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I want to show that if $G$ is a group with more than one element, and that $G$ has no proper non-trivial subgroups. Prove that $|G|$ is prime. (Do not assume at the outset that |G| is finite).

My question is not that how to prove it. I am saying that suppose $|G|\geq 2$ possibly $|G|=\infty.$ By assumption the only subgroups of $G$ are $\{e\}$ and $G$, i.e., the trivial groups. Let $a$ be non-identity element in $G$. Consider $\langle a\rangle$. Then $\langle a\rangle=G.$ So $G$ is cyclic.

My question is, why can I say that $G=\langle a\rangle$. I know there are only two subgroups and $\langle a\rangle\neq e$ because $a\neq e$. Therefore we must have $G=\langle a\rangle$. But my problem is why cant I say that consider $a,b\in G$ and then we look at $\langle a,b\rangle$. And then I say $G=\langle a,b\rangle$ and then I cannot say that $G$ is cyclic, and then I will have problem proving question.

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    @DonAntonio Thank $f$or your comment. In beginning I thought it has nothing to do with my question, but I was wrong. That is what I need. Thanks.2012-11-01

2 Answers 2

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Resuming all the above, together with my comment and, of course, what you did:

Take any $\,1\neq g\in G\,$ (exists such an element since $\,|G|>1\,$), then $\,\langle\,g\,\rangle=G\,$ , otherwise $\,G\,$ has a non-trivial subgroup, and we already know $\,G\,$ is cyclic:

1) It can't be the order of $\,g\,$ is infinite, otherwise $\,G=\langle\,g\,\rangle\cong\Bbb Z\,$ , but then there're lots of non-trivial subgroups: $\,\langle\,g^n\,\rangle\cong n\Bbb Z\,\lneq\Bbb Z\cong G\,$ , and thus $\,G\,$ is cyclic and finite.

2) Supose finally that $\,|G|=ord(g)=n\,$ . If there exists $\,k\in\Bbb N\,\,,\,1 , s.t. $\,n=mk\,\,,\,m\in\Bbb N\,$ , then the order of $\,g^k\in G\,$ is more than $\,1\,$ * and at most* $\,m\,$ , since

$\left(g^k\right)^m=g^{mk}=g^n=1$

and

$1

And we have a nontrivial subgroup. Thus, no such $\,k\,$ can exist and this means $\,n\,$ is a prime number. $\;\;\;\;\;\;\;\;\square\,$

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    Thanks, I have also proved it using more or less the same argument, I have used fundamental theorem of cyclic groups.2012-11-01
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You can say $G=\langle a, b\rangle$ and $G=\langle a\rangle$, there is no contradiction because the second equality implies that $b\in \langle a\rangle$ and $\langle a, b\rangle=\langle a \rangle$.

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    You have that $G$=$$ it means $G$ is generated by $a$ and $b$. But can you conclude that$G$is not cyclic at this point?. No.. you CAN'T unless you show$a$more stronger statement that it can't be generated by$a$single element, then you can say that it is not cyclic.2018-09-04