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.
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I'm assuming you mean proper subgroup? Otherwise $[G:G]=1$, always. – SquirtleSquad Jun 13 '16 at 21:29
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Yes. when H is a proper. – Seongqjini Jun 13 '16 at 21:31
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1In addition to the previous answers: infinite simple groups fulfill this condition. This includes some finitely presented groups. – YCor Jun 17 '16 at 22:43
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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...
David C. Ullrich
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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).
