# Group presentation question

1. Jan 31, 2008

### ehrenfest

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. Jan 31, 2008

### NateTG

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

3. Jan 31, 2008

### ehrenfest

See the EDIT.

4. Jan 31, 2008

### 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 $<x,y:x=y>$ is clearly equivalent to $\mathbb Z$.

5. Feb 1, 2008

### ehrenfest

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. Feb 1, 2008

### NateTG

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.

7. Feb 1, 2008

### 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.