Hadamard Product over Group forms Group

Theorem
Let $$\left({G, \circ}\right)$$ be a group whose identity is $$e$$.

Let $$\mathcal {M}_{G} \left({m, n}\right)$$ be a $m \times n$ matrix space over $$\left({G, \circ}\right)$$.

Then $$\left({\mathcal {M}_{G} \left({m, n}\right), +}\right)$$, where where $$+$$ is matrix addition, is also a group.

Proof
As $$\left({G, \circ}\right)$$, being a group, is also a monoid, it follows from Matrix Space Semigroup under Addition‎ that $$\mathcal {M}_{G} \left({m, n}\right)$$ is also a monoid.

As $$\left({G, \circ}\right)$$ is a group, it follows from Negative Matrix that all elements of $$\mathcal {M}_{G} \left({m, n}\right)$$ have an inverse.

The result follows.