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

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Homework Help Overview

The discussion revolves around the injective homomorphism from the group of integers, Z, into the group presentation . The original poster seeks to demonstrate the impossibility of such an injection, particularly focusing on the implications of group presentations and relators.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the nature of homomorphisms and relators, questioning whether a homomorphism must map relators to relators. There is also discussion about the equivalence of certain group presentations and their implications for injectivity.

Discussion Status

Participants are actively engaging with the problem, raising questions about the definitions and properties of group presentations. Some guidance has been offered regarding the relationship between the groups involved, but no consensus has been reached on the specific injective homomorphism in question.

Contextual Notes

There is an emphasis on the original poster's confusion regarding the nature of the groups and the implications of their presentations. The discussion reflects a learning process about group theory concepts without resolving the central question.

ehrenfest
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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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Fiddling around with the relation in the presentation will make it obvious.
 
See the EDIT.
 
Ah, so you're trying for an injective homomorphism G \rightarrow \mathbb Z.
Basically this is the same as saying that G is isomorphic to a (not necessarily proper) subgroup of \mathbb Z.

N.B.: Regarding the bit about relators - consider that &lt;x,y:x=y&gt; is clearly equivalent to \mathbb Z.
 
NateTG said:
Ah, so you're trying for an injective homomorphism G \rightarrow \mathbb Z.
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>.
 
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.
f: \mathbb{Z} \rightarrow G
f(n)=x^n
or
g(n)=y^n
look like they are such homomorphisms.
 
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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