1

Let $(S, \circ)$ be a semigroup.

Let $a \in S$.

It's a straightforward exercise to show that the left and right regular representations $\lambda_a$ and $\rho_a$ with respect to $a$ are permutations if $a$ is invertible.

However, it is not so obvious to prove the other way around.

That is:

"Suppose $\lambda_a$ and $\rho_a$ are permutations.

Then $(S, \circ)$ has an identity element and $a$ has an inverse."

So far I have that:

$\lambda_a$ and $\rho_a$ are injective iff $a$ is cancellable

As $\lambda_a$ is a permutation (and thus a bijection) there exists a mapping $\phi_\lambda$ such that $\forall x \in S: \phi_\lambda (\lambda_a (x)) = x$ and so $\phi_\lambda (a \circ x) = x$

and a similar mapping $\phi_\rho$ for $\rho_a$.

But I don't see how to get from $\forall x \in S: \phi_\lambda (a \circ x) = x$ and $\forall x \in S: \phi_\rho (x \circ a) = x$ to the fact that $\phi_\lambda$ and $\phi_\rho$ are themselves necessarily left and right regular representations of some element $a^{-1} \in S$.

Probably something conceptually simple, but I'm not seeing it.

This question appears in Seth Warner's 1965 "Modern Algebra", problem $7.13$.

Prime Mover
  • 5,005

1 Answers1

1

Since right-multiplying by $a$ is a permutation, there is some $g\in S$ such that $g\circ a=a$. For all $b\in S$, we have $(b\circ g)\circ a=b\circ(g\circ a)=b\circ a$. Since right-multiplying by $a$ is a permutation, we can cancel $a$ on the right, leaving $b\circ g=b$. Thus $g$ is a right identity. Analogously, $g$ is a left identity, and thus an identity. Since right-multiplying by $a$ is a permutation, there is $h$ such that $h\circ a=g$. Thus $a$ is left-invertible. Analogously, $a$ is right-invertible, and thus invertible.

joriki
  • 238,052