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?
 A: 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 $p$. 
A: 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. 
A: 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 $G$.
