# 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 $L$.

## 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.