0
$\begingroup$

What is the topology on the group $G =\prod_{n=1}^{\infty} \mathbb{Z}/(p^n)$ can be given so that it becomes a topological group ?

Here $\mathbb{Z}/(p^n)$ is finite cyclic group.

Is it the $I-$adic topology for ideal $I$ or is it the Zarisky topology?

But in that case we need ring structure.

I am confused. Is there product topology?

$\endgroup$
  • 1
    $\begingroup$ Yes, see here. $\endgroup$ – Dietrich Burde Aug 19 at 14:04
  • 3
    $\begingroup$ The default topology would be the product topology of the discrete topologies. The result will not be discrete. $\endgroup$ – Thomas Andrews Aug 19 at 14:08
  • 4
    $\begingroup$ There's no "the" topology. There are infinitely many different topologies on your $G$ that turn it into a topological group. What exactly are you trying to do here? What's your context? $\endgroup$ – freakish Aug 19 at 14:21
  • 2
    $\begingroup$ Each group $\mathbb{Z}/p^n$ is discrete. I did write this above. $\endgroup$ – Tyrone Aug 20 at 9:15
  • 2
    $\begingroup$ @BijanDatta, A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. Since $\mathbb{Z}/p^n$ is a group it also can be endowed with discrete metric $\endgroup$ – M. A. SARKAR Aug 20 at 11:03

Your Answer

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

Browse other questions tagged or ask your own question.