# Dehn's algorithm for word problem for surface groups

For some $g \geq 2$, let $\Gamma_g$ be the fundamental group of a closed genus $g$ surface and let $S_g=\{a_1,b_1,\ldots,a_g,b_g\}$ be the usual generating set for $\Gamma_g$ satisfying the surface relation $[a_1,b_1]\cdots[a_g,b_g]=1$. Dehn proved the following famous theorem.

Theorem : Let $w$ be a reduced word in the generators $S_g$ such that $w$ represents the identity in $\Gamma_g$. Then $w$ contains a subword $r_1$ of length $k \geq 2g+1$ such that there exists a word $r_2$ of length $4g-k$ with $r_1 r_2$ a cyclic permutation of $[a_1,b_1]\cdots[a_g,b_g]$.

This leads to Dehn's algorithm for solving the word problem in a surface group. Namely, start with a word $w$ which is reduced. If $w$ does not contain a subword $r_1$ as in the Theorem, then $w$ does not represent the identity in $\Gamma_g$. Otherwise, we can replace the subword $r_1$ of $w$ with $r_2$, shortening $w$. After possibly reducing $w$, we do the above again. We stop if we have reduced $w$ to the empty word.

There are many sources for Dehn's algorithm; for instance, Stillwell's book on geometric topology and combinatorial group theory gives a proof using elementary topology, and Lyndon-Schupp give a proof based on small cancellation theory. However, many sources say that Dehn's original proof used hyperbolic geometry. I've heard people tell me that he studied the usual tessalation of the hyperbolic plane by regular $4g$-gons and applied something like Gauss-Bonnet; however, I've never managed to reconstruct this kind of proof. Can someone either spell it out or give a (modern) reference?

• Its been a long time but I recall Dehn's proof being essentially identical to the small cancellation theory proof. Dehn is using the Cayley graph in the hyperbolic plane as he had no need for a general setup. – Ryan Budney Jul 17 '13 at 5:28
• One little remark: you remplace $r_1$ by $r_2^{-1}$ to keep the stuff correct ... Of course you understand it but to keep things clean :) – Olga Jan 19 '17 at 19:56
• Another remark: shouldn't $r_j$ be the words in the alphabet of $a_j, b_j$ but also their inverses ?... – Olga Jan 19 '17 at 20:03