Let $P$ be a probability on a finite group $G$, ${ P^{ (n)}}$ $n$-fold convolution of $P$ on $G$. Under mild condition, ${ P^{ (n)}}$ at $n \to \infty $ converges to the uniform probability on the group $G$. If $A = \left\{ { g \in G,\;P\left( g \right) \ne 0} \right\}$ be the carrier of the probability $P$, then ${ A^n} = \left\{ { { a_1} \cdot ... \cdot { a_n},\;\;{ a_1},...,{ a_n} \in A} \right\}$ be the carrier of probability ${ P^{ (n)}}$. One of necessary and sufficient conditions for the mentioned convergence is: sequence ${ A^n}$ at $n \to \infty $ stabilizes on $G$, i.e. ${ A^k}$ = ${ A^{ k + 1}} = ... = G$ for a natural number $k$. In other words, product of some multipliers equal to $A$ is $G$. The carrier $A$ is in general case any nonempty subset of group $G$. In the paper we find a condition under which product of some subsets of $G$ is $G$.
Categories: math.GR math.PR