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?

posted over 2 years ago

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2y ago

Prime Mover‭

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