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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?

W.Scott
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    How do you conclude that $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}/\mathbb{Z}$? – carmichael561 May 24 '17 at 22:11
  • Because $n\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ as a group. – W.Scott May 24 '17 at 22:13
  • It is true that $\mathbb{Z} \cong n \mathbb{Z}$; there is only one infinite cyclic group up to isomorphism (and no cyclic group of cardinality larger than $\mathbb{N}$) – Kaj Hansen May 24 '17 at 22:14
  • You have an isomorphism $f: \mathbb{Z}\to n\mathbb{Z}$. How do you get an isomorphism $\mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}/\mathbb{Z}$ from $f$? If you carefully write out your argument I imagine the error will become clear. – carmichael561 May 24 '17 at 22:17

4 Answers4

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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 $$\mathbf Z/ n\mathbf Z\simeq n\mathbf Z/ n^2\mathbf Z.$$

Bernard
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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 $\mathbb{Z}/n\mathbb{Z}$

ziggurism
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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.

Noah Schweber
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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

CopyPasteIt
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