Magma Subset Product with Self

Theorem
Let $\left({S, \circ}\right)$ be a groupoid.

Let $T \subseteq S$.

Then $\left({T, \circ}\right)$ is a groupoid iff $T \circ T \subseteq T$, where $T \circ T$ is the subset product of $T$ with itself.

Proof
By definition, $T \circ T = \left\{{x = a \circ b: a, b \in T}\right\}$.


 * Let $\left({T, \circ}\right)$ be a groupoid.

Then $T$ is closed, that is, $\forall x, y \in T: x \circ y \in T$.

Thus $x \circ y \in T \circ T \implies x \circ y \in T$.


 * Now suppose $T \circ T \subseteq T$.

Then $x \circ y \in T \circ T \implies x \circ y \in T$.

Thus $T$ is closed, and therefore $\left({T, \circ}\right)$ is a groupoid.