So I'm working with algebra and have to deal with this problem. I can't see the logic to start the problem. I know the definition of a group and cyclic group. Any suggestion would be so great! I'm using the book from Hungerford.
Problem: Prove that if $G$ is a group with $|G|\geq2$ and no subgroups except for the two trivial ones (i.e. $\{e_G\}$ and $G$) then $G$ is cyclic and of prime order.
On
As $G\ne\{e_G\}$, you can pick $a\in G\setminus\{e_G\}$. Then the subgroup $\langle a\rangle$ generated by $a$ is $\ne\{e_G\}$, hence $G=\langle a\rangle$ is cyclic. The order of $a$ must be either $1$, or a prime, or composite, or infinite.
Remains only the option that the order is prime.
On
One doesn't need to assume $G$ is finite: every infinite group has proper nontrivial subgroups. Indeed, take $g\in G$, $g\ne1$. If $\langle g\rangle$ is proper, we're done; otherwise $\langle g\rangle=G$; in this case we have $$ \{1\}\subsetneq\langle g^2\rangle\subsetneq\langle g\rangle=G $$ Suppose now $G$ is finite. Take $g\in G$, $g\ne 1$. If $\langle g\rangle\ne G$, we're done. Otherwise $\langle g\rangle=G$. If the order $n$ of $g$ is composite, say $n=ab$, with $1<a<n$, then the order of $g^a$ is less than $n$ and greater than $1$. Therefore $\{1\}\subsetneq\langle g^a\rangle\subsetneq\langle g\rangle=G$.
Thus the assumption can only hold for $|G|$ a prime and in this case it indeed holds, because subgroups can only have order $1$ or $|G|$.
If $|G| \geq 2$, then $G$ has some prime factor $p$. By Cauchy's theorem, $G$ has an element $g$ with order $p$.
Consider the subgroup generated by $g$. This subgroup is a cyclic group of $p$ elements. But since the only nontrivial subgroup is $G$ itself, we conclude that this subgroup is indeed $G$, and so $G$ is cyclic with prime order.