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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.

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

2 Answers 2

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Any group is a subgroup of itself.

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    $\begingroup$ If we disallow this trivial case, the statement is wrong: see math.stackexchange.com/questions/261145/… $\endgroup$ Oct 3, 2013 at 17:46
  • $\begingroup$ Is the statement still true if disallow trivial groups? $\endgroup$
    – Surya
    Oct 3, 2013 at 17:58
  • $\begingroup$ @Surya Do you mean trivial cases? That is what Trevor Wilson saying. The trivial group is finite... $\endgroup$
    – user1729
    Oct 3, 2013 at 18:05
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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.

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  • $\begingroup$ However, all known examples are infinitely presented. Which is interesting. $\endgroup$
    – user1729
    Oct 3, 2013 at 18:18
  • $\begingroup$ 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? $\endgroup$ Oct 3, 2013 at 18:24
  • $\begingroup$ 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. $\endgroup$
    – user1729
    Oct 3, 2013 at 18:35
  • $\begingroup$ I remember well, that I found the Riemann Hypothesis as an exercise in Serge Lang's "Complex Analysis". $\endgroup$ Oct 3, 2013 at 20:19

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