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Why doesn't this presentation embed?

  1. Sep 20, 2008 #1

    tgt

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    How to show <G,t|t^{-1}kt=k, k in K> does not embed into <G,t|->?

    Where K is a subgroup of an arbitrary group G.
     
  2. jcsd
  3. Sep 21, 2008 #2

    morphism

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    I admit I don't know much about this stuff, but what exactly does <G,t|-> mean? Is it the free group generated by the set [itex]G \cup \{t\}[/itex]? And how are you defining the embedding of presentations?
     
  4. Sep 21, 2008 #3

    tgt

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    If G=<X|R> where X is the generating set for G and R are the relations in G then <G,t|-> is <X,t|R>.

    Embedding means there exists an injective homomorphism.
     
  5. Sep 21, 2008 #4

    morphism

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    I still don't understand. What is "-", the empty relation? If so then isn't what I asked true: <G,t|-> is the free group generated by [itex]G \cup \{t\}[/itex]?
     
  6. Sep 21, 2008 #5

    tgt

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    I never said you were wrong. I like to only use the term free group for groups generated from a basis only. What I said in the previous post has no mistakes either.
     
  7. Sep 22, 2008 #6

    tgt

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    Correction, you were actually wrong. The free group generated by G U {t} will be greater. i.e take G={a,b} with the relations a=1, bb=1. Those relations will hold in <G,t|-> but not in the free group. In fact <G,t|-> is the free product of G with {t}.
     
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