How can I prove that, in this group, elements of this form are distinct? In the group $G=\langle x,y,z\mid xy=yx,zx=x^2z,zy=yz\rangle$, how can I prove that the elements of the form $x^iy^jz^k$ are all distinct?
As Arturo is asking I'm editing this question:  I want to understand what is a set of normal forms. At the beginning I thought that it was $x^iy^jz^k$, but they made me notice that is not so obvious how $z^{-1}x$ can be written in this form. Actually I think I proved that it cannot be, in fact: if $z^{-1}x=x^iy^jz^k$ then conjugating by $z$ we obtain $xz^{-1}=zx^iy^jz^{k-1}=x^{2i}y^jz^k$ and so $z^{-1}=x^{2i-1}y^jz^k$ and so $1=x^{2i-1}y^jz^{k+1}$, I also proved that if $x^iy^jz^k=1$ then $i=j=k=0$, and so we have $2i-1=0$ that has no solution in $\mathbb{Z}$. So could you tell me how can I compute a set of normal forms? (they don't have to be too complicated because I'm asked to draw the cayley graph)
 A: Consider the semidirect product $$G' = \langle X^i,Y \rangle \rtimes_{\varphi} \langle Z \rangle $$ where $i$ ranges over $\mathbb{Z}[1/2]$, powers of $X$ multiply in the obvious way and they commute with $Y$ (which has infinite order). The element $Z$ also has infinite order, and the action of $Z$ is $X^i Y^j \to X^{2i} Y^j$ (here $i \in \mathbb{Z}[1/2], j \in \mathbb{Z})$.  Then $X, Y, Z$ satisfy the given relations (with the map $x \mapsto X$, etc., of course), and by construction, all $X^i Y^j Z^k \ne 1$ for $(i,j,k) \ne (0,0,0)$, hence $x^i y^j z^k \ne 1$.
Edit: As Jack Schmidt points out in the comments, the original construction with $i \in \mathbb{Z}$ didn't work because the action of $Z$ is not an automorphism (only even powers of $X$ would be in the image).  
Edit 2: As Jack Schmidt points out in the comment to Manolito's answer, this can be fixed by letting $i$ range over $\mathbb{Z}[1/2]$ (rationals whose denominator is a power of 2) instead of just $\mathbb{Z}$.  The map $G \to G'$ is surjective because $x^z$ has image $X^{1/2}$, $x^{z^k}$ has image $x^{1/2^k}$, etc., so $G'$ is a quotient of $G$.
Edit 3: In fact, we can prove that $G \cong G'$.  It is sufficient to show that every element of $G$ has a normal form $$(x^a)^{z^b} z^c y^d$$ where $a$ is an odd integer and $b, c, d$ are arbitrary integers.  These elements are distinct because the image of the above element in $G'$ is $X^{a/2^b} Z^c Y^d$, and the decomposition of an element of $\mathbb{Z}[1/2]$ into the form $a/2^b$, where $a$ is odd, is unique. 
Take any word in $x,y,z$ and their inverses.  First of all, since $y$ commutes with everything, we can just pull all the $y$'s out to the right.  From now on I'll assume we just have $x$'s and $z$'s.
To avoid a lot of double superscripts, I'm going to write $x^{[a]}$ for $x^{z^a}$.  Notice that the original relation $zx = x^2z$ can be written as $$x^{[-1]} = x^2$$ and also that we have the law $$(x^{[a]})^{[b]} = x^{[a+b]}.$$  The above identities continue to hold if $x$ is replaced by any integer power of $x$. 
So we have a word in $x$'s and $z$'s (and their inverses).  By repeatedly applying the identity $$z^{-a} x = x^{[a]} z^{-a}$$ we can pull the $z$'s out to the right as well, and we are just left with a product of elements of the form $x^{[a]}$.  We must prove that such a product has the form $(x^b)^{[a]}$.
First of all, we have $$(x^2)^{[a]} = (x^{[-1]})^{[a]} = x^{[a-1]}$$ and the above is still valid if $x$ is replaced by any integer power of $x$, hence by iterating $$(x^{2^k})^{[a]} = x^{[a-k]}$$ for integers $k \ge 0$.  Now the general product $$x^{[a]} x^{[b]} = x^{[a]} x^{[a-(a-b)]} = x^{[a]} (x^{2^{a-b}})^{[a]} = (x^{1+2^{a-b}})^{[a]}$$ valid for $a-b \ge 0$, has the desired form.  If $a < b$, we rewrite $[a] = [b-(b-a)]$ instead, and do a similar calculation.  The even more general product $(x^m)^{[a]} (x^n)^{[b]}$ is similar.
So elements of $G$ have the normal form $(x^a)^{[b]} y^c z^d$, and we just need to show that we can make $a$ odd.  This follows from the identity $$(x^{2a})^{[b]} = ((x^a)^2)^{[b]} = ((x^a)^{[-1]})^{[b]} = (x^a)^{[b-1]}$$ which we can repeatedly apply to remove all the powers of 2 from $a$. 
A: let's do it the intuitive way: 
If you take some time to look at the relations, you will notice that every word in $x, y, z$ can be put into the form $x^iy^jz^k$, as $x$ and $y$ commute, and so do $y$ and $z$. Though $x$ and $z$ do not commute, strictly speaking, the relator $zx=x^2z$ lets you move all the $x$ to the left, doubling their exponents. Technically speaking, every word has a unique normal form $x^iy^jz^k$, and obviously, two such normal forms with different exponents are distinct. 
From the relators you can also deduce that the group is the direct product of the infinite free group $\langle y \rangle$ (isomorphic to $\mathbb{Z}$) with the group $\langle x, z \mid zx=x^2z \rangle$. 
Hope this helps. 
