If $H$ is a subgroup of $Z_{n}$, prove that there exist $m$ a positive integer dividing $n$ such that $H\cong \mathbb{Z}_{n/m}$ I consider the application $f$ from $H$ to $\mathbb{Z}_{n/m}$ how $[m]_{n}$ to $[1]_{n/m}$ , where $\langle [m]_{n}\rangle = H$. But I don't know if $f$ is right.

Any subgroup of a cyclic group is cyclic.   By Lagrange's theorem the order of $H$ divides $n$.  $\therefore n=m\mid H\mid\implies \mid H\mid=\frac nm\implies H\cong\Bbb Z_{\frac nm}$.
Alternatively, $\langle k\rangle =H$, where $k=\min\{t\gt0\mid t\in H\}$.
Or you could use $m$, as you began to do.  That is, $\langle m\rangle =H$, where $m=\frac nd$.  
