Product of Semigroup Element with Left Inverse is Idempotent
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Theorem
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$.
Let $x \in S$ such that $\exists x_L: x_L \circ x = e_L$, that is $x$ has a left inverse with respect to the left identity.
Then:
- $\paren {x \circ x_L} \circ \paren {x \circ x_L} = x \circ x_L$
That is, $x \circ x_L$ is idempotent.
Proof
\(\ds \paren {x \circ x_L} \circ \paren {x \circ x_L}\) | \(=\) | \(\ds x \circ \paren {x_L \circ x} \circ x_L\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ e_L \circ x_L\) | Definition of Left Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ x_L\) | Definition of Left Identity |
$\blacksquare$
Also see
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Semigroups: Exercise $5$