If G be a non-trivial group with no non-trivial proper subgroup, show that G is a group of prime order.
Attempt:
Given $G\neq \{e\}$ and let $a\in G, a\neq e$. Then $H=<a>$ is a subgroup of G. Then by the hypothesis, $H= \{e\} ~or~ G$. But $H\neq <a>, ~as ~a\neq e$. So G is generated by $a$.
I can show also that G is finite (if G is infinite, considering group $<a^2>$).
Let if possible, $o(G)=n=composite=kl, ~k,l>1$ are intergers. As $G=<a>$, $o(a)=n$. As $<a^k>$ is a subgroup of G, $(a^k)^l=a^{kl}=a^n=e\implies o(a^k)=l<n$, i.e $<a^k>$ is a subgroup of order $<n$, but G cannot have a non-trivial subgroup. So $l=1 ~or~ l=n$. How to conclude the rest?
The problem is in the shaded portion only. Please correct it.