In several works of Martin Bridson there is a definition of an asynchronously combable group. It looks like this is a rather convenient notion used for estimating Dehn functions, for proving that groups are automatic and for other things. It is only natural to ask how should a group look like if it is so 'bad' that it is not even asynchronously combable? Here are the relevant definitions.
Let $\Gamma$ be a group with finite generating set $\mathcal A$ and $d(x,y)$ be the associated word metric. A combing (normal form) for $\Gamma$ is a set of words $\{\sigma_\gamma\mid\gamma\in\Gamma\}$ in the letters $\mathcal A^\pm$ such that $\sigma_\gamma=\gamma$ in $\Gamma$. Let $$ R=\{\rho\colon\mathbb N\to\mathbb N\mid\rho(0)=0;\,\rho(n+1)\in\{\rho(n),\rho(n)+1\}\,\forall n;\,\rho\text{ unbounded }\}. $$ Given eventually constant paths in the Cayley graph $p_1,p_2\colon\mathbb N\to(\Gamma,d)$ we define $$ D(p_1,p_2)=\min_{\rho,\rho'\in R}\big\{\max_{t\in\mathbb N}\{d(p_1(\rho(t)),p_2(\rho'(t))\}\big\}. $$ Given a combing $\sigma$ for $\Gamma$, the asynchronous width of $\sigma$ is the function $\Phi_\sigma\colon \mathbb N\to\mathbb N$ defined by $$ \Phi_\sigma(n)=\max\{D(\sigma_g,\sigma_h)\mid d(1,g),d(1,h)\leq n;\,d(g,h)=1\}. $$ Here we treat words $\sigma_g$ as paths in the Cayley graph, by continuously extending the graph distance function $d$, which gives all edges length $1$, and assuming that the path $\sigma_g(t)$ is constantly equal to $g$ for $t>\operatorname{length}_{\mathcal A^\pm}(\sigma_g)$.
A finitely generated group $\Gamma$ is said to be asynchronously combable if there exists a combing $\sigma$ for $\Gamma$ and a constant $K>0$ such that $\Phi_\sigma(n)\leq K$ for all $n\in\mathbb N$.
The question now is: what is a simple example of a finitely presented group for which the asynchronous width function $\Phi_\sigma(n)$ is unbounded?