Left-associative elements of loop form subgroup
Suppose is a loop. Then, the left nucleus of , i.e., the set of left-associative elements of , is nonempty and forms a subgroup of . This subgroup is sometimes termed the left kernel of or the left-associative center of .
- 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.
- Monoid where every element is right-invertible equals group, which in turn uses equality of left and right inverses in monoid
Given: A loop with identity element . is the set of left-associative elements of .
To prove: is a subgroup of . More explicitly, is a group with identity element .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||is a monoid with identity element .||Fact (1)||has identity element||Follows directly from Fact (1).|
|2||For any , there is a right inverse of in , i.e., an element such that .||is a loop|
|3||For any , the right inverse constructed in Step (2) is in . In other words, for any , we have .||is a loop, so any equation has a unique solution for .||Step (2)|| Using left-associativity of , we get:|
Thus, we get: .
Since the equation has a unique solution, we get that .
|4||is a monoid with identity element in which every element has a right inverse.||Steps (1), (3)||Step-combination direct|
|5||is a group with identity element , completing the proof.||Fact (2)||Step (4)||Step-fact combination direct.|