Compact totally disconnected group topology on $\mathbb{Z}$

Question

Let $A$ be an infinite cyclic group. Does there exist a compact totally disconnected topology on the underlying set of $A$ such that multiplication and inversion are continuous?

I was thinking of why $A$ can not be a profinite (i.e. compact Hausdorff totally disconnected) group. The answer is negative, see here. One can ask weaker questions then. $A$ can be locally compact Hausdorff totally disconnected (take the discrete topology), it can be compact (take the topology that has only two opens sets). However I am not able to answer the above question.

Answer 1

No. A totally disconnected space is $T_1$ (thx to Eric Wofsey's comment) (components, and thus points, are closed) and in a topological group this implies $T_2$ (and Tychonoff) A (locally) compact countable Hausdorff space has an isolated point (by Baire's theorem) and as a topological group topology is homogeneous, all points are then isolated, so the group is discrete, and can only be finite.

Answer 2

No.

Every countable locally compact group is discrete.

Indeed, every locally compact space is a Baire space. If $G$ is a countable locally compact group, it follows from Baire (applied to singletons) that at least one singleton has non-empty interior. So there is an isolated point. Since left-translations are self-homeomorphisms, one deduces that every point is isolated, that is, $G$ is discrete.

Corollary: every countable compact group is finite.

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PS: the above assumes Hausdorff. At the opposite, every non-Hausdorff group topology on $\mathbf{Z}$ is compact (since the closure of $\{1\}$ then has a closure of finite index). Namely, it's obtained by pulling back the discrete topology from a finite quotient.