Can an abelian group have a minimal group topology?

Question

In the abstract of this paper, it is said that a minimal group topology on an abelian group is not Hausdorff.

Suppose $G$ is an abelian group and $\mathcal T$ is a minimal group topology on $G$ and let $N$ be the intersection of all neighborhoods of the identity element and let $\pi: G\to \frac{G}{N}$ be defined by $\pi(x)=Nx$. Then the set of all $\pi[F]$ where $F$ is a neighborhood of the identity element of $G$ is base of neighborhoods for a Hausdorff group topology on $\frac{G}{N}$ which has to be minimal. But $\frac{G}{N}$ is abelian and by the claim above cannot have a minimal Hausdorff group topology. Therefore an abelian group cannot have a minimal group topology.

Where am I wrong in this reasoning?

Answer 1

The paper that you mentioned is referring to a theorem proved by D. Remus in his dissertation:

Theorem: If $G$ is abelian and $\tau$ is a minimal group topology on $G$, then there exists a subgroup $N$ such that $G/N \cong \mathbb{Z}/p\mathbb{Z}$ ($p$ prime) and $\{N\}$ is a fundamental system of neighborhoods of $e$ in $\tau$.

Note that the only group topologies in $\mathbb{Z}/p\mathbb{Z}$ are the discrete and the indiscrete topologies, so there is in fact a Hausdorff minimal group topology on $\mathbb{Z}/p\mathbb{Z}$. But according to the theorem, these are the only examples of such a thing.

I don't have access to Remus' dissertation but I believe his article On the structure of the lattice of group topologies, Results Math. 6 (1983), presents the same results without the proofs. In this article he writes:

...the following results are proved for abelian groups... The atoms of $Tg(G)$, are the topologies induced by normal subgroups $N$ for which the factor group $G/N$ is a simple finite group.