$G$ is isomporphic to $\mathbb{Z}_{p}$ for a prime $p$ Prove that if $G$ is a finite group with no nontrivial, proper subgroups and $|G|>1$ then $G$ is isomporphic to $\mathbb{Z}_{p}$ for a prime $p$.
My attempt:  I know that $\mathbb{Z}_{p}$ for some prime $p$ will contain a prime number of elements. And, by Lagrange's Theorem (or a corollary to it), since $G$ is a finite group with no proper subgroups and with more than one elements, $G$ must have prime order. (otherwise, there would exist a subgroup with order that divides the order of $G$).
But I'm not sure how to make a solution out of these facts.  I know the formal definition of isomorphism, that $f(a)f(b)=f(ab)$, but trying to use that gets me even more confused. I think there are some properties or theorems that I can use but I can't figure out what.
 A: Pick $x \in G$ then $\langle x \rangle$ is a subgroup of $G$. Since $G$ has no nontrivial subgroups, either $\langle x \rangle = \{1\}$ or $\langle x \rangle = G$. If $\langle x \rangle = \{1\}$ then $x= 1$, if this happens for all $x$, then $G$ is the trivial group. If $\langle x \rangle = G$ and $|G|$ were composite, by Cauchy's Theorem applied to abelian groups, it would have subgroups for every $n$ dividing $|G|$, since it has no proper subgroups it has no divisors, so $|G|$ must be prime (alternatively,  $x^a$ generates $G$ for all $a$, otherwise, it would have a proper subgroup. This means $|G|$ is coprime to every integer $a<|G|$, therefore, it is prime). Now, $G$ is cyclic and of prime order, and all cyclic groups of the same order are isomorphic. 
A: If $G$ has no nontrivial subgroups then $G=\langle g\rangle$ is cyclic (For $g\neq e$, $\langle g\rangle$ is a nonzero subgroup which must be $G$). Take the epimorphism $\varphi:\mathbb{Z}\to G$ given by $n\mapsto g^n$.
Then by isomorphism theorem $\mathbb{Z}_n\cong G$ for some $n$. And as you have seen, $n$ must be prime.
