Congruence lattices forcing nilpotency

Description

We start from the following results, which are consequences of theorems by Freese, Hobby, and McKenzie. \begin{quote} Let $\mathbf{A}$ be a finite algebra in a congruence modular variety such that $\mathrm{Con} (\mathbf{A})$ has a $(0,1)$-sublattice $\mathbb{L}$ that is simple, complemented, and has at least three elements. Then $\mathbf{A}$ is abelian. \end{quote} Similarly if $\mathbb{L}$ has no $2$-element homomorphic image, then $\mathbb{A}$ is solvable. We derive a similar condition for nilpotency and investigate what could be converses to these results.

Period	27 May 2016
Event title	AAA 92
Event type	Conference
Location	Czech RepublicShow on map