Equivalence of definitions of inverse property loop

This article gives a proof/explanation of the equivalence of multiple definitions for the term inverse property loop
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a loop $(L,*)$.

1. Existence of left and right inverses: There exist bijective maps $\lambda ,\rho :L\to L$ such that $\lambda (a)*(a*b)=(b*a)*\rho (a)=b\ \forall \ a,b\in L$.
2. Existence of two-sided inverses: There exists a bijective map ${}^{-1}:L\to L$ such that $a^{-1}*(a*b)=(b*a)*a^{-1}=b$ for all $a,b\in L$.

Related facts

• Equality of left and right inverses in monoid

Proof

(1) implies (2)

It suffices to show that $\lambda =\rho$; then we can take the inverse map to equal that.

Given: A loop $(L,*)$ with a map $\lambda$ bijective maps $\lambda ,\rho :L\to L$ such that $\lambda (a)*(a*b)=(b*a)*\rho (a)=b\ \forall \ a,b\in L$.

To prove: For any $x\in L$, $\lambda (x)=\rho (x)$.

Proof: Consider the product $(\lambda (x)*x)*\rho (x)$.

Putting $a=b=x$ and using the property of $\rho$, this simplifies to $\lambda (x)$. On the other hand, we know that $\lambda (x)*x=\lambda (x)*(x*e)=e$ where $e$ is the identity element, so the expression $(\lambda (x)*x)*\rho (x)$ simplifies to $\rho (x)$. Thus, $\lambda (x)=\rho (x)$.

(2) implies (1)

We can take both $\lambda$ and $\rho$ as the inverse map.