We prove that if a recursively presented group admits a (nonempty) subshift of finite type with nonzero Medvedev degree then it fails to have the strong topological Rokhlin property. This result simplifies a known criterion and provides new examples of recursively presented groups without this property.

翻译：我们证明：若一个递归可呈现群允许具有非零Medvedev度的（非空）有限型子移位，则该群不具备强拓扑Rokhlin性质。此结果简化了已知的判定准则，并为不具备该性质的递归可呈现群提供了新的例证。