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? 



How many "steps" does one need to obtain a group from a semigroup
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](https://math.berkeley.edu/~gbergman/papers/u_d_u_d.pdf). 


