Does distributivity implies commutativity of one operation

Question

Suppose there is a set $S$, equipped with two binary operations, $*$ and $@$, such that S is closed and associative under both the operations. There exist inverses and identity with respect to both the operations. $*$ is distributive under $@$. (Both the operations are not commutative).
Let $a,b,c,d\in S$, and $(a@b)=\alpha$ and $(c@d)=\beta$.

$(a@b)*(c@d)=\alpha *(c@d)=(\alpha *c)@(\alpha *d) = ((a@b)*c)@((a@b)*d)=(a*c)@(b*c)@(a*d)@(b*d)\;\;----(1)$

Also $(a@b)*(c@d)=(a@b)*\beta=(a*\beta)@(b*\beta) = (a*(c@d))@(b*(c@d))=(a*c)@(a*d)@(b*c)@(b*d)\;\;----(2)$

Comparing (1) and (2). We conclude that $(b*c)@(a*d)=(a*d)@(b*c)$ using identity, inverses and associativity (cancellation property).
This implies that $@$ is commutative operation in S.

This a contradiction. That surprised me.

So the question is if the operations are not commutative then they can't be distributive always because then one of the operation comes out to be commutative? If one of them is commutative then only there is possibility of showing distributive property?

So like distributivity and commutativity are connected?

Answer 1

It takes more than distributivity: the existence of a $*$ identity and inverses is also crucial to your argument.

To see this, define operations $\otimes$ and $\oplus$ on $\Bbb R$ by $x\oplus y=x$ and $x\otimes y=y$ for all $x,y\in\Bbb R$. You can easily verify that both operations are associative and that

$$x\otimes(y\oplus z)=y=(x\otimes y)\oplus(x\otimes z)$$

and

$$(y\oplus z)\otimes x=x=(y\otimes x)\oplus(z\otimes x)\,,$$

i.e., that $\otimes$ distributes over $\oplus$. Clearly neither operation is commutative, however.

Answer 2

You're likely running into triviality problems. Note that your conditions are vastly stronger than even a ring, notably that "multiplication" has an identity and inverse.

In any ring, $0$ (the additive identity) must necessarily absorb under multiplication, and therefore cannot have a multiplicative inverse (look at $(a+(-a))\times b$). The proper definition of a ring, of course, looks at the multiplicative group of units, of which $0$ is never one.

You necessarily use cancellation of multiplication, but that is precisely the operation that need not exist in a non-trivial structure.