Infinite group must have infinite subgroups.

Prove that an Infinite group must have subgroup with infinite elements.

I know that if group was cyclic order of the generator is infinite and there are infinite number of divisors.

• Do you mean that it must have an infinite number of subgroups or subgroups which are infinite? Oct 3, 2013 at 18:03
• @DonAntonio infinite number of subgroups. Edited the question. Oct 3, 2013 at 18:05
• @Surya I think you should ask this as a separate question, so as not to nullify Trevor Wilson's answer. Oct 3, 2013 at 18:07
• (But for a hint: Consider the subgroup generated by each element. When do these subgroups precisely coincide?) Oct 3, 2013 at 18:07
• @user1729 ok edited it back. Oct 3, 2013 at 18:21

2 Answers

Any group is a subgroup of itself.

• If we disallow this trivial case, the statement is wrong: see math.stackexchange.com/questions/261145/… Oct 3, 2013 at 17:46
• Is the statement still true if disallow trivial groups? Oct 3, 2013 at 17:58
• @Surya Do you mean trivial cases? That is what Trevor Wilson saying. The trivial group is finite... Oct 3, 2013 at 18:05

There are infinite groups with no infinite proper subgroups. Even if the group is abelian, as show the $p$-component of $\mathbb{Q}/ \mathbb{Z}$ (that is the set of elements whose order is a power of $p$), for any prime number $p$.

Tarski monsters provide examples in the non abelian case.

• However, all known examples are infinitely presented. Which is interesting. Oct 3, 2013 at 18:18
• May be you can edit the question to include your remark. Or better ask the question independently : Is there a finitely presented infinite group in which every proper subgroup is finite? Oct 3, 2013 at 18:24
• When I said "all known examples" I meant all known examples (not just those known by me). It is a relatively famous open problem, related to Burnside's problem. See the discussion following Problem 1.1 of this list of problems. Oct 3, 2013 at 18:35
• I remember well, that I found the Riemann Hypothesis as an exercise in Serge Lang's "Complex Analysis". Oct 3, 2013 at 20:19