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Let $G$ be the group presented by $$G=\bigl\langle x,y,z\;\bigm|\; xy=yx,\;zxz^{-1}=x^my^n,\;zyz^{-1}=x^py^q\bigr\rangle.$$

I would like to prove that for every $m,n,p,q$, $G$ is solvable.

What is my idea: call $d=mq-np$. First suppose that $d\neq0$. I think that in this case $G\cong (\mathbb{Z}[\frac{1}{d}])^2\rtimes_\varphi\mathbb{Z}$ (that is solvable), where $\varphi(1)(1,0)=(m,n)$ and $\varphi(1)(0,1)=(p,q)$; but I have some problems to find $\alpha,\beta$ such that $\alpha^d=x$ and $\beta^d=y$, could you help me?

Now suppose $d=0$, if $m=n=p=q=0$ then $G=\mathbb{Z}$ and it's solvable. Otherwise suppose $m\neq0$, I don't know how to continue now, I proved that in this case $x^p=y^m$ I don't know if this helps, any idea?

EDIT: in the second case we can also suppose $p\neq0$, in fact if $p=0$ then $q=0$ and so $zyz^{-1}=1$ that implies $y=1$ and so $G$ becomes $\langle x,z|zxz^{-1}=x^m\rangle$ that is isomorphic to $\mathbb{Z}[\frac{1}{m}]\rtimes\mathbb{Z}$ that is solvable.

EDIT EDIT: now suppose $n=0$ then $q=0$ so the group becomes $\langle x,y,z|xy=yx,zxz^{-1}=x^m,zyz^{-1}=x^p\rangle$ but I proved that $x^p=y^m$ and so in this case $G\cong(\mathbb{Z}[\frac{1}{m}])^2\rtimes\mathbb{Z}$ that is solvable, so we can assume $n,m,p,q\neq0$ (I really don't know if this is useful).

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    Look at the subgroup normally generated by $x$ and $y$, perhaps writing $x_r=z^{r}xz^{-r}$ and $y_r=z^{r}yz^{-r}$. The relations of this subgroup look like $[x_r,y_r]=1$, $x_{r+1}=x_r^my_r^n$, and $y_{r+1}=x_r^py_r^q$. Now note that you can check if a group is abelian *locally*, as in, checking two elements at a time. So for example, $x_s$ commutes with $x_t$ when $s, by applying those relators over and over again.2012-02-21
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    @SteveD: are you sure that $x_{r+1}=x_r^my_r^n$? because to me it seems to be $x_{r-1}=x_r^my_r^n$, am I right? (the same for $y$).2012-02-21
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    Depends on how you write it. I usually write conjugation to mean $z^{-1}xz$, but you wrote it above as $zxz^{-1}$. So your relation in your group is the same as $x_1=x_0^my_0^n$.2012-02-21
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    ah right right, for some reason I read $x_r=z^{-r}xz^r$.2012-02-21
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    ok I got it, $^N=$, where $R_1,\ldots,R_k$ are possible relations that I don't care about; but with only the explicit relations I proved that this is abelian and so we have an abelian series $1\triangleleft^N\triangleleft G$, am I right?2012-02-21
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    Yes! In fact, a Reidemeister-Schreier rewriting will show there are no $R_1$, $R_2$, etc. But that doesn't effect the result.2012-02-21
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    @SteveD: Maybe I'm missing something...but surely $\langle x, y\rangle\unlhd G$, as $x^z\in\langle x, y\rangle$ and $y^z\in \langle x, y\rangle$. As $[x, y]=1$ you have that $\langle x, y\rangle$ is abelian and you're, basically, done...2012-02-21
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    @user1729: You need to check that $\langle x,y\rangle$ is also closed under conjugation by $z^{-1}$, which it may not be.2012-02-21

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