# Does distributivity implies commutativity of one operation

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?

• Commented Jan 3, 2021 at 6:25
• Commented Jan 3, 2021 at 6:33

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.

• $\otimes$ and $\oplus$ does not have unique identity. But in my example, I assume that $*$ and $@$ have unique identity.
– Iti
Commented Jan 3, 2021 at 6:27
• @Iti: That is precisely my point: you assumed far more than distributivity and associativity, and my example shows that those extra assumptions are essential. It isn’t just distributivity that got you commutativity: it was a whole bucket full of assumptions. Commented Jan 3, 2021 at 6:29
• Ok, I understand your point. I have used more assumptions than just distributivity, which leads to commutativity of $@$. Thank you very much.
– Iti
Commented Jan 3, 2021 at 6:33
• @Iti: You’re very welcome. Commented Jan 3, 2021 at 6:35
• @Iti: Probably because we normally think of fields as rings with extra properties, just as we normally think of rings as groups with extra properties. Commented Jan 3, 2021 at 7:04

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.

• May you please explain what are yo meant by non-trivial structure. I have not studied abstract algebra much. I know only group theory.
– Iti
Commented Jan 3, 2021 at 6:24
• You can have things like "groups with [insert wonderful property here]", and that may prove incredible theorems. Until someday it's proven that the only group that satisfies all these properties... is the trivial group. This is IMO rather likely when you require the existence of a full multiplicative inverse over the set - I think you're getting the trivial, 1-element ring only (and then the theorem that it's commutative is rather vacuous). Commented Jan 3, 2021 at 8:33