Proving that a group with no proper subgroups must be of prime order without using Lagrange theorem and under other restrictions. This is further to a question that I asked earlier. Based on suggestions/hints in the comments, I propose this proof for proving that a group $\displaystyle G$ with no proper subgroup is of prime order. I am not allowed to use Lagrange's theorem, Cauchy theorem, concept of order of an element of a group, isomorphisms.
I proceed as follows:
It's already been proven in my last post here that $\displaystyle G$ must be cyclic. If $\displaystyle G=\{e\}$, then the given statement is wrong as $\displaystyle 1$ is not a prime.
So let's consider the non-trivial case when $\displaystyle ord( G)  >1$.
Of course $\displaystyle G$ can be considered finite, if not finite then talking about order of $\displaystyle G$ in terms of prime does not make sense. So let order of $\displaystyle G=m >1$ and suppose on the contrary that $\displaystyle m$ is not a prime. It follows that $m$ can be written as product of $\displaystyle p$ and $\displaystyle q$ such that
\begin{equation}
m=pq,\ \ ( p\geq 2) \land ( q\geq 2) \tag{1}
\end{equation}
Let $a\in G$ be a non-identity element.
Either $\displaystyle a^{p} =1$ or $\displaystyle a^{p} \neq 1$.
Case 1: $\displaystyle a^{p} =1$
Let's consider the set $\displaystyle \{a,a^{2} ,a^{3} ,\cdots ,a^{p} \}$, which is clearly a proper subgroup of $\displaystyle G$ so this case is not possible.
Case 2: $\displaystyle a^{p} \neq 1$
Let's consider the set $\displaystyle S=\{a^{pi} :\ i\in \mathbb{Z} \}$, which is clearly a subgroup of $\displaystyle G.$ In$\displaystyle A=\{a^{p} ,a^{2p} ,\cdots ,a^{mp} ,\ a^{( m+1) p} \}$ at least two elements must be the same as $\displaystyle G$ is finite. For some $\displaystyle a^{ip} \in A$ there exists $\displaystyle a{^{\ }}^{jp} \in A$ such that
\begin{equation}
a^{ip} =a{^{\ }}^{jp} ,\ j >i\geq 1 \tag{2}
\end{equation}
It follows that $\displaystyle a^{( j-i) p} =e=a^{kp}$, where $\displaystyle k=j-i\geq 1$ and $\displaystyle k\leq m+1-j< m\Longrightarrow 1\leq k< m$
Now for any $\displaystyle i >k$, we have by Euclid's division lemma that $\displaystyle i=kq+r$, where $\displaystyle 0\leq r< k$ and therefore $\displaystyle a^{pi} =a^{ri}$ and $\displaystyle S$ boils down to the set $\displaystyle S=\{a^{p} ,a^{2p} ,\cdots ,a^{( k-1) p} ,e\}$, which is clearly a subgroup of order $\displaystyle k < m$ and therefore $\displaystyle S$ is a proper subgroup of $\displaystyle G$, which is not possible.
Therefore by contradiction our assumption in $\displaystyle ( 1)$ is wrong and therefore $\displaystyle m$ must be a prime.
Is my proof correct? Thanks.
 A: There are few gaps in the proof.
First you are not given that $G$ is finite. So you have to prove that $G$ cannot be infinite before you prove that $G$ must be of prime order.
Second, in Case 2, $G$ is finite does not imply that there exist two identical elements in $A$. In fact this is because $A$ is a subset of $G$ and $|A|=m+1>m=|G|$.
Third, in Case 2, it may happen that $i=1$ and $j=m+1$, which implies that $k=j-i=m$.
A: It is noted that the proof in OP is not complete yet and has some errors also which I'll try to fix up in this answer.
The statement to be proven is this: If a group has no proper subgroups then the group is cyclic of order $\displaystyle p$.
The group will have concept of order $\displaystyle p$ (finite) only if the group is also finite. So first let's prove that $\displaystyle G$ is finite.
Claim: $\displaystyle G$ is finite.
Proof: Suppose that $\displaystyle G$ is infinite group. Let $\displaystyle g\in G$ be a non-identity element and we consider the subset $\displaystyle A_{g} =\{g^{i} :i\in \mathbb{Z} \}$.
It is plain that all elements in $\displaystyle A_{g}$ are distinct for if any two elements in $\displaystyle A_{g} \ $are same (say $\displaystyle a^{i} =a^{j}$ then $\displaystyle a^{i-j} =1$, where $\displaystyle i >j$ and then it can be verified that $\displaystyle \{1,a,a,\cdots ,a^{i-j-1} \}$ is a subgroup of $\displaystyle G$.
It follows that $\displaystyle G=A_{g}$ and similarly, $\displaystyle G=A_{g^{2}}$.
Claim: $\displaystyle g\notin A_{g^{2}}$.
Proof: Suppose on the contrary that $\displaystyle \exists i\in \mathbb{Z} -\{0\}$ such that $\displaystyle g=g^{2i}$. It follows that $\displaystyle g^{2i-1} =1\Longrightarrow \{1,g,g^{2} ,\cdots ,g^{2i-2} \}$ is a proper subgroup of $\displaystyle G$, which is a contradiction.
Therefore $\displaystyle g\notin A_{g^{2}}$, which is a contradiction as $\displaystyle A_{g^{2}} =G$. Therefore, our assumption that $\displaystyle G\ $is infinite is not correct. It follows that $\displaystyle G$ is finite.
Having suitably dealt with the first part, it remains to prove that $\displaystyle G$ is cyclic and has order $\displaystyle p$. It has already been proven in my earlier post (linked in the OP) that $\displaystyle G$ is cyclic.
Suppose on the contrary that $\displaystyle G$ is of composite order $\displaystyle t$. Let $\displaystyle t=mn$, where both $\displaystyle m,n$ are at least equal to $\displaystyle 2.$
Let $\displaystyle g\in G$ be a non-identity element. Consider the set $\displaystyle A=\{g,g^{2} ,\cdots ,g^{mn} ,\ g^{mn+1} \}$. Clearly at least two elements of $\displaystyle A$ must be the same (if not then $\displaystyle G$ has $\displaystyle mn+1$ elements resulting immediately in contradiction). Let $\displaystyle g^{i} =g^{j}$ for some $\displaystyle i,j$ such that $\displaystyle 1\leq i< j\leq mn+1\Longrightarrow 1\leq j-i\leq mn$
It follows that $\displaystyle g^{k} =1$, where $\displaystyle k=j-i$.
Now $\displaystyle k\neq s$ for any $\displaystyle s< mn$ for if $\displaystyle k=s< mn$  then $\displaystyle \{1,g,g^{2} ,\cdots ,g^{s-1} \}$ is a proper subgroup which is contradiction so $\displaystyle k$. So $\displaystyle k=mn$. And it follows that
\begin{equation}
a^{mn} =1 \tag{1}
\end{equation}
Let's consider the set $\displaystyle \langle g^{m} \rangle =\{g^{mi} :\ i\in \mathbb{Z} \}\ $. Now if $\displaystyle i >n$, then by Euclid's division lemma, we must have $\displaystyle q$ and $\displaystyle r$ such that $\displaystyle i=nq+r$, where $\displaystyle r< n$ and it follows that $\displaystyle a^{mi} =a^{r}$ (using (1)).
For negative $\displaystyle i\in \mathbb{Z}$, we add $\displaystyle mn$ to the power so that $\displaystyle \langle g^{m} \rangle =\{g^{m} ,g^{2m} ,\cdots ,g^{( n-1) m} ,\ g^{nm} \}$. It is clear that this subgroup has only $\displaystyle n$ elements ($\displaystyle < mn)$, which is a contradiction as $\displaystyle G$ can't have a proper subgroup.
Therefore, we conclude by contradiction that $\displaystyle G\ ( \neq \{e\})$ is of prime order.
