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Does an infinite group whose every non-trivial subgroup is also infinite exist? If yes, what can be an example of such a group?

also,

Does an infinite group whose every non-trivial subgroup is finite exist? If yes, what can be an example of such a group?

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  • $\begingroup$ The free group on 2 elements? It is the group made of words over the alphabet $a,b,a^{-1},b^{-1}$, with concatenation as multiplication. $\endgroup$ Nov 19, 2013 at 6:11
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    $\begingroup$ @Ranveer, in the second part of the question, I think you want the first occurrence of "finite" to be "infinite", no? $\endgroup$
    – Doc
    Nov 19, 2013 at 6:23
  • $\begingroup$ Now this is the final question. I'm really sorry for the typos and thanks @Doc for pointing out the mistake. $\endgroup$
    – Ranveer
    Nov 19, 2013 at 6:30
  • $\begingroup$ just edit out the "also" in "also finite" and you're good. $\endgroup$
    – Doc
    Nov 19, 2013 at 6:49
  • $\begingroup$ For abelian examples, see math.stackexchange.com/questions/261145 $\endgroup$ Jan 12, 2017 at 23:11

2 Answers 2

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Any group without non-trivial elements of finite order has this property (and the converse also holds). One example is $\mathbb{Z}$.

For your second question, one example is the group $\{x \in \mathbb{C} : \exists n \in \mathbb{N}\: \: x^{2^n} = 1\}$ under multiplication.

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  • $\begingroup$ A very nice example that is also easy to understand. Well done! $\endgroup$
    – Doc
    Nov 19, 2013 at 16:20
  • $\begingroup$ Why are all the subgroups finite? $\endgroup$
    – Topology
    Sep 26, 2016 at 0:17
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There exist infinite groups all of whose proper subgroups are finite. Look here: http://en.wikipedia.org/wiki/Pr%C3%BCfer_group

The Tarski Monster is another example .. http://en.wikipedia.org/wiki/Tarski_monster_group

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  • $\begingroup$ The point of the Tarski monster group is that it is finitely generated, while the Prufer groups are not (just so you know the motivation...). $\endgroup$
    – user1729
    Nov 19, 2013 at 9:55

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