# Discrete abelian subgroups

## Main Question or Discussion Point

Quick question:

Suppose I have a (transitive) R^n action on a manifold M. If the isotropy group of R^n is discrete, does that mean that it is automatically isomorphic to Z/kZ, with 0<=k<=n?

Basically, my discrete subgroup is a lattice then, right?

Thanks!

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Office_Shredder
Staff Emeritus
Gold Member
The isotropy group is a subgroup of Rn so certainly can't have elements of finite order.

It is true that discrete subgroups of Rn have to look like Zk for some k

The isotropy group is a subgroup of Rn so certainly can't have elements of finite order.
Could you elaborate on that? Dummy here. So you are saying the isotropy group of R^n cannot be discrete?

I believe that Office_Shredder is incorrect. He seems to be thinking about subgroups of the translations group of $R^n$. Since that group is itself isomorphic to R^n, its subgroups can be discrete, but they can't have elements of finite order (elements such that a^n=1 for some n), aside from the identity. Z/kZ (better known as $Z_k$) is finite and all its elements have finite order.

I also have to question whether you're using the term isotropy group correctly.

I also have to question whether you're using the term isotropy group correctly.
Why? By isotropy group at some point m\in M, I mean the {g\in G | g(m)=m}. In my result I find that dim(iso-group)=0, so it is discrete. As G=G^n, that means the isotropy group is a discrete (abelian) subgroup of G, which I kind of hoped to be isomorphic to Z^k.

My group is isomorphic to R^n (group operation is addition). The left R-action acts on n factors, which I write as R^n.

Office_Shredder
Staff Emeritus