Group whose all subgroups have infinite index Is there a group $G$ satisfying the following conditions?
If $H$ is a proper subgroup of $G$ , then $[G:H]$ has infinite index.
I guess $\mathbb{Q}$ is such group.
 A: Yes, $\Bbb Q$ is such a group.
Say $H$ is a proper subgroup, and assume $H\ne\{0\}$. There exists $x\in H$ and a prime $p$ so that  $x/p\notin H$. Now, writing $H/n=\{r/n:r\in H\}$, it follows that $$H, H/p, H/p^2\dots$$is a strictly increasing sequence of subgroups.
So now we just have to show that $\{0\}$ has infinite index...
A: You can generalize David's answer by looking at the class of divisible groups, and you can look at this post: Subgroups of finite index in divisible group, for more on that.
There are more exotic groups, with a stronger property: infinite groups where every subgroup has cardinality less than the whole group. Some examples are:


*

*Prüfer groups: every subgroup is finite and cyclic (this is an example of a divisible group)

*Tarski Monsters: finitely generated groups where every proper subgroup is order $p$ for some prime. These are known to exist for every prime number sufficiently large (and there are continuum many non-isomorphic examples)

*Shelah constructed groups with cardinality $\aleph_1$, but every proper subgroup has cardinality $\leq \aleph_0$ in the paper On a problem of Kurosh, Jonsson groups, and applications. (so an uncountable group where every proper subgroup is countable).
