# Does every group have a finite index subgroup?

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.

• What about $(\mathbb{Q},+)$?: math.stackexchange.com/questions/1724728/… – crskhr Oct 17 '17 at 15:31
• There are finitely presented infinite simple groups, which cannot have such a subgroup. – Derek Holt Oct 17 '17 at 15:51
• @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 $G$. – user1729 Oct 17 '17 at 16:12
• @user1729 But the point of my comment was to point out that there exist finitely groups with no proper subgroups of finite index. – Derek Holt Oct 17 '17 at 16:20
• @DerekHolt Sorry, I missed the second paragraph! – user1729 Oct 19 '17 at 8:22

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.