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?

1$\begingroup$ Sounds a bit broad... I've never seen this problem specifically addressed, but plenty of examples can be given. $\endgroup$ – YCor May 10 '18 at 0:37

1$\begingroup$ @YCor Like what? $\endgroup$ – R W May 10 '18 at 0:54

$\begingroup$ 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". $\endgroup$ – YCor May 10 '18 at 7:30

$\begingroup$ Not very useful, indeed :) $\endgroup$ – 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 $MM^{1}MM^{1}\cdots M^{(1)^{n1}}$, 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 grouprepresentability of restricted relation algebras, which will appear in Algebra Universalis this year. It is also available on his website.