beginning that it almost follows form the definition, because it holds for any two non trivial items g,h from G,H respectively that g*h is not the same as h*g.
     
  7. Dec 15, 2011 #6

    Dick

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    Yes, it's almost trivial. g*h is not equal to h*g if g and h are not the identities of their respective groups.
     
  8. Dec 15, 2011 #7
    So is there actually something to prove here?
     
  9. Dec 15, 2011 #8

    Dick

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    Not much. gh is not equal to hg because g isn't equal h and neither is the identity.
     
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