Matrix Space Semigroup under Hadamard Product

Theorem
Let $\map {\mathcal M_S} {m, n}$ be the matrix space over a semigroup $\struct {S, \circ}$.

Then the algebraic structure $\struct {\map {\mathcal M_S} {m, n}, +}$, where $+$ is matrix entrywise addition, is also a semigroup.

If $\struct {S, \circ}$ is a commutative semigroup then so is $\struct {\map {\mathcal M_S} {m, n}, +}$.

If $\struct {S, \circ}$ is a monoid then so is $\struct {\map {\mathcal M_S} {m, n}, +}$.

Proof
$\struct {S, \circ}$ is a semigroup and is therefore closed and associative.

As $\struct {S, \circ}$ is closed, then so is $\struct {\map {\mathcal M_S} {m, n}, +}$ from Matrix Entrywise Addition is Closed.

As $\struct {S, \circ}$ is associative, then so is $\struct {\map {\mathcal M_S} {m, n}, +}$ from Matrix Entrywise Addition is Associative.

Thus if $\struct {S, \circ}$ is a semigroup then so is $\struct {\map {\mathcal M_S} {m, n}, +}$.

If $\struct {S, \circ}$ is commutative, then so is $\struct {\map {\mathcal M_S} {m, n}, +}$ from Matrix Entrywise Addition is Commutative.

Thus if $\struct {S, \circ}$ is a commutative semigroup then so is $\struct {\map {\mathcal M_S} {m, n}, +}$.

Let $\struct {S, \circ}$ be a monoid, with identity $e$.

Then from Zero Matrix is Identity for Matrix Entrywise Addition, $\struct {\map {\mathcal M_S} {m, n}, +}$ also has an identity and is therefore also a monoid.