What is this group called?

Let $X$ denote a set. There's a corresponding group obtained by taking the group freely generated by $X^2$ and then quotienting out by the following families of relations:

1. $(x,x) = 1$

2. $(x,y)(y,z) = (x,z)$

3. $(x,y)(y,x) = 1$

4. For each quadruple $(x,y,x',y')$ such that $\{x,y\} \cap \{x',y'\} = \emptyset$, we have: $$(x,y)(x',y') = (x',y')(x,y)$$

Question. What is this group called?

My motivation for considering this group is that it acts on the set $(\{0,1\}^\mathbb{N})^X,$ of $X$-many binary streams, by interpreting $(x,y)$ as the act of taking the first digit of stream $x$, removing it from $x$, and appending it to the beginning of $y$. Each of the four families above can be explained in these terms; for example $(x,x)=1$ is saying that if I take the first digit of stream $x$, and put it back on stream $x$, nothing changes.

Solution 1:

Not a complete answer, but too long for a comment:

The group you described has a neat equivalent characterization:

It can be described by the presentation $\langle a_1, a_2, ... , a_{|X| - 1}| [a_i, a_j] = [a_k, a_j] \forall i \neq j, j \neq k, k \neq i\rangle$

Proof:

Suppose, $X = \{0, ... , n\}$ ($|X| = n + 1$). Let's define $a_i := (0, i)$. That results in $(i, 0)$ being equal to $a_i^{-1}$ and for $i \neq j, j \neq 0, i \neq 0$ $(i, j) = a_i^{-1}a_j$. Now this change of notation already covers the first three rules, making them redundant.

Now, let's deal with the fourth rule. Without the loss of generality we can assume that one of the following cases takes place:

1)$x' = 0$, $x, y$ and $y' = z$ are pairwise distinct and non-zero. Then $(0, z)(x, y) = (x, y)(0, y')$, which turns into $a_{z}a_{x}^{-1}a_y = a_{x}^{-1}a_{y}a_{z}$ which results in $a_ya_za_y^{-1} = a_xa_za_{x}^{-1}$, or equivalently $[a_y, a_z] = [a_x, a_z]$

2)$x, y, x' = z$ and $y' = t$ are pairwise distinct and non-zero. Then $(x, y)(z, t)=(z, t)(x, y)$ turns into $a_x^{-1}a_ya_z^{-1}a_t = a_z^{-1}a_ta_{x}^{-1}a_y$ which is $(a_ya_z^{-1}a_y^{-1})(a_ya_ta_y^{-1})=(a_xa_z^{-1}a_x^{-1})(a_xa_ta_x^{-1})$ which is a direct corollary of case 1

Thus $[a_y, a_z] = [a_x, a_z]$ for all distinct $x$, $y$ and $z$ are the only non-trivial relations required.