## August 10, 2020

### The Group With No Elements

#### Posted by John Baez Maybe people who are all excited about the “field with one element” should start at the beginning and think a bit about the “group with no elements”.

I’m just kidding, but I did learn something fun today from Michael Kinyon’s response to a puzzle posed by Barbara Fantechi. Namely: there’s a nice way to define a group as a nonempty set with some structure and properties. This makes the nonemptiness condition seem silly and tacked-on… and if you drop it, you get a slight generalization of the definition of group, which allows there to be an empty group.

It goes like this. A group is the same as a nonempty set with an associative binary operation $\cdot : G \times G \to G$ called multiplication such that for any $g \in G$, the operations of left and right multiplication by $g$ are bijections.

Why? Clearly every group has such a multiplication, so the challenge is to prove the converse, and the key step is to get ahold of the identity element. So let’s do that.

For starters, let’s define left division by $g$ to be the inverse of left multiplication by $g$, and write $g\backslash h$ to mean $h$ left divided by $g$. Similarly, define right division by $g$ to be the inverse of right multiplication by $g$, and write $h/g$ to mean $h$ right divided by $g$.

Saying that left division by $g$ is the inverse of left multiplication by $g$ amounts to this:

$g \cdot (g\backslash h) = h, \qquad g\backslash(g \cdot h) = h$

Similarly, saying right division by $g$ is the inverse of right multiplication by $g$ amounts to this:

$(h/g) \cdot g = h, \qquad (h \cdot g)/g = h$

Now, how do we get the identity element $1 \in G$? If $G$ is nonempty, we can choose any element and try taking $1 = g/g$. But why is this independent of our choice? Also, it’s equally natural to try $1 = h\backslash h$ for some $h \in H$. Why is this the same thing, independent of our choice of $h$?

We can settle all these questions simultaneously. If $g,h \in G$ we have

$g \cdot h = g \cdot (g\backslash g) \cdot h$

Dividing on the left by $g$ we get

$h = (g\backslash g) \cdot h$

Dividing on the right by $h$ we get

$h/h = g\backslash g$

for all $g,h \in G$. Done!

So, if $G$ is nonempty, we can set $1 = g\backslash g = g/g$ for any $g \in G$. This is independent of our choice of $g \in G$, and it obeys

$1 \cdot g = (g/g) \cdot g = g$

and also

$g \cdot 1 = g \cdot (g\backslash g) = g$

It’s now easy to get inverses: whenever you have a monoid where every element $g$ has both a left inverse (here $1/g$) and a right inverse (here $g\backslash 1$), they must be equal, so we can take either one to be $g^{-1}$.

The fun part, for me, is that we can say

Definition. A possibly empty group is a set $G$ with an associative binary operation $\cdot : G \times G \to G$ called multiplication together with two more binary operations:

• left division: $\quad \backslash : G \times G \to G$
• right division: $\quad / : G \times G \to G$

such that

$g \cdot (g\backslash h) = h, \qquad g\backslash (g \cdot h) = h$ $(h/g) \cdot g = h, \qquad (h \cdot g)/g = h$

for all $g, h \in G$.

There’s an obvious concept of morphism between possibly empty groups — a map preserving all three operations. The category of possibly empty groups is equivalent to the category of groups with a new initial object thrown in, “the empty group”, and no morphisms from any groups to this one.

Note that there’s a Lawvere theory for possibly empty groups! Its category of nonempty algebras is equivalent to the category of groups.

I doubt this stuff is very important, but I’m surprised I hadn’t heard about it earlier, and a bit surprised that there’s a Lawvere theory whose algebras are the usual groups and also one extra ‘empty group’.

In hindsight it’s quite reminiscent of how a torsor of a group is defined as a nonempty set on which that group acts freely and transitively.

Here’s a puzzle that could help sharpen my intuitions. I haven’t solved it:

Puzzle. Prove or disprove this conjecture: there is no Lawvere theory whose category of algebras is the category of monoids together with one more algebra, the empty set.

(Note that this one extra algebra will be initial, with no morphisms from any other algebra to it.)

I bet this conjecture is true.

Posted at August 10, 2020 11:34 PM UTC

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### Re: The Group With No Elements

Your conjecture is false. The trick is to replace the nullary operation defining the unit element of a monoid by a “constant” unary operation. The Lawvere theory for “possibly empty monoids” is generated by a binary operation $m \colon A \times A \to A$ and a unary operation $e \colon A \to A$ subject to the following four axioms: (1) $m\circ (m \times 1) = m \circ (1 \times m) \colon A \times A \times A \to A$ ($m$ is associative), (2) $e \circ pr_1 = e \circ pr_2 \colon A \times A \to A$ ($e$ is constant), (3) $m \circ (1 \times e) = pr_1 \colon A \times A \to A$ ($e$ is a right unit), and (4) $m \circ (e \times 1) = pr_2 \colon A \times A \to A$ ($e$ is a left unit).

A similar description of “possibly empty groups” is given on the first page of Freyd’s Algebra valued functors in general and tensor products in particular.

Posted by: Alexander Campbell on August 11, 2020 2:58 AM | Permalink | Reply to this

### Re: The Group With No Elements

Nice! Now I’ll completely reverse my position and guess that for any Lawvere theory $T$ with nullary operations, we can create a Lawvere theory $T'$ where those nullary operations are replaced by unary operations, together with new equations like your equation (2) saying those unary operations are ‘constant’, together with equations mimicking those obeyed by the original nullary operations… with the result being that the algebras of $T'$ are just those of $T$ together with one new algebra: the empty set.

Posted by: John Baez on August 11, 2020 3:20 AM | Permalink | Reply to this

### Replacing constants with unary function symbols

At the last TACL (Topology, Algebra and Categories in Logic) in Nice, I had this same idea of wondering how much the category of models of a theory would be enlarged by replacing constant symbols with unary function symbols. I even had the opportunity to ask Gromov if he had any insight into it, but apparently it wasn’t something he’d ever had cause to think about.

A context which makes this a lot more interesting is to consider models in toposes other than Set. The idea of a distinguished element seems natural in Set, because every non-empty set has a global element. But in most toposes, pointed objects are uncommon. If we want to define a group object in a category of sheaves over a space, for example, the insistence on a global element forces the group to have global support, whereas with a unary operation we can distinguish between trivial groups on subspaces and the trivial group on the whole space.

Another flavour of this is when objects naturally come equipped with elements, such as in a topos of actions of a monoid or group. Such a topos comes with a forgetful functor to Set, but a global element of an M-set is not any element, but specifically an element on which M acts trivially. This is fine (we can still have interesting groups in a lot of cases), but seems like a somewhat artificial restriction in the context of this blog post.

If nothing else, I hope this illustrates how the category of models of a theory can be enlarged much more than just “adding the empty model” in other contexts :)

Posted by: Morgan Rogers on August 11, 2020 12:05 PM | Permalink | Reply to this

### Re: The Group With No Elements

In Functorial Semantics of Algebraic Theories (http://www.tac.mta.ca/tac/reprints/articles/5/tr5.pdf around p63-65) Lawvere gives two different definitions for the constants of a Lawvere theory $L$: an expressible constant is an expression in no variables (i.e. an element of $L(1, 0)$) and a definable constant is an expression in one variable (i.e. an element of $L(1, 1)$) which is unchanged under substituting that variable for a different variable.

He then shows that for any Lawvere theory, either the set of definable constants is isomorphic to the set of expressible constants, or there are no expressible constants. So Lawvere theories either “hide” their constants, or they don’t!

He goes on to say

Remark. Evidently one could complete; (or deplete) algebraic theories with regard to expressibility of constants; however, there seems to be no need to do so. As an example,the algebraic theory of groups can be ‘presented’ (see Section 2) in two ways, one involving a single binary operation $x, y \to x \cdot y^{-1}$ as generator, and the second involving a 0-ary generator $e$, a unary generator $x \to \x^{-1}$, and a binary generator $x, y \to x \cdot y$.This actually gives two theories, for in the first case no constant is expressible and in the second case the (only) constant $e$ is expressible. These two theories also give rise to different algebraic categories (see Chapter III), for according to the first theory the empty set is a group, whereas to the second it is not.

I think (though I haven’t checked the details) that what he means is that if you take a Lawvere theory $L$, and create a new category $L'$ by simply deleting every morphism into $0$, then $L'$ is in fact a Lawvere theory with the same definable constants but no expressible constants. I’m pretty sure $L'$ will then have the same models as $L$, except that it will also have an empty model.

### Re: The Group With No Elements

Isn’t Lawvere’s first presentation, with the single binary operation, a possibly more economical version of John’s three binary operations? It seems also to allow the empty group.

Posted by: L Spice on August 14, 2020 4:09 PM | Permalink | Reply to this

### Re: The Group With No Elements

Lurking in the background is the fact that there are exactly two trivial Lawvere theories, or a bit more generally, exactly two trivial monads on $Set$.

Let me explain what I mean. It’s a nice little exercise on monads to show that if $T$ is a monad on $Set$ and there exists at least one $T$-algebra with at least two elements, then the unit map

$\eta_X: X \to T(X)$

is injective for every set $X$. In that case, the free $T$-algebra on a set $X$ of generators contains a copy of $X$; the generators don’t get identified. So there are $T$-algebras of arbitrarily large cardinality.

But what about monads $T$ for which there isn’t an algebra with at least two elements? Well, that means that every $T$-algebra has either one element or no elements. I’m calling a monad with such a property “trivial”, I’m saying that there are exactly two such monads, and I’m saying that they’re both finitary (i.e. correspond to Lawvere theories).

Again this is a nice little exercise, but as it’s more relevant to your post, John, I’ll do it here.

Let $T$ be a trivial monad, in my sense. For every set $X$, the set $T(X)$ carries the structure of a $T$-algebra, namely the free $T$-algebra on $X$. So, $|T(X)| \leq 1$ for every set $X$. But we havea map $\eta_X: X \to T(X)$, so if $X$ is nonempty then so is $T(X)$. Thus, $T(X)$ has exactly one element for each nonempty set $X$.

That only leaves the question of whether $T(\emptyset)$ has $0$ elements or $1$ element. And both possibilities occur:

• There is a unique monad $T_0$ on $Set$ such that $T_0(X) = \begin{cases} \emptyset &\text{if } X = \emptyset\\ 1 &\text{otherwise}. \end{cases}$ Algebraically, it is the theory presented by no operations and the single equation $x = y.$ An algebra for $T_0$ is a set in which all elements are equal, i.e., a set with at most one element.

• There is a unique monad $T_1$ on $Set$ such that $T_1(X) = 1$ for all $X$. Algebraically, it is the theory presented by a single constant $c$ and the single equation $x = c.$ An algebra for $T_1$ is a set with a distinguished element to which all other elements are equal, i.e., a set with exactly one element.

Why am I talking about this here? Because the difference between $T_0$ and $T_1$ is the simplest example of the general phenomenon you’re talking about. A $T_1$-algebra is like a group; a $T_0$-algebra is like a possibly empty group.

Posted by: Tom Leinster on August 11, 2020 11:53 AM | Permalink | Reply to this

### Re: The Group With No Elements

Actually, this trivial monad $T_0$ is the key to answering your question:

guess that for any Lawvere theory $T$… we can create a Lawvere theory $T'$… [such that] the algebras of $T'$ are just those of $T$ together with one new algebra: the empty set.

Your guess is right, and there’s a simple formula: $T' = T_0 \times T$.

So, for instance, there’s a Lawvere theory whose algebras are “possibly empty complex Lie algebras”; there’s another one whose algebras are “possibly empty rings”, and so on.

In more detail: for any category $A$ with finite products, and any two endofunctors $S, T$ of $A$, one can define a new endofunctor $S \times T$ by

$(S \times T)(a) = S(a) \times T(a).$

Now $End(A)$ is a monoidal category under composition, and one can check that

$\times: End(A) \times End(A) \to End(A)$

is a lax monoidal functor. (There’s only one sensible thing that the coherence maps can be. It wasn’t screamingly obvious to me that the coherence diagrams would commute, but I checked and they do. What’s a good abstract explanation of this?)

Any lax monoidal functor between monoidal categories carries monoids in the domain to monoids in the codomain. In our case, this means that if $S$ and $T$ have the structure of monads then $S \times T$ acquires the structure of a monad.

Now take $A = Set$, take $S$ to be the trivial monad $T_0$ on $Set$, and take any monad $T$ on $Set$. Write $T'$ for the monad $T_0 \times T$. Then as a functor,

$T'(X) = \begin{cases} \emptyset &\text{if } X \cong \emptyset\\ T(X) &\text{otherwise}. \end{cases}.$

The monad structure on $T'$ is the obvious one: on nonempty sets it’s the same as that of $T$, and on the empty set it’s the only thing possible. So a $T'$-algebra structure on a nonempty set $X$ is just a $T$-algebra structure on $X$, and the empty set also has a unique $T$-algebra structure.

In other words, the algebras for $T'$ are exactly the algebras for $T$ together with the empty set (which is a $T'$-algebra in a unique way). That’s what you wanted.

Note that this construction works for arbitrary monads on $Set$, not necessarily finitary. For instance, if we take $T$ to be the powerset monad, which isn’t finitary, then the $T$-algebras are the complete lattices (with join-preserving maps as the homomorphisms), and a $T'$-algebra is a lattice that is either complete or empty. (Weird!) But if $T$ is finitary then so is $T'$, so the operation $T \mapsto T'$ corresponds to an operation on Lawvere theories.

Posted by: Tom Leinster on August 11, 2020 3:14 PM | Permalink | Reply to this

### Re: The Group With No Elements

What’s a good abstract explanation of this?

For any monoidal category $C$ (such as $End(A)$), the diagonal functor $C\to C\times C$ is strong monoidal. Therefore, by doctrinal adjunction, if it has a right adjoint (i.e. if $C$ has cartesian products, which are pointwise in a functor category), that right adjoint is lax monoidal.

Posted by: Mike Shulman on August 11, 2020 4:16 PM | Permalink | Reply to this

### Re: The Group With No Elements

Ah excellent, thanks.

For the sake of being explicit, let me pick out the part of the doctrinal adjunction story that we need here and say it directly. I’m just repeating back to you what you said to me :-)

Given an adjunction between monoidal categories, there’s a natural one-to-one correspondence between colax monoidal structures on the left adjoint and lax monoidal structures on the right adjoint. (It’s given by taking mates.)

Here, we start with a category $A$ with finite products, we take the monoidal category $C = End(A)$, and we observe that $C$ also has finite products, defined pointwise. Hence we have an adjunction

$\Delta: C \leftrightarrows C \times C: \times$

($\Delta$ left adjoint to $\times$). As you say, $\Delta$ is strong monoidal, and in particular colax monoidal. The correspondence above then gives us a lax monoidal structure on $\times$. And it follows that one can take products of monads on $A$ in the way I described.

Posted by: Tom Leinster on August 11, 2020 5:11 PM | Permalink | Reply to this

### Re: The Group With No Elements

Tom wrote:

Actually, this trivial monad $T_0$ is the key to answering your question:

guess that for any Lawvere theory $T$… we can create a Lawvere theory $T'$… [such that] the algebras of $T'$ are just those of $T$ together with one new algebra: the empty set.

Your guess is right, and there’s a simple formula: $T' = T_0 \times T$.

Cool! Here $\times$ the pointwise product of endofunctors that happen to be finitary monads, made into a finitary monad. But since finitary monads on Set correspond to Lawvere theories, we could also try to understand this $\times$ construction more directly as a monoidal structure on the category of Lawvere theories, right?

Posted by: John Baez on August 13, 2020 12:24 AM | Permalink | Reply to this

### Re: The Group With No Elements

Right! I think $\times$ is actually the product in the category of Lawvere theories. In any case, it’s given by

$(K \times L)(n, m) = K(n, m) \times L(n, m)$

for Lawvere theories $K$ and $L$.

It’s easy to make the translation from monads to Lawvere theories. For a Lawvere theory $L$, an element of $L(n, m)$ is to be thought of as an $m$-tuple of $n$-ary operations, whereas for a monad $T$, an element of $T(n)$ is to be thought of as a single $n$-ary operation. So if $L$ is the Lawvere theory corresponding to a finitary monad $T$, then $L(n, m) = T(n)^m$.

Hence, if we also have a Lawvere theory $K$ corresponding to a finitary monad $S$, then the Lawvere theory $K \times L$ corresponding to the monad $S \times T$ is given by

\begin{aligned} (K \times L)(n, m)& = (S \times T)(n)^m\\ & = (S(n) \times T(n))^m\\ & = S(n)^m \times T(n)^m\\ & = K(n, m) \times L(n, m). \end{aligned}

That’s the formula I started with.

The other stuff I said about monads translates easily too. The Lawvere theory $L_0$ corresponding to my trivial monad $T_0$ is given by

$L_0(n, m)= \begin{cases} \emptyset &\text{if }  m = 0 \lt n\\ 1 &\text{otherwise}. \end{cases}$

The algebras for $L_0$ are the sets with at most one element. Given a Lawvere theory $L$, the resulting theory $L' = L_0 \times L$ (corresponding to $T'= T_0 \times T$) is given by

$L'(n, m)= \begin{cases} \emptyset &\text{if }  m = 0 \lt n\\ L(n, m) &\text{otherwise.} \end{cases}$

The algebras for $L'$ in $Set$ are the algebras for $L$ together with the empty set. In other words, $L'$ is the theory of “possibly empty $L$-algebras”.

I don’t know what the algebras for $L'$ are in an arbitrary finite product category. Indeed, my proof of my statement about algebras for $L'$ in $Set$ is that it follows from the stuff about monads; I haven’t tried to think it through directly in terms of Lawvere theories.

Posted by: Tom Leinster on August 13, 2020 10:05 AM | Permalink | Reply to this

### Re: The Group With No Elements

What changes when going from trying to define possibly empty groups to trying to define possibly empty loops (i.e. not necessarily associative possibly empty groups)?

### Re: The Group With No Elements

Probably nothing, but the proof of the identity $1 = g\backslash g = h/h$ in a non-empty group will have to change, because in a loop $g \cdot (g\backslash g) \cdot h$ is not well defined due to the lack of associativity. I suppose we have to first show that $(g \cdot (g\backslash g)) \cdot h = g \cdot ((g\backslash g) \cdot h)$ before carrying with the rest of the proof.

### Re: The Group With No Elements

Dropping associativity gives you possibly-empty quasigroups. Is there a (reasonable) definition of “loop” different from “quasigroup with identity” that would allow the empty quasigroup to qualify?

If one allows semigroups and quasigroups to be empty but requires that monoids and loops have an actual identity element, then a group is the same as an associative loop is the same as a monoid with division while a possibly-empty group is the same as an associative quasigroup is the same as a semigroup with division. So you get a little boolean lattice of algebraic theories corresponding to subsets of {associativity, division, identity} and all eight theories are distinct, which seems nice to me.

(Also, whose bright idea was it that semigroups and quasigroups, adding a condition that makes them more group-like should change the name to something without “group” in it?)

Posted by: Nick Olson-Harris on August 11, 2020 6:55 PM | Permalink | Reply to this

### Re: The Group With No Elements

And you could then add commutativity to the identity and have a boolean lattice from [associativity, division, commutativity, identity] and define possibly empty abelian groups as a magma with associativity, division, and commutativity. Then you could define a possibly empty ring as a set $S$ with two binary operations $+$ and $\cdot$ such that $\cdot$ distributes over $+$, $S$ is an possibly-empty abelian group with respect to $+$, and a (possibly empty) semigroup with respect to $\cdot$. And then you could define possibly empty division rings as possibly empty rings with division, and possibly empty fields as possibly empty commutative division rings. So you can have a field with no elements and a field with two elements, but no field with one element!

### Re: The Group With No Elements

Actually, in this definition, the ring with one element would be a possibly empty division ring, as nowhere in the definition does it say the additive and multiplicative identities are different - because the definition does not use either additive nor multiplicative identities. And because such a ring is commutative it would be a possibly empty field as well. Hmmph. Perhaps there is a way to define actual possibly empty division rings and possibly empty fields without using identities, as opposed to possibly empty possibly trivial division rings and fields.

### Re: The Group With No Elements

And then you could define possibly empty division rings as possibly empty rings with division

(Preview always chokes on plain ASCII apostrophes, so I am awkwardly stripping them out.)

Doesn’t one have to take some care with this definition? A division ring only has division by non-0, so you would need some way to refer to 0, at which point you’re not speaking about possibly empty rings any more. Maybe the division condition is that $\forall x, y, z,\, x y \ne x \implies \exists w, x w = z$?

Posted by: L Spice on August 14, 2020 4:16 PM | Permalink | Reply to this

### Re: The Group With No Elements

Yeah, non-zero division; I realised a few hours after writing the above comment that the way I defined division in the comment above ensures that the resulting category of air quotes ‘division rings’ is simply the interval category consisting of the empty ring and the trivial ring. This is probably why the non-zero requirement for division is stipulated in the first place. In fact, division is not required, only that the ring be a (not necessarily associative) domain, for the non-zero aspect to apply.

### Re: The Group With No Elements

If the empty group is the new initial group, then what is the new coproduct in the new category of groups?

The ordinary coproduct still satisfies the universal property, so everything still works out exactly as before (which might not sound very exciting, but it’s also reassuring that nothing unexpected happens in this very-slightly-enlarged category). Moreover, the pushout of a span whose apex is the trivial (1-element) group still coincide with coproducts, for the same reason that in a category with non-trivial subterminal objects, pulling pack two morphisms whose codomain is subterminal is the same as taking the product: we can always extend the span/cospan whose colimit/limit is being taken to a span/cospan involving the initial/terminal object, and the uniqueness of this extension ensures that the universal property of the colimit/limit is unchanged.

Posted by: Morgan Rogers on August 12, 2020 5:46 PM | Permalink | Reply to this

### Re:

Since the empty group is also (trivially) commutative, we could add commutativity to the axioms and talk about the category of possibly empty abelian groups, and the empty group is probably the initial object in that category as well. So presumably the direct sum of abelian groups would still be the coproduct of the “slightly enlarged” category of abelian groups for the reasons you give.

### Re: The empty abelian group

By the way, we can add commutativity to the axioms of a possibly empty group, but there’s an equivalent and more efficient definition of a possibly empty abelian group:

Definition. A possibly empty abelian group is a set $A$ with a commutative associative operation $+ \colon A \times A \to A$ and an operation $- \colon A \times A \to A$ such that

$(a + b) - b = a = (a - b) + b$

The category of possibly empty abelian groups is not an abelian category because the initial object (the trivial group $0$) is not the terminal object (the empty group $\empty$). Also, the product of possibly empty abelian groups doesn’t match the coproduct, since

$0 \times \empty = \empty$

while

$0 + \empty = 0$

(Don’t just believe me, check them using the category-theoretic definitions and make sure I’m right!)

Posted by: John Baez on August 13, 2020 12:13 AM | Permalink | Reply to this

### Re: The empty abelian group

I had a think after writing that comment and for abelian groups, the direct sum and product are the same for finite abelian groups, and since the empty abelian group is finite, this would make $0 \oplus \emptyset = \emptyset$ using the direct sum (read: direct product) rather than equal to 0 as should be for a coproduct. But what operation would replace it?

### Re: The empty abelian group

As for your axioms of the possibly empty abelian group, you could get rid of the associativity requirement and end up with the axioms of a commutative quasigroup.

### Re: The empty abelian group

I had a think after writing that comment and for abelian groups, the direct sum and product are the same for finite abelian groups, and since the empty abelian group is finite, this would make $0 \oplus \emptyset = \emptyset$ using the direct sum (read: direct product) rather than equal to $0$ as should be for a coproduct. But what operation would replace it?

A notational comment:

I believe the term ‘direct sum’ and symbol $\oplus$ is most appropriate for a biproduct, which is both a product and coproduct in a compatible way. I think it’s confusing to say ‘direct sum’ and write $\oplus$ when you mean the categorical product, as you evidently do here. The fact that you say “direct sum (read: direct product)” shows how confusing this is. The concepts of product and coproduct make sense in any category, while the concepts of ‘direct sum’ and ‘direct product’ are traditionally used in certain particular categories, so when we encounter a brand-new category like ‘possibly empty abelian groups’ it’s good to use the concepts of product and coproduct.

I think maybe you agree with me that in the category of possibly empty abelian groups, the product of $0$ and $\emptyset$ is $\emptyset$.

Posted by: John Baez on August 13, 2020 5:04 PM | Permalink | Reply to this

### Re: The Group With No Elements

Sorry for the possibly dumb question, but why is the set G closed under the right and left divisions you defined?

Posted by: Bruno on August 13, 2020 9:12 PM | Permalink | Reply to this

### Re: The Group With No Elements

If you have a group $G$, you can define right and left division operations

$g/h = g \cdot h^{-1} , \qquad h\backslash g = h^{-1} \cdot g$

and check that they obey the axioms I listed:

$g \cdot (g\backslash h) = h, \qquad g\backslash (g \cdot h) = h$

$(h/g) \cdot g = h, \qquad (h \cdot g)/g = h$

In my post, I sketch the proof that conversely, any set $G$ with an associative product together with left and right division operations obeying these axioms is either a group or the empty set.

Posted by: John Baez on August 13, 2020 10:46 PM | Permalink | Reply to this

### Re: The Group With No Elements

Do these axioms indicate a Grothendieck construction for general semigroups that yields a possibly empty group?

Embedding semigroups in groups is covered by the classic of Clifford and Preston. Obviously cancellation ($a b=a c \Rightarrow b=c$) is necessary, and for a commutative cancellative semigroup the ordered-pair construction works. But Mal’cev gave an example of a semigroup that does not embed in any group. In general it seems the best you can do is to take the group generated by the elements of the semigroup with its multiplication table as the relations. Presumably the corresponding construction is to take the possibly-empty-group generated in the same way, and I suppose that the only difference between the free group and the free possibly-empty-group generated by a set $X$ comes when $X$ is empty and so the free group is the one-element group and the free possibly-empty-group is the empty possibly-empty-group. The empty semigroup would embed quite happily into either of those as a semigroup.