# Left-associative elements of loop form subgroup

From Groupprops

## Contents

## Statement

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 .

## Related facts

## Facts used

- 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

## Proof

**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 .

**Proof**:

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: Similarly, 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. |