In an (R,R)-bimodule on which elements act the same from the left or right, ring elements commute up to action

Let R be a ring and let M be an (R,R)-bimodule such that r \cdot m = m \cdot r for all m \in M and r \in R. Prove that for all r,s \in R and m \in M, rs \cdot m = sr \dot m. (That is, the assumption that R is commutative in the definition of an R-algebra is a rather natural one.)


Let r,s \in R and m \in M. Then we have rs \cdot m = r \cdot (s \cdot m) = r \cdot (m \cdot s) = (m \cdot s) \cdot r = m \cdot sr = sr \cdot m, as desired.

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