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## 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!

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!