Prove that every subgroup of a cyclic group is cyclic. Do this by working with exponents, and use the description of the subgroups of $\mathbb{Z}^{+}$.
My first thought on this problem is that if $G = \langle a \rangle$ is cyclic, then this is by definition the smallest subgroup containing $a$, so by definition, if a subgroup $H \subset G$ contains $a$, then $H = G$. Is that correct? So a subgroup will contain powers of $a$, but not necessarily $a$ itself.
Here is my attempt.
Let $G$ be cyclic generated by $a$, and $H \subset G$ a subgroup. Define $S = \{n \in \mathbb{Z} \mid a^n \in H\}$. I claim that $S \subset \mathbb{Z}$ is a subgroup. If $n,m \in S$, then we have $a^n a^m = a^{n+m} \in H$ by closure, so $n+m \in S$. $a^0 = e \in H$ since $H$ is a subgroup. If $a^n \in H$, then $\left(a^n\right)^{-1} = a^{-n} \in S$ since $-n \in \mathbb{Z}$. So $S$ is a subgroup of $\mathbb{Z}$, so we have $S = \mathbb{Z} b$ for some $b \geq 0$, as all subgroups of $\mathbb{Z}$ are of this form. If $S = \{0\}$, then only $a^0 \in H$ and $H$ is the trivial subgroup, which is certainly cyclic. Suppose $S \neq \{0\}$, then $S = \mathbb{Z} b$ for some $b > 0$ and $H$ is non-trivial. Furthermore, $a^n \in H$ if and only if $n = kb$ for some $k \in \mathbb{Z}$. That is, $$H = \{a^{kb} \mid k \in \mathbb{Z} \} = \{(a^b)^k \mid k \in \mathbb{Z}\} = \langle a^b \rangle.$$ So $H$ is cyclic, generated by $a^b$.
How does this look? I think I could shorten the argument by not distinguishing whether $S$ is trivial because it doesn't quite matter. If $S$ is trivial, $S = 0 \mathbb{Z}$, so any $a^n \in H$ is still of the form $a^{k \cdot 0}$ for $k \in \mathbb{Z}$, which is the identity for all $k$.
How does this look?