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.
 A: 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 $m=\frac nd$.  
A: If I understand your solution correctly, you start with a subgroup $H$ of $Z_n$.  Then $H = \left<[m]_n\right>$, that is, the subgroup generated by (the equivalence class of) $m$ in $Z_n$.  You construct $f \colon H \to Z_{m/n}$ by sending $[m]_n$ to $[1]_{m/n}$.
So my questions are:


*

*How do you know $H$ is cyclic?  Maybe you've learned that as a theorem already, or maybe this problem is a lemma in proving that theorem.

*What is $f$ on other elements of $H$?  You've only defined it on one element.  

*How do you know that $f$ is a homomorphism?

*How do you know that $f$ is bijective?
I think you're on the right track, but these are some details that need to be fleshed out.
