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.
 A: 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.
A: 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.
A: For $\mathbb{Z}$, use division algorithm to show that any subgroup $H\le \mathbb{Z}$ is of the form $\langle n \rangle$ where $n$ is the smallest positive element of $H$.
Let $G$ be a cyclic group with generator $a$, then define a group homomorphism $f:\mathbb{Z}\to G$ where $1\mapsto a$. This map is surjective since $\langle a\rangle = G$. Therefore $G\cong \mathbb{Z}/\ker f$. But $\ker f$ being a subgroup of $G$ is in the form $\langle n\rangle$, and we conclude that $G\cong \mathbb{Z}/\langle n\rangle$. If $n = 0$, then $G\cong \mathbb{Z}$. If $n = 1$, then $G\cong 0$. If $n$ is anything else, then $G\cong \mathbb{Z}_n$. These are up to isomorphism only possible cyclic groups.
For $\mathbb{Z}_n$, use Lagrange's theorem. Its possible subgroups are $\langle d\rangle \cong\mathbb{Z}_{n/d}$ where $d|n$. 
A: *

*If your proof is true, you proofed that subgroup $H$ of group $\langle a \rangle$ is always $\{a^i \mid 0 \le i < k\}$, for some $k$. That's False.

*The induction is false, I don't finally understand the try, but if you add an element $x$ to group $H$, and not add $x^{-1}$, $H \cup \{x\}$ is not a group.
A: No, you can't do induction in that way. Why should $H=\{g^i:0\le i<k\}$?
The cleanest proof exploits the knowledge of the subgroups of $\mathbb{Z}$ and the homomorphism theorems.


*

*If $H$ is a subgroup of $\mathbb{Z}$ (with respect to addition, of course), then $H=n\mathbb{Z}$ for a unique integer $n\ge0$.

*If $N$ is a normal subgroup of $G$, then the subgroups of $G/N$ are exactly those of the form $H/N$, where $H$ is a subgroup of $G$ containing $N$.
Now, with this knowledge, consider a cyclic group $G$; then there is a surjective homomorphism $\varphi\colon\mathbb{Z}\to G$, so its kernel $\ker\varphi=n\mathbb{Z}$ for a unique $n\ge0$.
Therefore, we can reduce to the case where $G=\mathbb{Z}/n\mathbb{Z}$.
A subgroup of $\mathbb{Z}/n\mathbb{Z}$ is of the form $H/n\mathbb{Z}$, where $H$ is a subgroup of $\mathbb{Z}$ containing $n\mathbb{Z}$; thus $H=m\mathbb{Z}$ where $m$ divides $n$. Since $m\mathbb{Z}$ is obviously cyclic, also its quotient $m\mathbb{Z}/n\mathbb{Z}$ is cyclic.
A: We consider the surjective homomorphism $\pi : \mathbf{Z} \rightarrow \mathbf{{Z}/{nZ}}$.
Let $H$ a subgroup of $\mathbf{{Z}/{nZ}}$, by applying the homomorphism $\pi^{-1}(H)$ we obtain by properties of homomorphism a subgroup of $(\mathbf{Z},+)$.
However, we know that the subgroups of $(\mathbf{Z},+)$ are exactly the $n\mathbf{Z}$ which contain one generator. They are cyclic !
No we apply the homomorphism $\pi(\pi^{-1}(H))$. Hence, we obtain a subgroup of $(\mathbf{Z/nZ},+)$ this time. By properties of homomorphism $H$ is cyclic.
