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Does every group have a finite index subgroup (besides the whole group)? I suspect the answer is no, but I haven't found an example.

What about finitely presented or generated groups? Or groups with torsion?

I tried something like taking a presentation and only selecting some of the generators, but I don't this works even if $G$ has torsion, since the other generators could interact in a complicated way. Perhaps if we could find a minimal presentation of a group with torsion containing (as a generator) a torsion element, then this would work, but I'm not sure why such a minimal presentation should exist.

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    $\begingroup$ What about $(\mathbb{Q},+)$?: math.stackexchange.com/questions/1724728/… $\endgroup$ – crskhr Oct 17 '17 at 15:31
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    $\begingroup$ There are finitely presented infinite simple groups, which cannot have such a subgroup. $\endgroup$ – Derek Holt Oct 17 '17 at 15:51
  • $\begingroup$ @Derek You don't need finitely presented (or even finitely generated). If $H$ has finite index in $G$ then there are only finitely many conjugates of $H$. Then intersecting these gives a normal subgroup which is also of finite index in $G$. $\endgroup$ – user1729 Oct 17 '17 at 16:12
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    $\begingroup$ @user1729 But the point of my comment was to point out that there exist finitely groups with no proper subgroups of finite index. $\endgroup$ – Derek Holt Oct 17 '17 at 16:20
  • $\begingroup$ @DerekHolt Sorry, I missed the second paragraph! $\endgroup$ – user1729 Oct 19 '17 at 8:22
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The condition that a group has a nontrivial finite index subgroup is equivalent to the condition that it has a nontrivial finite quotient, and also equivalent to the condition that it acts nontrivially on some finite set (exercise).

As stated in the comments, $\mathbb{Q}$ is an easy example of a group with no nontrivial finite quotients (exercise), the point being that quotients of divisible groups are divisible but no nontrivial element of a finite group is divisible. If you want an example which is torsion then take $\mathbb{Q}/\mathbb{Z}$, which is still divisible.

Even assuming finitely presented does not help you: the Higman group is a finitely presented group with no nontrivial finite quotients.

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