Matrix Space Semigroup under Hadamard Product

Theorem
Let $\mathcal M_S \left({m, n}\right)$ be the matrix space over a semigroup $\left({S, \circ}\right)$.

Then the algebraic structure $\left({\mathcal M_S \left({m, n}\right), +}\right)$, where $+$ is matrix entrywise addition, is also a semigroup.

If $\left({S, \circ}\right)$ is a commutative semigroup then so is $\left({\mathcal M_S \left({m, n}\right), +}\right)$.

If $\left({S, \circ}\right)$ is a monoid then so is $\left({\mathcal M_S \left({m, n}\right), +}\right)$.

Proof
$\left({S, \circ}\right)$ is a semigroup and is therefore closed and associative.

As $\left({S, \circ}\right)$ is closed, then so is $\left({\mathcal M_S \left({m, n}\right), +}\right)$ from Properties of Matrix Entrywise Addition.

As $\left({S, \circ}\right)$ is associative, then so is $\left({\mathcal M_S \left({m, n}\right), +}\right)$ from Properties of Matrix Entrywise Addition.

Thus if $\left({S, \circ}\right)$ is a semigroup then so is $\left({\mathcal M_S \left({m, n}\right), +}\right)$.

If $\left({S, \circ}\right)$ is commutative, then so is $\left({\mathcal M_S \left({m, n}\right), +}\right)$ from Properties of Matrix Entrywise Addition.

Thus if $\left({S, \circ}\right)$ is a commutative semigroup then so is $\left({\mathcal M_S \left({m, n}\right), +}\right)$.

Let $\left({S, \circ}\right)$ be a monoid, with identity $e$.

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