Structure under Left Operation is Semigroup
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Theorem
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
Then $\struct {S, \gets}$ is a semigroup.
Proof
We need to verify the semigroup axioms:
A semigroup is an algebraic structure $\struct {S, \circ}$ which satisfies the following properties:
\((\text S 0)\) | $:$ | Closure | \(\ds \forall a, b \in S:\) | \(\ds a \circ b \in S \) | |||||
\((\text S 1)\) | $:$ | Associativity | \(\ds \forall a, b, c \in S:\) | \(\ds a \circ \paren {b \circ c} = \paren {a \circ b} \circ c \) |
By the nature of the right operation, $\struct {S, \to}$ is closed:
- $\forall x, y \in S: x \gets y = y \in S$
whatever $S$ may be.
Hence Semigroup Axiom $\text S 0$: Closure holds.
From Right Operation is Associative, $\gets$ is associative.
Hence Semigroup Axiom $\text S 1$: Associativity holds.
So $\struct {S, \gets}$ is a semigroup.
$\blacksquare$