Subsemigroup of Cancellable Mappings is Subgroup of Invertible Mappings

Theorem

Let $S$ be a set.

Let $S^S$ denote the set of mappings from $S$ to itself.

Let $\CC \subseteq S^S$ denote the set of cancellable mappings on $S$.

Let $\MM \subseteq S^S$ denote the set of invertible mappings on $S$.

Then:

the subsemigroup $\struct {\CC, \circ}$ of $\struct {S^S, \circ}$ coincides with the subgroup $\struct {\MM, \circ}$ of $\struct {S^S, \circ}$

where $\circ$ denotes composition of mappings.

Proof

From Set of Invertible Mappings forms Symmetric Group, we have that $\struct {\MM, \circ}$ is a group.

Hence, by definition, $\struct {\MM, \circ}$ is a subgroup of $\struct {S^S, \circ}$.

Recall from Bijection iff Left and Right Inverse that a mapping is invertible if and only if it is a bijection.

By definition, a cancellable mapping is a mapping both left cancellable and right cancellable.

From Injection iff Left Cancellable and Surjection iff Right Cancellable, a cancellable mapping is both an injection and a surjection.

That is, a mapping is cancellable mapping if and only if it is a bijection.

That is, $\struct {\CC, \circ}$ is exactly the same as $\struct {\MM, \circ}$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.10 \ \text{(a)}$