Let $m$ and $n$ be two given positive integers such that $m<n$. Then what are the necessary and sufficient conditions for the groups $(\mathbb Z_m,+_m)$ and $(\mathbb Z_n,+_n)$ to be homomorphic under the map $\phi \colon \mathbb Z_n \to \mathbb Z_m$ such that $\phi(x)$ is the remainder when $x$ is divided by $m$, for each $x$ in $\mathbb Z_n$.

I know that if one of these integers is a divisor of the other, then the groups are homomorphic.

  • 3
    $\begingroup$ By homomorphic you mean there exists a non-trivial homomorphism between them? $\endgroup$ – Tobias Kildetoft Oct 11 '13 at 8:34
  • 2
    $\begingroup$ What do you mean when you say two groups are homomorphic? $\endgroup$ – user61527 Oct 11 '13 at 8:34
  • $\begingroup$ Oh I'm sorry! Yes, of course, I mean that there exists a non-trivial homomorphism between them. In fact, I would like the conditions under whcih the remainder map becomes a homorphism form the group with the larger order to one with the smaller order. Can I make some alterations to my questions? $\endgroup$ – Saaqib Mahmood Oct 11 '13 at 8:46
  • 1
    $\begingroup$ $\mathbb{Z}_n$ is a homomorphic image of $\mathbb{Z}_m$ if and only if $(m) \subset (n)$, which is equivalent to $n \mid m$. It's easy to see if you consider the canonical map from $\mathbb{Z}$ to both. $\endgroup$ – Daniel Fischer Oct 11 '13 at 8:49
  • $\begingroup$ I'd suggest to rephrase this as “when is that map ... a homomorphism”. $\endgroup$ – Carsten S Oct 11 '13 at 9:21

The definition of $\phi$ is not right: there is no Euclidean division defined on $\def\Z{\mathbf Z}\Z/n\Z$ (some elements like the class of $n-1$ have inverses, in which case exact division by them is possible, but in the contrary case no useful notion of division with remainder exists). Instead you mean to take a representative in $\Z$ of a class in $\Z/n\Z$, and apply Euclidean division by$~m$ to that representative. This defines a map $\Z/n\Z\to\Z/m\Z$ only if the remainder of that division is independent of the choice of that representative. You can easily figure out when this happens, and if it does the resulting map is easily seen to be a homomorphism.

  • $\begingroup$ Then what about the map $\phi \colon Z_6 \to Z_2$ defined by the remainder of $x$ when divided by $2$, as in the division algorithm? In this case $2$ divides $6$. Can't we define such a map of $Z_9$ into $Z_2$, for example, by taking the former to be represented by the set $\{0, 1, 2, \ldots, 8\}$? $\endgroup$ – Saaqib Mahmood Oct 11 '13 at 9:45
  • 1
    $\begingroup$ The point of my answer is that no operation of Euclidean division by $2$ is defined in $\Z/6\Z$ (you can of course define one in whatever crazy way you like, but then you are on your own to prove any properties). In this case choosing a representative and then doing Euclidean division gives a result independent of the choice, and allows defining a morphism $\Z/6\Z\to\Z/2\Z$. In the case of $\Z/9\Z$ the element $2$ is invertible (in the ring $\Z/9\Z$), so division by it is defined, but with no place for a remainder. Trying to define a nonzero morphism $\Z/9\Z\to\Z/2\Z$ does not work. $\endgroup$ – Marc van Leeuwen Oct 11 '13 at 10:16
  • 1
    $\begingroup$ If as you suggest you circumvent to problem of showing that the choice of representative does not matter (since it does matter for $\Z/9\Z$ when dividing by $2$) by forcing a particular choice of representative, then you will run into trouble showing the map defined using that is a homomorphism of groups. Try to prove it is a homomorphism, this is the best way to see where you need the freedom to choose representatives. $\endgroup$ – Marc van Leeuwen Oct 11 '13 at 10:24

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.