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Let $G$ be a non-trivial group with no non-trivial proper subgroup. Prove that $G$ cannot be infinite group.

A hints is given as order of G is not infinite since $a$ and $a^{-1}$ are only generators. I know that infinite cyclic group has only two generators. But don't know how to prove the problem.

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  • $\begingroup$ What about simplifying the double negation by just asking "Prove that any infinite group contains a nontrivial proper subgroup"? $\endgroup$ – J.-E. Pin Jan 11 '16 at 17:02
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Take $1 \neq g \in G$. Then $\langle g \rangle \subseteq G$ is a subgroup. Since it is not the trivial subgroup and there are no proper subgroups it is equal to $G$. Thus $G$ is cyclic. Assume $G$ is infinite. Then the group $\langle g^2 \rangle$ is a proper subgroup.

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  • $\begingroup$ I understand. Very good reply. I am unable to understand the last line "Then the group $\langle g^2 \rangle$ is a proper subgroup." Also what next $\endgroup$ – user1942348 Jan 11 '16 at 17:21
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    $\begingroup$ Well, $\langle g^2 \rangle$ is by definition a subgroup. Since $\operatorname{ord}(g) = \infty$ by assumption, $g^2 \neq 1$ and thus it is not the trivial group. Furthermore we have $g \notin \langle g^2 \rangle$, since otherwise we had $g = g^{2k}$ for some $k \in \mathbb{Z}$ and thus $1 = g^{2k - 1}$ and therefore $g$ had finite order. This yields that $\langle g^2 \rangle$ is a proper subgroup of $G$ which contradicts the assumption that $G$ does not have proper subgroups. $\endgroup$ – Paul K Jan 11 '16 at 17:26
  • $\begingroup$ @menag Would you please elaborate "$⟨g^2⟩ $ is by definition a subgroup". Would you please give me a outline to prove this. Thanks. $\endgroup$ – rama_ran Nov 21 '16 at 4:27
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Let $G$ be an infinite group, and $g \in G$ a non trivial element.

  • What happens if $g$ has finite order?
  • If $g$ has infinite order, then consider $\langle g^2\rangle$
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  • $\begingroup$ Please write some more lines about $<g^2>$. $\endgroup$ – rama_ran Nov 21 '16 at 5:06
  • $\begingroup$ @rama_ran it will be a non-trivial subgroup of $G$ $\endgroup$ – Mathmo123 Nov 21 '16 at 11:28
  • $\begingroup$ You wrote "What happens if g has finite order" . Would you please elaborate a bit? $\endgroup$ – rama_ran Nov 21 '16 at 11:40
  • $\begingroup$ @rama_ran you need to show that $G$ has a subgroup. If $g$ has finite order, can you see a way of showing that $\endgroup$ – Mathmo123 Nov 21 '16 at 11:58

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