# $\mathcal{B}_3$ modulo the normal closure of $\mathbb{Z} \times \mathbb{Z}$

Consider the braid group $\mathcal{B}_3$ on three strands. It is known that $\mathcal{B}_3 = \langle x,y | xyx = yxy \rangle$ and that the center $Z\mathcal{B}_3$ is infinite cyclic generated by $(xy)^3$.

(1) What is the group $\mathcal{B}_3$ modulo the normal closure of $\langle x,(xy)^3 \rangle$ ?

(2) What is the index of $\langle x,(xy)^3 \rangle$ in $\mathcal{B}_3$ ?

It is not hard to show that $\langle x,(xy)^3 \rangle$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$.

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Are any powers of $y$ in $H=\langle x, (xy)^3\rangle$? That would imply that the index is infinite. I don't remember how that nifty representation of $\mathcal{B}_3$ goes again, so can't prove it now. – Jyrki Lahtonen Jul 1 '11 at 21:53

$\langle x, y | xyx= yxy \rangle / \langle \langle x, (xy)^3 \rangle \rangle = \langle x,y | xyx=yxy, x=1, (xy)^3=1 \rangle=\langle y | y=y^2, y^3=1 \rangle = 1$.
The structure of the braid group $\mathcal{B}_3$ is well understood. It is easily seen to be isomorphic to the group $\langle a,b \mid a^2 = b^3 \rangle$. If you factor out the central subgroup generated by $a^2$, you get the free product of cyclic groups of orders 2 and 3, of which the commutator subgroup has index 6 and is free of rank 2. So $\mathcal{B}_3$ has a subgroup of index 6 which is a direct product of a free group of rank 2 and an infinite cyclic group. Hence it is clearly not virtually abelian. So the abelian subgroup $\langle x,(xy)^3 \rangle$ must have infinite index.