TITLE: If $G$ has no proper subgroup, then $G$ is cyclic of prime order
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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.