# If $G$ has no proper subgroup, then $G$ is cyclic of prime order

This is something I'm supposed to be able to prove for an upcoming test, but I can't find anything to help me prove this in my notes or the chapter, which is on cosets and Lagrange's theorem. If all I start with is the group having no proper subsets, then that means any subset is the whole group. By Lagrange's theorem, that means the index is $1$, but that doesn't get me anywhere. Where do I start?

• Do you know Cauchy's theorem? Commented Nov 24, 2014 at 2:16

I'll give the proof of it without using Cauchy's theorem:

1. $$G$$ is cyclic If not, the generating set of $$G$$ must have at least two elements. Let $$a$$ and $$b$$ be elements which satisfy $$a\notin \langle b\rangle$$ and $$b\notin \langle a\rangle$$, then $$\langle a\rangle$$ and $$\langle b\rangle$$ are distinct proper subgroups of $$G$$, a contradiction.

2. $$G$$ has a prime order If $$G$$ has infinite order, then it has proper subgroup (consider $$2\Bbb{Z}$$ in $$\Bbb{Z}$$.) so $$G$$ is finite. If $$|G|=rs$$ with $$r,s>1$$ and $$G=\langle a\rangle$$, then $$\langle a^s\rangle$$ is a proper subgroup of $$G$$.

• This is really nice! Cauchy's theorem seems overkill when you put it this way. Commented Nov 24, 2014 at 2:24
• Ok, this makes sense to me and sounds good, but how would I justify the first sentence. If $G$ is not cyclic it must be generated by two elements. It makes sense but what theorem or property proves that it is true? Commented Nov 24, 2014 at 2:41
• @chris We assume that $G$ is not cyclic, so $G$ is not generated by one element. Commented Nov 24, 2014 at 2:49
• Ho do you know that such an $a$ and such a $b$ exist? Commented Oct 20, 2016 at 18:07
• @Jerrywu Excluding trivial cases, of course :) Commented Nov 19, 2021 at 19:54

Here is a simple proof via Cauchy's theorem, proceeding by contrapositive.

Let $G$ be a group of composite order $n = pk$, $p$ prime. By Cauchy's theorem, $G$ has an element of order $p$, hence a cyclic proper subgroup of order $p$.

• This may be a stupid question, but how can we be sure there is a cyclic proper subgroup of order $p$? Is it because the element with order $p$ generates a cyclic group containing just itself and the identity? Commented Nov 24, 2014 at 2:29
• Any element $x\in G$ of order $\alpha$ generates a cyclic group of order $\alpha$. That is $\langle x \rangle = \{1, x, x^{2}, x^{3}, \ldots, x^{\alpha-1}\}$. Commented Nov 24, 2014 at 2:36

I'm answering a 5-year old question because I had questions on this one. If G has no nontrivial subgroups then any of it's elements would generate it. If the order were composite, say $$o(G)=nm \Rightarrow$$ pick $$a\neq e$$ so that $$a^{nm}=e=(\underbrace{a^{n}}_{b})^{m}=b^m\Rightarrow o(b)\leq m$$.

Either $$a^{n}=e$$ or $$b^{m}=e$$. In either case, $$G$$ would have a nontrivial proper subgroup.