Left-associative elements of loop form subgroup

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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

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


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.


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.