If a group $\displaystyle G$ has no proper subgroups then it is to be proven that $\displaystyle G$ is cyclic.
I am not allowed to use Lagrange's theorem or the concept of order of an element in a group.
I proceeded as follows:
If $\displaystyle G=\{e\}$, then the theorem is trivially true.
So let's consider the case when $\displaystyle G\neq \{e\}$. It follows that $\displaystyle G$ has at least one non-identity element $\displaystyle a\in \ G$.
Now I claim that $\displaystyle G\setminus \langle a\rangle =\emptyset $.
Proof: On the contrary, suppose that $\displaystyle G\setminus \langle a\rangle \neq \emptyset $.
Let $\displaystyle x\in G\setminus \langle a\rangle $. It is known that $ $$\displaystyle \langle a\rangle \subset G$ is a subgroup of $\displaystyle G$ however since $\displaystyle (b\in G)\land (b\notin \langle a\rangle)$, it follows that $\displaystyle \langle a\rangle $ is a proper subgroup. This contradicts the given hypothesis that $\displaystyle G$ has no proper subgroups.
Therefore, $\displaystyle G\setminus \langle a\rangle =\emptyset $. It follows that $\displaystyle G =( G\setminus \langle a\rangle ) \cup \langle a\rangle =\emptyset \cup \langle a\rangle =\langle a\rangle $.
This proves that $\displaystyle G$ is cyclic.
Is my proof correct? Thanks.