Magma Subset Product with Self

Theorem

Let $\struct {S, \circ}$ be a magma.

Let $T \subseteq S$.

Then $\struct {T, \circ}$ is a magma if and only if $T \circ T \subseteq T$, where $T \circ T$ is the subset product of $T$ with itself.

Proof

By definition:

$T \circ T = \set {x = a \circ b: a, b \in T}$

Necessary Condition

Let $\struct {T, \circ}$ be a magma.

Then $T$ is closed.

That is:

$\forall x, y \in T: x \circ y \in T$

Thus:

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

$\Box$

Sufficient Condition

Let $T \circ T \subseteq T$.

Then:

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

That is, $T$ is closed.

Therefore $\struct {T, \circ}$ is a magma by definition.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets