Non-trivial finite/infinite subgroups of infinite groups

Question

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?

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. — zarathustra, Nov 19 '13 at 06:11
@Ranveer, in the second part of the question, I think you want the first occurrence of "finite" to be "infinite", no? — Doc, Nov 19 '13 at 06:23
Now this is the final question. I'm really sorry for the typos and thanks @Doc for pointing out the mistake. — Ranveer, Nov 19 '13 at 06:30
For abelian examples, see http://math.stackexchange.com/questions/261145 — Bart Michels, Jan 12 '17 at 23:11

Arthur · Accepted Answer · 2013-11-19T06:29:27.203

8

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.

edited Nov 19 '13 at 06:29

answered Nov 19 '13 at 06:12

Arthur

A very nice example that is also easy to understand. Well done! – Doc Nov 19 '13 at 16:20
Why are all the subgroups finite? – Topology Sep 26 '16 at 00:17

score 5 · Answer 2 · answered Nov 19 '13 at 06:39

5

There exist infinite groups all of whose proper subgroups are finite. Look here: http://en.wikipedia.org/wiki/Pr%C3%BCfer_group

answered Nov 19 '13 at 06:39

Doc

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...). – user1729 Nov 19 '13 at 09:55

2 Answers2