6

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.

Reader
  • 1,353
  • If there exists $,x\in G;;,;x\notin\langle,g,\rangle,$ ,then $,\langle,g,\rangle\neq G,$ – DonAntonio Oct 31 '12 at 22:45
  • 1
    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 subgroups...ie. you need to prove that ANY group - if not of prime order - has a proper nontrivial subgroup. – amWhy Oct 31 '12 at 23:33
  • group of order $\geq 2$ with no proper, nontrivial subgroups $\implies$ group is of prime order $$\iff$$ group not of prime order (order $\geq 2$) $\implies$ group has at least one proper, nontrivial subgroup. – amWhy Oct 31 '12 at 23:39
  • @amWhy My problem isnot how to prove it, my problem is written in the above mentioned comment. – Reader Nov 01 '12 at 05:50
  • 1
    @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 01 '12 at 08:35

2 Answers2

6

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. $\;\;\;\;\;\;\;\;\square\,$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • Thanks, I have also proved it using more or less the same argument, I have used fundamental theorem of cyclic groups. – Reader Nov 01 '12 at 16:34
  • Sorry (or not) for requesting you help after you experiences, but could you explain me the 'then' in this part? $1<k<n\Longrightarrow 1\lneq\langle,g^k,\rangle \lneq\langle,g,\rangle=G$ – xtreyreader Oct 29 '20 at 14:11
  • It's just that the order of $;g^k;$ is $;m;$ , which is less than $;n;$ . – DonAntonio Oct 29 '20 at 16:08
  • Why does the order of $g^k$ need to be more than 1? – beginner Feb 18 '24 at 19:49
  • @beginner Otherwise the order of $;g^k;$ is one and thus $;g^k=1;,;;k<n;$ , contradicting the assumption that $;ord(g)=n,(>k);$ . – DonAntonio Feb 23 '24 at 19:15
2

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 $\langle a, b\rangle=\langle a \rangle$.

  • 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 ? –  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 01 '12 at 06:59
  • @Reader: yes, but they can also be generated by more than one element. Any group is generated by all its elements ! –  Nov 01 '12 at 20:16
  • @ QiL There can be many generators in a cyclic group, but when we write <a,b> we mean that only a or b alone is not enough to generate the whole group. And in this case group is not cyclic, because it is not generated by a single element. – Reader Nov 01 '12 at 20:40
  • @Reader: no, this means $G$ is generated by $a$ and $b$, nothing more. In your post, when you say $G=\langle a, b\rangle$, you can't prove none of $a$ or $b$ generate $G$. –  Nov 01 '12 at 20:46
  • @ QiL Thanks. If I say that G=<a,b>, then since $a\in G$ so there is also a subgroup $$ in G, so must be equal to G, because G has no proper subgroups. Hence G is cyclic. – Reader Nov 01 '12 at 20:51
  • Why did I answer ? –  Aug 08 '13 at 22:42
  • Unless there is a very good reason, we prefer not to arbitrarily delete good content: the answers could well help other users in the future. – robjohn Aug 09 '13 at 23:10
  • You have that $G$=$<a,b>$ 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. – Uncool Sep 04 '18 at 12:37