Left-associative elements of loop form subgroup

Statement

Suppose $(L,*)$ is an algebra 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 algebra loop form subgroup

Facts used

1. Left-associative elements of magma form submagma

Proof

Given: An algebra loop $(L,*)$ with identity element $e$. $S$ is the set of left-associative elements of $L$.

To prove: $S$ is a subgroup of $L$.

Proof: By fact (1), $S$ is closed under $*$. Also, $e$ is clearly in $S$. Since all elements of $S$ are left-associative, $S$ itself satisfies associativity, so $S$ is a submonoid of $L$.

We now show that if $a\in S$, and $b$ is the right inverse of $a$ in $L$, then $b\in S$. For any $c,d\in S$, we have:

$a*(b*(c*d))=(a*b)*(c*d)=e*(c*d)=c*d$.

Similarly, we have:

$a*((b*c)*d)=(a*(b*c))*d=((a*b)*c)*d)=(e*c)*d=c*d$.

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:

$b*(c*d)=(b*c)*d$.

Thus, $b$ is left-associative.

Thus, the right inverse of every element in $S$ is in $S$. Thus, $S$ 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 $a\in S$ with right inverse $b$. $b$ has right inverse $a'$. Then, by associativity in $S$, $a=a*e=a*(b*a')=(a*b)*a'=e*a'=a'$. Thus, $b$ and $a$ are two-sided inverses of each other, completing the proof.