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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.