Left Module induces Right Module over same Ring iff Actions are Commutative

Theorem
Let $\struct {R, +_R, \times_R}$ be a ring.

Let $\struct {G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.

Let $\circ': G \times R \to G$ be the binary operation defined by:
 * $\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$

Then $\struct {G, +_G, \circ'}$ is a right module over $\struct {R, +_R, \times_R}$ :
 * $\forall \lambda, \mu \in R: \forall x \in G: \paren {\lambda \times_R \mu} \circ x = \paren {\mu \times_R \lambda} \circ x$

Necessary Condition
Let $\struct {G, +_G, \circ'}$ be a right module over $\struct {R, +_R, \times_R}$.

Then:

Sufficient Condition
Let the scalar multiplication $\circ$ satisfy:
 * $\forall \lambda, \mu \in R: \forall x \in G: \paren {\lambda \times_R \mu} \circ x = \paren {\mu \times_R \lambda} \circ x$

It needs to be shown that $\struct {G, +_G, \circ'}$ satisfies the right module axioms.

Let $\lambda, \mu \in R, x \in G$.

Then:

Let $\lambda \in R, x, y \in G$.

Then:

Let $\lambda, \mu \in R, x \in G$.

Then:

Also see

 * Right Module induces Left Module over same Ring iff Actions are Commutative