infinite group that maps onto dihedral group The group is generated by $y_i$, $i=0, ...,p-1$
with relations
$y_0y_1=y_1y_2=...=y_{p-1}y_0$
$y_0y_2=y_1y_3=...=y_{p-1}y_1$
$\vdots$
$y_0y_{p-1}=y_1y_0=...y_{p-1}y_{p-2}$
I have run into this group and wondering if it is familiar. If you add the relations $y_i^2$ you get the dihedral group of order $2p$.
 A: Here is the argument for arbitrary $p>3$. Note that from the first line of equalities we get 
$y_2=y_1^{-1}y_0y_1$, $y_3=(y_0y_1)^{-1}y_1(y_0y_1)$ and that all $y_i$ are in the subgroup generated by $y_1, y_2$. From the first equality of the second line we get that 
$y_0y_2=y_1y_3$, that is $y_0y_1^{-1}y_0y_1=y_1(y_0y_1)^{-1}y_1(y_0y_1)$. After cancelation, we get $y_0^2=y_1^2$. Hence $y_1^2$ is in the center. The factor-group satisfies $y_0^2=y_1^2=1$, that is the factor-group over that central cyclic subgroup is a dihedral group (generated by two involutions).  Now for simplicity assume that $p$ is even. The case when $p$ is odd is similar. Let $n=p/2+1$. Then from the first line of equalities we get $y_{p-1}=(y_0y_1)^{-n}y_1(y_0y_1)^n$. Now the last term on the first line gives 
$y_{p-1}y_0=y_0y_1$, so $(y_0y_1)^{-n}y_1(y_0y_1)^ny_0=y_0y_1$. This means $(y_0y_1)^{n+1}=
(y_1y_0)^{n+1}$. Hence the dihedral factor-group is finite.  Thus your group is a cyclic central extension of a finite dihedral group. 
A: OK, here is an argument for $p$ odd (I have been [wrongly] assuming $p$ was a prime this whole time).
First, note that the abelianization is $\mathbb{Z}$; this shows is that if $K=[G,G]$, then your group is the semidirect product $K\rtimes \mathbb{Z}$.  The first half of the Reidemeister-Schreier theorem shows that $K$ is generated by $b_{i,n}$, where $b_{i,n}=y_0^ny_iy_0^{-n-1}$. The relations are all of the form $y_0^ny_iy_jy_{j+k}^{-1}y_{i+k}^{-1}y_0^{-n}$, where all the first indices are interpreted modulo $p$.  If we allow $b_{0,n}=1$ for all $n$, then these relations are all rewritten as $b_{i,n}b_{j,n+1}b_{j+k,n+1}^{-1}b_{i+k,n}^{-1}$.  If we think of all the generators in an array, where the columns are indexed by $i$ mod $p$, and the rows indexed by $n\in\mathbb{Z}$, then these relations allow us to "cancel" most of the generators, so that $K$ is generated by the $b_{1,n}$.  But it is also easy to see that $b_{1,n}$ lies in the subgroup generated by $b_{1,n+1}$.  Since $G$ is polycyclic (Mark showed this in his answer, when he showed $G$ is a quotient of the Klein bottle group), $K$ is finitely generated, and this gives us enough to conclude $K$ is cyclic.  Now in writing $G=K\rtimes\mathbb{Z}$, the $\mathbb{Z}$ factor is generated by $y_0$.  Quotienting out by $y_0^2$ gives the dihedral group $D_{2p}$, and thus $K$ must have order $p$, so that $G=C_p\rtimes\mathbb{Z}$.
