Let V be the variety of algebras (A,\ \cdot) satisfying



Answered question


Let V be the variety of algebras (A, ) satisfying the identitie
xxx and (xy)z(zy)x
Let W be the subvariety of V defined by the additional identity y(xy)
Determine FW(x, y). Write out a Cayley table.

Answer & Explanation



Beginner2021-12-16Added 30 answers

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 bf{F}W(x, y) is
{x, y, (xy), (yx)}.
More precisely, Bergmans
Shannon Hodgkinson

Shannon Hodgkinson

Beginner2021-12-17Added 34 answers

Step 1
To save typing, I'll write xy and xy and 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) (xy)(zw)=(xz)(yw)
2) x(yz)=(xy)(xz)
3) (yz)x=(yx)(zx)
4) y(xy)=(yx)y
5) (yx)x=xy
For completeness, I'll give proofs of these five identities inside the spoiler blocks.
1) (xy)(zw)=((zw)y)x=((yw)z)x=(xz)(yw)
2) x(yz)=(×)(yz)=(xy)(xz)
3) (yz)x=(yz)(×)=(yx)(zx)
4) y(xy)=(yx)(yy)=(yx)y
5) (yx)x=(×)y=xy
Ok, now we add the additional defining identity of W:6. y(xy)=x. Note that in conjunction with identity 4 above, we also have 7. (yx)y=x
Our task is to understand FW(a, b), 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 FW(a, b) 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 (n+1) are the generators and the products of two terms at Level (n).
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 =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:
I've hidden the derivations in the spoiler blocks below.
a(ab)=a((aa)b)=a((ba)a)=ba by 6. Similarly, b(ba)=ab.
b(ab)=a by 6. Similarly, a(ba)=b.
(ab)a=b by 7. Similarly,

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