Group with more than one element and with no proper, nontrivial sub groups must have prime order. 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. 
 A: 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 $\,k\in\Bbb N\,\,,\,1<k<n\,$ , 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 
$$1<k<n\Longrightarrow 1\lneq\langle\,g^k\,\rangle \lneq\langle\,g\,\rangle=G$$
And we have a nontrivial subgroup. Thus, no such $\,k\,$ can exist and this means $\,n\,$ is a prime number. $\;\;\;\;\;\;\;\;\square\,$
A: 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 $\langle a, b\rangle=\langle a \rangle$. 
