Let $G \neq \lbrace e \rbrace$ a group such that their only subgroups are the trivial ones. Prove that $G$ is a finite subgroup with prime order.

I don't know how to prove that $G$ is finite. If it's divided into two cases, whether cyclic or not, I can prove the case where $G$ is not cyclical. Any hint?


If $g$ is infinite, take $a\in G\setminus\{e\}$. Then $\langle a\rangle$ is a subgroup of $G$ which is not $\{e\}$. Therefore $G=\langle a\rangle$. But, since $G$ is infinite, $G$ is an infinite cyclic grouo and therefore $G\simeq(\mathbb{Z},+)$. But this group has non-trivial subgroups; take $2\mathbb Z$, for instance.

Therefore, $G$ is finite. Can you do the rest?

  • $\begingroup$ Yeah! Thanks for the help! $\endgroup$ – Corrêa Mar 23 '18 at 0:47

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