Product is Left Identity Therefore Left Cancellable
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Theorem
Let $\struct {S, \circ}$ be a semigroup.
Let $e_L \in S$ be a left identity of $S$.
Let $a \in S$ such that:
- $\exists b \in S: b \circ a = e_L$
Then $a$ is left cancellable in $\struct {S, \circ}$.
Proof
Let $x, y \in S$ be arbitrary.
Then:
\(\ds a \circ x\) | \(=\) | \(\ds a \circ y\) | Semigroup Axiom $\text S 0$: Closure | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b \circ \paren {a \circ x}\) | \(=\) | \(\ds b \circ \paren {a \circ y}\) | Semigroup Axiom $\text S 0$: Closure | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {b \circ a} \circ x\) | \(=\) | \(\ds \paren {b \circ a} \circ y\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e_L \circ x\) | \(=\) | \(\ds e_L \circ y\) | by hypothesis | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) | Definition of Left Identity |
The result follows by definition of left cancellable.
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.11$