$$\mathcal{P} = \{\Pr(G) : G \text{ a finite group}\},$$

namely the following.

Joseph's conjectures:

J1. All limit points of $\mathcal{P}$ are rational.

J2. $\mathcal{P}$ is well ordered by >.

J3. $\{0\}\cup\mathcal{P}$ is closed.

Earlier this month I uploaded a preprint to the arxiv which proves the first two of these conjectures, and yesterday I gave a talk at the algebra seminar in Oxford about the proof. While preparing the talk I noticed that some aspects of the proof are simpler from an ultrafinitary perspective, basically because ultrafilters can be used to streamline epsilon management, and I gave one indication of this perspective during the talk. In this post I wish to lay out the ultrafinitary approach in greater detail.

Throughout this post we fix a nonprincipal ultrafilter $u\in\beta\mathbf{N}\setminus\mathbf{N}$, and we let $\mathbf{R}^*$ be the ultrapower $\mathbf{R}^\mathbf{N}/u$, where two elements of $\mathbf{R}^\mathbf{N}$ are considered equivalent iff they are equal in $u$-almost-every coordinate. The elements of $\mathbf{R}^*$ are called "nonstandard reals" or "hyperreals". There is a principle at work in nonstandard analysis, possibly called Łoś's theorem, which asserts, without going into the finer details, that all "first-order" things that you do with reals carry over in a natural way to the field of hyperreals, and everything more-or-less works just how you'd like it to. For instance, if $r$ and $s$ are hyperreals then $r<s$ naturally means that the inequality holds in $u$-almost-every coordinate, and similarly the field operations of $\mathbf{R}$ extend naturally, and with these definitions $\mathbf{R}^*$ becomes a totally ordered field. We will seldom spell out so explicitly how to naturally extend first-order properties in this way.

An ultrafinite group $G$ is an ultraproduct $\prod_u G_i = \prod_{i=1}^\infty G_i/u$ of finite groups. Its order $|G| = (|G_i|)/u$ is a nonstandard natural number, and its commuting probability $\Pr(G) = (\Pr(G_i))/u$ is a nonstandard real in the interval $[0,1]$. Joseph's first two conjectures can now be stated together in the following way.

Somewhat similarly, Joseph's third conjecture can be stated in an ultrafinitary way as follows: For every ultrafinite group $G$ there is a finite group $H$ such that $\text{st}\Pr(G)=\Pr(H)$. Here $\text{st}$ is the standard part operation, which maps a finite hyperreal to the nearest real. When phrased in this way it resembles a known result about compact groups. Every compact group $G$ has a unique normalised Haar measure, so we have a naturally defined notion of commuting probability $\Pr(G)$. However every compact group $G$ with $\Pr(G)>0$ has a finite-index center, so $\Pr(G)=\Pr(H)$, where $H$ is the finite group $G/Z(G)$. This is a theorem of Hofmann and Russo. Nevertheless, I find Joseph's third conjecture rather hard to believe.Theorem:The commuting probability of every ultrafinite group $G$ has the form $q+\epsilon$, where $q$ is a standard rational and $\epsilon$ is a nonnegative infinitesimal.

For us the most important theorem about commuting probability is a theorem of P. Neumann, which states that if $\epsilon>0$ then every finite group $G$ such that $\Pr(G)\geq\epsilon$ has a normal subgroup $H$ such that $|G/H|$ and $|[H,H]|$ are both bounded in terms of $\epsilon$. To prove the above theorem we need the following "amplified" version:

Here if $G = \prod_u G_i$ we say that $S\subset G$ isTheorem (Neumann's theorem, amplified, ultrafinitary version):Every ultrafinite group $G$ has an internal normal subgroup $H$ such that $[H,H]$ is finite and such that almost every pair $(x,y)\in G^2$ such that $[x,y]\in[H,H]$ is contained in $H^2$.

*internal*if $S$ is itself an ultraproduct $\prod_u S_i$ of subsets $S_i\subset G_i$, and "almost every" needs little clarification because the set of pairs in question is an internal subset of $G^2$. (Otherwise we would need to introduce Loeb measure.)

**Proof:**If $\text{st}\Pr(G)=0$ the theorem holds with $H=1$, so we may assume $\text{st}\Pr(G)>0$. By Neumann's theorem $G$ has an internal normal subgroup $K_0$ of finite index such that $[K_0,K_0]$ is finite. Since $G/K_0$ is finite there are only finitely many normal subgroups $K\leq G$ containing $K_0$ and each of them is internal, so we may find normal subgroups $K,L\leq G$ containing $K_0$ such that $[K,L]$ is finite, and which are maximal subject to these conditions.

Suppose that a positive proportion of pairs $(x,y)\in G^2$ outside of $K\times L$ satisfied $[x,y]\in[K,L]$. Then we could find $(x,y)\in G^2\setminus (K\times L)$, say with $x\notin K$, such that for a positive proportion of $(k,l)\in K\times L$ we have $[xk,yl]\in[K,L]$. After a little commutator algebra one can show then that for a positive proportion of $l\in L$ we have $[x,l]\in[K,L]$, or in other words that the centraliser

$$ N_0 = C_{L/[K,L]}(x) = C_{L/[K,L]}(\langle K,x\rangle)$$

of $x$ in $L/[K,L]$ has finite index. But this implies that the largest normal subgroup contained in $N_0$, namely

$$ N = C_{L/[K,L]}(K'),$$

where $K'$ is the normal subgroup of $G$ generated by $K$ and $x$, also has finite index. Since certainly $K\leq C_{K'/[K,L]}(L)$ a classical theorem of Baer implies that

$$[K'/[K,L], L/[K,L]] = [K',L]/[K,L]$$

is finite, and hence that $[K',L]$ is finite, but this contradicts the maximality of $K$ and $L$.

Hence almost every pair $(x,y)\in G^2$ such that $[x,y]\in[K,L]$ is contained in $K\times L$, and thus also in $L\times K$, so the theorem holds for $H=K\cap L$.$\square$

Now let $G$ be any ultrafinite group $G$ and let $H$ be as in the theorem. Then

$$\Pr(G) = \Pr(H) + \epsilon,$$

where $\epsilon$ is nonnegative and infinitesimal. Thus it suffices to show that $\Pr(H)$ is a standard rational whenever $[H,H]$ is finite. Note in this case that Hall's theorem implies that the second centre

$$Z_2(H) = \{h\in H : [h,H]\subset Z(H)\}$$

has finite index. One can complete the proof using a little duality theory of abelian groups, but the ultrafinite perspective adds little here so I refer the reader to my paper.

The other thing I noticed while preparing my talk is that the best lower bound I knew for the order type of $\mathcal{P}$, $\omega^2$, is easy to improve to $\omega^\omega$, just by remembering that $\mathcal{P}$ is a subsemigroup of $(0,1]$. In fact the order type of a well ordered subsemigroup of $(0,1]$ is heavily restricted: it's either $0$, $1$, or $\omega^{\omega^\alpha}$ for some ordinal $\alpha$. This observation reduces the possibilities for the order type of $\mathcal{P}$ to $\{\omega^\omega,\omega^{\omega^2}\}$. I have no idea which it is!