Solution to the Henckell--Rhodes problem: finite $F$-inverse covers do exist

Abstract

For a finite connected graph $\mathcal{E}$ with set of edges $E$, a finite $E$-generated group $G$ is constructed such that the set of relations $p=1$ satisfied by $G$ (with $p$ a word over $E\cup E^{-1}$) is closed under deletion of generators (i.e.~edges). As a consequence, every element $g\in G$ admits a unique minimal set $\mathrm{C}(g)$ of edges (the \emph{content} of $g$) needed to represent $g$ as a word over $\mathrm{C}(g)\cup\mathrm{C}(g)^{-1}$. The crucial property of the group $G$ is that connectivity in the graph $\mathcal{E}$ is encoded in $G$ in the following sense: if a word $p$ forms a path $u\longrightarrow v$ in $\mathcal{E}$ then there exists a $G$-equivalent word $q$ which also forms a path $u\longrightarrow v$ and uses only edges from their content; in particular, the content of the corresponding group element $[p]_G=[q]_G$ spans a connected subgraph of $\mathcal{E}$ containing the vertices $u$ and $v$. As an application it is shown that every finite inverse monoid admits a finite $F$-inverse cover. This solves a long-standing problem of Henckell and Rhodes.Comment: 46 pages, 14 figures, new result Cor. 2.7 included, several inaccuracies removed, more details include