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:

$$\mathrm{\forall }a,b,c,d\in S:(a\circ b)\circ (c\circ d)=(a\circ c)\circ (b\circ d)$$

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 almost 3 years ago

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3y ago

Prime Mover‭

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