We give a necessary and sufficient condition, in terms of a certain reflection principle, for every unconditionally closed subset of a group G to be algebraic. As a corollary, we prove that this is always the case when G is a direct product of an Abelian group with a direct product (sometimes also called a direct sum) of a family of countable groups. This is the widest class of groups known to date where the answer to the 63 years old problem of Markov turns out to be positive. We also prove that whether every unconditionally closed subset of G is algebraic or not is completely determined by countable subgroups of G.According to Markov [6], a subset S of a group G is called:(a) elementary algebraic if there exist an integer n > 0, a 1 , . . . , a n ∈ G and ε 1 , . . . , ε n ∈ {−1, 1} such that S = {x ∈ G : x ε 1 a 1 x ε 2 a 2 . . . a n−1 x εn = a n }, (b) algebraic if S is an intersection of finite unions of elementary algebraic subsets of G, (c) unconditionally closed if S is closed in every Hausdorff group topology of G.Since the family of all finite unions of elementary algebraic subsets of G is closed under finite unions and contains all finite sets, it is a base of closed sets of some T 1 topology Z G on G, called the Zariski topology of G. (This topology is also known under the name verbal topology, see [1].) The family of all unconditionally closed subsets of G coincides with the family of closed subsets of a T 1 topology M G on G, namely the infimum (taken in the lattice of all topologies on G) of all Hausdorff group topologies on G.