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Group presentation question

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that you cannot injectively take Z into <x,y:xyx=x> by a homomorphism.

    EDIT:the presentation should be <x,y:xyx=y>
    2. Relevant equations



    3. The attempt at a solution

    I am new to group presentation, can someone just give me a hint? Z has the presentation <x: no relations>. Is it true that any homomorphism has to take a relator to a relator? Then it is clearly impossible since Z has only one relator...
     
    Last edited: Jan 31, 2008
  2. jcsd
  3. Jan 31, 2008 #2

    NateTG

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    Fiddling around with the relation in the presentation will make it obvious.
     
  4. Jan 31, 2008 #3
    See the EDIT.
     
  5. Jan 31, 2008 #4

    NateTG

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    Ah, so you're trying for an injective homomorphism [itex]G \rightarrow \mathbb Z[/itex].
    Basically this is the same as saying that [itex]G[/itex] is isomorphic to a (not necessarily proper) subgroup of [itex]\mathbb Z[/itex].

    N.B.: Regarding the bit about relators - consider that [itex]<x,y:x=y>[/itex] is clearly equivalent to [itex]\mathbb Z[/itex].
     
  6. Feb 1, 2008 #5
    Not really. I am trying to show that there is no injective homomorphism from Z to <x,y:xyx=y>. That is, I want to show that you cannot find any copies of Z in <x,y:xyx=y>.
     
  7. Feb 1, 2008 #6

    NateTG

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    Hmm, I'm missing something then.
    [tex]f: \mathbb{Z} \rightarrow G[/tex]
    [tex]f(n)=x^n[/tex]
    or
    [tex]g(n)=y^n[/tex]
    look like they are such homomorphisms.
     
  8. Feb 1, 2008 #7
    Maybe you're right. I wanted to use this as a step in a different problem, but maybe I need to find another step.
     
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