Left-associative elements of loop form subgroup: Difference between revisions

Latest revision as of 20:17, 20 June 2012

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

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 $(L, *)$ 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.

@@ Line 5: / Line 5: @@
 ==Related facts==
-* [[Right-associative elements of algebra loop form subgroup]]
+* [[Right-associative elements of loop form subgroup]]
 ==Facts used==
-# [[uses::Left-associative elements of magma form submagma]]
+# [[uses::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]].
+# [[uses::Monoid where every element is right-invertible equals group]], which in turn uses [[uses::equality of left and right inverses in monoid]]
 ==Proof==
-'''Given''': An algebra loop <math>(L,*)</math> with identity element <math>e</math>. <math>S</math> is the set of left-associative elements of <math>L</math>.
+'''Given''': A loop <math>(L,*)</math> with identity element <math>e</math>. <math>S</math> is the set of left-associative elements of <math>L</math>.
-'''To prove''': <math>S</math> is a subgroup of <math>L</math>.
+'''To prove''': <math>S</math> is a subgroup of <math>L</math>. More explicitly, <math>(S,*)</math> is a [[group]] with identity element <math>e</math>.
-'''Proof''': By fact (1), <math>S</math> is closed under <math>*</math>. Also, <math>e</math> is clearly in <math>S</math>. Since all elements of <math>S</math> are left-associative, <math>S</math> itself satisfies associativity, so <math>S</math> is a submonoid of <math>L</math>.
+'''Proof''':
-We now show that if <math>a \in S</math>, and <math>b</math> is the right inverse of <math>a</math> in <math>L</math>, then <math>b \in S</math>. For any <math>c,d \in S</math>, we have:
+{| class="sortable" border="1"
+! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
-<math>a * (b * (c * d)) = (a * b) * (c * d) = e * (c * d) = c * d</math>.
+|-
+| 1 || <math>(S,*)</math> is a monoid with identity element <math>e</math>. || Fact (1) || <math>(L,*)</math> has identity element <math>e</math> || || Follows directly from Fact (1).
-Similarly, we have:
+|-
+| 2 || For any <math>a \in S</math>, there is a right inverse of <math>a</math> in <math>L</math>, i.e., an element <math>b \in L</math> such that <math>a * b = e</math>. || || <math>L</math> is a loop || ||
-<math>a * ((b * c) * d) = (a * (b * c)) * d = ((a * b) * c) * d) = (e * c) * d = c * d</math>.
+|-
+| 3 || For any <math>a \in S</math>, the right inverse <math>b</math> constructed in Step (2) is in <math>S</math>. In other words, for any <math>c,d \in S</math>, we have <math>(b * c) *d  = b * (c * d)</math>. || || <math>L</math> is a loop, so any equation <math>a * x = z</math> has a unique solution for <math>x</math>. || Step (2) || Using left-associativity of <math>a</math>, we get:<math>a * ((b * c) * d) = (a * (b * c)) * d = ((a * b) * c) * d) = (e * c) * d = c * d</math><br>Similarly, <math>a * (b * (c * d)) = (a * b) * (c * d) = e * (c * d) = c * d</math><br>Thus, we get: <math>a * (b * (c * d)) = a * ((b * c) * d) = c * d</math>.<br>Since the equation <math>a * x = c * d</math> has a unique solution, we get that <math>b * (c * d) = (b * c) * d</math>.
-Thus, we get:
+|-
+| 4 || <math>(S,*)</math> is a monoid with identity element <math>e</math> in which every element has a right inverse. || || || Steps (1), (3) || Step-combination direct
-<math>a * (b * (c * d)) = a * ((b * c) * d) = c * d</math>.
+|-
+| 5 || <math>(S,*)</math> is a group with identity element <math>e</math>, completing the proof. || Fact (2) || || Step (4) || Step-fact combination direct.
-Since the equation <math>a * x = c * d</math> has a unique solution, we get that:
+|}
-<math>b * (c * d) = (b * c) * d</math>.
-Thus, <math>b</math> is left-associative.
-Thus, the right inverse of every element in <math>S</math> is in <math>S</math>. Thus, <math>S</math> is a monoid in which every element has a right inverse. We now want to show that every element has a two-sided inverse.
-Suppose <math>a \in S</math> with right inverse <math>b</math>. <math>b</math> has right inverse <math>a'</math>. Then, by associativity in <math>S</math>, <math>a = a * e = a * (b * a') = (a * b) * a' = e * a' = a'</math>. Thus, <math>b</math> and <math>a</math> are two-sided inverses of each other, completing the proof.