Creating a Topological group from modulo multiplication Group. If I were to create a Topology out of the Modulo 3 Multiplication group $\mathbb{Z}_3$, what elements would it consist of and why? So $\mathbb{Z}_3 = \{0,1,2\}$ as a group over modulo 3. What are the elements of this group as a Topology?
Note: I edited out what I thought it would look like and why as it was pointed out that it was confusing, and I can not figure out how to better explain my methodology.
Thanks, 
Brian
 A: Assume you're not giving it the discrete topology, and you don't consider the trivial topology (i.e. only the total and the empty set as topology, the coarsest topology that exists).  Then there is an open set with two elements, say $\{a,b\}$.  Clearly, you can order them so that $a+1=b$ ((modulo $3$, of course).
Translating the above by the group action you get that $\{0,1\}, \{1,2 \}, \{2,0\}$ are all open sets, and so this is precisely the discrete topology, after you intersect them in pairs.
Thus, the only possibility is the discrete topology, if your topology is not to be the trivial topology.
You can play this trick on any finite cyclic group, and you'll get that every topology, including the trivial topology, follows from taking the initial topology of an epimorphism
$\phi:C_n\twoheadrightarrow G,$ where $G=C_m$ is discrete cyclic of order $m$ and $m|n,$ or $0$ -that is, $$T=\{\phi^{-1}(U): U \mbox{ open in } G\}.$$  The case of the trivial topology would be to consider the trivial group $\{0\}$.
Thus, the 'atom' open set that generates the whole topology is precisely $Ker\ \phi$, which is $\{0\}$ when $\phi=Id.$
A: Maybe a totally different approach would be to consider your elements as rotations of $\frac{2 \pi}{3}$ acting on a regular 3-gon i.e. an equilateral triangle?
