Is there a non-trivial topological group structure of $\mathbb{Z}$?

Question

More specificaly, is there a haussdorf non-discrete topology on $\mathbb{Z}$ that makes it a topological group with the usual addition operation?

Answer 1

Yes. Take, for example, the subgroups $p^k\mathbb{Z}$, for $k>0$ and a fixed prime $p$, as a basis of neighborhoods of the identity.

Answer 2

There is a topology on $\mathbb Z$ which has the set of all arithmetic sequences as a basis. It shows up in the topological proof of the infinitude of primes, cf. [H. Fürstenberg, On the Infinitude of Primes, Amer. Math. Monthly 62 (1955), 353]

Answer 3

There is a huge number of such topologies. Let $G$ be any discrete abelian group. By $\widehat G$ we denote the family (in fact, a group) of its characters, that is of homomorphisms from $G$ to the unit circle group $\Bbb T$. By Pontrjagin duality theory, the family $\widehat G$ separate points of $G$ (that is for each two distinct elements $g,h\in G$ there exists a character $\chi\in\widehat G$ such that $\chi(a)\ne\chi(b)$). Thus a diagonal product $\Delta\{\chi:\chi\in\widehat G\}:G\to\prod \{\Bbb T_\chi: \chi\in\widehat G\}$ is an injective homomorphism of $G$ into a compact group. In endows $G$ with a (totally bounded) group topology (called Bohr topology on the group $G$). In particular, if $G$ is infinite then its Bohr topology is indiscrete. On the other hand, according to this answer, $G$ admits continuum many (pairwise transversal) group topologies, none of which is totally bounded. The case of $G=\Bbb Z$ is espectially simple because it admits an injective homomorphism into each topological group with an element of infinite order. As a more concrete example, in this MSE answer I proposed for each $\kappa\le\frak c$ an injective homomorphism of the group $\Bbb Z$ onto a dense subroup of a compact group $\Bbb T^k$.