Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let A be finite commutative group say $(Z_m)^h$. I will say that $S \subset A$ is an orbit if exist group $H$ which acts on A such that $S$ is an orbit of $H$.

Can one give a simple characterization of all orbits of $(Z_m)^h$?

By action on $A$ I mean automorphisms of a group A.

share|improve this question
    
Does it possible to say something inteligent about how orbits are look like? –  Klim Efremenko Oct 15 '10 at 6:38

1 Answer 1

up vote 3 down vote accepted

The abelian group in question is the product of its Sylow-$p$ subgroups, which are preserved by automorphisms. Therefore the orbits in it are the products of orbits in the Sylow $p$-subgroups. Therefore, we may consider the case where $m=p^k$ for some prime $p$ and some natural number $k$.

I can answer this question for maximal orbits (orbits under the full automorphism group). I think the more general questions may not have a nice answer.

In $(Z_{p^k})^h$, there are precisely $k+1$ orbits of the full automorphism group, represented by $e, pe, \ldots, p^ke$, where $e=(1,0,\ldots,0)$. The orbit of $p^i e$ consists of those vectors where the gcd of entries divides $p^i$, but not $p^{i+1}$ (except for $i=k$, where the orbit is just the element $0$).

For a general finite abelian group, this problem was solved more than a 100 years ago by Miller, and also discussed by Birkhoff and Baer. For the exact references, as well as a modern treatment see http://arxiv.org/abs/1005.5222.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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