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.
Best Answer
I would abandon induction as it does not help you for the infinite case. Let $g$ be the generator of $G$. Suppose $H$ is not cyclic. Then there exists $j$ and $k$ relatively prime where $g^j$ and $g^k$ are in $H$. But since $j$ and $k$ are relatively prime there exists $a,b$ such that $aj+bk = 1$. Hence $(g^j)^a+(g^k)^b = g$, so $g$ is in $H$. Hence $H=G$ which is cyclic, contradiction.