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This a followup question on the first answer to this post: Why the term and the concept of quotient group?

In the first answer, Lahtonen says that for the quotient group Z/10Z, one can "equate 9 with 99", etc.

Let's take Z/3Z to make the question shorter. The quotient group has three elements: (...,0,3,6...),(,1,4,7...),(,2,5,8...) Each element is an infinite set of integers, specifically, the integers divisable by 3 plus an offset that is 0,1, or 2.

From my understanding of the answer by Lahtonen (which I'm not questioning, I've seen the same idea stated somewhere else) it should be that these cosets are somehow the same as equating 3 and 0, or 5 and 2.

The point is lost on me. In what sense does "5==2" relate to the 3 cosets? The latter are infinite, and have more structure than the single integers.

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  • Like, pretty much all algebra book I read get into equivalence relation and equivalent class before they get into coset, as it's part of the standard proof of Lagrange theorem. Using an element of an equivalent class to represent the whole class is the usual thing. Of course, one might want to use the notation $[5]$ to refer to the coset and $5$ for the element, but that's cube sum, and get really tiring one you have a bunch of quotient and end up having to write something like $[[[5]]]$. – Gina Jan 05 '14 at 09:50

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When we say $5$ and $2$ are equivalent here (or congruent $\mod3$) what we mean is that they belong to the same congruence class. The congruence classes are specifically those partitions of $\mathbb{Z}$ (i.e. cosets) that make up $\mathbb{Z}/3\mathbb{Z}$. Notice both $2$ and $5$ belong to $[2]_3=\{\dots,2,5,8,\dots\}$.

Notice the similarity with the typical idea of say taking $p$ objects and partitioning into groups of $q$ objects; we end up with $p/q$ such partitions. The idea behind quotients in algebra is of course very much the same.

Following @TheNumber23's great reply, notice that aggregating elements into partitions is making the individual elements "indistinguishable". If you can understand this, the isomorphism theorems will be a piece of cake.

obataku
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Let $G$ be a group and $N$ be normal in $G$. So $G/N$ is in fact studying elements outside of $G$. Just think $N$ is now our identity element. Elements in $N$ are no longer interesting. Furthermore two elements are equivalent if they differ by an element in $N$. Taking the quotient $\mathbb{Z}/3\mathbb{Z}$ is not saying $2==5$ all the time. It is more saying $2$ and $5$ differ by something in $3\mathbb{Z}$.

TheNumber23
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