### 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.

## 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.