I think you may follow: http://www.prometej.ba/Uploads/Datoteke/topbook.pdf is very interesting or playing in topological fundamentals.

A nice question. Remark that it is open since 2009. See Question 6.3 of

Lydia Außenhofer, M.J. Chasco, Dikran Dikranjan, Xabier Domínguez,

Locally minimal topological groups, arXiv:0912.2946v2

published in final form as

Lydia Außenhofer, M.J. Chasco, Dikran Dikranjan, Xabier Domínguez,

Locally minimal topological groups 1,

Journal of Mathematical Analysis and Applications,

Volume 370, Issue 2, 2010, Pages 431-452,

https://doi.org/10.1016/j.jmaa.2010.04.026.

(http://www.sciencedirect.com/science/article/pii/S0022247X10003197)

(this is a correction of my previous comment in which ResearchGate robot substituted the correct reference by a similar one that exists on ResearchGate)

I think every locally minimal topology on the group is discrete. By way of contradiction, we assume every neighbourhood of 0 is infinite. Suppose there is a closed U witnesses that every closed subgroup contained in U is minimal and K is a maximal subgroup contained in U, then K is closed and infinite. Indeed, if K were finite, then there would be a neighbour V of 0 s.t. K+V+V is contained in U and V∩K=0, then K+ is in U for any nonzero x in V. This contradicts to the maximality.

So K is a countable infinite minimal group of exponent 2, this is impossible.

Dear Dekui Peng ,

One moment in your reasoning remains unclear for me. Namely, the sentence "if K were finite, then there would be a neighbourhood V of 0 s.t. K+V+V is contained in U". There would be no problem in the case of open U, but I don't see why this should be true for your closed U. Say, a closed $U$ may be of the form of a "big" neighborhood W and just one point y that lies "far" outside of W. Then for the set K = {0, y} there may be no neighborhood V of 0 such that K+V+V is contained in U.

Best, Vladimir

Dear @Vladimir Kadets,

Thanks a lot for pointing this mistake.

We must let K be a maximal subgroup contained in the interior of U, then K is infinite and the closure of K is in U, then it works.

I am a bit confused by this answer: Protasov and Peng gave (completely different, yet both in ZFC) proofs that such a topology need not exist. Maybe it is all a question of misunderstanding, I guess.

Dear Michael Hrušák

I am a little bit confused with your answer. It looks like the topology in your example (the neighborhood base of 0=emptyset is an ultrafilter) is locally MAXIMAL (i.e. there is no STRONGER non-discrete topology on any neighborhood of zero element). Why it should be locally MINIMAL?

Also, every neighborhood of 0 contains 0. So maybe neighborhoods of 0 in your example should be unions of {0} and elements of the ultrafilter.

Sorry, if I misunderstood something.

Also, by now I don't see problems with the argument given by Dekui Peng

(after the minor correction that he made).

Best, Vladimir

This question was answered by (first) Igor Protasov and a few days later by Dekui Peng (who was not aware of Protasov's) answer.