Let $F\_n$ be the free group on $n$ generators. Of course, every finitely-presentable group $G$ is a finite colimit of copies of $F\_n$, where $n$ is allowed to vary. But is $G$ a finite colimit of copies of $F\_2$?


Of course, because $F\_{2n}$ is a finite coproduct of copies of $F\_2$, we have that any finitely-presentable group $G$ is a finite colimit of finite colimits of copies of $F\_2$ -- a "2-fold" finite colimit of copies of $F\_2$. But I'm curious about the 1-fold case. 


To make the question a bit more concrete, let's unwind what it means to be a finite colimit of copies of $F\_2$:



> 
> Let $G$ be a group. Then $G$ is a finite colimit of copies of copies of $F\_2$ if and only if $G$ admits a presentation of the following description:
> 
> 
> * There are $2n$ generators coming in pairs $x\_1,y\_1, \dots x\_n, y\_n$;
> * There is a finite set of generating relations, each of the form $w(x\_i,y\_i)=v(x\_j,y\_j)$, where $w,v$ are group words and $1 \leq i \leq j \leq n$.
> 
> 
> 


So for example, $x\_1y\_1^2x\_1^{-1} = y\_2^{-1}x\_2$ is a permissible generating relation (with $i=1,j=2$) but $x\_1 x\_2 = x\_3$ is not a permissible generating relation because only 2 different subscripts are allowed to appear in a permissible generating relation. So my question is:


**Question:** Let $G$ be a finitely-presented group.


* Is $G$ a finite colimit of copies of $F\_2$?
* Equivalently, does $G$ admit a presentation of the above form?


**Edit:**


The form of the presentation can be constrained even further, to look like this:


* There is a finite set of generating relations, coming in pairs each of the form $x\_i = w(x\_j,y\_j)$, $y\_i=v(x\_j,y\_j)$, where $w,v$ are group words and $1 \leq i, j \leq n$.


Other variations are possible too; I'm not sure what the most convenient description to work with might be.



Is every finitely-presentable group a finite colimit of copies of $F_2$?
I believe the answer is yes. Assume, by way of contradiction, that some finitely presented group cannot be so expressed. Then we can choose such a group $G$ where for any generating set of the form $x\_1,y\_1, x\_2,y\_2,\ldots, x\_n,y\_n$ the number of non-permissible relations needed to define $G$ (together with some finite number of permissible relations) is minimized; say those non-permissible relations are $w\_1=1, w\_2=1,\ldots, w\_m=1$. Write $w\_1=z\_1z\_2\cdots z\_p$ where each $z\_j\in \{x\_1^{\pm 1},y\_1^{\pm 1},\ldots, x\_n^{\pm 1},y\_n^{\pm 1}\}$, where we may also assume that $p$ has been minimized. Note that since $w\_1=1$ is not permissible, we must have $p\geq 3$.


Add new generators $x\_{n+1},y\_{n+1},x\_{n+2},y\_{n+2}$. The relations $x\_{n+1}=z\_{p}$, $y\_{n+1}=z\_{p-1}$, $x\_{n+2}y\_{n+2}=1$ are permissible. The relation $x\_{n+2}y\_{n+1}x\_{n+1}=1$ is also permissible (since it is equivalent to $x\_{n+2}=x\_{n+1}^{-1}y\_{n+1}^{-1}$). Asserting these relations still gives us the same group (since our new relations merely tell us how to write the new generators in terms of the old ones). The relation $w\_1=1$ is equivalent to $z\_1z\_2\cdots z\_{p-2}y\_{n+2}=1$, but it is now shorter, a contradiction.


