Set of all Self-Maps under Composition forms Semigroup
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Theorem
Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let the operation $\circ$ represent composition of mappings.
Then the algebraic structure $\struct {S^S, \circ}$ is a semigroup.
Proof
Let $f, g \in S^S$.
As the domain of $g$ and codomain of $f$ are the same, the composition $f \circ g$ is defined.
By the definition of composition, $f \circ g$ is a mapping from the domain of $g$ to the codomain of $f$.
Thus $f \circ g: S \to S$, so $f \circ g \in S^S$.
Since this holds for all $f, g \in S^S$, $\struct {S^S, \circ}$ is closed.
By Composition of Mappings is Associative, $\circ$ is associative.
Since $\struct {S^S, \circ}$ is closed and $\circ$ is associative:
- $\struct {S^S, \circ}$ is a semigroup.
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $4$. Groups: Exercise $7$
- 1968: Ian D. Macdonald: The Theory of Groups ... (previous) ... (next): $\S 1$: Some examples of groups: Theorem $1.12$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \alpha$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 29$. Semigroups: definition and examples: $(2)$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Remark