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.


Here's a pretty direct way to do this. Choose any presentation by generators $x_1,\ldots,x_n$ and relations $r_1,\ldots,r_m$; say $r_i = z_{i,1} \cdots z_{i,k}$ for $z_{i,1},\ldots,z_{i,k} \in \{x_1^{\pm 1}, \ldots, x_n^{\pm 1}\}$. Firstly, we may assume all $r_i$ have length $k = 3$: the new variables $$x_{i,j} = z_{i,1} \cdots z_{i,j}$$ for $0 \leq j \leq k$ are subject only to the relations \begin{align*} x_{i,0} = e = x_{i,k}, & & & & & & x_{i,j} = x_{i,j-1}z_{i,j} & & (1 \leq j \leq k). \end{align*} If $k < 3$, we can eliminate the variable $z_{i,1}$ at the expense of replacing all $z_{i,1}^{\pm 1}$ by $z_{i,2}^{\mp 1}$ (if $k = 2$) or $e$ (if $k = 1$) in the other relations, so we may assume all $r_i$ have length exactly $3$. Then introduce new variables $x_{n+1},\ldots,x_{n+m}$ as well as variables $y_1,\ldots,y_{n+m}$, subject to the relations \begin{align*} x_{n+i} &= z_{i,1},& & 1 \leq i \leq m,\\ y_i &= e & & 1 \leq i \leq n,\\ y_{n+i} &= z_{i,2}, & & 1 \leq i \leq m,\\ x_{n+i}y_{n+i} &= z_{i,3}^{-1}, & & 1 \leq i \leq m,\\ \end{align*} This gives a presentation of the desired form. $\square$