Abstract
An algebraic proof is given of the classical fact that two different concentric circular annuli A ( r ) A(r) and A ( s ) A(s) are conformally inequivalent, where A ( r ) = { z ∈ C : 1 > | z | > r } A(r) = \{ z \in {\mathbf {C}}:1 > \left | z \right | > r\} . Indeed, it is shown that the covering groups of these annuli are not elementarily equivalent in the context of PSL ( 2 , R ) {\text {PSL}}(2,{\mathbf {R}}) . Considering the universal covering surface as U U , the upper half-plane, the covering group of a bounded plane domain is naturally contained in PSL ( 2 , R ) {\text {PSL}}(2,{\mathbf {R}}) as the group of covering transformations.
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