Left Module Does Not Necessarily Induce Right Module over Ring

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 $\circ’ : G \times R \to R$ 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 not necessarily a right module over $\struct {R, +_R, \times_R}$

Proof
Proof by Counterexample

Let $\struct {S, +_S, \times_S}$ be a ring with unity

Let $0_S, 1_S$ be the zero and unity of $S$ respectively.

Let $\struct {\map {\mathcal M_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.

Let $R = \map {\mathcal M_S} 2$.

From Ring of Square Matrices over Ring is Ring, $\struct {R, +, \times}$ is a ring.

Let $G = \set {\begin{bmatrix} x & 0_S \\ y & 0_S \end{bmatrix} : x, y \in S }$

Lemma 1

 * $G$ is a left ideal of $\struct {R, +, \times}$.

From Leigh.Samphier/Sandbox/Left Ideal is Left Module over Ring, $\struct {G, +, \times}$ is a left module over $\struct {R, +, \times}$.

Let $\mathbf A = \begin{bmatrix} 1_S & 0_S \\ 0_S & 0_S \end{bmatrix}, \mathbf B = \begin{bmatrix} 0_S & 0_S \\ 1_S & 0_S \end{bmatrix}, \mathbf C = \begin{bmatrix} 1_S & 0_S \\ 1_S & 0_S \end{bmatrix}$

Then:

And:

Hence for $\mathbf A, \mathbf B \in R, \mathbf C \in G$:
 * $ \paren {\mathbf A \times \mathbf B} \times \mathbf C \neq \paren {\mathbf B \times \mathbf A} \times \mathbf C$

Let $\circ : G \times R \to R$ be the binary operation defined by:
 * $\forall \mathbf \Lambda \in R: \forall \mathbf X \in G: \mathbf X \circ \mathbf \Lambda = \mathbf \Lambda \times \mathbf X$

From Left Module induces Right Module over same Ring iff Actions are Commutative, $\struct {G, +, \circ}$ is not a right module over $\struct {R, +, \times}$

Also see

 * Leigh.Samphier/Sandbox/Right Module Does Not Necessarily Induce Left Module over Ring