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

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.

• What about simplifying the double negation by just asking "Prove that any infinite group contains a nontrivial proper subgroup"? – J.-E. Pin Jan 11 '16 at 17:02

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.

• 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 – user1942348 Jan 11 '16 at 17:21
• 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. – Paul K Jan 11 '16 at 17:26
• @menag Would you please elaborate "$⟨g^2⟩$ is by definition a subgroup". Would you please give me a outline to prove this. Thanks. – rama_ran Nov 21 '16 at 4:27

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