Is $\mathbb{Z}$ isomorphic to $n\mathbb{Z}$ for every $n\neq 0$? It is clearly the case that the map $f:\mathbb{Z} \rightarrow n\mathbb{Z}$ defined as $f(x) = nx$ is a group isomorphism.
But if that is the case, we also have $\mathbb{Z}/n\mathbb{Z}$ isomorphic to $\mathbb{Z}/\mathbb{Z}$ isomorphic to $\{0\}$, but this is wrong because we know $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to the cyclic group $\mathbb{Z}_n$. What did I do wrong?
 A: In the isomorphism $\mathbf Z\simeq n\mathbf Z$, you simply forgot that $n\mathbf Z$ is mapped onto  $n^2\mathbf Z$, so that what you can deduce is
$$\mathbf Z/ n\mathbf Z\simeq n\mathbf Z/ n^2\mathbf Z.$$
A: A quotient group $G/H$ depends not only on $G$ and the isomorphism class of the abstract group $H$, but it also matters how $H$ sits inside $G$. $H\cong K$ does not imply $G/H\cong G/K$.
Although $\mathbb{Z}$ and $n\mathbb{Z}$ are isomorphic, they sit inside $\mathbb{Z}$ differently. In particular, the identity map $\mathbb{Z}\to\mathbb{Z}$ is onto, so the quotient is a singleton, while the map $a\mapsto na$ is not. There are many integers which are not multiples of $n$. Each one belongs to a coset corresponding to a nontrivial element of the quotient $\mathbb{Z}/n\mathbb{Z}$
A: You are assuming that if $H_1\cong H_2$ are isomorphic subgroups of $G$, then $G/H_1$ is isomorphic to $G/H_2$. But this isn't true: a quotient group $G/H$ depends on the way $H$ lives inside $G$, not just what $H$ looks like on its own.
A: Yes $n\mathbb{Z}$ 'looks like' $\mathbb{Z}$, but why? It is useful to be able to ground abstract math with real world experience.
Imagine you made a ruler thousands of years ago, and you marked l-unit-of-measure with a tick, Later you say, heck, let me 'tick divide' my unit of measure into 5 sub-divisions. Well, you might as well admit it, you can look at 1/5 now as your unit of measure.
Next step - rational numbers!
If you want to see a picture and more discussion, see 5-tick per Unit ruler
