Cyclic subgroup generated by $a$ is the smallest subgroup that contains $a$

Question

We know that for any element $a\in G$, $\langle a\rangle$ generates a cyclic subgroup $H$ of $G$. Theorem 5.17 from Fraleigh A first course in abstract algebra says that every subgroup containing $a$ contains $H$. I honestly do not understand the proof that he provides. Could someone explain it with extreme detail?

Answer 1

Let $L$ be a subgroup of $G$ such that $a\in L$. Since $L$ is a group and $a\in L$, then $e\in L$ and all the powers of $a$ and $a^{-1}$ belong to $L$. But, since$$\langle a\rangle=\{e\}\cup\left\{a^n\,\middle|\,n\in\mathbb{N}\right\}\cup\left\{(a^{-1})^n\,\middle|\,n\in\mathbb{N}\right\},$$this means that $\langle a\rangle\subset L$.

Answer 2

well, suppose that $a \in H$. let $a^n \in \langle a\rangle$ with $n>0$. Then $a^n= a\cdot \cdots a$, so by closure under the group operation, $a^n \in H$ as well. if $n<0$, the argument is the same, since $a^{-1} \in H$.

Answer 3

Every subgroup containing $a$ will also contain all integral powers $a^n$ of $a$ with $n\in \mathbb{Z}$ by the subgroup criterion (which says in particular that with $a,b\in S$ also $ab,a^{-1}\in S$). But these integral powers are just given by $H$. So it will contain $H$.