With ZFC, is there an infinite group $G$ such that there is no nontrivial nondiscrete topology on $G$ with the functions $G\times G\to G,~~ (a,b) \mapsto ab$ and $G\to G,~~ a\mapsto a^{1}$ continuous?
There is a large literature about this, see "nontopologizable groups". These are, by definition, groups for which the only Hausdorff group topology is discrete. There are various examples, the first of which were obtained by Olshanskii and Shelah (see here for references) around 1980.
An observation is that for a group $G$, we have the equivalence between:
 the only group topologies on $G$ are the discrete and the indiscrete ones
 $G$ is simple and nontopologizable.
Shelah's example being simple, it therefore answers your question.


$\begingroup$ By default I expect that yes; you can check if necessary. $\endgroup$ – YCor May 10 '14 at 21:18

$\begingroup$ I do not access the main papers, but a survey by Dikranjan and Megrelishvili says: " The problem of existence of infinite nontopologizable groups was raised by Markov. The first example of such a group, under the assumption of CH, was given by Shelah [185], later Hesse [118] eliminated the use of CH in his argument.". [118] is German even if I can find it. "Eleminated CH"!. CH most probably is Continuum hypothesis. Does it mean that it is ZFC ?! $\endgroup$ – Minimus Heximus May 10 '14 at 21:25

$\begingroup$ Also I wonder if CH Elimination resulted in simpleness elimination! G. Luk ́acs's paper says nothing about CH for existence of nontopologizible. $\endgroup$ – Minimus Heximus May 10 '14 at 21:41

4$\begingroup$ You're right, Shelah may use a transfinite induction based on CH. I guess that these various constructions are not related. Anyway the more recent examples in arxiv.org/abs/1210.7895 are simple (and finitely generated!) and not based on CH. $\endgroup$ – YCor May 10 '14 at 21:52