I got this problem from my group theory's lecture.
If $m,n>1$ and $m$ is a multiple of $n$, then
$$ {\mathbb{Z}_{m}}/{\mathbb{Z}_{n}} \cong \mathbb{Z}_{\frac{m}{n}} $$
My attempt: Define a function $f:\mathbb{Z}_{m}\rightarrow{\mathbb{Z_{\frac{m}{n}}}}$ such that for every $z\in{\mathbb{Z}} \text{ } \text{ } f([z]_{m})=[z]_{\frac{m}{n}}$, and testing specific cases in $n$ and $m$ I've found that $Ker(f) \cong \mathbb{Z}_{n}$, to show this in the general case the problem is reduced to prove for every $n$ and $m$ multiple of n
$$ \{ a <m \text{ } | \text{ } a|\frac{m}{n} \}$$
must have $n$ elements. But I can't find a way to prove this. Any hint?
To prove that two groups are isomrphic, you need to give an isomorphism (an invertible homomorphism), your map is a good choice , but you need to verify that it is well defined (it does not depend on the representative of classe, i.e., $a\cong b[m]\implies a\cong b[\frac{m}{n}]$), then it is homomorphism ,surjective and that $\ker(f)\cong \mathbb{Z}_{n}$,more specifically $\ker(f)=\langle[\frac{m}{n}]\rangle$ (one inclusion is trivial$\langle[\frac{m}{n}]\rangle\subset \ker(f)$ and u can prove the other direction using surjectivity by applying the first isomorphism theorem).