A subset $X$ of an Abelian group $G$ with zero element $0$ is said to be independent if, given any $n\in \mathbb N$, any pairwise distinct $x_1,\dots, x_n\in A$, and any $k_1,\dots, k_n\in \mathbb Z$, we have $k_1\cdot x_1=\dots= k_n\cdot x_n= 0$ whenever $k_1\cdot x_1 +\dots +k_n\cdot x_n=0$. In other words, $X\subset G$ is independent if the natural homomorphism $\bigoplus_{x\in X}\langle x\rangle \to G$ is injective (here $\langle x\rangle$ denotes the subgroup of $G$ generated by $x$). We say that $X$ is a basis of $G$ if $X$ is independent and $\langle X\rangle =G$.

We consider independent subsets of Hausdorff Abelian topological groups of prime exponent $p$. It is well known that any such group $G$ is a direct sum of copies of the cyclic group $\mathbb Z/p\mathbb Z$ of order $p$ and hence can be treated as a vector space over the field $\mathbb F_p$. Therefore, $G$ has a basis $E$. Thus, on any Abelian topological group $G$ of prime exponent $p$ with basis $E$, there exists the natural topology induced by the Tychonoff product topology of $\prod_{e\in E}\langle e\rangle$. We refer to this topology as the product topology on $G$ associated with $E$.

A subset $X$ of $G$ is said to be topologically independent if, given any $n\in \mathbb N$, any pairwise distinct $x_1,\dots, x_n\in X$, any $k_1,\dots, k_n\in \mathbb Z$, and any neighborhood $U$ of $0$, there exists a neighborhood $V$ of $0$ such that $k_1\cdot x_1, \dots, k_n\cdot x_n \in U$ whenever $k_1\cdot x_1 +\dots +k_n\cdot x_n\in V$. Clearly, any topologically independent set is independent, but the converse is not true: it is known that if $X\subset G$ is topologically independent, then the topology of $H=\langle X\rangle $ is coarser than the product topology on $H$ associated with the basis $X$ of $H$.

Recall that the intersection of the kernels of continuous characters of a topological group is called the von Neumann kernel of $G$ and denoted by $n(G)$; a group $G$ with $n(G)= G$ is said to be minimally almost periodic and a group $G$ with trivial $n(G)$ is said to be maximally almost periodic. It is easy to see that an Abelian topological group $G$ of prime exponent is maximally almost periodic if and only if there exists a basis $E$ of $G$ such that the product topology on $G$ associated with $E$ is coarser than the original topology of $G$, i.e., $E$ is topologically independent. There exist examples of minimally almost periodic Abelian groups of any prime exponent. However, any infinite topological Abelian group of prime exponent contains an infinite maximally periodic subgroup (in other words, any such group contains an infinite topologically independent set). This is one of the main results of the report.

The second main result is that any countable topological Abelian group of prime exponent has a closed discrete basis. Moreover, any countable-dimensional topological vector space over a finite field or over a complete second-countable valued field (such as $\mathbb R$ or $\mathbb C$) has a closed discrete basis. For uncountable-dimensional spaces, this is not true.