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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 $2p$.

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If I am not mistaken, all of these groups are quotients of the Klein bottle group. In particular, they are polycyclic, so you can find out much about them. Plugging a bunch of these into GAP, it would seem only one more relation is necessary. –  Steve D May 11 '11 at 23:01
    
In fact, after doing a couple more examples I would guess that your group is always $C_p\rtimes\mathbb{Z}$, with $\mathbb{Z}$ acting by inversion. –  Steve D May 11 '11 at 23:49
    
I might be missing something but I don't get this. If p=3, I get the braid group -- in agreement in this case. –  Menton May 12 '11 at 0:22
    
No, these groups are always solvable, so certainly there is no braid group. –  Steve D May 12 '11 at 22:30
    
I will have more time tonight, but I am fairly confident what I said in the above comments is accurate. –  Steve D May 12 '11 at 22:34
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2 Answers 2

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.

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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 $G=C_p\rtimes\mathbb{Z}$.

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