0
$\begingroup$

It is well known (and asked countless times as I gathered while searching an answer for this question) that $\mathbb{Z}/n\mathbb{Z}$ $\cong$ $\mathbb{Z}_n$. However, what I am trying to do is to infer the structure of $\mathbb{Z}/\mathbb{Z}_n$ (if this is even a valid structure that is, as I am not even sure that $\mathbb{Z}_n$ is a valid normal subgroup of $\mathbb{Z}$ under normal addition: details below).

Obviously taking a quotient group is not the same thing as division as we are used to (as in multiplying by the multiplicative inverse), but the separation (partition) into "equal" groups of the original supergroup does resemble similar concepts. I was wondering if this structure will also be isomorphic to $n\mathbb{Z}$ as if we simply switched the denominator like in regular division. I simply cannot comprehend how to partition $\mathbb{Z}$ by the cosets of $\mathbb{Z}_n$.

And this ultimately is my problem. If I take regular (nonmodular addition) clearly if I take the coset where I add 1 to, say, $\mathbb{Z}_5$, this overlaps with $\mathbb{Z}_5$, and does not parition $\mathbb{Z}$. But then I also consider how nonmodular addition will not make $\mathbb{Z}_5$ a well defined binary structure in the first place. So if we consider just modular addition we arrive that it is its only coset.

So did I answer my own question? The only reason that we can't have cosets of $\mathbb{Z}_n$ in the context of a supergroup $\mathbb{Z}$ is because of the fact that their operations are incompatible, and would not even make sense, as the incompatibility would not make it a proper subgroup. I am curious as to the relationship here as it has never been brought up in any introductory algebra class.

$\endgroup$
  • $\begingroup$ I'm not sure I understand the question, but it seems as if you are having trouble with the idea of a quotient group. A quotient group of $G$ is not a subgroup of $G$, and thus the notion of $\mathbb{Z}/\mathbb{Z}_n$ does not make sense. If $\mathbb{Z}_n$ was a subgroup of $\mathbb{Z}$, then $\mathbb{Z}$ would have torsion elements, which it does not! $\endgroup$ – TomGrubb Jul 3 '17 at 21:36
  • $\begingroup$ Yes I figured that I answered my own question, because they are incompatible. And I never once said the quotient group was a subgroup of G, I was referring the group we were taking quotients of. However, I believe that there is not a way to describe $\mathbb{Z}_n$ as a normal subgroup of $\mathbb{Z}$, and thus not a valid thing to take quotients of. $\endgroup$ – Charlie Tian Jul 3 '17 at 21:38
  • $\begingroup$ also not sure what a torsion element is either $\endgroup$ – Charlie Tian Jul 3 '17 at 21:39
  • 1
    $\begingroup$ @CharlieTian: A torsion element is one whose product with a non-zero element equals zero, e.g., in $ Z_6 , 2,3,4$ are torsion, since $2(3)=3(2)=4(3)=0(Mod6)$ $\endgroup$ – MSIS Jul 3 '17 at 22:24
  • $\begingroup$ Would you define a torsion as nonzero? Because then 0 is also a torsion element. This sounds very similar to the concept of zero divisors $\endgroup$ – Charlie Tian Jul 6 '17 at 15:02
0
$\begingroup$

I think you are having notation issues. The notation $\mathbb{Z}/n\mathbb{Z}$ always means the integers modulo $n$. The notation $\mathbb{Z}_n$ is sometimes used sometimes for the integers modulo $n$, but sometimes it is not. See the section of the wikipedia on notation.

However the way you are using it means the integers modulo $n$. Same object, different notation.

$\endgroup$
  • $\begingroup$ Yes this is what I mean, and the textbook I have followed (Fraleigh) uses this frequently as the integers mod $n$ $\endgroup$ – Charlie Tian Jul 3 '17 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.