Right Inverse for All is Left Inverse

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$ such that:

$\forall x \in S: \exists x_R: x \circ x_R = e_R$

That is, every element of $S$ has a right inverse with respect to the right identity.


Then $x_R \circ x = e_R$, that is, $x_R$ is also a left inverse with respect to the right identity.


Proof

Let $y = x_R \circ x$. Then:

\(\displaystyle y \circ e_R\) \(=\) \(\displaystyle y \circ \paren {y \circ y_R}\) Definition of Right Inverse Element
\(\displaystyle \) \(=\) \(\displaystyle \paren {y \circ y} \circ y_R\) as $\circ$ is associative
\(\displaystyle \) \(=\) \(\displaystyle y \circ y_R\) Product of Semigroup Element with Right Inverse is Idempotent: $y = x_R \circ x$
\(\displaystyle \) \(=\) \(\displaystyle e_R\) Definition of Right Inverse Element


So $x_R \circ x = e_R$, and $x_R$ behaves as a left inverse as well as a right inverse with respect to the right identity.

$\blacksquare$


Also see


Sources