Endomorphisms on an entropic structure whose pointwise product is the identity automorphism - entropic idempotent structure?
Context: self-study from Warner's "Modern Algebra (1965): Exercise 16.27.
Let $\alpha$ and $\beta$ be endomorphisms of an entropic structure $(S, \odot)$ such that $\alpha \odot \beta$ is the identity automorphism, and let $\otimes$ be the operation on $S$ defined by: $$x \otimes y = \alpha (x) \odot \beta (y)$$ for all $x, y \in S$. Then $(S, \otimes)$ is an entropic idempotent algebraic structure.
An entropic structure is an algebraic structure $(S, \circ)$ such that $\circ$ is an entropic operation, that is:
$$\forall a, b, c, d \in S: (a \circ b) \circ (c \circ d) = (a \circ c) \circ (b \circ d)$$
Idempotence is trivial, but I'm having trouble proving entropicness.
Here is my attempt:
$(w \otimes x) \otimes (y \otimes z)$
$= (\alpha (w \otimes x) ) \odot (\beta (y \otimes z) )$
$= (\alpha (\alpha (w) \odot \beta (x))) \odot (\beta (\alpha (y) \odot \beta (z)))$
$= (\alpha (\alpha (w) ) \odot \alpha (\beta (x))) \odot (\beta (\alpha (y)) \odot \beta (\beta (z)))$ (justified by homomorphism)
$= (\alpha (\alpha (w) ) \odot \beta (\alpha (y))) \odot (\alpha (\beta (x)) \odot \beta (\beta (z)))$ (as $\odot$ is entropic)
$= (\alpha (w) \otimes \alpha (y)) \odot (\beta (x) \otimes \beta (z))$
If we can say that $\alpha (w) \otimes \alpha (y) = \alpha (w \otimes y)$ then we are home and dry, but I can't see why we can make that leap, as it is far from obvious that $\alpha$ and $\beta$ are endomorphisms on $(S, \otimes)$ in the same way that they are on $(S, \odot)$. Or are they? And why?
What is it I'm missing here?
1 answer
$(S,\otimes)$ is not entropic. The following is an example, that is, a counterexample, to the claim of the exercise.
$(\mathbb Z_3,\odot)$ where $\odot$ is defined as $x\odot y=x+2y$, is entropic: $$\begin{align} (x\odot y)\odot(z\odot w) & = (x+2y)+2(z+2w) \\ & = x+2y+2z+4w \\ & = (x+2z)+2(y+2w) \\ & = (x\odot z)\odot(y\odot w). \end{align}$$
Define $\alpha(x):=x^2$ and $\beta(x):=2(x-x^2)$. Then $$\begin{align} (\alpha\odot\beta)(x) & = \alpha(x)+2\beta(x) \\ & = x^2+4(x-x^2) \\ & = x^2+x-x^2 \\ & = x. \end{align}$$
Construe $\otimes$ as $x\otimes y:=\alpha(x)\odot\beta(x)$. Given $x,y,z,w\in \mathbb Z_3$, $$\begin{align} & (x\otimes y)\otimes(z\otimes w) \\ & = \alpha(\alpha(x)+2\beta (y))+2\beta(\alpha(z)+2\beta(w)) \\ & = 2w^4+2w^3+2w^2z^2+w^2+wz^2+w+x^4+x^2y^2+2x^2y+y^4+y^3+y^2+2z^4+z^2 \\ & = 2w^2+2w^2+2w^2z^2+w^2+wz^2+w+x^2+x^2y^2+2x^2y+y^2+y^2+y^2+2z^2+z^2 \\ & = 2w^2+2w^2z^2+wz^2+w+x^2+x^2y^2+2x^2. \end{align}$$
In conclusion, $(\mathbb Z_3,\otimes)$ is not entropic. For example, $$(1\otimes2)\otimes(0\otimes0)=0 \neq 1=(1\otimes0)\otimes(2\otimes0).$$
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