# How many “steps” does one need to obtain a group from a semigroup

Given a countable group $G$ and a generating subsemigroup $S\subset G$, let us consider the increasing sequence of alternating products of $S$ and $S^{-1}$ beginning with $S$ $$A_1=S, A_2=S S^{-1}, A_3=S S^{-1}S, A_4=S S^{-1}S S^{-1}, A_5=S S^{-1}S S^{-1}S, \dots \;,$$ and put $$\kappa(S) = \min \{k: G=A_k \} \;.$$ Obviously, $\kappa(S)=1$ if and only if $S=G$, and $\kappa(S)=\infty$ if $S$ is the semigroup of positive words in a free group $G$. One can show that if $G$ is nilpotent, then $\kappa(S)= 2$ for any generating subsemigroup $S\neq G$, and, more generally, that $\kappa(S)=2$ for any generating subsemigroup iff $G$ does not contain a free subsemigroup. What else is known?

• Sounds a bit broad... I've never seen this problem specifically addressed, but plenty of examples can be given. – YCor May 10 '18 at 0:37
• @YCor Like what? – R W May 10 '18 at 0:54
• Take any group (there are quite many); take a generating subsemigroup (there can be quite many)... no idea where to start with. This is what I mean by "too broad". – YCor May 10 '18 at 7:30
• Not very useful, indeed :) – R W May 10 '18 at 9:04

Marek Kuczma asked in 1980 whether for every positive integer $n$ there exists a subsemigroup $M$ of a group $G$ such that $G$ is equal to the $n$-fold product $MM^{-1}MM^{-1}\cdots M^{(-1)^{n-1}}$, but not to any proper initial subproduct of the product.

George Bergman proved that the answer is affirmative for all $n$. The result (and sundry generalizations) appear in Submonoids of groups, and group-representability of restricted relation algebras, which will appear in Algebra Universalis this year. It is also available on his website.