If $k$ divides $n$ then there's only one subgroup of $G$ of order $k$ 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 $\,r&gt;0\;$ 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 $\,K\,$.
Suppose $h\in H \leq G$, for $|H|=k$. What can the order of $h$ be? Is it in the group generated by $g^{g/k}$?