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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?

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  • $\begingroup$ to better frame the question, can you point to a property that $G$ and $H$ have, but $G\times H$ does not have? $\endgroup$
    – Ittay Weiss
    Mar 14, 2013 at 9:04
  • $\begingroup$ @IttayWeiss one option: "simple" $\endgroup$
    – Alexander Gruber ♦
    Mar 14, 2013 at 9:09
  • $\begingroup$ @AlexanderGruber simple what? $\endgroup$
    – Ittay Weiss
    Mar 14, 2013 at 9:10
  • $\begingroup$ @IttayWeiss If $G$ and $H$ are simple groups $G\times H$ isn't. (I don't know whether simplicity is a "thing" in magmas in general, but groups are magmas, so this is one example.) $\endgroup$
    – Alexander Gruber ♦
    Mar 14, 2013 at 9:11
  • $\begingroup$ I'm not sure that is what OP had in mind, hence my question. $\endgroup$
    – Ittay Weiss
    Mar 14, 2013 at 9:12

3 Answers 3

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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 $A\times B$.

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  • $\begingroup$ An identity being a statement like $\forall a,b \in A(ab=ba)$? Are we allowed to use "OR" and "There exists"? $\endgroup$
    – goblin GONE
    Mar 14, 2013 at 10:48
  • $\begingroup$ No, only $\forall$. $\endgroup$
    – Boris Novikov
    Mar 14, 2013 at 11:28
  • $\begingroup$ Conversely, the direct factors, being homomorphic images of the product, inherit all equational properties of the product. $\endgroup$
    – James
    Mar 16, 2013 at 2:25
  • $\begingroup$ @James: Certainly, but I only answered thr question. $\endgroup$
    – Boris Novikov
    Mar 16, 2013 at 6:49
  • $\begingroup$ @Boris: The question also says "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?" (I haven't written anything about that in my answer yet, either.) $\endgroup$
    – Tara B
    Mar 18, 2013 at 13:08
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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.

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  • $\begingroup$ @user18921: I like your question and could say more about it, but it is quite broad. Could you please let me know in which direction it would be most helpful to extend my answer? $\endgroup$
    – Tara B
    Mar 15, 2013 at 23:30
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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 $H$

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  • $\begingroup$ Helpful reminder of key definitions! +1 $\endgroup$
    – amWhy
    Mar 15, 2013 at 1:32
  • $\begingroup$ @amWhy: Thanks a lot. :-) $\endgroup$
    – Mikasa
    Mar 15, 2013 at 5:26

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