Show that the order of group $G$ is $1$ or a prime number 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. 
 A: 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?).
A: For an element $a$ of order a number $n$, the element $a^k$ has order $\dfrac n{\gcd(n,k)}$. so if the only subgroups of $G$ are the trivial subgroups $\{1_G\}$ and $G$ it means that, for any $k$, $\gcd(n,k)$  is $1$ or $n$. This is possible only if the only divisors of $n$ are $1$ and $n$, i.e. if $n$ is prime.
Conversely, if $n$ is prime, $G$ is cyclic and its only subgroups, by Lagrange's theorem, are the trivial subgroups.
A: Your proof is incorrect at the point where you use Lagrange's theorem. True, the theorem claims that the order of a subgroup must divide the order of the group, but there is no claim to the converse, i.e. that, if something divides the order of the group, it must be an order of a subgroup. For example, a convenient link from Wikipedia shows that there is a group of order $12$ with no subgroup of order $6$.
For a hopefully better proof, let $n=|G|$ and we need to prove that either $n=1$ or $n$ is prime. Suppose $n\gt 1$, so $G$ is nontrivial, and there is an element $a\in G, a\ne e$. Look at the (cyclic) group generated by $a$: $\langle a\rangle=\{a^k\mid k\in\mathbb N\}$. Because $a\ne e$, $\langle a\rangle$ is nontrivial, so it must then coincide with the whole $G$ (i.e. $G$ is cyclic). But then, if $n=|G|=|\langle a\rangle|$ was composite, say $n=mk$, then look at the subgroup generated by $a^k$: $|\langle a^k\rangle|=m$ so $G$ would have another subgroup of order $m, 1\lt m\lt n$ - contradiction! 
