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In every Modern Algebra book I've read, I've seen that the groups $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}_n$ are isomorphic, but not equal. I understand the difference between "isomorphic" and "equal," but this particular example raises a couple of questions for me.

I know the first group consists of cosets $k + n\mathbb{Z}$, and the second group consists of equivalence classes $[k]_n$, but isn't it true that $k + n\mathbb{Z} = [k]_n$ for every integer $k$? (They are both sets containing the same elements from $\mathbb{Z}$.) And if so, then can't we say that $\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$? Thanks in advance!

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    $\begingroup$ Presumably, one could define $\mathbb{Z}_n$ with only $n$ symbols, no reference at all to these other integers. Then the equality you put for $[k]_n$ would just be the isomorphism, not an equality. I've seen $\mathbb{Z}_2$ introduced this way. $\endgroup$ – Artimis Fowl May 19 '17 at 22:43
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    $\begingroup$ If you define $\mathbb Z_n$ in terms of equivalence classes, you are correct. The "intuitive" definition of $\mathbb Z_n$ as consisting of $\{0,1,2,\dots,n-1\}$ with "clock arithmetic" would not be equal. $\endgroup$ – Thomas Andrews May 19 '17 at 22:45
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    $\begingroup$ Cosets are equivalence classes, so there's no difference between your two definitions. In my opinion, the notation $\mathbb{Z}_n$ should be avoided because it can lead to confusion when $n$ is prime, since $\mathbb{Z}_p$ also means the $p$-adic integers. $\endgroup$ – carmichael561 May 19 '17 at 22:47
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    $\begingroup$ @Kaynex Isomorphism is not the same as equality. $\endgroup$ – Jair Taylor May 19 '17 at 23:07
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    $\begingroup$ @d4rk_1nf1n1ty Your question isn't unimportant, but the distinction between isomorphism and equal are. You're in a class where the objective is to become abstract. The structure of groups are important, their labels are not. When you're comparing the rotations of a polygon to the invarients of a field extension, it suddenly doesn't matter what "equal" means. But isomorphism means everything. $\endgroup$ – Kaynex May 19 '17 at 23:27
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Suppose we have a surjective homomorphism $h: G \to H$. The distinction, in my mind, between $\Bbb Z/n\Bbb Z$ and $\Bbb Z_n$ is the same as the distinction between $G/\text{ker }h$ and $H$.

Clearly, these are isomorphic groups (they have the same algebraic properties), but they are distinct in their "set structure" (the quotient group elements are elements of the power set of $G$, and the homomorphic image elements are just set elements (singletons) (of $H$)). In some areas of math (cough, topology) it can be crucial to keep track of "which level of set-construction" you are in.

Here is a similar example (from vector spaces) to illustrate the "ontological" versus "abstract" distinction.

The $x$-axis is a (normal) subgroup of the additive (vector addition) group of the Euclidean plane. We can form the coset space of all horizontal lines, which is isomorphic to, but surely not equal to, the $y$-axis.

As algebraic objects, we "abstract away" the particulars of how a given group arises, as we are usually only interested in its properties. However, the group may arise in a particular context we are studying, where its basis of definition may matter. One should not, for example, confuse the congruence class modulo $n$ of the integer $k$, with the integer $k$ itself.

This is the nature of algebraic morhpsims in general-they "filter away" information. In some sense, the quotient object has "more information to keep track of" (in $\Bbb Z/5\Bbb Z$, for example, one has to perform a (simple) mental calculation to resolve $7 + 5\Bbb Z$ to $2 + 5\Bbb Z$), whereas the homomorphic image has discarded the excess baggage.

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  • $\begingroup$ This seems to be the idea I was getting from everyone's comments above. I really like how you stated that the strict definition can matter, though--as @Kaynex stated--if you're only concerned about the structure, the symbols are (practically) irrelevant. $\endgroup$ – d4rk_1nf1n1ty May 21 '17 at 2:56
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Let's consider another example, and see if we can tease out what we want from our definition of equality and isomorphic.

One group is the rotations and reflections of a square which leave it in the same place it started (though the points inside may be moved about). The other group is generated by $2$ elements $a,b$, and follows the relations: $a^4 = e$, $b^2 = e$, and $ab = ba^3$. It turns out these two groups are isomorphic - we can identify $b$ with horizontal reflection, and $a$ with rotation by $\frac{\pi} 2$.. $b$ with vertical reflection and a with $-\frac{\pi} 2$. Further, if I draw a heart inside the square, I know what it means to reflect or rotate that heart. I can immediately apply the group operators for the first group. But what does it mean to apply $a$ to that heart?

Let's also look at a case where we can agree $2$ objects are equal. For example, $1 + 1 = 2$. The first sentence would have made just as much sense had I written "we can agree 1+1 objects are equal." (though linguistically horrific). The two representations, $2$ vs $1 + 1$, are interchangeable everywhere, and there's no new ambiguity from using one or the other. By this, I'd suspect that one would not say that the two groups I mentioned above are equal, but everyone would agree they're isomorphic.

Now, to return to the question for $\mathbb{Z}_n$ and $\mathbb{Z}/n\mathbb{Z}$. We agree they're certainly isomorphic, now we want to decide if they're equal. But first, a note on definitions. here and here both use $\mathbb{Z}_n$ and $\mathbb{Z}/n\mathbb{Z}$ for two different algebraic objects! Worse yet, they both note a third meaning for the notation $\mathbb{Z}_n$: the p-adics. Hence, to decide if they are different, we first need to agree on a definition for both. If we use yours, the first two of these links seem to suggest they're equal - but I can easily imagine a text defining the finite cyclic groups as $\mathbb{Z}_n$ and the equivalence classes as $\mathbb{Z}/n\mathbb{Z}$. Now, some sentances to consider (with ones definition of choice for $\mathbb{Z}_n$):

$[0]_5$ is a principle ideal. VS $0 + 5\mathbb{Z}$ is a principle ideal. (similarly for 'additive subgroup of $\mathbb{Z}$', or 'maximal ideal', etc)

$[k]_n$ is an equivalence class. VS $k + n\mathbb{Z}$ is an equivalence class.

$|k + n \mathbb{Z}|$ is infinite. VS $|[k]_n|$ is infinite.

$7$ is a member of $k + n \mathbb{Z}$. VS $7$ is a member of $[k]_n$

A few of these make me a bit nervous/uncomfortable - there seems like there could be some ambiguity between sending $[k]_n$ to $k + n \mathbb{Z}$ vs $-k + n \mathbb{Z}$ - as additive groups, these both look fine.

To answer this more concretely, one would need to know who asserted the two symbols were different, and their definitions of the two groups (as opposed to the most memorable definitions you've found).

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  • $\begingroup$ This is an excellent answer. I wish I could accept 2 answers, but David got to it first... :P $\endgroup$ – d4rk_1nf1n1ty May 21 '17 at 2:58
  • $\begingroup$ I agree with d4rk's assessment, and the comparative statements that make Artimis "nervous" are quite to the point. +1 from me. $\endgroup$ – David Wheeler May 21 '17 at 16:52
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It could be that $\mathbb{Z}_n$ is the multiplicative group of $n^\text{th}$ roots of unity, meaning the set of all complex numbers $z$ of length 1 with $z^n=1$, using multiplication as the group operation . The map $H: r\mapsto e^{2\pi ir}$, where $i = \sqrt{-1}$, goes from the additive group of real numbers onto the unit circle (as a multiplicative group, with $H$ a homomorphism) with image $Z_n$ and the kernel $H^{-1}(1) = n\mathbb{Z}$ (the additive group of integer multiples of $n$), so $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}_n$ by a familiar theorem of group theory.

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  • $\begingroup$ So what you're saying is that there can be varying definitions for $\mathbb{Z}_n$? (As long as those definitions give rise to the same group structure.) $\endgroup$ – d4rk_1nf1n1ty May 20 '17 at 0:38
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No , I am suggesting that Z_n is the group multiplicative group of nth roots of unity as I stated above . Surely the books you referred to give a definition for Z_n . How can they mention a group without telling you what it's elements are .If not take my definition ,it's right .Also it is important because is gives a natural connection between the multiplication and addition operations . You can do the same with the additive group of R (real numbers ) and the multiplicative group of the Positive Real numbers via x ---> e^x . Here no quotient group is involved and they are infinite groups . Again in your example Z/nZ and Z_n are specific groups with definitions given or known . Of course all this assumes that C (the field of complex numbers and R (real numbers) are fixed throughout ,Of course they have many different constructions better forgotten.(sometimes )

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    $\begingroup$ This should not be a separate answer. I would edit it into the previous answer, probably in a new paragraph saying Added: It could also be a comment replying to the OPs comment on your answer, but it is a bit long for that. $\endgroup$ – Ross Millikan May 20 '17 at 4:35
  • $\begingroup$ no ,I am answering the dark_infinity's question on my previous post which is definitive and stands alone .It would be confusing to combine them,sorry $\endgroup$ – StuartMN May 20 '17 at 4:58

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