zakinutuzi

2021-12-15

Let $V$ be the variety of algebras satisfying the identitie
$x\cdot x\approx x$ and $\left(x\cdot y\right)\cdot z\approx \left(z\cdot y\right)\cdot x$
Let $W$ be the subvariety of V defined by the additional identity $y\cdot \left(x\cdot y\right)\approx$
Determine . Write out a Cayley table.

Neunassauk8

Step 1
Bergman asks to derive a number of identities from the defining identities for $W$, and these identities help to show that the universe of is
.
More precisely, Bergmans

Shannon Hodgkinson

Step 1
To save typing, I'll write $x\cdot y$ and $xy$ and $\approx$ as $=$.
Bergman gives in part (a) a list of identities which follow from the defining identities of $V$ These identities are very helpful in solving part (b)!
1) $\left(xy\right)\left(zw\right)=\left(xz\right)\left(yw\right)$
2) $x\left(yz\right)=\left(xy\right)\left(xz\right)$
3) $\left(yz\right)x=\left(yx\right)\left(zx\right)$
4) $y\left(xy\right)=\left(yx\right)y$
5) $\left(yx\right)x=xy$
For completeness, I'll give proofs of these five identities inside the spoiler blocks.
1) $\left(xy\right)\left(zw\right)=\left(\left(zw\right)y\right)x=\left(\left(yw\right)z\right)x=\left(xz\right)\left(yw\right)$
2) $x\left(yz\right)=\left(×\right)\left(yz\right)=\left(xy\right)\left(xz\right)$
3) $\left(yz\right)x=\left(yz\right)\left(×\right)=\left(yx\right)\left(zx\right)$
4) $y\left(xy\right)=\left(yx\right)\left(yy\right)=\left(yx\right)y$
5) $\left(yx\right)x=\left(×\right)y=xy$
Ok, now we add the additional defining identity of . Note that in conjunction with identity 4 above, we also have 7. $\left(yx\right)y=x$
Our task is to understand , the free algebra in $W$ on two generators a and b. I'm using a and b for the generators so as not to be confused with the variables x and y used in the identities above. Let's try to find all its elements.
Recall that every element of the free algebra is an equivalence class of terms in the generators a and b. The terms can be built up in levels, where the terms at Level 0 are the generators and the terms at Level $\left(n+1\right)$ are the generators and the products of two terms at Level $\left(\le n\right)$.
Level 0: a, b
Level 1: a, b, aa, ab, ba, and bb.
Note that we can eliminate aa and bb since they are redundant: $aa=a$ and $\mathbf{=}b$
Now it turns out that any term at Level 2 (any product of a, b, ab, and ba) can be shown to be equivalent to a term at Level 1, using the identities above. They multiply according to the following Cayley table:
$\begin{array}{ccccc}& a& b& ab& ba\\ a& a& ab& ba& b\\ b& ba& b& a& ab\\ ab& b& ba& ab& a\\ ba& ab& a& b& ba\end{array}$
I've hidden the derivations in the spoiler blocks below.
$a\left(ab\right)=a\left(\left(aa\right)b\right)=a\left(\left(ba\right)a\right)=ba$ by 6. Similarly, $b\left(ba\right)=ab$.
$b\left(ab\right)=a$ by 6. Similarly, $a\left(ba\right)=b$.
$\left(ab\right)a=b$ by 7. Similarly,

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