Product of Semigroup Element with Right Inverse is Idempotent

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Theorem

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

Let $x \in S$ such that $\exists x_R: x \circ x_R = e_R$, i.e. $x$ has a right inverse with respect to the right identity.


Then:

$\paren {x_R \circ x} \circ \paren {x_R \circ x} = x_R \circ x$

That is, $x_R \circ x$ is idempotent.


Proof

\(\displaystyle \paren {x_R \circ x} \circ \paren {x_R \circ x}\) \(=\) \(\displaystyle x_R \circ \paren {x \circ x_R} \circ x\) as $\circ$ is associative
\(\displaystyle \) \(=\) \(\displaystyle x_R \circ e_R \circ x\) Definition of Right Inverse Element
\(\displaystyle \) \(=\) \(\displaystyle x_R \circ x\) Definition of Right Identity

$\blacksquare$


Examples

$2 x$ and $\dfrac x 2$ Mappings on Integers

Let $\struct {\Z^\Z, \circ}$ be the semigroup defined such that:

$\Z$ is the set of all mappings on the integers.
$\circ$ denotes composition of mappings.


Let $\rho, \sigma \in \Z^\Z$ such that:

$\forall x \in \Z: \map \rho x = \begin{cases} \dfrac x 2 & : x \text { even} \\ 0 & : x \text { odd} \end{cases}$
$\forall x \in \Z: \map \sigma x = 2 x$


Then:

$\rho$ is a right inverse for $\sigma$

but:

$\rho$ is not a left inverse for $\sigma$


As a result:

$\paren {\sigma \circ \rho}^2 = \sigma \circ \rho$


Also see


Sources