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

[ehrenfest]

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...

 

[NateTG]

Fiddling around with the relation in the presentation will make it obvious.

 

[ehrenfest]

See the EDIT.

 

[NateTG]

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.

 

[ehrenfest]

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>.

 

[NateTG]

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.

 

[ehrenfest]

Maybe you're right. I wanted to use this as a step in a different problem, but maybe I need to find another step.