1
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.