If I have a cyclic finite group $G$ that only has the subgroups $\{id\}$ and $G$. How can I show that the order of $G$ is either $1$ or a prime number?
My first thought is to Lagrange's Theorem, stating that the order of all subgroups of a group must divide the order of the group. Then since there are only two subgroups, and one only contains the identity then the number of elements in $G$ must be either $1$ or $p$.
From that, I'd reason that $p$ has to be a prime number since $G$ itself must be the other subgroup.
I don't feel this is the full proof though.
Best Answer
Firstly, note that every element has order $p$ for some fixed prime $p$ (why?). Next, note that if $g\in G\setminus\{1\}$ then for every $h\in G$ we have that $h\in\langle g\rangle$ (that is, the smallest sub group of $G$ containing $g$, which is cyclic, also contains $h$) (why?). The result follows (why?).