Let $G$ be a cyclic group of order $n$ generated by, say, $g$. Show that if $k$ divides $n$, then there is one, and only one, subgroup of $G$ of order $k$, and that this is generated by $g^{n/k}$.
I can see it being true if we take $n$ to be prime. The $k$ is either $1$ or $n$, thus if $k=1$, then the subgroup of order $1$ will be generated by $e$ and if $k=n$, then the subgroup will be $G$ itself generated by $g$ which is true since $n$ is prime.
How about in a case where $n$ is not prime?
Suppose $\,k\mid n\implies n=dk\;$:
If $\;G=\langle g\rangle\;$ is ciclyc then clearly $\,K:=|\langle g^d\rangle|=k\;$ .
Suppose now we have $\;1\ne H\le G\;,\;\;|H|=k \;$ and let $\,r>0\;$ be the minimal natural number s.t. $\,g^r\in H\;$ , then
$$1=(g^r)^k=g^{rk}\implies n\mid rk\implies rk=ns=dks\implies r=ds\implies s\mid d$$
and from here we have
$$g^r=(g^d)^s\in K$$
and from $\,|H|=||K|\;$ we get equality and thus uniqueness of $\,K\,$.
Hint: We know that $G=\left\{g^0,g^1,g^2,\dots,g^{n-1}\right\}$. Let $H\le G$ be a non-trivial subgroup ($H\ne \left\{g^0\right\}$) and let $k>0$ be the minimal positive number such that $g^k\in H$. What can you say about $k$ and $g^k$?
Suppose $h\in H \leq G$, for $|H|=k$. What can the order of $h$ be? Is it in the group generated by $g^{g/k}$?
