Is there a universal group in NFU?

Question

This is a follow-up to is there a universal group.

I'm curious about whether this holds in NFU (+ pairing, infinity, and choice).

I haven't seen a reference for pairing before, but I'm going to make $(\cdot, \cdot)$ a binary function symbol that takes two arguments of type level $n$ and returns a result also of type level $n$, for the purposes of assessing stratification.

Other than how $(\cdot, \cdot)$ interacts with stratification and the defining property of pairs, $[\forall x y z w]((x, y) = (z, w) \leftrightarrow (x = z \land y = w))$, it has no other properties.

I'm wondering whether all the groups in NFU embed into the group of all bijections.

We know from this answer that NFU can be a little annoying to work with, mostly for reasons related to stratification. Here's the defining formula for the group of all bijections, superscripts are type annotations, intended to make it easier to verify that the formula is stratified.

$F(x)$ expresses that $f$ is a binary relation.

$$ F(x^4) \;\;\text{is}\;\; [\forall z^3 \in x^4][\exists a^3][\exists b^3](z^3 = (a^3, b^3)) $$

$\varphi(x)$ expresses that $x$ is a bijection of the space.

$$ \varphi(x^4) \;\; \text{is}\;\; \\ F(x^4) \land \\ [\forall z^3][\exists! a^3 \in x^4][\exists b^3](a^3 = (b^3, z^3)) \land \\ [\forall z^3][\exists! a^3 \in x^4][\exists b^3](a^3 = (z^3, b^3)) $$

The conjuncts are intended to mean:

1. $x$ is a binary relation
2. For every element $z$ whatsoever, there exists a unique pair $a$ in $x$, such that the second element of $a$ is z.
3. For every element $z$ whatsoever, there exists a unique pair $a$ in $x$, such that the first element of $a$ is z.

Here is my proposed universal group $G$, the set of all bijections of the entire space.

$$ G = \{ x : \varphi(x) \} $$

$G$ exists by stratified comprehension. Does every group embed into it?

My initial thoughts are that every group must because every group is isomorphic to a subgroup of its automorphism group, and those automorphisms can be extended to bijections of the space (at least conceptually).

I'm uncomfortable with working intuitively with NFU though, so I'm wondering whether every group does indeed embed into $G$ (and also whether $G$ is even well-defiend)

Answer 1

Here is a failed attempt at proving that every group is a subgroup of $G$.

I also have not yet reached the "post-rigorous stage" of working with NFU; I feel like I roughly understand the "true meaning" of stratification and what it entails, but I'm not yet comfortable working fluently without explicitly checking everything.

Let $V$ be the universal set (that is, the set of all objects; in other words, the union of all sets), and let $V!$ denote the set of all bijections of $V$. I think it should be uncontroversial that $V!$ can be made into a group; call that group $G$.

Now, suppose that we have any group $H$. We wish to show that $H$ is a subgroup of $G$. Specifically, let's assume that we have a set $H$; a singleton subset $\{e\}$ of $H$; and a function $f : (H \times H) \to H$ satisfying the group axioms, with $e$ as the identity element.

Let $\tbinom{H}{1}$ denote the set of all subsets of $H$ of cardinality $1$. We can construct the "curried" version of $f$ as follows:

$$f' : \tbinom{H}{1} \to (H \to H)$$ $$f' = \{ (X, j) : X = \{x\}, \ j \in H^H, \ [\forall y \in H] (j(y) = f(x, y)) \}$$

(For stratification, note that $x$ and $y$ appear at the first level; $X$, $j$, $H$, and $f$ appear at the second level; and $H^H$ appears at the third level. The resulting function $f'$ can be thought of as being at the third level.)

Next, we can define a "promotion function" $p : (H \to H) \to (V \to V)$ as

$$ p(j)(x) = \begin{cases} j(x) & x \in H \\ x & \text{otherwise.} \end{cases} $$

(We have $x$ at the first level, $j$ and $H$ (and $V$, but it doesn't matter) at the second level, $p$ at the third level. Technically, we're treating $H$ as a constant and so its level doesn't matter either, but if we do handle the level of $H$ correctly, so to speak, that should mean that the entire construction we're performing right now can be performed by a single function.)

If $j$ is a bijection, then $p(j)$ is a bijection as well.

At this point, we can compose these functions to get an "embedding function" $m : \tbinom{H}{1} \to (V \to V)$, which takes any singleton subset of $H$ and returns an element of $G$.

After writing all of that, it finally occurs to me that this is not what we wanted. We wanted to have a function $H \to (V \to V)$, which takes any element of $H$ and returns an element of $G$. Oops.

A priori, the set $\tbinom{H}{1}$ is not even a group. However, it should be straightforward to take the group structure on $H$ and use it to define a group structure on $\tbinom{H}{1}$. Having done that, it should also be possible to show that our function $m$ is, in fact, an embedding of groups.

The good news is that we've successfully shown that the group $\tbinom{H}{1}$ is a subgroup of $G$. The bad news is that I don't believe there's any particular useful relationship between the group $H$ and the group $\tbinom{H}{1}$. I'm quite sure that the two groups are not isomorphic in general. I'm also pretty sure that $H$ is not even guaranteed to be isomorphic to a subgroup of $\tbinom{H}{1}$; after all, not every set $S$ is equinumerous with a subset of the set $\tbinom{S}{1}$, and I would expect the same to be true for groups. (Most famously, the universal set $V$ is strictly larger than $\tbinom{V}{1}$.)

So, what we have shown is that every group whose elements are all singletons is a subgroup of $G$, but we haven't shown the same for groups that are so large (or so "strange") that they're not isomorphic to any group whose elements are all singleton.

Only after writing all of the above do I realize this: all of the elements of $G$ are sets, and all sets can be mapped onto different elements of $G$, so the cardinality of $G$ is the cardinality of $\mathcal{P}(V)$, the set of all sets. This means that $G$ is strictly smaller than $V$, and so if there exists any group structure at all on $V$, then that group is not a subset of $G$.

So that leaves us with a few possibilities:

• There is no universal group.
• There is a group structure on $V$, but there is a universal group (which, for the above reason, cannot be $G$).
• There is no group structure on $V$, and $G$ is in fact a universal group.
• There is no group structure on $V$, and $G$ fails to be a universal group for some other reason, but there is nevertheless a universal group.