Regular Representations wrt Element are Permutations then Element is Invertible
Theorem
Let $\struct {S, \circ}$ be a semigroup.
Let $\lambda_a: S \to S$ and $\rho_a: S \to S$ be the left regular representation and right regular representation with respect to $a$ respectively:
\(\ds \forall x \in S: \, \) | \(\ds \map {\lambda_a} x\) | \(=\) | \(\ds a \circ x\) | |||||||||||
\(\ds \forall x \in S: \, \) | \(\ds \map {\rho_a} x\) | \(=\) | \(\ds x \circ a\) |
Let both $\lambda_a$ and $\rho_a$ be permutations on $S$.
Then there exists an identity element for $\circ$ and $a$ is invertible.
Proof
We have that $\rho_a$ is a permutation on $S$.
Hence:
\(\ds \exists g \in S: \, \) | \(\ds a\) | \(=\) | \(\ds \map {\rho_a} g\) | |||||||||||
\(\ds \) | \(=\) | \(\ds g \circ a\) | Definition of Right Regular Representation |
Then we have:
\(\ds \forall b \in S: \, \) | \(\ds \paren {b \circ g} \circ a\) | \(=\) | \(\ds b \circ \paren {g \circ a}\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||||
\(\ds \) | \(=\) | \(\ds b \circ a\) | from above | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b \circ g\) | \(=\) | \(\ds b\) | Right Cancellable iff Right Regular Representation Injective |
which demonstrates that $g$ is a right identity for $\circ$.
In the same way, we have that $\lambda_a$ is also a permutation on $S$.
Hence:
\(\ds \exists g \in S: \, \) | \(\ds a\) | \(=\) | \(\ds \map {\lambda_a} g\) | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ g\) | Definition of Left Regular Representation |
Then we have:
\(\ds \forall b \in S: \, \) | \(\ds \paren {a \circ g} \circ b\) | \(=\) | \(\ds a \circ \paren {g \circ b}\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||||
\(\ds \) | \(=\) | \(\ds a \circ b\) | from above | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds g \circ b\) | \(=\) | \(\ds b\) | Left Cancellable iff Left Regular Representation Injective |
which demonstrates that $g$ is a left identity for $\circ$.
So, by definition, $g$ is an identity element for $\circ$.
Again, we have that $\rho_a$ is a permutation on $S$, and so:
\(\ds \exists h \in S: \, \) | \(\ds g\) | \(=\) | \(\ds \map {\rho_a} h\) | |||||||||||
\(\ds \) | \(=\) | \(\ds h \circ a\) | Definition of Right Regular Representation |
and that $\lambda_a$ is also a permutation on $S$, and so:
\(\ds \exists h \in S: \, \) | \(\ds g\) | \(=\) | \(\ds \map {\lambda_a} h\) | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ h\) | Definition of Right Regular Representation |
So $h$ is both a left inverse and a right inverse for $a$.
Hence by definition $h$ is an inverse for $a$.
Hence $a$ is invertible by definition.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.13$
- joriki (https://math.stackexchange.com/users/6622/joriki), Left and Right Regular Representations are Permutations therefore Identity and Inverses Exist, URL (version: 2022-03-01): https://math.stackexchange.com/q/4393494