There is a lemma that says 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
Best Answer
Given $a\in G$, $\langle a\rangle$ is defined to be the smallest subgroup of $G$ containing $a$ as an element. Knowing that $\langle a\rangle=\langle b\rangle$ isn't enough to conclude that $a=b$, though. For example, we always have $\langle a\rangle=\langle a^{-1}\rangle,$ no matter the order of $a$.