Determine group G as cyclic 
If $G$ has no proper subgroup, prove that $G$ is cyclic of order $p$, where $p$ is a prime number.  

I know that since $G $is a group with no proper subgroups, $g \in G$ is not just the identity. I don't know where to go from there.
 A: Examine the cyclic subgroup generated by some $g \in G$, where $g$ is not the identity.
A: You don't need to assume $G$ is finite.
Proposition. If $G$ is a group which has no nontrivial proper subgroups, then either $G$ is the trivial group or $G$ is cyclic of prime order.
Proof. If $G$ is the trivial group, we are done. If $G$ is not the trivial group, let $g\in G$ be any element other than the identity. Then $\langle g\rangle$ is a nontrivial subgroup of $G$, and therefore must equal all of $G$ by hypothesis. Thus, $G$ is cyclic.
If $g^2=1$, then $\langle g\rangle =\{1,g\} = G$, so $G$ is cyclic of order $2$ and we are done. If $g^2\neq 1$, then $\langle g^2\rangle$ is a nontrivial subgroup of $G$, so $G=\langle g^2\rangle = \langle g\rangle$, hence there exists $k$ such that $g = (g^2)^k$. Thus, $g^{2k-1}=1$, which proves that $g$ is of finite order. Thus, $G$ is finite cyclic.
Let $n$ be the order of $g$. If $a|n$, $0\lt a\lt n$, then $\langle g^a\rangle = G$ (since it is a nontrivial subgroup). Therefore, $g\in \langle g^a\rangle$, so there exists b such that $g = (g^{a})^b = g^{ab}$. Therefore, $g^{ab-1} = 1$, so $n|ab-1$. Since $a|n$, then $a|ab-1$, hence $a|-1$, so $a=\pm 1$.
That is, the only divisors of $n$ are $\pm 1$ and $\pm n$, so $n$ is prime. $\Box$
A: In view of the original poster's request for further explanation after an answer that got five up-votes, here are further comments.
Let $g$ differ from the identity $e$.  Look at $g, g^2, g^3, \ldots$.  Since the group is finite, eventually you reach $g^m=\text{some earlier term in this sequence}= g^\ell$.  So $\ell<m$. Since $g^m=g^\ell$, you get $g^{m-1}=g^{\ell-1}$ unless $\ell=0$, and that means the $m$th term wasn't the first term equal to some earlier term; the $(m-1)$th term is an earlier one.  So if it's the first one, then $\ell=0$.  So the sequence is $g, g^2, g^3, \ldots,g^{m-2},g^{m-1},g^m$, and the last term is $e$.  This is then a subgroup.  But there are no proper subgroups besides the trivial one, so you've got the whole group, and $m=n$.
That gets you a cyclic group; now you need to prove that $n$ is prime.  Suppose it's not, so that $n=jk$ and $j, k$ are smaller numbers than $n$.  Then $g^k, g^{2k}, g^{3k},\ldots,g^{jk}=e$ is a subgroup.  But there are no proper subegroups, so the assumption that $n$ is not prime is refuted.
A: HINT $\ $ Any noncyclic group has a proper subgroup generated by any  non-identity element. Any infinite cyclic group $\rm\:\left<g\right>\:$  has the proper subgroup $\rm\:\left<g^2 \right>\:.\:$ A finite cyclic group $\rm\:\left<g\right>\:$ of composite order $\rm\:nk\:,\ n,k > 1\:,\:$ has proper subgroup $\rm\:\left<g^n\right>\:.\:$ What remains are cyclic groups of prime order.
