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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.

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

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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.

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