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I am wondering whether the following property of two binary operations $\diamond$ and $\star$ has a name. I haven't seen it listed in overviews of properties of binary operations, and I wouldn't know how to search for it directly:

$$(a \diamond b) \star (c \diamond d) = (a \star c) \diamond (b \star d)$$

This is trivially true if $\diamond = \star$ is associative and commutative. It is also true for any associative and commutative binary operation with an inverse $\diamond = +$, and the binary operation induced by the inverse $\diamond = -$.

If I had to come up with a name for this, I would say that the two operations are orthogonal, or maybe mutually commutative.

Are there any canonical, less trivial examples of this? (The context in which I myself came across this property would take some work to define properly.)

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    $\begingroup$ You might want to use [terminology]. $\endgroup$
    – soupless
    Dec 23, 2021 at 10:24
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    $\begingroup$ Are you familiar with the Eckmann–Hilton argument? It basically says whenever both you binary operations have a unit, then your property implies that they are in fact equivalent and both associative and commutative... $\endgroup$
    – G. Blaickner
    Dec 23, 2021 at 10:41
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    $\begingroup$ @G.Blaickner I wasn't. From a quick glance at the Wikipedia page, it seems that that would make for a nice answer as to why this property isn't of general interest and hasn't received a name. (NB $-$ in the example above is not unital; in my original more complicated example one of the operations is partial and not unital either.) $\endgroup$
    – oulenz
    Dec 23, 2021 at 10:50
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    $\begingroup$ In category theory, for the partially defined operations of horizontal and vertical composition of natural transformations, it's called the interchange law. The same terminology is also used here. It doesn't seem to be used much outside of category theory, but see exercises II.5.4-II.5.6 of Mac Lane, Categories for the Working Mathematician $\endgroup$
    – Calum Gilhooley
    Dec 23, 2021 at 13:44
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    $\begingroup$ @CalumGilhooley Thanks! My original example is actually also for categories, to be precise (skeletal) categories with something like a product or coproduct. The first (partial) binary operation is composition, and the other sends two morphisms to the induced morphism between the (co)products of their sources and targets. $\endgroup$
    – oulenz
    Dec 23, 2021 at 18:40

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Let me convert my comment to an answer:

Lets suppose that $M$ is a set equipped with two binary operations $\ast:M\times M\to M$ and $\circ:M\times M\to M$. Then the Eckmann-Hilton argument says that whenever you binary operations fulfils the kind of twisting property $$(a\circ b)\ast (c\circ d)=(a\ast c)\circ (b\ast d)$$ for all $a,b,c,d\in M$ and both binary operations are unital, meaning that there exists an element $1_{\ast}\in M$ such that $1_{\ast}\ast a=a\ast 1_{\ast}=a$ for all $a\in M$ and there exists an element $1_{\circ}\in M$ such that $1_{\circ}\circ a=a\circ 1_{\circ}=a$ for all $a\in M$, then it follows that $\ast=\circ$ and both $(M,\ast)$ and $(M,\circ)$ are abelian monoids, i.e. $\ast$ and $\circ$ are associative and commutative.

The proof is actually quite simple: First of all, observe that $1_{\ast}=1_{\circ}.$ since $$1_{\circ }=1_{\circ }\circ 1_{\circ }=(1_{\ast }\ast 1_{\circ })\circ (1_{\circ }\ast 1_{\ast })=(1_{\ast }\circ 1_{\circ })\ast (1_{\circ }\circ 1_{\ast})=1_{\ast }\ast 1_{\ast }=1_{\ast }.$$

The statement then follows from

$$ a\circ b=(1\ast a)\circ (b\ast 1)=(1\circ b)\ast (a\circ 1)=b\ast a=(b\circ 1)\ast (1\circ a)=(b\ast 1)\circ (1\ast a)=b\circ a$$

The claim that $\ast$ and $\circ$ are associative can be proven similarly.

This is probably the reason why your property does not really have a name in the literature, although you can say that $$(a\circ b)\ast (c\circ d)=(a\ast c)\circ (b\ast d)$$ means that $\ast$ is a homomorphism for the other operation $\circ$ if you want.

The Theorem of Eckmann-Hilton might not be true for the general case if your operations do not admit units, however, there are some variations, i.e. similar statements when assuming other properties of your binaries. One of them, for example, asserts that whenever your binary operations are commutative and indempotent $(a\ast a=a$ and $b\circ b=b$), then the operations coincide, since

$$a\ast b=(a\ast b)\circ (a\ast b)=(a\ast b)\circ (b\ast a)=(a\circ b)\ast (b\circ a)=(a\circ b)\ast (a\circ b)=a\circ b.$$

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