subgroups of finite cyclic group

Let $n$ be the order of a finite cyclic group $G$.  For every positive divisor (http://planetmath.org/Divisibility) $m$ of $n$, there exists one and only one subgroup of order $m$ of $G$.  The group $G$ has no other subgroups.

Proof.  If $g$ is a generator of $G$ and  $n=mk$,  then $g^k$ generates the subgroup $⟨g^k⟩$, the order of which is equal to the order of $g^k$, i.e. equal to $m$.  Any subgroup $H$ of $G$ is cyclic (see http://planetmath.org/node/4097this entry).  If  $|H|=m$,  then $H$ must have a generator of order $m$; thus apparently  $H=⟨g^{\pm k}⟩=⟨g^k⟩$.