Left-associative elements of loop form subgroup

Statement

Suppose $(L,*)$ is a loop. Then, the left nucleus of $L$, i.e., the set of left-associative elements of $L$, is nonempty and forms a subgroup of $L$. This subgroup is sometimes termed the left kernel of $L$ or the left-associative center of $L$.

Related facts

• Right-associative elements of loop form subgroup

Facts used

1. Left-associative elements of magma form submagma: Further, this submagma is a subsemigroup, and if the whole magma has a neutral element, it has the same neutral element and becomes a monoid.
2. Monoid where every element is right-invertible equals group, which in turn uses equality of left and right inverses in monoid

Proof

Given: A loop $(L,*)$ with identity element $e$. $S$ is the set of left-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 right inverse of $a$ in $L$, i.e., an element $b\in L$ such that $a*b=e$.		$L$ is a loop
3	For any $a\in S$, the right inverse $b$ constructed in Step (2) is in $S$. In other words, for any $c,d\in S$, we have $(b*c)*d=b*(c*d)$.		$L$ is a loop, so any equation $a*x=z$ has a unique solution for $x$.	Step (2)	Using left-associativity of $a$, we get:$a*((b*c)*d)=(a*(b*c))*d=((a*b)*c)*d)=(e*c)*d=c*d$ Similarly, $a*(b*(c*d))=(a*b)*(c*d)=e*(c*d)=c*d$ Thus, we get: $a*(b*(c*d))=a*((b*c)*d)=c*d$. Since the equation $a*x=c*d$ has a unique solution, we get that $b*(c*d)=(b*c)*d$.
4	$(S,*)$ is a monoid with identity element $e$ in which every element has a right 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.