Equivalence of definitions of inverse property loop

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

(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.