Suppose that $m,n \in \mathbb{Z}$ and $m$ divides $n$. Show that $\frac{\mathbb{Z}_n}{\mathbb{Z}_m} \cong \mathbb{Z}_\frac{n}{m}$ Suppose that $m,n \in \mathbb{Z}$ and $m$ divides $n$. Show that $$\frac{\mathbb{Z}_n}{\mathbb{Z}_m} \cong \mathbb{Z}_\frac{n}{m}$$ I try to use the third isomorphism theorem to show but I don know how to apply it here. Anyone can guide me ?
 A: Hint: First, note that, strictly speaking, $\mathbb Z_m$ is not a subgroup of $\mathbb Z_n$ (unless $n=m$), and what is really meant here is to first identify a certain subgroup of $\mathbb Z_n$ that is isomorphic to $\mathbb Z_m$. If you first make this precise, a very natural choice of a function $f:\mathbb Z_n\to \mathbb Z_{\frac{n}{m}}$ will spring to mind, prompting an application of the (first) isomorphism theorem.
General note: instead of trying to think which of the isomorphism theorems to use, look for the 'evident homomorphism' and apply the (first) isomorphism theorem to it. More often than not, it is easier to find naturally occurring homomorphisms. The second and third isomorphism theorems are immediate consequences of the first one (so in a sense, there is only one isomorphism theorem). 
A: You can see this problem via another point of view. For any $m$ which divides $n$ we have $$n\mathbb{Z}\lhd m\mathbb{Z}$$ and by letting $$\varphi:m\mathbb{Z}\to \mathbb{Z}/\frac nm\mathbb{Z}$$ with $$\varphi(k)=\big[{\frac1m\cdot k}\big]$$ one could check; $\varphi$ is an epimorphism and so $\ker(\varphi)=n\mathbb{Z}$. And therfore according to first isomorphism theorem, $$m\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/\frac nm\mathbb{Z}$$ This can lead you to have the conclusion.
A: Define $\phi: \mathbb{Z}_n \to \mathbb{Z}_{n}$ such that
$$\phi(x)=mx$$
And consider $\ker \phi$ and $\operatorname{Im}\phi$, and apply First isomorphism theorem.
A: Hint: A quotient of a cyclic group is cyclic--compare orders.
A: Consider $\mathbb{Z}_{n}$ as the set $\{0, \cdots, n-1\}$ with addition defined modulo $n$. If we have $\mathbb{Z}_{6} = \{0,1,2,3,4,5\}$, we can identify a copy of $\mathbb{Z}_{3}$ inside $\mathbb{Z}_{6}$ as $\{0,2,4\}$. If we mod out by this subgroup, we get the cosets $\{0,2,4\}$ and $\{1,3,5\}$, which we can identify with $\mathbb{Z}_2$.
Let us use this example to motivate the general construction. Given $\mathbb{Z}_n$, the subgroup generated by $\frac{n}{m}$ has order $m$. The quotient subgroup is still cyclic, and is generated by $[1]$, which has order $\frac{n}{m}$, so the quotient is $\mathbb{Z}_{\frac{n}{m}}$.
