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A finitely generated group is synchronously combable if it admits a combing of a group with the synchronous fellow-traveller property. Without a regular-language assumption, this is more general than an automatic group. A bounded-rung strip argument gives a finite group presentation, and finite-state row shortening gives a recursive exponential filling bound and a soluble word problem for a group.

Work with a finite symmetric alphabet $\Sigma =A^{\pm 1}$. If $w=a_1\cdots a_m$, its path $\hat{w}$ starts at $1$, visits $a_1\cdots a_j$ at integer time $j$, travels at unit speed along the intervening generator edges, and stays at its endpoint for $t\ge m$.

Two such paths synchronously $K$-fellow travel ifAn asynchronous version permits increasing time reparametrizations. Checking all integer times suffices after a bounded enlargement of $K$ for the edge interiors.

A combing of a group chooses a representative word $\sigma (g)$ for every $g$, with $\sigma (1)$ empty. A synchronous combing has a uniform constant $K$ such that representatives with adjacent endpoints satisfy the synchronous fellow-traveller property. A group admitting this is a synchronous combable group. The formal language need not be regular or computable. A regular language of representatives satisfying the synchronous condition instead gives an automatic group. This distinction is relevant to the finite-state automata request in part (iii); the present argument uses the synchronous meaning throughout. The definitions and distinction are also made explicitly in Combings of Groups and the Grammar of Reparameterization, Section 1.

Two paths satisfy the synchronous fellow-traveller property with constant $K$ if $d(p(t),q(t))\le K$ for all equal times, with completed paths held at their endpoints. Checking integer times suffices up to increasing $K$ by at most two for unit-speed edge paths. A synchronous combing of a group requires a uniform constant for all adjacent endpoints.