If $X$ is a self-distributive algebra, then define $x^{[n]}$ for all $n\geq 1$ by letting $x^{[1]}=x$ and $x^{[n+1]}=x*x^{[n]}$. The motivation for this question comes from the following fact about self-distributivity on one generator.

Theorem: Suppose that $X$ is a self-distributive algebra generated by one element. Then for all $x,y\in X$, there are $m,n$ with $x^{[m]}=y^{[n]}$.

Define the Fibonacci terms by letting $t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$.

An algebra $(X,*,1)$ is said to be a reduced permutative algebra if $x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$ for all $x,y,z\in X$ and where for all $x,y\in X$, there is some $n$ where $t_{n}(x,y)=1$. We say that a reduced permutative algebra $(X,*,1)$ is critically simple if whenever $\simeq$ is a congruence, either $\simeq$ is the trivial congruence or $x\simeq 1$ for some $x\neq 1$.

Suppose that $(X,*,1)$ is a finite reduced permutative algebra and $x,y\in X$ and suppose that $p\geq 0$. Then does there a reduced permutative algebra $(Y,*,1)$, natural numbers $m,n$ and a homomorphism $\phi:Y\rightarrow X$ and $x',y'\in Y$ where $\phi(y')=y,\phi(x')=x$ and $x'^{[m]}=y'^{[n]}$ and where $x'^{[m+p]}\neq 1$? What if $(X,*,1)$ is critically simple?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.