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I am just looking for a bit of help understanding this concept: Let $$\mathbb Z/6\mathbb Z=\{[0],[1],[2],[3],[4],[5]\}$$be the set of equivalence classes.

But what elements does $\mathbb Z/6\mathbb Z$ under addition modulo $6$ have? Is it again the same as $\mathbb Z/6\mathbb Z$ or is it $\mathbb (Z/6\mathbb Z,+6)=\{0,1,2,3,4,5\}$

I am having trouble differentiating between the set and the set under the group operation.

Shaun
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student626
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    As with most things in mathematics, it depends. However, if you define $\mathbb{Z}/6\mathbb{Z}$ to be a quotient group, then the elements of that group are equivalence classes. Period. We might choose to represent those classes using certain distinguished elements, but they are still equivalence classes. – Xander Henderson Feb 01 '22 at 16:53
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    $\Bbb Z_n ={0,1,2,\ldots,n-1}$ is an isomorphic (ring) representation of $\Bbb Z/n\Bbb Z,$ using normal-form choices as reps of each coset (here the choice is the least nonnegative element in the coset, i.e. the remainder $\bmod n$ of every coset element). Though it is often convenient to blur the distinction between the two, it is essential to keep in mind that they are different (isomorphic) structures. See here for further discussion. – Bill Dubuque Feb 01 '22 at 17:15
  • Be careful interpreting the claims in your accepted answer - see my comment there. – Bill Dubuque Feb 01 '22 at 17:29

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You can write $[n]$ to underlie that you're working with equivalence classes (and this is the "correct" way to write the elements of a quotient) but to simplify the notation, sometimes you can also just write $0,1,\cdots,5$ for the classes $[0],\cdots,[5]$ if there's no risk of confusion with actual integers and remember that, for example, $3+5=2$ in this case. The result is a different, although equivalent ring. As Bill Dubuque points out, we choose $0,1,\cdots,5$ as representatives because they are the "minimal" representatives, meaning that every other $n\in\mathbb Z$ can uniquely be written as $6q+r$, with $0\leq r\leq 5$.

Alessandro
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    Technically they are not the "same thing", i.e. same ring. Rather they are isomorphic rings - see my comment on the question for further detail. Also the choice of rep is "motivated by a formal argument", here remainder reps in Euclidean domains (see the post linked in said comment). One should be careful making claims like those lest they (conceptually) mislead beginners. – Bill Dubuque Feb 01 '22 at 17:26
  • I expressed myself poorly. You're right that the two rings are just isomorphic and not the same. Also, when I said that the choice is not motivated by a "formal argument", I meant that choosing different representatives is also valid to give the ring structure. It won't give the same ring, but isomorphic rings for sure. – Alessandro Feb 01 '22 at 18:03
  • My point is that while experienced users may be able to guess correct interpretations of those fuzzy claims, many beginners may wrongly interpret them as meaning something false and misleading, and this may inhibit their learning process. I highly recommend that you edit your answer to remedy this. – Bill Dubuque Feb 01 '22 at 18:08
  • @BillDubuque, done, hopefully now the answer is less "fuzzy" – Alessandro Feb 01 '22 at 18:20