---
abstract: 'A pseudofinite group satisfying the uniform chain condition on centralizers up to finite index has a big finite-by-abelian subgroup.'
address: 'Université Lyon 1; CNRS; Institut Camille Jordan UMR 5208, 21 avenue Claude Bernard, 69622 Villeurbanne-cedex, France'
author:
- 'Frank O. Wagner'
date: '1/9/2015'
title: 'Pseudofinite $\widetilde{\mathfrak M_c}$-groups'
---

[^1]

Introduction {#introduction .unnumbered}
============

We generalize the results of Elwes, Jaligot, MacPherson and Ryten [@ER08; @EJMR11] about pseudofinite superstable groups of small rank to the pseudofinite context, possibly of infinite rankl

Rank
====

A [*dimension*]{} on a theory $T$ is a function $\dim$ from the collection of all interpretable subsets of a monster model to $\R^{\ge0}\cup\{\infty\}$ satisfying

-   [*Invariance:*]{} If $a\equiv a'$ then $\dim(\p(x,a))=\dim(\p(x,a'))$.

-   [*Algebraicity:*]{} If $X$ is finite, then $\dim(X)=0$.

-   [*Union:*]{} $\dim(X\cup Y)=\max\{\dim(X),\dim(Y)\}$.

-   [*Fibration:*]{} If $f:X\to Y$ is a interpretable surjection whose fibres have constant dimension $r$, then $\dim(X)=\dim(Y)+r$.

For a partial type $\pi$ we put $\dim(\pi)=\inf\{\dim(\p):\pi\proves\p\}$. We write $\dim(a/B)$ for $\dim(\tp(a/B))$.It follows that $X\subseteq Y$ implies $\dim(X)\le\dim(Y)$, and $\dim(X\times 
Y)=\dim(X)+\dim(Y)$. Moreover, any partial type $\pi$ can be completed to a complete type $p$ with $\dim(\pi)=\dim(p)$. A dimension is [*additive*]{} if it satisfies

-   [*Additivity:*]{} $\dim(a,b/C)=\dim(a/b,C)+\dim(b/C)$.

Additivity is clearly equivalent to [*fibration*]{} for type-definable maps.Examples for an additive dimension include:

1.  coarse pseudofinite dimension (in the expansion by cardinality comparison quantifiers, in order to ensure invariance);

2.  Lascar rank, $SU$-rank or -rank, possibly localised at some $\emptyset$-invariant family of types;

3.  for any ordinal $\alpha$, the coefficient of $\omega^\alpha$ in one of the ordinal-valued ranks in (2) above.

In any of the examples, we put $\dim(X)=\infty$ if $\dim(X)>n$ for all $n<\omega$.Note that additivity in the above examples follows from the Lascar inequalities for the corresponding rank in examples (2) and (3) and from [@Hr12] in example (1). Let $G$ be a type-definable group with an additive dimension, and suppose $0<\dim(G)<\infty$. A partial type $\pi(x)$ implying $x\in G$ is [*wide*]{} if $\dim(\pi)=\dim(X)$. It is [*broad*]{} if $\dim(\pi)>0$. An element $g\in G$ is [*wide*]{}/[*broad*]{} over some parameters $A$ if $\tp(g/A)$ is. We say that $\pi$ is [*negligible*]{} if $\dim(\pi)=0$.An additive dimension is invariant under definable bijections.Let $f$ be an $A$-definable bijection, and $a\in\text{dom}(f)$. Then $\dim(f(a)/A,a)=0$ and $\dim(a/A,f(a))=0$, whence $$\dim(a/A)=\dim(a,f(a)/A)=\dim(f(a)/A).\qquad\qed$$ If $\dim(a/A)<\infty$, we say that $a\indd_AB$ if $\dim(a/A)=\dim(AB)$.It follows that $\indd$ satisfies transitivity. Moreover, for any partial type $\pi$ there is a comple type $p\supseteq\pi$ (over any set of parameters containing dom$(\pi)$) with $\dim(p)=\dim(\pi)$, so $\indd$ satisfies extension. Let $\dim$ be an additive dimension. If $\dim(a/A)<\infty$ and $\dim(b/A)<\infty$, then $a\indd_Ab$ if and only if $b\indd_Aa$, if and only if $\dim(a,b/A)=\dim(a/A)+\dim(b/A)$.Obvious.

\[generics\] Let $G$ be a type-definable group with an additive dimension, and suppose $0<\dim(G)<\infty$. If $g$ is wide over $A$ and $h$ is wide over $A,g$, then $gh$ and $hg$ are wide over $A,g$ and over $A,h$.As dimension is invariant under definable bijections $$\dim(gh/A,g)=\dim(h/A,g)=\dim(G).$$ The other statements follow similarly, possibly using symmetry.

Big abelian subgroups
=====================

\[centralizer\] Let $G$ be a pseudofinite group, and $\dim$ an additive dimension on $G$. Then there is an element $g\in G\setminus\{1\}$ such that $\dim(C_G(g))\ge\frac13\dim(G)$.Suppose first that $G$ has no involution. If $G\equiv\prod_I G_i/\U$ for some family $(G_i)_I$ of finite groups and some non-principal ultrafilter $\U$, then $G_i$ has no involution for almost all $i\in I$, and is soluble by the Feit-Thompson theorem. So there is $g_i\in G_i\setminus\{1\}$ such that $\langle g_i^{G_i}\rangle$ is commutative. Put $g=[g_i]_I\in 
G\setminus\{1\}$. Then $\langle 
g^G\rangle$ is commutative and $g^G\subseteq C_G(g)$. As $g^G$ is in definable bijection with $G/C_G(g)$, we have $$\dim(C_G(g))\ge\dim(g^G)=\dim(G/C_G(g))=\dim(G)-\dim(C_G(g)).$$ In particular $\dim(C_G(g))\ge\frac12\dim(G)$.

Now let $i\in G$ be an involution and suppose $\dim(C_G(i))<\frac13\dim(G)$. Then $$\dim(i^G)=\dim(G/C_G(i))=\dim(G)-\dim(C_G(i))>\frac23\dim(G)$$ and there is $j=i^g\in i^G$ with $\dim(j)\ge\frac23\dim(G)$. Note that $$\dim(C_G(j))=\dim(C_G(i)^g)=\dim(C_G(i))<\frac13\dim(G).$$ Then $$\dim(j^Gj)=\dim(G/C_G(j))=\dim(G)-\dim(C_G(j))>\frac23\dim(G)$$ and there is $h=j^{g'}j\in j^Gj$ with $\dim(h/j)\ge\frac23\dim(G)$. Note that $h$ is inverted by $j$. By additivity, $$\dim(j/h)=\dim(j,
h)-\dim(h)\ge\dim(h/j)+\dim(j)-\dim(G)>\frac13\dim(G).$$ If $H=\{x\in G:h^x=h^{\pm1}\}$, then $H$ is an $h$-definable subgroup of $G$, and $C_G(h)$ has index two in $H$. Since $j\in H$, we have $$\dim(C_G(h))=\dim(H)\ge\dim(j/h)>\frac13\dim(G).\qed$$

\[abelian\] Let $G$ be a pseudofinite -group, and $\dim$ an additive dimension on $G$. Then $G$ has a definable broad finite-by-abelian subgroup $Z$. More precisely, $Z=\Z(C)$ where $C$ is a minimal broad centralizer (up to finite index) of a finite tuple.By the  condition, there is a broad centralizer $C$ of some finite tuple, such that $C_C(g))$ is not broad for any $g\in C\setminus\Z(C)$. Put $Z=\Z(C)$, a definable finite-by-abelian normal subgroup of $C$. If $Z$ is broad, we are done. Otherwise $\dim(Z)=0$, and $$\dim(C/Z)=\dim(C)-\dim(Z)=\dim(C).$$ For $g\in C\setminus Z$ we have $\dim(C_C(g))=0$, whence for $\bar g=gZ$ we have $$\begin{aligned}\dim(\bar g^{C/Z})&=\dim(g^CZ/Z)\ge\dim(g^C)-\dim(Z)\\
&=\dim(C)-\dim(C_C(g))-\dim(Z)=\dim(C/Z).\end{aligned}$$ Hence for all $\bar g\in (C/Z)\setminus\{\bar1\}$ we have $$\dim(C_{C/Z}(\bar g))=\dim(C/Z)-\dim(\bar g^{C/Z})=0.$$ As $C$ and $Z$ are definable, $C/Z$ is again pseudofinite, contradicting Proposition \[centralizer\].

Theorem \[abelian\] holds in particular for any pseudofinite -group with the pseudofinite counting measure. Note that the broad finite-by-abelian subgroup is defined in the pure group, using centralizers and almost centres. Moreover, the -condition is just used in $G$, not in the section $C/Z$.

\[corTh\] A superrosy pseudofinite group with $\Ut(G)\ge\omega^\alpha$ has a definable finite-by-abelian subgroup $A$ with $\Ut(A)\ge\omega^\alpha$.A superrosy group is . If $\alpha$ is minimal with $\Ut(G)<\omega^{\alpha+1}$ and we put $$\dim(X)\ge n\quad\text{if}\quad\Ut(X)\ge\omega^\alpha\cdot n,$$ then $\dim$ is an additive dimension with $0<\dim(G)<\infty$. The assertion now follows from Theorem \[abelian\].

For any $d,d'<\omega$ there is $n=n(d,d')$ such that if $G$ is a finite group without elements $(g_i:i\le d')$ such that $$|C_G(g_i:i<j):C_G(g_i:i\le j)|\ge d$$ for all $j\le d'$, then $G$ has a subgroup $A$ with $|A'|\le n$ and $n\,|A|^n\ge|G|$.If the assertion were false, then given $d,d'$, there were a sequence $(G_i:i<\omega)$ of finite groups satisfying the condition, such that $G_i$ has no subgroup $A_i$ with $|A_i'|\le i$ and $i\,|A_i|^i\ge|G_i|$. But any non-principal ultraproduct $G=\prod G_n/\U$ is a pseudofinite -group, and has a definable subgroup $A$ with $A'$ finite and $\dim(A)\ge\frac1n\dim(G)$ for some $n<\omega$. Unravelling the definition of the pseudofinite counting measure (and possibly increasing $n$) we get $|A_i'|\le n$ and $n\cdot|A_i|^n\ge|G_i|$ for almost all $i<\omega$, a contradiction for $i\ge n$.

Pseudofinite -groups of dimension $2$
=====================================

\[dim2\] Let $G$ be a pseudofinite -group, and $\dim$ an additive integer-valued dimension on $G$. If $\dim(G)=2$, then $G$ has a broad definable finite-by-abelian subgroup $N$ whose normalizer is wide. Note that since $\dim$ is integer-valued, a broad subgroup has dimension at least $1$. By Corollay \[corTh\], if $C$ is a minimal broad centralizer (up to finite index), then $A=\Z(C)$ is broad and finite-by-abelian.

If $A$ is commensurate with $A^g$ for all $g\in G$, then commensurativity is uniform by compactness. So by Schlichting’s Theorem there is a normal subgroup $N$ commensurate with $A$. But then $\Z(N)$ contains $A\cap N$, has finite index in $N$ and is broad; since it is characteristic in $N$, it is normal in $G$ and we are done.

Suppose $g\in\N_G(H)$ is such that $A$ is not commensurable with $A^g$. Then clearly $C$ cannot be commensurable with $C^g$, and $$\dim(A\cap A^g)\le\dim(C\cap C^g)=0.$$ Hence $$\dim(AA^g)\ge\dim(AA^g/(A\cap A^g))=\dim(A/(A\cap 
A^g)+\dim(A^g/(A\cap A^g))=2\,\dim(A).$$ As $A$ is broad and $\dim(G)=2$, we have $\dim(A)=1$.

Choose some $d$-independent wide $a,b_0,c_0$ in $A$ over $g$. Then $\dim(a^gb_0/g,c_0)=2$, and $$2\ge\dim(c_0a^gb_0/g)\ge\dim(c_0a^gb_0/g,c_0)=\dim(a^gb_0/g,
c_0)=\dim(a^gb_0/g)=2$$ and $c_0\indd_gc_0a^gb_0$. Thus $c_0$ is wide in $A$ over $g,c_0a^gbc_0$. Similarly, $b_0$ is wide in $A$ over $g,c_0a^gbc_0$.

Choose $d,b_1,c_1\equiv_{c_0a^gb_0,g}a,b_0,c_0$ with $d,b_1,c_1\indd_{c_0a^gb_0,g}a,b_0,c_0$. Then $c_0a^gb_0=c_1d^gb_1$, whence $$a^gb=cd^g,$$ where $c=c_0^{-1}c_1$ and $b=b_0b_1^{-1}$. Moreover, $$\dim(b/a,g)\ge\dim(b_0b_1^{-1}/a,b_0,c_0,g)\ge\dim(b_1/a,b_0,c_0,
g)=\dim(b_1/c_0a^gb_0,g)=1,$$ so $b$ is wide in $A$ over $g,a$. Similarly, $c$ is wide in $A$ over $g,d$.

Let $x$ and $y$ be two d-independent wide elements of $C_A(a,b,c,d)$ over $a,b,c,d,g$. Then they are d-independent wide in $A$, and for $z=xgy$ we have $$\dim(z/a,b,c,d,g)=2,$$ so $z$ is wide in $G$ over $g,a,b,c,d$. Moreover $$a^zb=a^{xgy}b=a^{gy}b^y=(a^gb)^y=(cd^g)^y=c^yd^{gy}=cd^{xgy}=cd^z.$$ Choose $z'\in G$ with $z'\equiv_{a,b,c,d}z$ and $z'\indd_{a,b,c,d}z$, and put $r=z'^{-1}z$. Then $r$ is wide in $G$ over $g,a,b,c,d,z$, and $$a^zb^r=a^{z'r}b^r=(a^{z'}b)^r=(cd^{z'})^r=c^rd^{z'r}=c^rd^z.$$ Hence $$c^{-1}a^zb=d^z=c^{-r}a^zb^r.$$ Putting $b'=bb^{-r}$ and $c'=cc^{-r}$, we obtain $$a^zb'=c'a^z,$$ where $a$ is wide in $A$. As $r\indd_{g,a,b,c,d}z$ we have $$\dim(z/a,b',c')\ge\dim(z/a,b,c,d,r)=\dim(z/a,b,c,d)=2,$$ and $z$ is wide in $G$ over $a,b',c'$. If $z''\equiv_{a,b',c'}z$ with $z''\indd_{a,b',c'}z$, then $a^{z''}b'=c'a^{z''}$, whence $$b'^{a^z}=c'=b'^{a^{z''}}$$ and $a^za^{-z''}=a'^z$ commutes with $b'$, where $a'=aa^{-z''z^{-1}}$. $\dim(b')\ge1$, $\dim(a'/b')\ge1$, $\dim(z/a',b')\ge2$ and $\dim(a'^z/b')\ge1$.If $\dim(b'/b)=0$, then $\dim(r/b,b')=\dim(r/b)=2$. Choose $r'\equiv_{b,b'}r$ with $r'\indd_{b,b'}r$. So $b^r=b'^{-1}b=b^{r'}$ and $r'r^{-1}\in C_G(b)$. Since $r'r^{-1}$ is wide in $G$ over $b$, so is $C_G(b)$, and it has finite index in $G$ by minimality. Thus $b\in\Z(G)$ and $\dim\Z(G)\ge1$, so we can take $N=\Z(G)$. Thus we may assume $\dim(b')\ge\dim(b'/b)\ge1$.

Next, note that $z$ and $z''z^{-1}$ are wide and $d$-independent over $a,b',c'$ by Lemma \[generics\]. An argument similar to the first paragraph yields $\dim(a'/b')\ge\dim(a'/a,b',z''z^{-1})\ge1$ and $\dim(a'^z/b')\ge1$.

To finish, note that $\dim(C_G(b'))\ge\dim(a'^z/b')\ge1$. If $\dim(C_G(b'))=2$, then $b'\in\Z(G)$ and $\dim(\Z(G))\ge\dim(b')\ge1$, so we are done again. Otherwise $\dim(C_G(b'))=1$. By the -condition there is a broad centralizer $D\le C_G(b')$ of some finite tuple, minimal up to finite index, and $\dim(D)=\dim(C_G(b'))=1$, whence $\dim(C_G(b')/D)=0$. Choose $z^*\in G$ with $z^*\equiv_{a',b'}z$ and $z^*\indd_{a',b'}z$, and put $h=z^{*-1}z$. Then $a'^{z^*}\in C_G(b')$, so $a'^z\in C_G(b',b'^h)$. Moreover, $h$ is wide in $G$ over $a',b'$ and $d$-independent of $z$, so $$\dim(C_G(b',b'^h))\ge\dim(a'^z/b',h)=\dim(a'^z/b')\ge1$$ and $\dim(C_G(b')/C_G(b',b'^h))=0$. It follows that $$\begin{aligned}\dim(D/D\cap D^h)&=\dim(D/(D\cap C_G(b'^h)))+\dim((D\cap 
C_G(b'^h))/(D\cap D^h))\\
&\le\dim(C_G(b')/C_G(b',b'^h))+\dim(C_G(b'^h)/D^h)=0,\end{aligned}$$ whence $\dim(D\cap D^h)=1$, and $D^h$ is commensurable with $D$ by minimality. Thus $$\dim(\N_G(D))\ge\dim(h/b')=2$$ and $\N_G(D)$ is wide in $G$. Since $D$ is finite-by-abelian by Corollary \[corTh\], we finish as before, using Schlichting’s Theorem.

\[soluble\] Let $G$ be a pseudofinite group whose definable sections are , and $\dim$ an additive integer-valued dimension on $G$. If $\dim(G)=2$, then $G$ has a definable wide soluble subgroup.By Theorem \[dim2\], there is a definable finite-by-abelian group $N$ such that $N_G(N)$ is wide. Replacing $N$ by $C_N(N')$, we may assume that $N$ is (finite central)-by-abelian. If $\dim(N)=2$ we are done. Otherwise $\dim(N_G(N)/N)=1$; by Corollary \[corTh\] there is a definable finite-by-abelian subgroup $S/N$ with $\dim(S/N)=1$. As above we may assume that $S/N$ is (finite central)-by-abelian, so $S$ is soluble. Moreover, $$\dim(S)=\dim(N)+\dim(S/N)=1+1=2,$$ so $S$ is wide in $G$.

A pseudofinite superrosy group $G$ with $\omega^\alpha\cdot2\le\Ut(G)<\omega^\alpha\cdot3$ has a definable soluble subgroup $S$ with $\Ut(S)\ge\omega^\alpha\cdot2$.Superrosiness implies that all definabel sections of $G$ are . We put $$\dim(X)=n\quad\Leftrightarrow\quad\omega^\alpha\cdot 
n\le\Ut(X)<\omega^\alpha\cdot(n+1).$$ This defines an additive dimension with $\dim(G)=2$. The result now follows from Corollary \[soluble\].

[99]{} Richard Elwes and Mark Ryten. , Math. Log. Quart. 54, No. 4, 374–386 (2008).

R. Elwes, E. Jaligot, H.D. Macpherson and M. Ryten, , Proc. London Math. Soc. (3) 103 (2011), 1049-1082.

E. Hrushovski. , J. Amer. Math. Soc. 25(1):189–243, 2012.

[^1]: Partially supported by ValCoMo (ANR-13-BS01-0006)
