Krohn-Rhodes complexity 1 decidable implies that complexity $n \geqslant 0$ is decidable

When decomposing a finite semigroup into a wreath product of groups and aperiodic semigroups, complexity measures the minimal number of groups that are needed. Determining an algorithm to compute complexity has been an open problem for almost 60 years. The main result of this paper proves that if it is decidable for a finite semigroup or finite automaton to have Krohn-Rhodes complexity 1, then it is decidable if a finite semigroup or finite automaton has complexity $n$ for all $n \geqslant 0$.