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

$\begingroup$ Of course in any of these cases, once you have your topology on Z, then it is useful to contemplate the completion. $\endgroup$– Gerald EdgarMar 17, 2010 at 0:04

1$\begingroup$ How about padic topology? $\endgroup$– Minimus HeximusMay 11, 2014 at 11:36
3 Answers
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.

$\begingroup$ Sorry, I had those examples and related ones in mind and wanted to prove they were the only ones. I'll think of a wellposed question and post it again. $\endgroup$ Mar 16, 2010 at 22:48
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]

$\begingroup$ This is a nice topology. Another description says that an integer is close to zero iff it is divisible by a large factorial. $\endgroup$ Mar 17, 2010 at 0:03

4$\begingroup$ This is a very natural topology, if I've got it right: I think it's the topology on $\mathbf{Z}$ induced from its inclusion into its completion (give the completion the profinite topology). Thought about this way, you see that an element is close to 0 iff it's in a lot of finite index subgroups. $\endgroup$ Mar 17, 2010 at 7:10
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 $\Bbb T^k$.