Wikipedia:Reference desk/Archives/Mathematics/2014 February 3

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February 3

Unnecessarily strong axiomatization of groups?

A magma may be defined with the axioms of

• a set G with a binary operation • with closure: ∀a,b ∈ G (a • b ∈ G)

A semigroup may be defined as an associative magma: with the axioms of

• a set G with a binary operation • with closure: ∀a,b ∈ G (a • b ∈ G)
• associativity: ∀a,b,c ∈ G ((a • b) • c = a • (b • c))

A group may be defined as a semigroup with identity and inverses: with the axioms of

• a set G with a binary operation • with closure: ∀a,b ∈ G (a • b ∈ G)
• associativity: ∀a,b,c ∈ G ((a • b) • c = a • (b • c))
• existence of an identity element: ∃e ∈ G (∀a ∈ G (e • a = a = a • e))
• existence of two-sided inverses: ∀a ∈ G (∃x ∈ G (a • x = e = x • a))

That is, a group is exactly a semigroup with the last two axioms added.

A quasigroup may be defined as a magma with left and right division: with the axioms

• a set G with a binary operation • with closure: ∀a,b ∈ G (a • b ∈ G)
• left division: ∀a,b ∈ G (∃!x ∈ G (a • x = b))
• right division: ∀a,b ∈ G (∃!y ∈ G (y • a = b))

The quasigroup article claims that "an associative quasigroup is either empty or is a group", so it would seem that a group can be defined as a nonempty quasigroup with associativity. Not every quasigroup is a loop, so the last two axioms of a quasigroup applied to a nonempty magma do not imply the last two axioms of a group in the absence of associativity. It seems to me however that for a magma even in the absence of associativity, the last two axioms of a group do imply left and right division (but I may be wrong: I notice a uniqueness requirement on division in a quasigroup missing from inverses in a group). If so, the axioms defining a nonempty quasigroup prior to adding associativity are strictly weaker than those of a group with the associativity axiom omitted.

My questions:

1. Am I correct in saying that the axioms defining a nonempty quasigroup are strictly weaker than those of a magma with identity and two-sided inverses?
2. If so, why is it usual to use the unnecessarily stronger set of axioms for a group?

—Quondum 02:31, 3 February 2014 (UTC)

Consider the operation defined by:

1 2 3 4 5

2 1 4 5 3

3 4 5 1 2

4 5 2 3 1

5 3 1 2 4

This has left and right division and therefore each element has and left and right identity, but they are not always the same, so I believe this is an example of a quasigroup which is not a magma with identity and two-sided inverses. As for your second question I have seen presentations that start with weaker axioms and it's an interesting problem to determine a "minimal" set of axioms, but as a practical issue verifying the axioms for a proposed group operation is often so trivial that having to do a few extra is not a problem. There are some constructions, such as semidirect product, where it's nice not to have to prove both halves of the inverse law, but such things aren't always covered anyway. --RDBury (talk) 06:43, 3 February 2014 (UTC)

In my first question, by "are strictly weaker" I also meant "are implied by", which is particularly relevant to being able to replace the axioms; as I understand you, you have shown the "do not imply" part. How I understand you is essentially that the typical axioms are easier to work with and understand, and being minimal in the axioms is not need normally. My interest derives from the idea of cleanly defining a class of objects that is exactly the groups plus the empty set, which may be useful in a context that I have in mind. So perhaps I'm less interested in the detail of the axioms than in whether a reasonable set of axioms allows the "empty group", and the statement that I quoted from the quasigroup article, if true, implies this. —Quondum 17:04, 3 February 2014 (UTC)

The statement from the quasigroup article is true: in a non-empty associative quasigroup, fix an element a, and let e be the unique element with $ae=a$. Then $aee=ae$, so $e^{2}=e$ (by uniqueness of divisors). Then for any element b, $(be)e=be$ and $e(eb)=eb$, so by uniqueness of divisors, $be=b$ and $eb=b$.

Then the existence of a two-sided inverse is standard: for every b, there is an element with $bx=e$, and an element with $yb=e$. Then $y=ye=ybx=ex=x$.--130.195.55.201 (talk) 19:00, 3 February 2014 (UTC)

Actually there are magmas with identity and two-sided inverse which are not quasigroups also. Consider

1 2 3

2 1 x

3 y 1

where x and y are replaced by any of 1, 2, or 3. --RDBury (talk) 22:34, 3 February 2014 (UTC)

You may have missed that associative was part of the premise of the statement.--130.195.55.150 (talk) 01:01, 4 February 2014 (UTC)

@RDBury: Neatly illustrated with a simple counterexample – existence of inverses does not imply division. That puts question 1 to bed with a negative.

@130.195.55.201: Thank you for the confirmation. —Quondum 06:00, 4 February 2014 (UTC)

There's a lot of stuff I haven't digested here, but addressing what I think may be your central point — in general, there is not a lot of interest in writing axioms that work for empty structures. In the usual setup, structures are simply required to be nonempty, period, regardless of what kind of structures they are. This corresponds to the fact that first-order logic in its standard form can prove, using no axioms at all, the statement $\exists x(x=x)$, so the "empty structure" is not a model for any first-order theory whatsoever.

Some people find this aesthetically unappealing, which as far as I can tell is the only justification for so-called free logic, which is weaker than first-order logic in that it cannot prove "something exists". I am not aware that this accomplishes anything at all beyond aesthetics, and it opens you up to lots of uninteresting trivial cases. --Trovatore (talk) 23:27, 3 February 2014 (UTC)

Trivial cases have real value when including them in a class simplifies the statement of related theorems, or serves as a terminal object in category theory. A nice historical prototype is the number zero. Exclusion of such cases is often because they (and the associated vacuously satisfied axioms) may be confusing to think about, but this periodically also stands in the way of insights. Zero, the empty string, the empty set... We cannot sensibly work with these categories without the empty cases. So I can't agree that excluding empty structures from consideration as a general strategy is even workable, and this has little to do with aesthetics. Considering the amount of buggy, convoluted software that I've fixed and simplified by treating the empty string as a regular limiting case instead of as a special case, I'd say that the avoidance of trivial cases is more about human psychology than about mathematical pragmatism or lack of application for trivial cases. —Quondum 06:00, 4 February 2014 (UTC)

Trivial cases in general, yes. Trying to make model theory work for the empty structure, no.

Zero is a useful object. The empty set is a useful object. The empty string is a useful object. Vacuous satisfaction is a pain to avoid. It's more painful to try to avoid vacuous satisfaction than it is to accept it.

Empty structures, on the other hand, I have never seen any use for at all. I don't think they "simplify the collection of related theorems", not at all. Unlike the four useful cases above, what they wind up doing is adding complications that turn out to have nothing interesting inside them, not unless you're interested in a certain class of philosophical questions (e.g. "can we know by pure logic that something exists"). That class of philosophical questions has never appeared to me to have an awful lot of content to it. I'm open to changing my mind on that point, but only if someone can show me the content, not just complain about logical axioms having existential content as a problem in itself. --Trovatore (talk) 08:26, 4 February 2014 (UTC)

Including trivial cases redefines old words, creating confusion. Multiplication by one or by zero is not literally multiplication. Multitasking usually means performing two or more tasks simultaneously, but to a mathematician it probably means: performing zero or more tasks simultaneously. Bo Jacoby (talk) 08:34, 4 February 2014 (UTC).

There's nothing wrong with trivial cases in general. There is something wrong with empty structures, in particular. Oh, not wrong in some deep a priori sense. Wrong in the same way that it's wrong to consider 1 to be a prime number, or {0} to be a field. That is, in the long run, you have to keep repeating "prime greater than 1" or some such, more often than you see any useful conceptual simplification. It's an empirical fact, not something you see in advance. --Trovatore (talk) 09:34, 4 February 2014 (UTC)

The guideline you give about going for reduced need to exclude cases is in general a good one. But to use the word "wrong" the way you do, even in the weak, qualified way you do here, seems overdone. And your examples don't really help: excluding 1 as a prime number is a very weak case: it does not simplify many cases (primarily unique factorization). And mentioning {0} as a field, I think that it is possible allow it naturally, without treating it as a special case, or, in general necessarily introducing the need for the exclusions you mention. It takes a little care, as trivial cases often do, but also sharpens one's thinking. And it is not even an empty structure. I would posit that a stronger empirical fact is that people unthinkingly exclude empty and trivial cases out of habit without good reason, contributing to precisely the problem that you describe. And as YohanN7 suggests below, considering such departures from the mainstream can be instructive. —Quondum 07:58, 5 February 2014 (UTC)

I repeat that I have just never seen anything they're good for. As far as I can tell they're utterly useless, a total waste of time. But hey, it's your time. Go for it. Let me know if you find anything interesting. --Trovatore (talk) 09:04, 5 February 2014 (UTC)

I want to clarify again what I mean by "they". Trivial structures are useful. Empty structures are not. --Trovatore (talk) 09:11, 5 February 2014 (UTC)

Your statement puzzles me. So the empty set is not useful? —Quondum 17:29, 5 February 2014 (UTC)

The empty set is very useful. But it's not a structure. --Trovatore (talk) 18:34, 5 February 2014 (UTC)

The empty set is a topological manifold (if that counts as having a structure (edit: nope, according to Trovatores link)). At least if you don't explicitly exclude it from the definition. (Introduction to Smooth Manifolds, John M Lee) YohanN7 (talk) 19:17, 5 February 2014 (UTC)

Despite its limited applicability, first-order logic and its constraints of convenience now apply to all of mathematics? – Hey, why do I have this feeling that I am an xkcd character? —Quondum 04:26, 6 February 2014 (UTC)

Well, while its not the only thing you want to know about a structure, you pretty much always want to know its first-order theory. For an empty structure, what would be its first-order theory, if there were one, is obvious: Any statement in prenex normal form that starts with a universal quantifier is true; any that starts with an existential quantifier is false.

But that isn't a first-order theory, because first-order logic can prove an existential statement without using any axioms at all, namely "something exists".

Is that a "constraint" of first-order logic? In some very technical sense, I suppose it is. But it's hard to imagine any that's more eminently reasonable. We know absolutely, apodeictically, that something exists. The mere asking of the question proves it, because the question is something.

Now we get to the question I'm not very interested in: Is that piece of knowledge something we know by pure logic, or does it have some sort of synthetic or empirical content? If you care about that, which I don't, much, then I suppose you might want to investigate free logic as a way of formulating the question without assuming the answer. But it seems like an awful lot of trouble, for very little reward. --Trovatore (talk) 08:42, 6 February 2014 (UTC)

There is one way of (seemingly) weakening the group axioms (as stated above) that is actually useful, in that it is instructive. Only one-sided identity and inverses is required. The task is then to show that the above axioms follow. Not hard, but not entirely trivial. YohanN7 (talk) 09:10, 4 February 2014 (UTC)

One caveat here, you have to get the correct combination of left/right identity and left/right inverse. There are are two combinations that work (I can never remember which) and two where non-groups exist. --RDBury (talk) 21:08, 4 February 2014 (UTC)

As I recall, it should be left/left (and then probably right/right), but I'm not sure. My ref for this is Mathematics of Classical and Quantum Physics by Byron and Fuller, but I don't have any of my books at hand a t m. Nonetheless, this "weakening" should have practical utility too. Given a multiplication table, do we have a group? Both for manual and computerized verification, the weakening is certainly useful. YohanN7 (talk) 12:19, 5 February 2014 (UTC)

Slightly off topic: To test a multiplication table for associativity, there is something called Light's test for associativity. A nice reference is Abstract Algebra by Pierre Antoine Grillet. YohanN7 (talk) 16:03, 5 February 2014 (UTC)

We have an article Light's associativity test. And Light's test for associativity now redirects there. —Tobias Bergemann (talk) 07:41, 6 February 2014 (UTC)