Homework Help: 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.