How to create multiplication table ("Cayley table") for an algebra or class of algebras?
You should do Part (a) of the problem before you try to do Part (b).
In Part (a), Bergman asks you to derive a number of identities from the defining identities for $\mathcal W$, and these identities help to show that the universe of $\mathbf F_{\mathcal W}(x,y)$ is $\{x, y, (x\cdot y), (y\cdot x)\}$. More precisely, Bergman's identities help to show that every word in $x$ and $y$ reduces to one of $x, y, (x\cdot y)$, or $(y\cdot x)$. To show that no further reductions are possible, it suffices to observe that if $\mathbb F_4=\mathbb F_2[\alpha]$ for $\alpha$ satisfying $x^2+x+1=0$, then the operation $x\cdot y:=\alpha x+(1-\alpha)y$ acting on the set $\mathbb F_4$ satisfies all the defining identities of $\mathcal W$, and the four words $x, y, x\cdot y$, and $y\cdot x$ have distinct interpretations on $\mathbb F_4$.
To show that every word in $x$ and $y$ reduces to some word in the set $\{x, y, (x\cdot y), (y\cdot x)\}$ one must show that the product of any two words in this set reduces to a word in this set.
For example, the product of $x$ and $(x\cdot y)$ (I mean the product $x\cdot (x\cdot y)$) reduces to $(y\cdot x)$. To see this, substitute $Y=x, X=(y\cdot x)$ in the defining identity $Y\cdot (X\cdot Y) = X$ to obtain $$\tag{E} x\cdot ((\underline{y}\cdot x)\cdot \underline{x})=(y\cdot x).$$ Now swap the underlined characters using the defining identity $(\underline{X}\cdot Y)\cdot \underline{Z} = (\underline{Z}\cdot Y)\cdot \underline{X}$ to obtain from (E) that $x\cdot ((x\cdot x)\cdot y)=(y\cdot x)$ holds. Now apply the defining identity $X\cdot X=X$ to reduce this to $x\cdot (x\cdot y)=(y\cdot x)$. This verifies the first sentence of this paragraph.
To save typing, I'll write $x\cdot y$ and $xy$ and $\approx$ as $=$.
You left out some important information here: Bergman gives in part (a) a list of identities which follow from the defining identities of $\mathcal{V}$. These identities are very helpful in solving part (b)!
- $(x y) (z w)= (x z) (y w)$
- $x (y z) = (x y) (x z)$
- $(y z) x= (y x) (z x)$
- $y (x y)= (y x) y$
- $(y x) x = x y$
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. $x(yz) = (xx)(yz) = (xy)(xz)$
3. $(yz)x = (yz)(xx) = (yx)(zx)$
4. $y(xy) = (yx)(yy) = (yx)y$
5. $(yx)x = (xx)y = xy$
Ok, now we add the additional defining identity of $\mathcal{W}$: 6. $y(xy) = x$. Note that in conjunction with identity 4 above, we also have 7. $(yx)y = x$.
Our task is to understand $\mathbf{F}_{\mathcal{W}}(a,b)$, the free algebra in $\mathcal{W}$ 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 $\mathbf{F}_{\mathcal{W}}(a,b)$ 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 $(\leq n)$.
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 $bb = b$.
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}{c|cccc} & a & b & ab & ba\\ \hline 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.
$a(ab) = a((aa)b) = a((ba)a) = ba$ by 6. Similarly, $b(ba) = ab$.
$b(ab) = a$ by 6. Similarly, $a(ba) = b$.
$(ab)a = b$ by 7. Similarly, $(ba)b = a$.
$(ab)b = ba$ by 5. Similarly, $(ba)a = ab$.
$(ab)(ab) = ab$. Similarly, $(ba)(ba) = ba$.
$(ab)(ba) = ((ab)b)((ab)a) = (ba)b = a$ by 2, 5, and 7. Similarly, $(ba)(ab) = b$.
It follows that if the terms $a$, $b$, $ab$, and $ba$ are distinct in the free algebra $\mathbf{F}_{\mathcal{W}}(a,b)$, then the table above is its Cayley table. To prove that the terms are distinct, it suffices to show that there is any algebra in $\mathcal{W}$ with generators $a$ and $b$ such that these four terms are distinct. One way to do this is to check that the Cayley table above actually describes an algebra in $\mathcal{W}$, i.e., that all three defining identities are satisfied. This is mechanical, but tedious.