# 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? – Alex Wertheim Nov 24 '14 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 element. Let $a$ and $b$ are elements which satisfies $a\notin \langle b\rangle$ 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. – Alex Wertheim Nov 24 '14 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? – chris Nov 24 '14 at 2:41
• @chris We assume that $G$ is not cyclic, so $G$ is not generated by one element. – Hanul Jeon Nov 24 '14 at 2:49
• Ho do you know that such an $a$ and such a $b$ exist? – philooooooo Oct 20 '16 at 18:07
• @HanulJeon I am unable to understand case 2 that you have written. what happened if |G|=composite=rs? How you have written '$\langle a^s\rangle$ is a proper subgroup of $G$. – user1942348 Apr 5 '17 at 17:56

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? – chris Nov 24 '14 at 2:29
• Any element $x\in G$ of order $\alpha$ generates a cyclic group of order $\alpha$. That is $\langle x \rangle = \{1, x, x^{2}, x^{3}, \ldots, x^{\alpha-1}\}$. – Alex Wertheim Nov 24 '14 at 2:36