Which properties are inherited by the Cartesian product of two sets equipped with a binary operation?

Question

Let $G$ and $H$ denote sets equipped with a binary operation (aka magmas). We can form the Cartesian product magma $G \times H$ in the obvious way. I'm interested in which properties of $G$ and $H$ transfer to $G \times H$. For instance, if both $G$ and $H$ are associative, then $G \times H$ is associative. Similarly with commutativity.

Is there a general principle that dictates which properties $G \times H$ will inherit?

And what about the other way around? For which properties does it hold that if $G \times H$ has that property, then either/both of $G, H$ must have it?

Answer 1

By Birkhof Theorem every variety of (universal) algebras is closed respectively direct product. So if an identity is true for $A$ and $B$, it is true for $A\times B$.

Answer 2

Some points I know:

• If $G$ and $H$ are semigroups, so is $G\times H$ since:

$$[(a,b)(c,d)](h,f)=(ac,bd)(h,f)=[(ac)h,(bd)f]=[a(ch),b(df)]=...(a,b)[(c,d),(h,f)]$$

• An element $(a,b)\in G\times H$ is idempotent iff $a$ is an idempotent in $G$ and $b$ is an idempotent in $H$.

• An element $(a,b)\in G\times H$ is a left (or right) identity element of $G\times H$ iff $a$ is a left (or right) identity element of $G$ and $b$ is a left (or right) identity element of $H$

Answer 3

I am quite certain that there is no currently known general rule which dictates exactly which properties $G\times H$ will inherit from $G$ and $H$, nor one which dictates which properties $G$ and $H$ inherit from $G\times H$. If there were, I would not have seen so many results on special cases in recent publications (not proved by referring to some general principle).

Here are some examples of properties not (necessarily) inherited by $G\times H$:

• Freeness
• Finite generation: This is inherited in magmas with identity, but not always otherwise. If $S$ and $T$ are f.g. semigroups, then $S\times T$ is f.g. iff either at least one of $S$ and $T$ is finite or $S^2=S$ and $T^2=T$. (This result is due to Ruskuc, Robertson and Wiegold.)
• Having word problem in a specific complexity class. For example the infinite cyclic group $\mathbb{Z}$ has context-free word problem, but $\mathbb{Z}\times \mathbb{Z}$ does not. Similarly, the semigroup $\mathbb{N}$ of natural numbers under addition has rational word problem, but $\mathbb{N}\times \mathbb{N}$ does not.

I'll post this for now, but might come back and add to it later, since I haven't by any means addressed everything in your question.