Left Inverse for All is Right Inverse
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Theorem
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that:
- $\forall x \in S: \exists x_L: x_L \circ x = e_L$
That is, every element of $S$ has a left inverse with respect to the left identity.
Then $x \circ x_L = e_L$, that is, $x_L$ is also a right inverse with respect to the left identity.
Proof
Let $y = x \circ x_L$. Then:
\(\ds e_L \circ y\) | \(=\) | \(\ds \paren {y_L \circ y} \circ y\) | Definition of Left Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds y_L \circ \paren {y \circ y}\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds y_L \circ y\) | Product of Semigroup Element with Left Inverse is Idempotent: $y = x \circ x_L$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e_L\) | Definition of Left Inverse Element |
So $x \circ x_L = e_L$, and $x_L$ behaves as a right inverse as well as a left inverse with respect to the left identity.
$\blacksquare$
Also see
- Left Identity while exists Left Inverse for All is Identity
- Right Identity while exists Right Inverse for All is Identity
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.4$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Semigroups: Exercise $5$