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.
Best Answer
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.