Right-associative elements of loop form subgroup

Statement

Suppose $(L,*)$ is an algebra loop. Then, the set of right-associative elements of $L$ is nonempty and forms a subgroup of $L$. This subgroup is termed the right kernel of $L$ or the right-associative center of $L$.

Definitions used

Right-associative element

An element $c\in L$ is termed right-associative if, for all $a,b\in L$, we have:

$(a*b)*c=a*(b*c)$.

Related facts

• Left-associative elements of loop form subgroup: Note that the proofs are identical and the facts can be deduced from each other using the opposite magma construction.

Facts used

1. Right-associative elements of magma form submagma
2. Monoid where every element is left-invertible equals group

Proof

Given: A loop $(L,*)$ with identity element $e$. $S$ is the set of right-associative elements of $L$.

To prove: $S$ is a subgroup of $L$. More explicitly, $(S,*)$ is a group with identity element $e$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$(S,*)$ is a monoid with identity element $e$.	Fact (1)	$(L,*)$ has identity element $e$		Follows directly from Fact (1).
2	For any $a\in S$, there is a left inverse of $a$ in $L$, i.e., an element $b\in L$ such that $b*a=e$.		$L$ is a loop
3	For any $a\in S$, the left inverse $b$ constructed in Step (2) is in $S$. In other words, for any $c,d\in S$, we have $(c*d)*b=c*(d*b)$.		$L$ is a loop, so any equation $x*a=z$ has a unique solution for $x$.	Step (2)	Using right-associativity of $a$, we get: $((c*d)*b)*a=(c*d)*(b*a)=(c*d)*e=c*d$. Similarly, $(c*(d*b))*a=c*((d*b)*a)=c*(d*(b*a))=c*(d*e)=c*d$. Thus, $((c*d)*b)*a=(c*(d*b))*a=c*d$. Thus, both $(c*d)*b$ and $c*(d*b)$ are solutions for $x$ to $x*a=c*d$, hence they are equal.
4	$(S,*)$ is a monoid with identity element $e$ in which every element has a left inverse.			Steps (1), (3)	Step-combination direct
5	$(S,*)$ is a group with identity element $e$, completing the proof.	Fact (2)		Step (4)	Step-fact combination direct.