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

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Of course in any of these cases, once you have your topology on Z, then it is useful to contemplate the completion. – Gerald Edgar Mar 17 '10 at 0:04
How about p-adic topology? – user47958 May 11 '14 at 11:36
up vote 12 down vote accepted

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.

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Sorry, I had those examples and related ones in mind and wanted to prove they were the only ones. I'll think of a well-posed question and post it again. – Cristos A. Ruiz Mar 16 '10 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]

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This is a nice topology. Another description says that an integer is close to zero iff it is divisible by a large factorial. – Gerald Edgar Mar 17 '10 at 0:03
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. – Kevin Buzzard Mar 17 '10 at 7:10

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