# Defining Multiplication on Strongly Homogeneous Commutative Semigroups

Let us call any commutative semigroup $$(S, +)$$ strongly homogeneous if it satisfies the following three properties:

P-1. Every endomorphism on $$S$$ is a bijection.

P-2. Any two endomorphisms on $$S$$ commute.

P-3. For every $$x, y \in S$$ there exist one and only one endomorphism $$\psi$$ on $$S$$ such that $$\psi(x) = y$$.

Is the following true?

Corresponding to any element $$1 \in S$$ there exist one a̶n̶d̶ ̶o̶n̶l̶y̶ ̶o̶n̶e̶ associative binary operation $$\times$$ satisfying, for all $$x,y,z \in S$$,

$$\tag 1 1 \times 1 = 1$$

$$\tag 2 x \times y = y \times x$$

$$\tag 3 x \times (y+z) = x \times y + x \times z$$

Moreover, for every $$x \ S$$ there exists $$y \in S$$ with $$x \times y = 1$$.

My Work

I sketched out a proof for a special case here, and I think the argument can be lifted, but I'm not completely confident in using this logic and would like verification/confirmation.

If the theory is false, a counter-example would be appreciated.

• Apart from the empty semigroup and the one-element semigroup, do you know any example of a strongly homogeneous commutative semigroup? – J.-E. Pin Jun 19 at 6:01
• @J.-E.Pin Yes - $(\Bbb R^{\gt 0}, 1, +)$. – CopyPasteIt Jun 19 at 11:30