There is a lemma that says t if a group $G$ has no proper nontrivial subgroups, then $G$ is cyclic. And here is the proof of the lemma:
Suppose $G$ has no proper nontrivial subgroups. Take an element $a$ in $G$ for which $a$ is not equal to $e$. Consider the cyclic subgroup $\langle a \rangle$. This subgroup contains at least e and a, so it is not trivial. But G has no proper subgroups, so it must be that $\langle a \rangle = G$. Thus $G$ is cyclic, by definition of a cyclic group.
But here i do not understand the following: Why must $\langle a \rangle$ be a subgroup of $G$? For every single element $a$ in $G$, if $\langle a \rangle$ is a subgroup of $G$, then every group should have at least as many subgroups as the number of its elements. I would appreciate any help. Thanks