Can Z be Injected into <x,y:xyx=y> by a Homomorphism?

In summary, the conversation discusses how to show that there is no injective homomorphism from the group Z to the group <x,y:xyx=y>. The group Z has the presentation <x: no relations>, making it impossible to find any copies of Z in <x,y:xyx=y>. The conversation also considers using homomorphisms f(n)=x^n or g(n)=y^n as a possible solution, but it is determined that this may not be applicable to the problem at hand.
  • #1
ehrenfest
2,020
1

Homework Statement


Show that you cannot injectively take Z into <x,y:xyx=x> by a homomorphism.

EDIT:the presentation should be <x,y:xyx=y>

Homework Equations


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...
 
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  • #2
Fiddling around with the relation in the presentation will make it obvious.
 
  • #3
See the EDIT.
 
  • #4
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].
 
  • #5
NateTG said:
Ah, so you're trying for an injective homomorphism [itex]G \rightarrow \mathbb Z[/itex].
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>.
 
  • #6
ehrenfest said:
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>.

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.
 
  • #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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