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Are there non-trivial topologies (neither discrete nor indiscrete) on the additive group of integers $\mathbb{Z}$, making it into a topological group. Could someone list them all, possibly with some details?

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May I ask why you posted this? thanks. –  awllower Jul 20 '11 at 14:32
    
Actually, I was trying to prove that with Fürstenburg's topology, $\mathbb{Z}$ is a topological group. Then this question came to my mind. –  Johnathon Jul 20 '11 at 15:09
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1 Answer 1

up vote 11 down vote accepted

For any set $S$ of primes, the embedding

$$\mathbb{Z} \to \prod_{p \in S} \mathbb{Z}_p$$

of $\mathbb{Z}$ into the corresponding product of rings of $p$-adic integers gives a nontrivial topology on $\mathbb{Z}$ generated by arithmetic progressions of length $d$ where all prime divisors of $D$ lie in $S$. All of these topologies are distinct. When $S$ is the entire set of primes, we get the topology induced from the profinite completion, notably used in Fürstenburg's proof of the infinitude of the primes.

More generally one can take the initial topology with respect to any collection of quotients $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ where $n$ is chosen from a set $S$ of positive integers closed under taking divisors and $\text{lcm}$s. (It is generated by arithmetic progressions of length chosen from $S$.)

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Thanks Yuan. Are these all the topologies making $\mathbb{Z}$ into a topologial group. –  Johnathon Jul 20 '11 at 15:08
    
Thanks Yuan. Your answer is very helpful. I think that there might be examples of finite topological groups with non-trivial topology. –  Johnathon Jul 20 '11 at 16:16
    
Continuing Quaochu Y's comment: we could easily rationalize demanding that a sane topology on $\mathbb Z$ should make the quotient maps to (discrete?!) $\mathbb Z/n$ continuous, which implies that the map factors through $\lim_n \mathbb Z/n=\hat{\mathbb Z}$, which discretizes $\mathbb Z$. Dropping various sets of primes dividing the $n$'s produces other effects. –  paul garrett Jul 20 '11 at 16:18
    
+1: Nice characterisation of Fürstenburg's topology! But how do we know that these topologies are compatible with the group operations? –  Zhen Lin Jul 20 '11 at 16:28
    
@Zhen: they're initial with respect to topologies on the finite quotients which are compatible with group operations (either discrete or indiscrete). –  Qiaochu Yuan Jul 20 '11 at 17:27
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