  In Question 19.35 of the Kourovka Notebook, M. H. Hooshmand asks whether,
given a finite group $G$ and a factorization $\mathrm{card}(G)= n_1\ldots n_k$,
one can always find subsets $A_1,\ldots,A_k$ of $G$ with
$\mathrm{card}(A_i)=n_i$ such that $G=A_1\ldots A_k;$ equivalently, such that
the group multiplication map $A_1\times\ldots\times A_k\to G$ is a bijection.
  We show that for $G$ the alternating group on 4 elements, $k=3$, and
$(n_1,n_2,n_3) = (2,3,2)$, the answer is negative. We then generalize some of
the tools used in our proof, and note an open question.
