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Q&A

Endomorphisms on an entropic structure whose pointwise product is the identity automorphism - entropic idempotent structure?

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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?

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2 comment threads

I don't know and right now can't see the answer, but the obvious point (which has probably already oc... (1 comment)
Could you please add the definition of an entropic structure? (2 comments)

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