Prove any subgroup of a cyclic group is cyclic. was just wondering if this is a valid proof for the aforementioned question? I am quite confident that it isn't, but not exactly sure why. Maybe I am missing the point of proofs by induction (amateur...).
Let $G$ be a cyclic group generated by $g$ and $H$ a subgroup of it. Since $H$ is a subgroup, $1_G\in H$ thus the trivial subgroup is cyclic. Proceeding by induction: $H=\lbrace g^i:0\leq i<k, \rbrace$ is true for case $k=1$.
So assume true for $n=k$ thus $H=\lbrace g^i:0\leq i<k \rbrace$ and $H\cup \lbrace g^{k}\rbrace=\lbrace g^i:0\leq i<k \rbrace \cup \lbrace g^{k}\rbrace=\lbrace g^i:0\leq i<k+1 \rbrace$ so since it is true for $n=k+1$ we have that it is true for all $n\in \mathbb{N}$.
Hopefully someone can point out the flaw, thanks.

Since $H$ is a subgroup of $G$ every element of $H$ can be expressed as $g^r$ for some $r\in \mathbb Z$. One possibility is that $H$ is trivial, hence cyclic. Else Let $m$ be the least positive value of $r$. Then $H$ is the cyclic group generated by $g^m$. Else there is an element $g^n\in H$ with $n=km+l$ and $1\leq l \lt m$, whence $g^l\in H$, which is a contradiction.
Least positive is justified by $g^r\in H \implies g^{-r}\in H$ - excluding the trivial group.
