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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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If there exists $\,x\in G\;\;,\;x\notin\langle\,g\,\rangle\,$ ,then $\,\langle\,g\,\rangle\neq G\,$ – DonAntonio Oct 31 '12 at 22:45
You can say $G = \langle a,b \rangle.$ It is a perfectly true statement- it just does not help to prove what you need to prove. – Geoff Robinson Oct 31 '12 at 22:48
@ Geoff Robinson I know I am saying $G=\langle a\rangle$, because I can then say that $G$ is cyclic. But I think this is strange, that I can choose between that I want $G$ to be cyclic or not. If $G=\langle a\rangle$, then it is cyclic. If $G=\langle a,b\rangle$ then it is not cyclic. Why cannot I see logic in this? – Reader Oct 31 '12 at 23:01
It turns out that all groups of prime order are cyclic and have no proper, non trivial subgroups, and, as you are trying to prove, if a group has no proper, nontrivial subgroup, it is of prime order (finite) (and therefore cyclic - you'll probably encounter that problem soon). What you'll need to show is that any non-cyclic group must have a proper subgroup, and that even cyclic groups (both finite and infinite) have proper you need to prove that ANY group - if not of prime order - has a proper nontrivial subgroup. – amWhy Oct 31 '12 at 23:33
@DonAntonio Thank for your comment. In beginning I thought it has nothing to do with my question, but I was wrong. That is what I need. Thanks. – Reader Nov 1 '12 at 8:35
up vote 1 down vote accepted

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



$$1<k<n\Longrightarrow 1\lneq\langle\,g^k\,\rangle \lneq\langle\,g\,\rangle=G$$

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

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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. – Reader Nov 1 '12 at 16:34

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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No I am not saying both thing I am only saying that G=⟨a,b⟩. And then I have a problem. I have that G is not cyclic. – Reader Oct 31 '12 at 23:06
@Reader: The group $\mathbb Z$ is generated by $1$ and $2$. But it is still cyclic, right ? – user18119 Oct 31 '12 at 23:16
But a cyclic group is a group that is generated by a single element? is it that <a,b> is also cyclic? Then everything is cyclic? – Reader Oct 31 '12 at 23:24
Z is also infinite cyclic group, and these are generated by two elements. But what about finite groups, they are generated by single element, right? – Reader Nov 1 '12 at 6:59
@Reader: yes, but they can also be generated by more than one element. Any group is generated by all its elements ! – user18119 Nov 1 '12 at 20:16

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