Profinite completions of free-by-free groups contain everything

Martin R. Bridson

Published 2023-12-11Version 1

Given an arbitrary, finitely presented, residually finite group $\Gamma$, one can construct a finitely generated, residually finite, free-by-free group $M_\Gamma = F_\infty\rtimes F_4$ and an embedding $M_\Gamma \hookrightarrow (F_4\ast \Gamma)\times F_4$ that induces an isomorphism of profinite completions. In particular, there is a free-by-free group whose profinite completion contains $\widehat{\Gamma}$ as a retract.