# On non-abelian dp-minimal groups

Atticus Stonestrom (Email<EMAIL_ADDRESS>
Abstract: Let $G$ be a dp-minimal group; we prove some consequences of several
different hypotheses on $G$. First, if $G$ is torsion-free, then it is
abelian. Second, if $G$ admits a distal f-generic type, then it is virtually
nilpotent; we prove this by equipping the quotient of $G$ by its FC-center in
this case with a valued group structure. Finally, if $G$ has the uniform chain
condition, for example if $G$ is stable, then $G$ is virtually solvable.

Acknowledgements: I would like to thank Caroline Terry for interesting
discussion of the Gowers inverse question, and Anand Pillay for his
encouragement and for feedback on an earlier version of this paper. After
completing the work in this paper, I learned that Eran Alouf and Frank Wagner
had independently proved that any stable dp-minimal group is virtually
solvable, and I would like to thank both for stimulating subsequent
discussion.

I am especially grateful to Frank Wagner for numerous very helpful comments,
most especially for pointing out a mistake in an earlier version of the paper.
I would also like to thank him for helpful references on MC-groups and in
particular for the proof of Fact 2.7.

Notation: Throughout $T$ will be an NIP theory with a monster model
$\mathfrak{C}$, and $G$ will be a group definable in $T$. We use
$\mathop{\mathchoice{\kern 5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower
3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\displaystyle{}}{\kern
5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox
to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\textstyle{}}{\kern 5.14375pt\hbox
to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern
5.14375pt\scriptstyle{}}{\kern 5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower
3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\scriptscriptstyle{}}}$
to denote non-forking independence. We use the ‘left-normed’ convention for
commutators, so that $[a,b]=a^{-1}b^{-1}ab$ and
$[a_{1},\dots,a_{n+1}]=[[a_{1},\dots,a_{n}],a_{n+1}]$. We use $\sim$ to denote
the relation of being conjugate in $G$. Conjugation is for us a right action,
with $a^{b}=b^{-1}ab$ for all $a,b\in G$. We use one piece of non-standard
notation: given sets $A_{1},\dots,A_{n}$, we write $[A_{1},\dots,A_{n}]$ to
denote the set of commutators $\\{[a_{1},\dots,a_{n}]:a_{i}\in A_{i}\\}$.

##  1 Introduction

‘Dp-minimality’, first introduced in [23] and [24] and then later isolated as
a notion in [18], is a kind of abstract ‘one-dimensionality’ condition which
has proved a very fruitful strengthening of the NIP, both from the perspective
of ‘pure’ model theory and from the perspective of its algebraic implications.
Here we are concerned with dp-minimal groups. The typical examples, including
for instance superstable groups of $U$-rank $1$,
$(\mathbb{Z}/4\mathbb{Z})^{\omega}$, and the multiplicative and additive
groups of $\mathbb{R},\mathbb{Q}_{p},$ and ACVFs, are all virtually abelian,
but there exist dp-minimal groups which are not virtually abelian [29], and it
is an open question, asked for example in [4], whether every dp-minimal group
is virtually nilpotent. This is known to be the case under several stronger
hypotheses. For instance, C-minimal valued groups are virtually nilpotent
[30], $\omega$-categorical dp-minimal groups are virtually abelian [7],
profinite dp-minimal groups are virtually abelian [6], and left-ordered dp-
minimal groups are abelian [8].

In this paper we investigate this question under several other hypotheses.
First we show that any torsion-free dp-minimal group is abelian; this is the
content of Section 4. We then show that any dp-minimal group $G$ which admits
a distal f-generic type is virtually nilpotent. This is deduced from a more
general analysis, in which we construct a valued group structure on the
quotient of $G$ by its FC-center in such a way that the structure inherited by
$G^{00}/Z(G^{00})$ is ‘plain’ (see Section 2.1.4); this is the content of
Section 5. Finally, we show that any dp-minimal group with the ‘uniform chain
condition’, ie in which there is no infinite descending chain of uniformly
definable subgroups, is virtually solvable111In private communication, Frank
Wagner has shared with us a proof that solvable abelian-by-(finite exponent)
groups with the intersection chain condition are virtually nilpotent, and it
accordingly follows from Fact 2.15 and 2.10 that dp-minimal ucc groups are
virtually nilpotent.; this applies for example to stable or more generally
rosy dp-minimal groups. In the stable case, this was independently proved in
unpublished work of Eran Alouf and Frank Wagner, via a different approach than
we use here.

##  2 Preliminaries

###  2.1 Groups

Let $(H,\cdot)$ be a group.

####  2.1.1 Radicals

Recall that the Fitting subgroup of $H$ is the subgroup
$F(H)\trianglelefteqslant H$ generated by all nilpotent normal subgroups of
$H$, and that the solvable radical of $H$ is the subgroup
$R(H)\trianglelefteqslant H$ generated by all solvable normal subgroups of
$H$. The following can be found for instance in [12]:

######  Fact 2.1.

If $F(H)$ is nilpotent, then it is a definable subgroup of $(H,\cdot)$. If
$R(H)$ is solvable, then it is a definable subgroup of $(H,\cdot)$.

We will freely use the following throughout the paper.

######  Remark 2.2.

If $H$ is virtually nilpotent (resp. virtually solvable), then the Fitting
subgroup (resp. solvable radical) of $H$ is nilpotent (resp. solvable). Thus,
by Fact 2.1, a definable group is virtually nilpotent (resp. virtually
solvable) iff it has a definable finite-index nilpotent (resp. solvable)
subgroup.

####  2.1.2 Commutators

Throughout this paper we will use the non-standard notation that, for subsets
$A_{1},\dots,A_{n}\subseteq H$, $[A_{1},\dots,A_{n}]$ denotes the set of
commutators $\\{[a_{1},\dots,a_{n}]:a_{i}\in A_{i}\\}$; when $A_{i}=\\{a\\}$
is a singleton we will just write
$[A_{1},\dots,A_{i-1},a,A_{i+1},\dots,A_{n}]$. The following is standard:

######  Lemma 2.3.

Suppose $A_{1},\dots,A_{n}\subseteq H$ are each fixed setwise under
conjugation by elements of $H$. Then, for any $k\in\\{0\\}\cup[n]$, the set of
$g\in H$ such that $[A_{1},\dots,A_{k},g,A_{k+1},\dots,A_{n}]=\\{1\\}$ is a
normal subgroup of $H$.

###### Proof.

Noting that $[A_{1},\dots,A_{k},g]=[g^{-1},[A_{1},\dots,A_{k}]^{g}]$, and that
$[A_{1},\dots,A_{k}]$ is fixed setwise under conjugation by elements of $H$,
it suffices to show the claim when $k=0$, ie to show that the set of $g\in H$
with $[g,A_{1},\dots,A_{n}]=\\{1\\}$ is a normal subgroup of $H$. This follows
by induction on $n$; in the case $n=1$ the group is just $C_{H}(A_{1})$.

For the inductive step, let $U$ denote the set of $u\in H$ with
$[u,A_{2},\dots,A_{n}]=\\{1\\}$ and let $V$ denote the set of $v\in H$ with
$[v,A_{1},\dots,A_{n}]=\\{1\\}$. Since
$[v^{h},A_{1},\dots,A_{n}]=[v,A_{1},\dots,A_{n}]^{h}$, $V$ is fixed setwise
under conjugation by elements of $H$. Now, by the inductive hypothesis, $U$ is
a normal subgroup of $H$. If $g\in V$, then $[g,A_{1}]\subseteq U$, whence
$[g^{-1},A_{1}]=[g,A_{1}]^{-g^{-1}}\subseteq U$, so that $g^{-1}\in V$.
Finally, if $g,h\in V$, then $[g,A_{1}],[h,A_{1}]\subseteq U$, whence
$[g,A_{1}]^{h}[h,A_{1}]\subseteq U$; since
$[gh,A_{1}]\subseteq[g,A_{1}]^{h}[h,A_{1}]$ we thus have $gh\in V$, as
desired. ∎

####  2.1.3 Torsion-free groups

We will need some facts about torsion-free groups; see for example 5.2.19 and
14.5.9 in [21].

######  Fact 2.4.

Suppose $H$ is nilpotent and torsion-free. Then $H/Z(H)$ is torsion-free.

######  Fact 2.5.

If $H$ is torsion-free and $[H:Z(H)]$ is finite, then $H$ is abelian.

The following is an easy computation which we put here for convenience.

######  Lemma 2.6.

Suppose $H$ is abelian-by-torsion and that $a\in H$ is a non-torsion element.
If $i,j\in\mathbb{Z}$ are such that $a^{i}\sim a^{j}$, then $|i|=|j|$.

###### Proof.

Let $g\in H$ be such that $(a^{i})^{g}=a^{j}$. For notational convenience, let
$f:H\to H$ denote the conjugation-by-$g$ map. Since $H$ is abelian-by-torsion,
there is some $n>0$ such that $f^{n}(a^{n})=a^{n}$. Let $b=a^{n}$; then $b$ is
still non-torsion, and we have $f(b)^{i}=f(a^{i})^{n}=a^{jn}=b^{j}$. By
induction on $k$ it follows that $f^{k}(b)^{i^{k}}=b^{j^{k}}$ for all
$k\geqslant 1$, since

$f^{k+1}(b)^{i^{k+1}}=f(f^{k}(b)^{i^{k}})^{i}=f(b^{j^{k}})^{i}=(f(b)^{i})^{j^{k}}=b^{j^{k+1}}.$

On the other hand, we have $b=f^{n}(b)$, and hence
$b^{i^{n}}=f^{n}(b)^{i^{n}}=b^{j^{n}}$. Since $b$ is non-torsion, thus
$i^{n}=j^{n}$, so that indeed $|i|=|j|$. ∎

####  2.1.4 MC-groups

Recall that $H$ is said to be an ‘MC-group’ if it satisfies the descending
chain condition on centralizers, ie if the sequence of subgroups
$C_{H}(a_{1})\supseteq C_{H}(a_{1},a_{2})\supseteq\dots$ stabilizes for every
sequence $(a_{i})_{i\in\omega}$ from $H$. We will use the following fact in
Section 6; thank you to Frank Wagner for the reference to [3].

######  Fact 2.7.

Suppose $H$ is a torsion MC-group. Then the solvable radical of $H$ is
solvable.

###### Proof.

The solvable radical is locally solvable, and (as a subgroup of $H$) it is
still a torsion MC-group. The result thus follows from Theorem 2.2 in [3]. ∎

####  2.1.5 Valued groups

Let $(\mathcal{P},\leqslant)$ be a totally ordered set with a greatest
element. A $\mathcal{P}$-valued group structure on $H$ is a a surjective map
$v:H\to\mathcal{P}$ satisfying the following:

  1. 1.

$v(a)=\max\mathcal{P}$ if and only if $a=1$.

  2. 2.

$v(a^{g})<v(b^{g})$ for all $g\in H$ and $a,b\in H$ with $v(a)<v(b)$.

  3. 3.

$v(a)=v(a^{-1})$ for all $a\in H$

  4. 4.

$v(ab)\geqslant\min\\{v(a),v(b)\\}$ for all $a,b\in H$.

Following [30], we will say that $(H,v)$ is ‘plain’ if we additionally have
$v([a,b])>\max\\{v(a),v(b)\\}$ for all $a,b\in H\setminus\\{1\\}$. The
following is proved in [30]:

######  Fact 2.8.

Suppose $(H,v)$ is a plain valued group of finite exponent. Then $H$ is
locally nilpotent.

###  2.2 Model theory

####  2.2.1 NIP groups

In this section assume only that $G$ is NIP. We will use several general facts
about NIP groups. First recall the Baldwin-Saxl theorem, proved in [2]: for
any $L$-formula $\phi(x,y)$, there is a constant $k$ such that, for all
$c_{1},\dots,c_{n}\in\mathfrak{C}^{y}$ such that each $\phi(x,c_{i})$ defines
a subgroup of $G$, there is a subset $s\subseteq[n]$ of size at most $k$ with
$\bigcap_{i\in s}\phi(x,c_{i})=\bigcap_{i\in[n]}\phi(x,c_{i})$. We isolate two
consequences of this:

######  Fact 2.9.

There is a constant $n$ such that, for any finite subset $B\subset
G(\mathfrak{C})$, there is $S\subseteq B$ of size at most $n$ with
$C_{G}(S)\subseteq C_{G}(b)$ for all $b\in B$. In particular, if
$(u_{i})_{i\in[n]}+(a_{i})_{i\in\omega}+(v_{i})_{i\in[n]}$ is indiscernible
and $W=\\{u_{i},v_{i}:i\in[n]\\}$, then $C_{G}(W)\subseteq C_{G}(a_{i})$ for
all $i\in\omega$.

######  Fact 2.10.

Suppose $G$ has the ‘uniform chain condition’, ie that it has no infinite
strictly descending chain of uniformly $\mathfrak{C}$-definable subgroups.
Then $G$ has the ‘intersection chain condition’; namely, for any formula
$\phi(x,y)$, and any (small or large) subset $B\subseteq\mathfrak{C}^{y}$ such
that $\phi(x,b)$ defines a subgroup of $G$ for every $b\in B$, there is a
finite subset $C\subseteq B$ such that $\bigcap_{c\in
C}\phi(x,c)\subseteq\phi(x,b)$ for every $b\in B$. In particular, $G$ is then
an MC-group.

We will also make extensive use of Shelah’s theorem on the existence of
$G^{00}$ [25]:

######  Fact 2.11.

There is a partial type $G^{00}(x)$, without parameters, such that $G^{00}$
type-defines a bounded index subgroup of $G$, and such that any bounded index
subgroup of $G$ type-definable with parameters contains $G^{00}$.

We will also need some of the theory of definably amenable NIP groups, as
developed in [13], [14], and [5]. Recall from [13] that $G$ is said to be
‘definably amenable’ if it admits a global translation-invariant Keisler
measure. Recall also from [5] that a $\mathfrak{C}$-definable subset
$D\subseteq G$ is said to be left (resp. right) ‘f-generic’ if, for any small
$M$ over which $D$ is defined, and for all $g\in G(\mathfrak{C})$, the formula
$x\in gD$ (resp. $x\in Dg$) does not fork over $M$. A partial type is said to
be left (right) f-generic if it implies only left (right) f-generic formulas.
With no further qualifer, f-generic means left f-generic. Now, from [13],
[14], [5], and [31], the following are equivalent:

  1. 1.

$G$ is definably amenable.

  2. 2.

$G$ admits a global type invariant under left-translation by elements of
$G^{00}(\mathfrak{C})$.

  3. 3.

$G$ admits a global f-generic type.

  4. 4.

The non-f-generic $\mathfrak{C}$-definable subsets of $G$ form an ideal.

  5. 5.

$G$ admits a global ‘strongly f-generic’ type, ie a global f-generic type
invariant over some small set.

Moreover, we have:

######  Fact 2.12.

A global type $p(x)\in S_{G}(\mathfrak{C})$ is f-generic if and only if it is
invariant under left translation by all elements of $G^{00}(\mathfrak{C})$. In
particular, if $p$ is f-generic, then, for any small $D\subset\mathfrak{C}$,
every element of $G^{00}(\mathfrak{C})$ is a difference of two realizations of
$p$, ie of form $ab^{-1}$ for some $a,b\models p|_{D}$.

Note that, if $p(x)\in S_{G}(\mathfrak{C})$ is strongly f-generic, then the
pushforward of $p_{x}\otimes p_{y}$ under the map $(x,y)\mapsto xy^{-1}$ is
(i) concentrated on $G^{00}$, and (ii) invariant under both left-translation
and right-translation by all elements of $G^{00}(\mathfrak{C})$. Thus one has:

######  Fact 2.13.

If $G$ is definably amenable, then $G$ has a global bi-f-generic type
concentrated on $G^{00}$.

####  2.2.2 Dp-minimality

There are numerous equivalent characterizations of dp-minimality; see for
example [9] for an introduction. A partial type $\pi(x)$ over a small
parameter set is said to be ‘dp-minimal’ if the following holds for some (any)
parameters $B$ over which $\pi$ is defined: for any infinite sequences $I,J$
mutually indiscernible over $B$, and any realization $a\models\pi(x)$, one of
$I$ and $J$ is indiscernible over $(a,B)$. Then $\pi(x)$ is not dp-minimal if
and only if there are $L(\mathfrak{C})$-formulas $\phi(x,y)$ and $\psi(x,z)$
and tuples $b_{i}\in\mathfrak{C}^{y}$ and $c_{j}\in\mathfrak{C}^{z}$ such
that, for all $i,j\in\omega$, the partial type

$\pi(x)\wedge\phi(x,b_{i})\wedge\psi(x,c_{j})\wedge\bigwedge_{k\neq
i}\neg\phi(x,b_{k})\wedge\bigwedge_{l\neq j}\neg\psi(x,c_{l})$

is consistent. We say that the definable group $G$ is dp-minimal if the
formula $x\in G$ is dp-minimal.

The following was noted in [27] for definable subgroups of dp-minimal groups,
and the analogue is true for invariant subgroups:

######  Fact 2.14.

Suppose $G$ is dp-minimal, and that $A,B\leqslant G$ are subgroups invariant
over some small model $M$. Then $A\cap B$ has bounded index in one of $A$ and
$B$.

###### Proof.

Suppose for contradiction that $A\cap B$ has unbounded index in both $A$ and
$B$. Then there are sequences $I=(a_{i})_{i\in\omega}$ in $A$ and
$J=(b_{j})_{j\in\omega}$ in $B$, each indiscernible over $M$, and such that
$a_{i}B\cap a_{j}B=Ab_{i}\cap Ab_{j}=\varnothing$ for all $i\neq j$. By Ramsey
and $M$-invariance of $A,B$ we may assume that $I$ and $J$ are mutually
indiscernible over $M$, and then by dp-minimality one of $I$ and $J$ must be
indiscernible over $(M,a_{0}b_{0})$. In the former case, there is an
automorphism fixing $(M,a_{0}b_{0})$ and taking $a_{0}$ to $a_{1}$, and since
$a_{0}^{-1}(a_{0}b_{0})\in B$ it follows by invariance that
$a_{1}^{-1}(a_{0}b_{0})\in B$. But then $a_{1}^{-1}a_{0}\in B$, a
contradiction. A symmetric argument, using that $(a_{0}b_{0})b_{0}^{-1}\in A$,
gives the result in the latter case. ∎

We will also use the following, from [27], [31], and [16] respectively:

######  Fact 2.15.

If $G$ is dp-minimal, then $G$ has a definable normal abelian subgroup
$A\trianglelefteqslant G$ such that $G/A$ has finite exponent. In fact, if $G$
is dp-minimal, then there is $n>0$ such that, for all $a,b\in G$, one of
$[a^{n},b]=1$ and $[a,b^{n}]=1$ holds.

######  Fact 2.16.

Any dp-minimal group is definably amenable.

######  Fact 2.17.

If $G$ is dp-minimal and $q(x)\in S_{G}(\mathfrak{C})$ is f-generic, then $q$
is $M$-invariant for every small $M\prec\mathfrak{C}$.

####  2.2.3 Invariant types in NIP theories

In this section assume only that $T$ is NIP. We will need several facts about
invariant types under this assumption. Recall from [23] that, if
$M\prec\mathfrak{C}$ is a small model and $q(x)\in S(\mathfrak{C})$ is a
global type, then $q(x)$ does not fork over $M$ if and only if it is
$M$-invariant. Thus invariant types are plentiful in the NIP context. There
are two distinguished classes we will be considering here.

The first is that of generically stable types, introduced in [22], then
investigated in [14] and in [32] (mostly with an NIP hypothesis) and in [19]
(without an NIP hypothesis). A global type $p(x)\in S(\mathfrak{C})$ invariant
over some small set is said to be ‘generically stable’ if for some (any) small
$B\subset\mathfrak{C}$ over which it is invariant, we have: for any linear
order $I$, any Morley sequence $(a_{i})_{i\in I}\models p^{\otimes I}|_{B}$,
and any $L(\mathfrak{C})$-formula $\phi(x,c)$, the set
$\\{i:a_{i}\models\phi(x,c)\\}$ is either finite or co-finite in $I$. With the
NIP hypothesis on $T$, the following conditions are all equivalent:

  1. 1.

$p(x)$ is generically stable.

  2. 2.

Any Morley sequence of $p(x)$ is totally indiscernible.

  3. 3.

$p(x)$ is the limit type of some totally indiscernible sequence.

  4. 4.

$p(x)$ is definable over and finitely satisfiable in any small model over
which it is invariant.

  5. 5.

For any global type $q(y)$ invariant over some small set, we have $p(x)\otimes
q(y)=q(y)\otimes p(x)$.

The following is proved in [14]:

######  Fact 2.18.

Suppose $p(x)\in S(\mathfrak{C})$ is $B$-invariant and generically stable.
Then $p(x)|_{B}$ is stationary, ie $p(x)$ is the unique global extension of
$p(x)|_{B}$ that does not fork over $B$.

The other class of invariant type we will use is that of distal types,
implicitly introduced in [28],111In an NIP theory, distal types are to distal
indiscernible sequences as generically stable types are to totally
indiscernible sequences. but first explicitly introduced in [26]. For the
definition, recall that, for a parameter set $B\subset\mathfrak{C}$ and
disjoint tuples of variables $x,y$, two types $p(x)\in S_{x}(B)$ and $q(y)\in
S_{y}(B)$ are said to be ‘weakly orthogonal’, denoted $p(x)\perp^{w}q(y)$, if
$p(x)\wedge q(y)$ determines a unique complete type in $S_{(x,y)}(B)$. Now a
$B$-invariant global type $p(x)\in S(\mathfrak{C})$ is said to be ‘distal over
$B$’ if the following holds: for any tuple $c$, if $I\models
p^{\otimes\omega}|_{(B,c)}$ then
$\operatorname{tp}(c/B,I)\perp^{w}p|_{(B,I)}$. It is proved in [26] that, if
$p(x)$ is a global type invariant over sets $B,C\subset\mathfrak{C}$, then
$p(x)$ is distal over $B$ if and only if it is distal over $C$. So one may
speak of a ‘distal type’ without ambiguity.

Distal types and generically stable types lie at precisely opposite ends of
the NIP spectrum, respectively capturing ‘purely order-like’ and ‘purely
stable-like’ behavior in NIP theories. In the dp-minimal case, every 1-type
falls into one of those two classes; see for example Proposition 9.13 in [26]:

######  Fact 2.19.

Suppose $B\subset\mathfrak{C}$ is a small set, and that $p(x)\in
S(\mathfrak{C})$ is a $B$-invariant global type such that $p(x)|_{B}$ is dp-
minimal. Then $p(x)$ is either generically stable or distal.

Fact 2.19 will be quite important for Section 6 of this paper. In the
meantime, let us point out one consequence here for convenience, which follows
quickly just from results in the literature:

######  Corollary 2.20.

Suppose $G$ is a dp-minimal group. Then either every global f-generic type of
$G$ is generically stable, or every global f-generic type of $G$ is distal.

###### Proof.

By Fact 2.17, any f-generic type of $G$ is invariant over a small set. Thus by
Fact 2.19 it suffices to show that, if $G$ has a generically stable f-generic
type, then every f-generic type of $G$ is generically stable. We sketch the
argument; suppose $p(x)\in S_{G}(\mathfrak{C})$ is f-generic and generically
stable, and let $M$ be a small model containing a representative of every
coset of $G^{00}$ and over which $p(x)$ is invariant. By Fact 2.12, every left
translate of $p(x)$ is of form $gp(x)$ for some $g\in G(M)$, whence every left
translate of $p(x)$ is generically stable and $M$-invariant and hence finitely
satisfiable in $M$. It follows that $G$ is fsg. By item (iii) of Corollary 3.3
in [15], it follows that a global type of $G$ is f-generic if and only if it
is generic; call this ($\ast$). By ($\ast$), $p$ is generic, so $G$ is
generically stable in the sense of Section 6 of [14], and by the results
proved there the global generic types of $G$ are precisely the translates of
$p$. Again by ($\ast$), it follows that the f-generic types are precisely the
translates of $p$, so the claim follows. ∎

##  3 Generic Nilpotence

In this section assume only that $G$ is NIP.

It is well known that, in a stable group, to check virtual nilpotence of class
at most $(c-1)$ it suffices to check that an independent $c$-tuple of
principal generics satisfies $[x_{1},\dots,x_{c}]=1$; see eg [33]. We point
out here an analogous fact in the NIP setting, which follows quickly from Fact
2.11; compare also de Aldama’s proof of the existence of nilpotent envelopes
in NIP groups [1].

######  Lemma 3.1.

Suppose there is a left or right f-generic type $p(x)\in
S_{G^{00}}(\mathfrak{C})$ invariant over some small model and such that
$p^{\otimes c}(x_{1},\dots,x_{c})\vdash[x_{1},\dots,x_{c}]=1$. Then $G$ is
virtually nilpotent of class at most $(c-1)$.

###### Proof.

By induction on $c$. In the case $c=0$, $G$ must be finite, so the result
follows. Now suppose the result holds for $c$, and that $p(x)\in
S_{G^{00}}(\mathfrak{C})$ is an $M$-invariant f-generic type satisfying
$p^{\otimes{(c+1)}}(x_{1},\dots,x_{c+1})\vdash[x_{1},\dots,x_{c+1}]=1$. Let
$q(y)$ be the pushforward of $p^{\otimes c}(x_{1},\dots,x_{c})$ under the map
$G^{c}\to G$, $(x_{1},\dots,x_{c})\mapsto[x_{1},\dots,x_{c}]$; then
$p(x)\otimes q(y)\vdash[x,y]=1$.

If $b\models q|_{M}$ and $a\models p_{(M,b)}$, then $a\in C_{G}(b)$; since
$\operatorname{tp}(a/b)$ is f-generic it follows that $C_{G}(b)$ must have
finite index in $G$. In particular, $G^{00}(x)\vdash x\in C_{G}(b)$, so by
compactness there is a $\varnothing$-definable set $D\supseteq G^{00}$ such
that $b\in C_{G}(D)$. Since $D$ is $\varnothing$-definable, thus $q(y)\vdash
y\in C_{G}(D)$.

Now, let $H=C_{G}(C_{G}(D))$. Then $H\supseteq D\supseteq G^{00}$; thus $H$
has finite index in $G$, so it suffices to show that $H$ is virtually
nilpotent of class at most $c$, and we moreover have $p(x)\vdash x\in H$ and
$q(y)\vdash y\in H$. Since we also have $q(y)\vdash y\in C_{G}(D)$, it follows
that $q(y)\vdash y\in Z(H)$.

Let $\bar{p}$ be the pushforward of $p$ under the quotient map $H\to H/Z(H)$;
this is well-defined since $p(x)\vdash x\in H$, and by the above we have
$\bar{p}{}^{\otimes c}(x_{1},\dots,x_{c})\vdash[x_{1},\dots,x_{c}]=1$. Since
$p$ is concentrated on $H^{00}$ and the pullback of a type-definable subgroup
of bounded index under a quotient map is type-definable and bounded index,
$\bar{p}$ is concentrated on $(H/Z(H))^{00}$. Moreover, the image of an
f-generic set under a quotient map is still f-generic, so $\bar{p}$ is
f-generic. By the inductive hypothesis applied to the interpretable group
$H/Z(H)$, it follows that $H/Z(H)$ is virtually nilpotent of class at most
$(c-1)$, and taking its pullback to $H$ gives the result. ∎

##  4 Torsion-free case

In this section assume that $G$ is dp-minimal. We will show that, if $G$ has a
global bi-f-generic type concentrated on $G^{00}$ whose realizations are non-
torsion, then $G$ is virtually nilpotent of class at most $2$. Moreover, if
$G$ is torsion-free, then $G$ is abelian.

######  Lemma 4.1.

Suppose that $I=(a_{i})_{i\in\omega}$ is a sequence in $G$ such that (i)
$a_{i}\not\sim a_{j}$ for all $i\neq j$, and (ii) $[a_{i},a_{j}]=1$ for all
$i,j$. Then, for every sequence $J=(b_{j})_{j\in\omega}$ in $G$, there is some
$i\neq j$ and $k\in\omega$ with $[a_{k}^{b_{i}},a_{k}^{b_{j}}]=1$.

###### Proof.

Suppose that $J$ is otherwise, so that $[a_{k}^{b_{i}},a_{k}^{b_{j}}]\neq 1$
for all $i\neq j$ and $k\in\omega$. By Ramsey and compactness, we may then
assume that $I$ and $J$ are mutually indiscernible sequences. Now, let $n$ be
given Fact 2.9. Pick $u_{i},v_{i}$ so that
$(u_{i})_{i\in[n]}+I+(v_{i})_{i\in[n]}$ is still $J$-indiscernible and let
$W=\\{u_{i},v_{i}:i\in[n]\\}$; then $I$ and $J$ are mutually indiscernible
over $W$. Moreover, $D:=C_{G}(W)$ is $W$-definable, and by condition (ii) and
Fact 2.9 we have $a_{k}\in D\subseteq C_{G}(a_{k})$ for all $k\in\omega$.

Now, by condition (i), we have $a_{i}\sim a_{0}^{b_{0}}$ iff $i=0$, so $I$ is
not indiscernible over $a_{0}^{b_{0}}$ and hence not over $(W,a_{0}^{b_{0}})$.
Thus, by dp-minimality, $J$ must be indiscernible over $(W,a_{0}^{b_{0}})$.
Since $a_{0}\in D$, we have $b_{0}\models a_{0}^{b_{0}}\in D^{y}$, and thus
also $a_{0}^{b_{0}}\in D^{b_{1}}$. But $D\subseteq C_{G}(a_{0})$, so that
$a_{0}^{b_{0}}\in C_{G}(a_{0})^{b_{1}}=C_{G}(a_{0}^{b_{1}})$, ie
$[a_{0}^{b_{0}},a_{0}^{b_{1}}]=1$, a contradiction. ∎

######  Lemma 4.2.

Suppose that $a\in G(\mathfrak{C})$ is non-torsion, and that
$J=(g_{i})_{i\in\omega}$ is an $a$-indiscernible sequence in $G$. Then
$[a^{g_{0}},a^{g_{1}}]=1$.

###### Proof.

Let $s\subseteq\omega$ be the set of all $k\in\omega$ such that
$[a^{kg_{0}},a^{kg_{1}}]\neq 1$; since $J$ is $a$-indiscernible, note that, if
$k\in s$, then $[a^{kg_{i}},a^{kg_{j}}]\neq 1$ for all $i\neq j$. On the other
hand, we have $[a^{i},a^{j}]=1$ for all $i,j$, and by Lemma 2.6 and Fact 2.15
we have $a^{i}\not\sim a^{j}$ for all $i\neq j\in\omega$, so it follows from
Lemma 4.1 that $s$ must be finite. On the other hand, letting $b=a^{g_{0}}$
and $c=a^{g_{1}}$, we have $[b^{k},c^{k}]=1$ for all $k\notin s$, and it
follows that $[b,c]=1$. (For example, we may find four pairwise coprime
numbers $p,q,r,s\in\omega\setminus s$. Then $b^{p}$ and $b^{q}$ each lie in
$C_{G}(c^{pq})$, so $b\in C_{G}(c^{pq})$, and likewise $b\in C_{G}(c^{rs})$.
Now $c^{pq}$ and $c^{rs}$ each lie in $C_{G}(b)$, so $c\in C_{G}(b)$, as
needed.) ∎

######  Lemma 4.3.

Let $M$ be a small model. Suppose that $p(x)\in S_{G}(\mathfrak{C})$ is an
$M$-invariant type whose realizations are non-torsion, and that $q(y)\in
S_{G^{00}}(\mathfrak{C})$ is an $M$-invariant right f-generic type. Then
$q(y)\otimes p(x)$ and $p(x)\otimes q(y)$ each imply $[x,x^{y}]=1$.

###### Proof.

To see that $q(y)\otimes p(x)$ implies $[x,x^{y}]=1$, let $a\models p|_{M}$
and let $J=(g_{i})_{i\in\omega}\models q^{\otimes\omega}|_{(M,a)}$. Then $J$
is $a$-indiscernible, so by Lemma 4.2 $[a^{g_{0}},a^{g_{1}}]=1$ and hence
$[a,a^{g_{1}g_{0}^{-1}}]=1$. But $q$ is right f-generic and concentrated on
$G^{00}$, so $g_{1}g_{0}^{-1}\models q|_{(M,a)}$, and the result follows.

To see that $p(x)\otimes q(y)$ implies $[x,x^{y}]=1$, let
$J=(g_{i})_{i\in\omega}\models q^{\otimes\omega}|_{M}$ and let $a\models
p|_{(M,J)}$. Since $p$ is $M$-invariant and $J$ is $M$-indiscernible, also $J$
is $(M,a)$-indiscernible. So again by Lemma 4.2 we have
$[a,a^{g_{1}g_{0}^{-1}}]=1$. Also, again by right f-genericity,
$g_{1}g_{0}^{-1}\models q|_{M}$, and now since $a\models
q|_{(M,g_{1}g_{0}^{-1})}$ the result follows. ∎

######  Lemma 4.4.

Suppose there is an $M$-invariant bi-f-generic type concentrated on $G^{00}$
whose realizations are non-torsion. Then $G$ is virtually nilpotent of class
at most $2$.

###### Proof.

Let $q(x)\in S_{G^{00}}(\mathfrak{C})$ be bi-f-generic with non-torsion
realizations, and let $(a_{i})_{i\in\omega}+(b_{i})_{i\in\omega}$ be a Morley
sequence of $q$ over $M$. By Lemma 4.3, we have $[a_{1},a_{1}^{b_{1}}]=1$ and
$[b_{1},b_{1}^{a_{1}}]=1$. In other words, $[a_{1},b_{1}]\in C_{G}(a_{1})\cap
C_{G}(b_{1})$, so that $\langle a_{1},b_{1}\rangle$ is nilpotent of class at
most $2$. Now, by dp-minimality, one of $(a_{i})_{i\in\omega}$ and
$(b_{i})_{i\in\omega}$ is indiscernible over $[a_{1},b_{1}]$. If
$(b_{i})_{i\in\omega}$ is, then, since $[a_{1},b_{1},b_{1}]=1$, we have
$[a_{1},b_{1},b_{2}]=1$, so the result follows by Lemma 3.1. So we may assume
instead that $(a_{i})_{i\in\omega}$ is indiscernible over $[a_{1},b_{1}]$.
Since $[a_{1},b_{1},a_{1}]=1$, thus also $[a_{1},b_{1},a_{2}]=1$; since $q$ is
bi-f-generic and concentrated on $G^{00}$, and $a_{2}\models q|_{(M,a_{1})}$,
we have $a_{2}^{a_{1}}\models q|_{(M,a_{1})}$, and since $b_{1}\models
q|_{(M,a_{1},a_{2})}$ it follows that $[a_{1},b_{1},a_{2}^{a_{1}}]=1$.

Now, again by dp-minimality, one of $(a_{i})_{i\in\omega}$ and
$(b_{i})_{i\in\omega}$ is indiscernible over $[b_{1},a_{1}^{b_{1}^{-1}}]$. If
$(b_{i})_{i\in\omega}$ is, then, since $[b_{1},a_{1}^{b_{1}^{-1}},b_{1}]=1$,
also $[b_{1},a_{1}^{b_{1}^{-1}},b_{2}]=1$, so that
$[b_{1},a_{1},b_{2}^{b_{1}}]=1$. Since $q$ is bi-f-generic and concentrated on
$G^{00}$, and $b_{2}\models q|_{(M,a_{1},b_{1})}$, we have
$b_{2}^{b_{1}}\models q|_{(M,a_{1},b_{1})}$, so it follows that
$[b_{1},a_{1},b_{2}]=1$. But then $[a_{1},b_{1},b_{2}]=1$ and so again by
Lemma 3.1 we are done. So we may assume that $(a_{i})_{i\in\omega}$ is
indiscernible over $[b_{1},a_{1}^{b_{1}^{-1}}]$. Since
$[b_{1},a_{1}^{b_{1}^{-1}},a_{1}]=1$, it follows that
$[b_{1},a_{1}^{b_{1}^{-1}},a_{0}]=1$, and hence that
$[b_{1},a_{1},a_{0}^{b_{1}}]=1$.

Altogether we have shown that, if the desired conclusion does not hold, then
$[a,c,b^{a}]=1$ and $[c,b,a^{c}]=1$ for all $(a,b,c)\models q^{\otimes
3}|_{M}$. By the Hall-Witt identity, we have

$[c,b,a^{c}][a,c,b^{a}][b,a,c^{b}]=1,$

so it follows that $[b,a,c^{b}]=1$. By one final application of bi-f-
genericity of $q$, we have $c^{b}\models q|_{(M,a,b)}$, so that $[b,a,c]=1$,
whence $[a,b,c]=1$ and the result follows from Lemma 3.1. ∎

######  Theorem 4.5.

If $G$ is torsion-free, then it is abelian.

###### Proof.

By Fact 2.16 and Fact 2.13, let $q(x)\in S_{G^{00}}(\mathfrak{C})$ be an
$M$-invariant bi-f-generic type, and let $(a,b)\models q^{\otimes 2}|_{M}$. As
in Lemma 4.4, we have that $\langle a,b\rangle$ is nilpotent of class at most
$2$, so that $[a^{k},b]=[a,b]^{k}=[a,b^{k}]$ for all $k\in\omega$. On the
other hand, by Fact 2.15, for some $n>0$, one of $[a^{n},b]=1$ and
$[a,b^{n}]=1$ holds, whence $[a,b]^{n}=1$. Since $G$ is torsion-free, thus
$[a,b]=1$, and by Lemma 3.1 it follows that $G^{00}$ is abelian.

Now, to show the desired result, it suffices by Fact 2.5 to show that
$G^{00}\subseteq Z(G)$. Thus fix arbitrary $g\models G$ with $g\neq 1$. Let
$N$ be any small model containing $(M,g)$, and let $c\models q|_{N}$. By Lemma
4.3 applied to the $N$-invariant type $\operatorname{tp}(g/N)$, we have
$[g,g^{c}]=1$. On the other hand, $c\models G^{00}$, and $G^{00}$ is a normal
abelian subgroup of $G$, so also $[c,c^{g}]=1$. Thus $\langle c,g\rangle$ is
nilpotent of class at most $2$, and so again since $G$ is torsion-free and by
Fact 2.15 it follows that $[c,g]=1$. So $c\in C_{G}(g)$, and by f-genericity
of $q$ it follows that $C_{G}(g)$ has finite-index in $G$, whence
$G^{00}\subseteq C_{G}(g)$. Since $g$ was arbitrary, indeed $G^{00}\subseteq
Z(G)$, and the result follows. ∎

##  5 Distal case

In this section, we will study dp-minimal groups admitting a distal f-generic
type; by Corollary 2.20, these are precisely those dp-minimal groups with no
generically stable f-generic type. Hence the class of such groups includes not
only distal dp-minimal groups, ie dp-minimal groups in which every invariant
type is distal, but also, for instance, dp-minimal $\mathcal{P}$-valued groups
whenever $(\mathcal{P},\leqslant)$ is a total order with no least element.

Many of the results we will prove here were proved for C-minimal valued groups
in [30]. In particular, the results of Section 5.3 and Theorem 5.21 were both
proved in the C-minimal case there. The main difference here is that we do not
start with a valued group structure on the group.

Throughout this section, assume that $G$ is a dp-minimal group admitting a
distal f-generic type. By Corollary 2.20, then every f-generic type of $G$ is
distal, and by Fact 2.13 and Fact 2.17 there is hence a bi-f-generic type
$q(x)\in S_{G^{00}}(\mathfrak{C})$ which is distal and invariant over any
small model; fix this type $q$ for the entire section.

###  5.1 Conjugacy classes are thin

Recall from [11] that a $\mathfrak{C}$-definable subset $X\subseteq G$ with
$1\in X=X^{-1}$ is called ‘thick’ if, for any sequence $(a_{i})_{i\in\omega}$
indiscernible over the parameters defining $X$, we have $a_{i}^{-1}a_{j}\in X$
for all $i,j\in\omega$. In this section we will show that, for any $u\in
G\setminus\\{1\\}$, the set $G\setminus(u^{G}\cup u^{-G})$ is thick.

######  Lemma 5.1.

Let $B\subset\mathfrak{C}$ be a parameter set. Suppose $u,g\models G$ are such
that $\operatorname{tp}(u/B)\perp^{w}\operatorname{tp}(g/B)$ and that
$\operatorname{tp}(u/B)\perp^{w}\operatorname{tp}(u^{g}/B)$, and that there is
some $n>0$ with $[g^{n},u]=1$. If $[g,u]\equiv_{B}u$, then $u=1$.

###### Proof.

Suppose $[g,u]\equiv_{B}u$. We prove by induction on $k$ that
$[g^{k},u]\equiv_{B}u$ for each $k>0$; taking $k=n$ will then give the result.
The base case $k=1$ is by hypothesis. For the inductive step, assume that
$[g^{k},u]\equiv_{B}u$. Then, since
$\operatorname{tp}(u/B)\perp^{w}\operatorname{tp}(g/B)$, it follows that
$[g^{k},u]^{g}\equiv_{B}u^{g}$. By the case $k=1$, we also have
$[g,u]\equiv_{B}u$. Now since
$\operatorname{tp}(u/B)\perp^{w}\operatorname{tp}(u^{g}/C)$, the $2$-tuple
$([g,u],u^{g})$ has the same type over $B$ as $([g,u],[g^{k},u]^{g})$ does.
But we have $u^{g}\cdot[g,u]=u$, and so the result follows from the identity
$[g^{k},u]^{g}[g,u]=[g^{k+1},u]$. ∎

######  Lemma 5.2.

Let $M$ be a small model, and suppose that $v\in G(M)$ and that $u\in
v^{G^{00}}$. If $J+g\models q^{\otimes(\omega+1)}|_{(M,u)}$, then
$\operatorname{tp}(u/M,J)\perp^{w}\operatorname{tp}(g/M,J)$ and
$\operatorname{tp}(u/M,J)\perp^{w}\operatorname{tp}(u^{g}/M,J)$.

###### Proof.

That $\operatorname{tp}(u/M,J)\perp^{w}\operatorname{tp}(g/M,J)$ follows
immediately from distality of $q$. Now, let $h\models G^{00}$ be such that
$u=v^{h}$, and let $J^{\prime}+g^{\prime}\models
q^{\otimes(\omega+1)}|_{(M,u,h)}$. We have
$(u,J^{\prime},g^{\prime})\equiv_{M}(u,J,g)$, so it suffices to show that
$\operatorname{tp}(u/M,J^{\prime})\perp^{w}\operatorname{tp}(u^{g^{\prime}}/M,J^{\prime})$.
Since $h\models G^{00}$ and $g^{\prime}\models q|_{(M,u,h,J^{\prime})}$, it
follows by left f-genericity of $q$ that $hg^{\prime}\models
q|_{(M,u,h,J^{\prime})}$. Since $v\in G(M)$, we thus have in particular that
$v^{g^{\prime}}\equiv_{(M,J^{\prime})}v^{hg^{\prime}}=u^{g^{\prime}}$. On the
other hand, again since $v\in G(M)$, and since
$\operatorname{tp}(u/M,J^{\prime})\perp^{w}\operatorname{tp}(g^{\prime}/M,J^{\prime})$,
we have
$\operatorname{tp}(u/M,J^{\prime})\perp^{w}\operatorname{tp}(v^{g^{\prime}}/M,J^{\prime})$,
so the result follows. ∎

######  Lemma 5.3.

Let $M$ be a small model, and suppose that $I=(a_{i})_{i\in\omega}$ is an
$M$-indiscernible sequence such that, for some $v\in G(M)$, we have
$a_{i}^{-1}a_{j}\in v^{G^{00}}$ for all $i<j$. Then $I$ is constant.

###### Proof.

By Ramsey and compactness, it suffices to show the result when $I$ has order
type $\mathbb{R}$, so replace $I$ by a sequence $(a_{i})_{i\in\mathbb{R}}$
with the same properties. Now, I claim that, for some $n>0$,
$q(x)\vdash[x^{n},a_{i}^{-1}a_{j}]=1$ for all $i,j\in\mathbb{R}$; call this
($\star$). If $q(x)\vdash x^{n}=1$ for some $n>0$ then the claim is clear.
Otherwise the realizations of $q(x)$ are non-torsion, whence by Lemma 4.4
$G^{00}$ is nilpotent of class at most $2$; in particular then
$[b^{n},c]=[b,c^{n}]$ for all $b,c\models G^{00}$ and all $n\in\omega$. Since
$a_{i}^{-1}a_{j}\models G^{00}$ and $q$ is concentrated on $G^{00}$, ($\star$)
then follows from Fact 2.15.

Let $J+g\models q^{\otimes(\omega+1)}|_{(M,I)}$, and let $u=a_{0}^{-1}a_{1}$.
By Lemma 5.2, we have
$\operatorname{tp}(u/M,J)\perp^{w}\operatorname{tp}(g/M,J)$ and
$\operatorname{tp}(u/M,J)\perp^{w}\operatorname{tp}(u^{g}/M,J)$; moreover, by
5.2 applied to the bi-f-generic distal type $q^{-1}$, we also have
$\operatorname{tp}(u/M,J)\perp^{w}\operatorname{tp}(g^{-1}/M,J)$ and
$\operatorname{tp}(u/M,J)\perp^{w}\operatorname{tp}(u^{g^{-1}}/M,J)$. Call
these facts ($\ast$).

Note also that, since $I$ is $M$-indiscernible and $q$ is $M$-invariant, $I$
is $(M,J,g)$-indiscernible. Thus $u\equiv_{(M,J,g)}a_{i}^{-1}a_{j}$ for all
$i<j\in\mathbb{R}$.

Now, $(a_{i})_{i\in[0,1]}$ and $(a_{j})_{j\in[3,4]}$ are mutually
indiscernible over $(M,a_{2},a_{5},J,g)$. By dp-minimality, one of them is
thus indiscernible over $(M,a_{2},a_{5},J,g,a_{0}ga_{3}^{-1})$. First suppose
$(a_{i})_{i\in[0,1]}$ is indiscernible over that set. Note that

$g^{-1}\cdot a_{0}^{-1}\cdot(a_{0}ga_{3}^{-1})\cdot
a_{5}=a_{3}^{-1}a_{5}\equiv_{(M,J)}u.$

Hence, by indiscernibility, also
$g^{-1}a_{1}^{-1}a_{0}ga_{3}^{-1}a_{5}\equiv_{(M,J)}u$, ie
$(a_{1}^{-1}a_{0})^{g}(a_{3}^{-1}a_{5})\equiv_{(M,J)}u$. But we have
$a_{1}^{-1}a_{0}\equiv_{(M,J,g)}u^{-1}$, hence
$(a_{1}^{-1}a_{0})^{g}\equiv_{(M,J)}u^{-g}$, and
$a_{3}^{-1}a_{5}\equiv_{(M,J)}u$. Since
$\operatorname{tp}(u^{-g}/M,J)\perp^{w}\operatorname{tp}(u/M,J)$, it follows
that $u^{-g}u\equiv_{(M,J)}u$, ie that $[g,u]\equiv_{(M,J)}u$. By Lemma 5.1,
using ($\ast$) and ($\star$), we must have $u=1$, so that indeed $I$ is
constant.

Now suppose $(a_{j})_{j\in[3,4]}$ is indiscernible over
$(M,a_{2},a_{5},J,g,a_{0}ga_{3}^{-1})$. As above, note that

$a_{2}^{-1}\cdot(a_{0}ga_{3}^{-1})\cdot a_{3}\cdot
g^{-1}=a_{2}^{-1}a_{0}\equiv_{(M,J)}u^{-1}.$

By indiscernibility, thus
$a_{2}^{-1}a_{0}ga_{3}^{-1}a_{4}g^{-1}\equiv_{(M,J)}u^{-1}$, ie
$(a_{2}^{-1}a_{0})(a_{3}^{-1}a_{4})^{g^{-1}}\equiv_{(M,J)}u^{-1}$. But we have
$a_{3}^{-1}a_{4}\equiv_{(M,J,g)}u$, hence
$(a_{3}^{-1}a_{4})^{g^{-1}}\equiv_{(M,J)}u^{g^{-1}}$, and
$a_{2}^{-1}a_{0}\equiv_{(M,J)}u^{-1}$. Since

$\operatorname{tp}(u^{g^{-1}}/M,J)\perp^{w}\operatorname{tp}(u/M,J),$

it follows that $u^{-1}u^{g^{-1}}\equiv_{(M,J)}u^{-1}$, so that
$u^{-g^{-1}}u\equiv_{(M,J)}u$, ie $[g^{-1},u]\equiv_{(M,J)}u$. By Lemma 5.1
applied to the bi-f-generic distal type $q^{-1}$, using ($\ast$) and
($\star$), we must have $u=1$, so that again $I$ is constant. ∎

######  Corollary 5.4.

Suppose $I=(b_{i})_{i\in\omega}$ is an indiscernible sequence such that
$b_{i}^{-1}b_{j}\sim b_{0}^{-1}b_{1}$ for all $i<j$. Then $I$ is constant.

###### Proof.

By Ramsey and compactness, we may find a model $M$ such that (i) $G(M)$
contains a representative of every coset of $G^{00}$, (ii) $I$ is
$M$-indiscernible, and (iii) for some $v\in G(M)$, we have $b_{i}^{-1}b_{j}\in
v^{G}$ for all $i<j$. By (i), there is some $\lambda\in G(M)$ such that
$b_{i}^{-1}b_{j}\in v^{G^{00}\lambda^{-1}}$ for all $i<j$. Let
$a_{i}=b_{i}^{\lambda}$. Then $(a_{i})_{i\in\omega}$ is $M$-indiscernible, and
$a_{i}^{-1}a_{j}\in v^{G^{00}}$ for all $i<j$; by Lemma 5.3,
$(a_{i})_{i\in\omega}$ must be constant, so $I$ is too. ∎

By considering a Morley sequence we hence obtain the following:

######  Corollary 5.5.

Let $M\prec\mathfrak{C}$, and suppose $p(x)\in S_{G}(\mathfrak{C})$ is
$M$-invariant with $p^{\otimes 2}(x,y)\vdash p(x^{-1}y)$ and $p^{\otimes
2}(x,y)\vdash x\sim y$. Then $p(x)\vdash x=1$.

###  5.2 Conjugacy classes determine types

In this section we will show that any type not concentrated on finitely many
conjugacy classes is uniquely determined by the $G^{00}$-conjugacy class of
any of its realizations.

######  Lemma 5.6.

Suppose $B\subset\mathfrak{C}$ is a small set of parameters and that $a\models
G$ begins a $B$-indiscernible sequence of non-conjugates. Then
$a^{g}\equiv_{B}a$ for all $g\models G^{00}$.

###### Proof.

By Ramsey and compactness, we may assume without loss of generality that $B=M$
is a model. I first claim that, for any $J+g\models
q^{\otimes(\omega+1)}|_{(M,a)}$, we have $a^{g}\equiv_{(M,J)}a$; it suffices
for this to show that $a^{g}\equiv_{(M,J)}a$ for some $J+g\models
q^{\otimes(\omega+1)}|_{(M,a)}$. To see this, let $I=(a_{i})_{i\in\omega}$ be
an $M$-indiscernible sequence with $a_{0}=a$ and $a_{i}\not\sim a_{j}$ for all
$i\neq j$. Let $J\models q^{\otimes\omega}|_{(M,I)}$ and let
$K=(g_{i})_{i\in\omega}\models q^{\otimes\omega}|_{(M,I,J)}$. Since $I$ is
$M$-indiscernible and $q$ is $M$-invariant, $I$ and $K$ are mutually
indiscernible over $(M,J)$, and thus by dp-minimality one of them is
indiscernible over $(M,J,a^{g_{1}})$. Since $a_{i}\sim
a^{g_{1}}=a_{0}^{g_{1}}$ if and only if $i=0$, $K$ must therefore be
indiscernible over $(M,J,a^{g_{1}})$, and hence there is some
$a_{*}\equiv_{(M,J)}a$ such that $a_{*}^{g_{0}}=a^{g_{1}}$. But now
$a^{g_{1}g_{0}^{-1}}=a_{*}$, so that $a^{g_{1}g_{0}^{-1}}\equiv_{(M,J)}a$.
Since $q$ is right f-generic and concentrated on $G^{00}$, and $g_{1}\models
q|_{(M,a,J,g_{0})}$, we have $g_{1}g_{0}^{-1}\models q|_{(M,a,J)}$, so the
claim follows.

Now, let $J+g\models q^{\otimes(\omega+1)}|_{(M,a)}$. By the claim,
$a^{g}\equiv_{(M,J)}a$, and by the claim applied to the bi-f-generic distal
type $q^{-1}$ also $a^{g^{-1}}\equiv_{(M,J)}a$. On the other hand, by
distality of $q$, we have
$\operatorname{tp}(a/M,J)\perp^{w}\operatorname{tp}(g/M,J)$, so it follows
that $a_{*}^{g}\equiv_{(M,J)}a\equiv_{(M,J)}a_{*}^{g^{-1}}$ for all
$a_{*}\equiv_{(M,J)}a$. In other words, if $r(x)=\operatorname{tp}(a/M,J)$,
then $g$ lies in the normalizer $N_{G}(r):=\\{h\in G:r(x)^{h}=r(x)\\}$.
$N_{G}(r)$ is an $(M,J)$-invariant subgroup of $G$; since
$\operatorname{tp}(g/M,J)=q|_{(M,J)}$ and $q$ is f-generic, it follows by Fact
2.12 that $G^{00}\subseteq N_{G}(r)$, so that $a^{h}\equiv_{(M,J)}a$ for all
$h\models G^{00}$. The result follows ‘a fortiori’. ∎

######  Corollary 5.7.

If $r(x)\in S_{G}(B)$ for some $B\subset\mathfrak{C}$ and $r$ is not
concentrated on finitely many conjugacy classes, then $r(x)^{g}=r(x)$ for all
$g\models G^{00}$. If $B=M$ is a model, then the same holds provided $r$ is
not concentrated on a single conjugacy class.

######  Corollary 5.8.

Suppose $p(x)\in S_{G}(\mathfrak{C})$ is an $M$-invariant type not
concentrated on any conjugacy class. If $(a_{i})_{i\in\omega}\models
p^{\otimes\omega}|_{M}$, then $(a_{i}^{g_{i}})_{i\in\omega}\models
p^{\otimes\omega}|_{M}$ for all $g_{i}\models G^{00}$.

###### Proof.

If $(b_{i})_{i\in\omega}$ is a Morley sequence of $p$ over $M$, then each
$b_{i}$ starts an $(M,b_{\neq i})$-indiscernible sequence of non-conjugates,
so by Lemma 5.6 $b_{i}^{g}\equiv_{(M,b_{\neq i})}b_{i}$ for all $g\models
G^{00}$. Thus by induction on $n\in\omega$ we have
$(a_{i}^{g_{i}})_{i<n}+(a_{i})_{i\geqslant n}\models p^{\otimes\omega}|_{M}$
for all $g_{i}\models G^{00}$, and the result follows. ∎

###  5.3 Commutators

In this section we will show that, for any $a\models G$, the set of
commutators $[a,G^{00}]$ is a group; this was proved for C-minimal valued
groups in [30].

######  Lemma 5.9.

Let $a\models G$ such that $q(x)\vdash[a,x]\neq 1$. Then $[a,q]$, ie the
pushforward of $q$ under the map $x\mapsto[a,x]$, is not concentrated on a
single conjugacy class.

###### Proof.

Suppose otherwise for contradiction that $[a,q](x)\vdash x\sim u$. Let $M$ be
any small model containing $a,u$, and let $(g_{i})_{i\in\omega}\models
q^{\otimes\omega}|_{M}$. Since $q$ is right f-generic and concentrated on
$G^{00}$, we have $g_{j}g_{i}^{-1}\models q|_{M}$ for all $i<j$ and hence
$[a,g_{j}g_{i}^{-1}]\sim u$ for all $i<j$. For $i\in\omega$, let
$b_{i}=a^{g_{i}}$, and consider the $M$-indiscernible sequence
$I=(b_{i})_{i\in\omega}$. For $i<j$, we have

$b_{i}^{-1}b_{j}=a^{-g_{i}}a^{g_{j}}=[a,g_{j}g_{i}^{-1}]^{g_{i}}\sim u.$

By Lemma 5.4, $I$ is constant, so that $a^{g_{0}}=a^{g_{1}}$ and hence
$g_{1}g_{0}^{-1}\in C_{G}(a)$. Since $g_{1}g_{0}^{-1}\models q|_{M}$, this
contradicts that $q(x)\vdash[a,x]\neq 1$. ∎

By Corollary 5.7, and since $[a,q]$ is just
$\operatorname{tp}(1/\mathfrak{C})$ if $q(x)\vdash[a,x]=1$, we get the
following:

######  Corollary 5.10.

For any $M\prec\mathfrak{C}$ and any $a\in G(M)$, we have
$[a,q|_{M}]=[a,q|_{M}]^{g}$ for all $g\models G^{00}$.

######  Lemma 5.11.

Let $a\models G$. Then $[a,g][a,q]=[a,q]$ for all $g\models G^{00}$, and
$[a,G^{00}][a,q|_{M}]\subseteq[a,G^{00}]$ for any small model $M$ with $a\in
G(M)$.

###### Proof.

First note the following general fact. Let $r\in S_{G}(M)$ be a complete type
over some model $M\prec\mathfrak{C}$ such that $a\in G(M)$ and such that
$[a,r]=[a,r]^{g}$ for all $g\models G^{00}$. Then, from the identity
$[a,g][a,c]^{g}=[a,cg]$, it follows that $[a,g][a,r]=[a,rg]$ for all $g\models
G^{00}$.

The desired result now follows; indeed by Corollary 5.10 and by the remark
above we have $[a,g][a,q|_{M}(x)]=[a,q|_{M}(x)g]$ for all $M\prec\mathfrak{C}$
with $a\in G(M)$ and all $g\models G^{00}$; this shows already that
$[a,G^{00}][a,q|_{M}]\subseteq[a,G^{00}]$. Furthermore, since $q$ is right
f-generic, we have $qg=q$ for all $g\in G^{00}(\mathfrak{C})$, so that
$q|_{M}(x)g=q|_{M}(x)$ whenever $G(M)\ni g$, and hence indeed
$[a,g][a,q]=[a,q]$. ∎

######  Lemma 5.12.

For any $a\models G$, the set of commutators $[a,G^{00}]$ is a type-definable
subgroup of $G^{00}$, and it is ‘strongly connected’, ie has no proper type-
definable subgroups of bounded index.

###### Proof.

Certainly $[a,G^{00}]$ is type-definable, and it is a subset of $G^{00}$ since
$G^{00}$ is a normal subgroup of $G$. Now, let $M$ be any small model
containing $a$. By Corollary 5.10, we have $[a,q|_{M}]^{g}=[a,q|_{M}]$ for all
$g\models G^{00}$; likewise, by Corollary 5.10 applied to the bi-f-generic
distal type $q^{-1}$, we also have $[a,q|_{M}^{-1}]^{g}=[a,q|_{M}^{-1}]$ for
all $g\models G^{00}$. Thus, for any $b\models q|_{M}$, since $q$ is
concentrated on $G^{00}$ we have

$\displaystyle[a,b^{-1}]$
$\displaystyle=[a,b]^{-b^{-1}}\in[a,q|_{M}]^{-b^{-1}}=[a,q|_{M}]^{-1}\text{
and}$ $\displaystyle[a,b]^{-1}$
$\displaystyle=[a,b^{-1}]^{b^{-1}}\in[a,q|_{M}^{-1}]^{b^{-1}}=[a,q|_{M}^{-1}],$

so that $[a,q|_{M}^{-1}]=[a,q|_{M}]^{-1}$, and it follows that
$[a,q|_{M}][a,q|_{M}^{-1}]$ is closed under taking inverses. Now, by Fact
2.12, we have $G^{00}\subseteq q|_{M}^{-1}(x)q|_{M}(x)$. Thus, for any
$g\models G^{00}$, there are $b,c\models q|_{M}$ with $g=b^{-1}c$, and then we
have

$[a,g]=[a,c][a,b^{-1}]^{c}\in[a,q|_{M}][a,q|_{M}^{-1}]^{c}=[a,q|_{M}][a,q|_{M}^{-1}],$

so that $[a,G^{00}]\subseteq[a,q|_{M}][a,q|_{M}^{-1}]$. Finally, by Lemma 5.11
applied to $q$ and to $q^{-1}$, we have

$[a,G^{00}][a,q|_{M}][a,q|_{M}^{-1}]\subseteq[a,G^{00}][a,q|_{M}^{-1}]\subseteq[a,G^{00}].$

Thus $[a,q|_{M}][a,q|_{M}^{-1}]=[a,G^{00}]$, and it is symmetric and closed
under multiplication; in other words, it is a group.

To see that $[a,G^{00}]$ is strongly connected, note that, by Lemma 5.11,
$[a,q]$ is invariant under left translation by all elements of
$[a,G^{00}(\mathfrak{C})]$. If $H\leqslant[a,G^{00}]$ is type-definable and of
bounded index in $[a,G^{00}]$, then, since $[a,q]$ is concentrated on
$[a,G^{00}]$, it must be concentrated on some right coset of $H$, and then
left-invariance of $[a,q]$ forces $H=[a,G^{00}]$. ∎

######  Corollary 5.13.

For any $a,b\models G$, one of $[a,G^{00}]\supseteq[b,G^{00}]$ and
$[a,G^{00}]\subseteq[b,G^{00}]$ holds.

###### Proof.

By Fact 2.14, $[a,G^{00}]\cap[b,G^{00}]$ has bounded index in one of
$[a,G^{00}]$ and $[b,G^{00}]$. Since the latter two groups are strongly
connected, $[a,G^{00}]\cap[b,G^{00}]$ must be equal to one of them, and the
result follows. ∎

######  Corollary 5.14.

For any type-definable set $X\subseteq G$, the set of commutators $[X,G^{00}]$
is a type-definable strongly connected subgroup of $G^{00}$.

###### Proof.

$[X,G^{00}]$ is equal to $\bigcup_{a\models X}[a,G^{00}]$, and by Lemma 5.12
and Corollary 5.13 this is the union of a chain of subgroups of $G^{00}$. So
indeed $[X,G^{00}]$ is a group, and it is certainly type-definable. To see
that it is strongly connected, suppose that $H\leqslant[X,G^{00}]$ is a
bounded-index type-definable subgroup. Then, for every $a\models X$,
$H\cap[a,G^{00}]$ has bounded index in $[a,G^{00}]$, hence must be equal to
$[a,G^{00}]$. So $H\geqslant[a,G^{00}]$ for all $a\models X$ and hence
$H=[X,G^{00}]$, as desired. ∎

The following now has an identical proof as Proposition 2.4.1 of [30]:

######  Lemma 5.15.

$[G^{00},G^{00}]$ has finite exponent.

###### Proof.

Fix $a,b\models G^{00}$; by Lemma 5.12, $[a,b,G^{00}]$ is now a normal
subgroup of $G^{00}$. Working in the quotient $G^{00}/[a,b,G^{00}]$, it
follows from the identities $[a^{k+1},b]=[a,b][a,b,a^{k}][a^{k},b]$ and
$[a,b^{k+1}]=[a,b^{k}][a,b][a,b,b^{k}]$ that $[a^{k},b]$ and $[a,b^{k}]$ are
each equivalent to $[a,b]^{k}$ modulo $[a,b,G^{00}]$ for each $k\in\omega$. By
Fact 2.15, there is some $n>0$ such that one of $[a^{n},b]$ and $[a,b^{n}]$ is
trivial, and it follows that $[a,b]^{n}\in[a,b,G^{00}]$.

Thus there is some $g\models G^{00}$ with $[a,b]^{n}=[a,b,g]$, so that
$[a,b]^{n+1}=[a,b]^{g}\sim[a,b]$. By Lemma 2.6 and Fact 2.15, it follows that
$[a,b]$ must be a torsion element. Since $a,b\models G^{00}$ were arbitrary,
thus every element of $[G^{00},G^{00}]$ has finite order, and so by
compactness $[G^{00},G^{00}]$ has finite exponent. ∎

######  Corollary 5.16.

If $G$ does not have finite exponent, then it is virtually nilpotent of class
$\leqslant 2$.

###### Proof.

If the realizations of $q(x)$ are non-torsion, then by Lemma 4.4 we are done,
so we may assume that $q(x)\vdash x^{k}=1$ for some $k>0$; we aim to show that
$G$ has finite exponent. By Lemma 5.15, let $m>0$ be the exponent of
$[G^{00},G^{00}]$. By Fact 2.12, every element of $G^{00}$ is a difference of
two realizations of $q$, and it follows that $G^{00}$ has exponent at most
$mk$: if $a,b\models G^{00}$ have order at most $k$, then $(ab)^{k}$ is
equivalent to $a^{k}b^{k}=1$ modulo $[G^{00},G^{00}]$, so that $(ab)^{k}$ lies
in $[G^{00},G^{00}]$, and hence $(ab)^{km}=1$. Thus indeed $G^{00}$ has
exponent at most $n:=mk$.

Now, since $G^{00}$ is a group, we have $G^{00}(x)\wedge
G^{00}(y)\vdash(xy^{-1})^{n}=1$, and so by compactness there is some
$\varnothing$-definable set $D\supseteq G^{00}$ such that $(ab^{-1})^{n}=1$
for all $a,b\models D$; in other words $c^{n}=1$ for all $c\in DD^{-1}$.

Now I claim that $G$ is torsion; by compactness the result will follow.
Otherwise, there is some non-torsion element $b\models G$. Since every element
of $DD^{-1}$ is torsion, we have $b^{k}\notin DD^{-1}$ for all $k\neq 0$, and
hence $b^{i}D\cap b^{j}D=\varnothing$ for all $i\neq j\in\omega$. Letting
$I=(a_{i})_{i\in\omega}$ be an indiscernible sequence realizing the EM-type of
$(b^{i})_{i\in\omega}$ over $\varnothing$, we then have $a_{i}D\cap
a_{j}D=\varnothing$ for all $i\neq j$; since $D\supseteq G^{00}$, the $a_{i}$
thus lie in pairwise distinct cosets of $G^{00}$. Since $I$ is indiscernible,
this contradicts that $G^{00}$ has bounded index in $G$, and the result
follows. ∎

###  5.4 Valued group structure

Suppose that $(H,\cdot)$ is an arbitrary group and that $N\trianglelefteqslant
H$ is a normal subgroup such that the collection $\mathcal{P}=\\{[a,N]:a\in
H\\}$ is a family of subgroups of $H$ totally ordered by the reverse inclusion
relation, which we will denote $\preccurlyeq$. Note that
$(\mathcal{P},\preccurlyeq)$ has a maximal element $\\{1\\}=[1,N]$. Now,
define a map $v:H/C_{H}(N)\to\mathcal{P}$ by $v(a)=[a,N]$; this is well-
defined since $[C_{H}(N)a,g]=\\{[a,g]\\}$ for all $a\in H$ and $g\in N$. By
definition, $v(a)\preccurlyeq v(b)$ if and only if $[a,N]\supseteq[b,N]$.

######  Lemma 5.17.

In the above circumstances, $(H,v)$ is a $\mathcal{P}$-valued group.

###### Proof.

For property (1), note that, for $a\in H$, we have $v(a)=\max\mathcal{P}$ if
and only if $[a,N]=\\{1\\}$, ie if and only if $a\in C_{H}(N)$. For property
(2), suppose that $v(a)\prec v(b)$ and that $g\in H$ is arbitrary. We have
$[a,N]\supsetneq[b,N]$, and hence $[a,N]^{g}\supsetneq[b,N]^{g}$. But $N$ is a
normal subgroup of $H$, so $[a,N]^{g}=[a^{g},N^{g}]=[a^{g},N]$, and likewise
$[b,N]^{g}=[b^{g},N]$, so that $v(a^{g})\prec v(b^{g})$ as desired. For
property (3), fix $a\in H$. We have $[a^{-1},N]=[N,a]^{a^{-1}}=[N,a]$. But
$[a,N]$ is by hypothesis a subgroup of $H$, hence in particular is closed
under inverses, and so $[a^{-1},N]=[a,N]$, whence $v(a)=v(a^{-1})$.

So we need only show property (4), ie that
$v(ab)\succcurlyeq\min\\{v(a),v(b)\\}$ for all $a,b\in H$. First suppose that
$v(b)=\min\\{v(a),v(b)\\}$, ie that $[a,N]\subseteq[b,N]$; we wish to show
that $[ab,N]\subseteq[b,N]$. By the identity $[ab,g]=[a,g]^{b}[b,g]$, we have
$[ab,N]\subseteq[a,N]^{b}[b,N]$. But $[a,N]^{b}\subseteq[b,N]^{b}=[b,N]$, so
we have $[ab,N]\subseteq[b,N][b,N]\subseteq[b,N]$, where the second inclusion
uses that $[b,N]$ is closed under multiplication. In other words
$v(ab)\succcurlyeq v(b)$, as desired.

Now suppose that $v(a)=\min\\{v(a),v(b)\\}$. Using that $v(c)=v(c^{-1})$ for
all $c\in H$, we have $v(ab)=v(b^{-1}a^{-1})$ and $v(b^{-1})=v(b)\succcurlyeq
v(a)=v(a^{-1})$. Thus $v(a^{-1})=\min\\{v(b^{-1}),v(a^{-1})\\}$ and so, by the
case in the previous paragraph, we have $v(b^{-1}a^{-1})\succcurlyeq
v(a^{-1})$. But this means $v(ab)\succcurlyeq v(a)$, as desired. ∎

Now, let us return to the group $G$. The subgroup $N=G^{00}$ is normal in $G$,
and by Lemma 5.12 and Corollary 5.13 the set
$\mathcal{P}=\\{[a,G^{00}]:a\models G\\}$ is a chain of subgroups of $G$. So,
by Lemma 5.17, the map $v:a\mapsto[a,G^{00}]$ makes $G/C_{G}(G^{00})$ into a
$\mathcal{P}$-valued group. Note that, by Fact 2.11, the group $C_{G}(G^{00})$
has a purely group-theoretic description; recall that the FC-center of $G$,
which we will denote $\mathrm{FC}(G)$, is the set of all $g\in G$ with
$[G:C_{G}(g)]<\infty$. This is a characteristic subgroup of $G$, and by Fact
2.11 we have $g\in\mathrm{FC}(G)$ if and only if $G^{00}\subseteq C_{G}(g)$.
So $C_{G}(G^{00})=\mathrm{FC}(G)$ and thus the map $v$ makes
$G/\mathrm{FC}(G)$ into a valued group.

We have $\mathrm{FC}(G)\cap G^{00}=C_{G}(G^{00})\cap G^{00}=Z(G^{00})$. Thus
$G^{00}/Z(G^{00})$ inherits the valued group structure from
$G/\mathrm{FC}(G)$.

######  Theorem 5.18.

$(G^{00}/Z(G^{00}),v)$ is a plain valued group.

###### Proof.

We need only show that $v$ is a plain valuation. Hence fix $a,b\in
G^{00}\setminus Z(G^{00})$; we wish to show that
$v([a,b])\succ\max\\{v(a),v(b)\\}$. Since $v([a,b])=v([b,a])$, we may assume
without loss that $v(a)=\max\\{v(a),v(b)\\}$, so that
$[a,G^{00}]\subseteq[b,G^{00}]$. We wish to show that $v([a,b])\succ v(a)$, ie
that $[a,b,G^{00}]\not\supseteq[a,G^{00}]$. To see this, let $M$ be any small
model and let $g\models q|_{(M,a,b)}$; it suffices to show that
$[a,g]\notin[a,b,G^{00}]$. Suppose otherwise for contradiction that
$[a,g]=[a,b,h]$ for some $h\models G^{00}$. Now,
$[a,g]\models[a,q]|_{(M,a,b)}$, and $b\models G^{00}$, so by Lemma 5.11 we
have $[a,b][a,g]\models[a,q]|_{(M,a,b)}$. But
$[a,b][a,g]=[a,b][a,b,h]=[a,b]^{h}$. Thus $[a,q]$ is concentrated on the
conjugacy class $[a,b]^{G}$, and by Lemma 5.9 we must have $q(x)\vdash x\in
C_{G}(a)$. But then, by f-genericity of $q$, $C_{G}(a)$ has finite index in
$G$, contradicting that $a\notin\mathrm{FC}(G)$. ∎

Now by Fact 2.8 we obtain the following:

######  Corollary 5.19.

$G^{00}$ is locally nilpotent.

###### Proof.

If $G$ has finite exponent, then by Fact 2.8 and Theorem 5.18
$G^{00}/Z(G^{00})$ is locally nilpotent, whence $G^{00}$ is too. So we may
assume that $G$ does not have finite exponent, and then it follows from
Corollary 5.16 that $G^{00}$ is nilpotent of class at most $2$, hence in
particular locally nilpotent. ∎

###  5.5 Local nilpotence to nilpotence

The following lemma and its proof work just with the assumption that $G$ is
dp-minimal; we do not need the existence of a distal f-generic type for it.

######  Lemma 5.20.

Let $M\prec\mathfrak{C}$ be any model, and suppose $p(x)\in
S_{G}(\mathfrak{C})$ is an $M$-invariant type with the property that, for some
(any) $(a,b)\models p^{\otimes 2}|_{M}$, the group $\langle a,b\rangle$ is
nilpotent of class at most $(n-1)$. Then there is some $\sigma\in S_{n}$ such
that $p^{\otimes
n}(x_{1},\dots,x_{n})\vdash[x_{\sigma(1)},\dots,x_{\sigma(n)}]=1$.

###### Proof.

Let $(a_{i})_{i\in\omega}+(u_{i})_{i\in\omega}+(b_{i})_{i\in\omega}$ be a
Morley sequence of $p$ over some model $M$. Write $I=(a_{i})_{i\in\omega}$ and
$J=(b_{i})_{i\in\omega}$ and $K=(u_{i})_{i\in\omega}$ for notational
convenience.

By hypothesis, $\langle a_{0},b_{1}\rangle$ is nilpotent of class at most
$(n-1)$. We will prove by backwards induction on $k$ that, for every
$k\in[n]\cup\\{0\\}$, and for all $c_{1},\dots,c_{k}\in\\{a_{0},b_{1}\\}$,
there is some $\tau\in S_{n-k}$ such that
$[c_{1},\dots,c_{k},u_{\tau(n-k)},\dots,u_{\tau(1)}]=1$. The base case $n=k$
is clear. Now, suppose we have shown the result for $k+1$, and fix
$c_{1},\dots,c_{k}\in\\{a_{0},b_{1}\\}$. $I$ and $J$ are mutually
indiscernible over $K$, and so by dp-minimality one of $I$ and $J$ is
indiscernible over $(K,[c_{1},\dots,c_{k}])$.

First suppose $I$ is indiscernible over that set. By the inductive hypothesis,
there is some $\nu\in S_{n-k-1}$ such that
$[c_{1},\dots,c_{k},a_{0},u_{\nu(n-k-1)},\dots,u_{\nu(1)}]=1$. By
indiscernibility of $I$ over $(K,[c_{1},\dots,c_{k}])$, it follows that
$[c_{1},\dots,c_{k},a_{1},u_{\nu(n-k-1)},\dots,u_{\nu(1)}]=1$. Now if we
define $\tau\in S_{n-k}$ by taking $\tau(n-k)=1$ and taking $\tau(i)=\nu(i)+1$
for $i\in[n-k-1]$ then we get the desired result by indiscernibility of
$a_{0}+a_{1}+u_{1}+\dots+u_{n-k}+b_{1}$.

Now suppose $J$ is indiscernible over that set. By the inductive hypothesis,
there is some $\nu\in S_{n-k-1}$ such that
$[c_{1},\dots,c_{k},b_{1},u_{\nu(n-k-1)},\dots,u_{\nu(1)}]=1$. By
indiscernibility of $J$ over $(K,[c_{1},\dots,c_{k}])$, it follows that
$[c_{1},\dots,c_{k},b_{0},u_{\nu(n-k-1)},\dots,u_{\nu(1)}]=1$. Now if we
define $\tau\in S_{n-k}$ by taking $\tau(n-k)=n-k$ and $\tau(i)=\nu(i)$ for
$i\in[n-k-1]$ then we get the desired result by indiscernibility of
$a_{0}+u_{1}+\dots+u_{n-k}+b_{0}+b_{1}$.

So the induction goes through, and when $k=0$ we get that
$[u_{\sigma(n)},\dots,u_{\sigma(1)}]=1$ for some $\sigma\in S_{n}$, giving the
desired result. ∎

Now we can obtain the main result.

######  Theorem 5.21.

$G$ is virtually nilpotent.

###### Proof.

By Corollary 5.19, $G^{00}$ is locally nilpotent, so any global type $p(x)$
concentrated on $G^{00}$ and invariant over some small set satisfies the
hypothesis of Lemma 5.20 for some $n>0$. In particular, there is some $n>0$
and some $\sigma\in S_{n}$ such that $q^{\otimes
n}(x_{1},\dots,x_{n})\vdash[x_{\sigma(1)},\dots,x_{\sigma(n)}]=1$. Fix any
sequence $(a_{1},\dots,a_{n})\models q^{\otimes n}|_{M}$. Since $q(x)$ is
concentrated on $G^{00}$, we have $q^{\otimes 2}(x,y)\vdash q(x^{-1}y)$ by
Fact 2.12, and so by Corollary 5.5 $q$ is not concentrated on any conjugacy
class. By Corollary 5.8, we thus have
$(a_{1}^{g_{1}},\dots,a_{n}^{g_{n}})\models q^{\otimes n}|_{M}$ for all
$g_{i}\models G^{00}$, and hence in particular
$[a_{\sigma(1)}^{G^{00}},\dots,a_{\sigma(n)}^{G^{00}}]=\\{1\\}$. For each
$k\in\\{0\\}\cup[n]$ and $l\in[n]$, define $A_{kl}=a_{l}^{G^{00}}$ if
$l\leqslant n-k$ and $A_{kl}=G^{00}$ otherwise. We prove by induction on $k$
that $[A_{k\sigma(1)},\dots,A_{k\sigma(n)}]=\\{1\\}$; we have already shown
the base case $k=0$.

Now, for the inductive step, suppose we have shown the result for $k$, and let
$i=\sigma^{-1}(n-k)$. Note that each $A_{kl}$ is setwise invariant under
conjugation by elements of $G^{00}$. Hence, by Lemma 2.3, the set of $g\models
G^{00}$ such that

$[A_{k\sigma(1)},\dots,A_{k\sigma(i-1)},g,A_{k\sigma(i+1)},\dots,A_{kn}]=\\{1\\}$

is a subgroup of $G^{00}$; call it $U$. By the inductive hypothesis,
$U\supseteq A_{k\sigma(i)}=a_{n-k}^{G^{00}}$, and hence in particular
$a_{n-k}\in U$. On the other hand, for $j\neq i$, $A_{k\sigma(j)}$ is either
equal to $G^{00}$ or of form $a_{l}^{G^{00}}$ for some $l<n-k$, and it follows
that $U$ is $(M,a_{<n-k})$-invariant. But $a_{n-k}\models q|_{(M,a_{<n-k})}$
and $a_{n-k}\in U(\mathfrak{C})$, so all the realizations of
$q|_{(M,a_{<n-k})}$ lie in $U(\mathfrak{C})$, and by Fact 2.12 it follows that
$G^{00}\subseteq U$ and hence $G^{00}=U$. Since $A_{(k+1),\sigma(i)}=G^{00}$
and $A_{(k+1),\sigma(j)}=A_{k\sigma(j)}$ for $j\neq i$, the inductive step
follows.

Taking $k=n$, we get $[G^{00},_{\ n-1}G^{00}]=\\{1\\}$, and the result
follows. (For example, we now have in particular that $q^{\otimes
n}(x_{1},\dots,x_{n})\vdash[x_{1},\dots,x_{n}]=1$, and so the result follows
by Lemma 3.1.) ∎

##  6 UCC case

In this section we will study dp-minimal groups with the ‘uniform chain
condition’ (ucc), ie with the property that they contain no infinite strictly
decreasing sequence of uniformly definable subgroups. The class of ucc groups
includes stable groups, groups definable in o-minimal structures, and more
generally all rosy groups; see eg [10].

After completing this project, we learned that Eran Alouf and Frank Wagner had
independently proved Theorem 6.7 in the case that $G$ is stable. Since then,
Frank Wagner has shared with us a proof that any solvable abelian-by-(finite
exponent) group with the intersection chain condition is virtually nilpotent.
By Fact 2.15, Fact 2.10, and Theorem 6.7, it follows that dp-minimal ucc
groups are in fact virtually nilpotent.

Throughout this section, assume that $G$ is dp-minimal.

We begin first with a remark that failures of ucc give rise to type-definable
subgroups of $G$ that have distal f-generic types, where we define
f-genericity for relatively definable subsets of a type-definable group in the
natural way.111Namely, suppose $O\leqslant G$ is a subgroup of $G$ type-
definable over some parameters. Given a $\mathfrak{C}$-definable set
$D\subseteq G$, let us say that $D$ is ‘f-generic for $O$’ if the following
holds: for some (any) small model over which $D$ is defined and $O$ is type-
defined, and for any $g\in O(\mathfrak{C})$, the partial type $x\in gD\cap O$
does not fork over $M$. Then say that a partial type is f-generic for $O$ if
it is concentrated on $O$ and implies only formulas that are f-generic for
$O$. We will not use this observation, but thought it worth making in light of
Section 5. (Note however that throughout this paper we have worked under the
assumption that $G$ is a definable group, so Theorem 5.21 does not apply to
type-definable dp-minimal groups with a distal f-generic type.)

######  Remark 6.1.

Suppose $G$ does not have ucc. Then there is an infinite type-definable
subgroup $O\leqslant G$ and a global distal type $p(x)\in S_{O}(\mathfrak{C})$
that is f-generic for $O$.

###### Proof.

By Ramsey and compactness, there is a formula $\phi(x,y)$ and an indiscernible
sequence $(b_{i})_{i\in\omega}$ such that (i) $\phi(x,b_{0})$ defines a
subgroup of $G$, and (ii) for each $i\in\omega$, $\phi(x,b_{i+1})$ defines a
proper subgroup of $\phi(x,b_{i})$. We will take $O$ to be the subgroup of $G$
type-defined by $\bigwedge_{i\in\omega}\phi(x,b_{i})$.

By an essentially identical argument to that of Section 4 in [31], the
$\mathfrak{C}$-definable subsets of $G$ not f-generic for $O$ form an ideal,
and hence the partial type $\pi(x)$ containing $\neg\psi(x,c)$ for every such
formula $\psi(x,c)$ is consistent. Letting $M$ be any small model containing
$\\{b_{i}:i\in\omega\\}$, then $\pi(x)\wedge x\in O$ is $M$-invariant and
consistent, hence does not divide over $M$, and so by NIP extends to a global
$M$-invariant type $p(x)\in S_{O}(\mathfrak{C})$ which is now f-generic for
$O$. We claim that $p(x)$ is not generically stable. To see this, by NIP, let
$q(y)\in S(\mathfrak{C})$ be the limit type of the sequence
$(b_{i})_{i\in\omega}$; this is an $M$-invariant type.

On the one hand, we have $q(y)\otimes p(x)\vdash\phi(x,y)$. Indeed, if
$a\models p|_{M}$, then $a\models O$, ie $a\models\phi(x,b_{i})$ for all
$i\in\omega$, and hence $q(y)\vdash\phi(a,y)$, as needed. On the other hand,
we have $p(x)\otimes q(y)\vdash\neg\phi(x,y)$. To see this, let $b\models
q|_{M}$. Then $(b_{i})_{i\in\omega}+b$ is indiscernible, and so $\phi(x,b)$
defines an infinite-index subgroup of $\phi(x,b_{i})$ for all $i\in\omega$. It
follows by compactness that $\phi(x,b)$ has unbounded index in $O$, whence
$\phi(x,b)$ is not f-generic for $O$. By f-genericity of $p$, thus
$p(x)\vdash\neg\phi(x,b)$, as needed.

So $p(x)\otimes q(y)\neq q(y)\otimes p(x)$, and thus $p(x)$ cannot be
generically stable. By Fact 2.19, it follows that $p(x)$ is distal, and we are
done.∎

Now we will turn to showing that, if $G$ has ucc, then it is virtually
solvable. First we begin with a general observation, which does not need a ucc
assumption.

######  Lemma 6.2.

Let $p(x)\in S_{G}(\mathfrak{C})$ be an unrealized generically stable type.
Then $p(x)$ has infinite left-stabilizer.

###### Proof.

Let $M$ be a small model over which $p$ is invariant, and let
$(a_{i})_{i\in\mathbb{Z}}+(b_{j})_{j\in\mathbb{Z}}$ be a Morley sequence of
$p$ over $M$. By dp-minimality, one of $(a_{i})_{i\in\mathbb{Z}}$ and
$(b_{j})_{j\in\mathbb{Z}}$ is indiscernible over $(M,a_{0}^{-1}b_{0})$. Since
$(a_{i})_{i\in\mathbb{Z}}+(b_{j})_{j\in\mathbb{Z}}$ is totally indiscernible,
and $a_{0}^{-1}b_{0}$ and $b_{0}^{-1}a_{0}$ are interdefinable, it follows
that $(a_{i})_{i\in\mathbb{Z}}$ and $(b_{j})_{j\in\mathbb{Z}}$ are each
indiscernible over $(M,a_{0}^{-1}b_{0})$. So let $f$ be an automorphism fixing
$(M,a_{0}^{-1}b_{0})$ pointwise and mapping $a_{i}$ to $a_{i+1}$ for each
$i\in\mathbb{Z}$. Let $b=b_{0}$ and $b^{\prime}=f(b_{0})$; then
$a_{0}^{-1}b=a_{1}^{-1}b^{\prime}$, ie $a_{1}a_{0}^{-1}b=b^{\prime}$, and
$(a_{0},a_{1},b)$ and $(a_{0},a_{1},b^{\prime})$ each realize $p^{\otimes
3}|_{M}$. Thus, letting $g=a_{1}a_{0}^{-1}$, we have $b\models p|_{(M,g)}$ and
$gb\models p|_{(M,g)}$.

Let $N$ be any small model containing $(M,g)$ and such that $b\models p|_{N}$.
Then $b\mathop{\mathchoice{\kern 5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower
3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\displaystyle{}}{\kern
5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox
to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\textstyle{}}{\kern 5.14375pt\hbox
to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern
5.14375pt\scriptstyle{}}{\kern 5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower
3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern
5.14375pt\scriptscriptstyle{}}}_{M}N$, whence $b\mathop{\mathchoice{\kern
5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox
to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\displaystyle{}}{\kern 5.14375pt\hbox
to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern
5.14375pt\textstyle{}}{\kern 5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower
3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\scriptstyle{}}{\kern
5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox
to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\scriptscriptstyle{}}}_{(M,g)}N$,
whence $gb\mathop{\mathchoice{\kern 5.14375pt\hbox
to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern
5.14375pt\displaystyle{}}{\kern 5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower
3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\textstyle{}}{\kern
5.14375pt\hbox to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox
to0.0pt{\hss$\smile$\hss}\kern 5.14375pt\scriptstyle{}}{\kern 5.14375pt\hbox
to0.0pt{\hss$\mid$\hss}\lower 3.48746pt\hbox to0.0pt{\hss$\smile$\hss}\kern
5.14375pt\scriptscriptstyle{}}}_{(M,g)}N$. But $gb\models p|_{(M,g)}$, and it
follows from Fact 2.18 that $gb\models p|_{N}$. So $b$ and $gb$ each realize
$p|_{N}$; since $g\in G(N)$, and $p$ is definable over $N$ and hence an heir
of $p|_{N}$, it follows that $gp=p$.

We thus have that $c_{0}^{-1}c_{1}$ left-stabilizes $p$ for any
$(c_{0},c_{1})\models p^{\otimes 2}|_{M}$, and since $p$ is unrealized the
claim follows. ∎

Now we can begin proving the main result.

######  Lemma 6.3.

Suppose $G$ has ucc. Then there is an infinite type-definable subgroup
$H\leqslant G$ such that (i) $H\leqslant O$ for every infinite
$\mathfrak{C}$-definable subgroup $O\leqslant G$, and (ii) $H(\mathfrak{C})$
left-stabilizes and right-stabilizes every unrealized generically stable type
concentrated on $G$: if $p(x)\in S_{G}(\mathfrak{C})$ is generically stable
and unrealized, and $g_{1},g_{2}\in H(\mathfrak{C})$, then $g_{1}pg_{2}=p$.
Moreover, $H$ is an intersection of $\varnothing$-definable subgroups of $G$,
and it is normal in $G$.

###### Proof.

For an $L$-formula $\phi(x,y)$, let $H_{\phi}$ be the intersection of all
instances of $\phi(x,y)$ defining an infinite subgroup of $G$. By Fact 2.10,
$H_{\phi}$ is an intersection of finitely many such instances, whence it is
definable and, by Fact 2.14, has finite index in some such instance; in
particular it must itself be infinite. Moreover, it is automorphism-invariant,
and hence $\varnothing$-definable.

Thus take $H$ to be the intersection of $H_{\phi}$ over all $L$-formula
$\phi(x,y)$. $H$ is type-definable and normal in $G$ by construction. It is
also an intersection of boundedly many infinite definable subgroups of $G$,
and again by Fact 2.14 it follows from compactness that $H$ is infinite. So we
need only show that $H(\mathfrak{C})$ left-stabilizes and right-stabilizes
every unrealized generically stable type concentrated on $G$; since $p$ is
generically stable if and only if $p^{-1}$ is, it suffices to show that
$H(\mathfrak{C})$ left-stabilizes every unrealized generically stable type
$p(x)\in S_{G}(\mathfrak{C})$.

So fix such $p$. By Lemma 6.2, $p(x)$ has infinite left-stabilizer. But $p(x)$
is a definable type, so for any $L$-formula $\phi(x,y)$ the
left-$\phi$-stabilizer $\mathrm{st}_{\phi}(p)$ is now an infinite
$\mathfrak{C}$-definable subgroup of $G$, and by construction of $H$ it
follows that $H\leqslant\mathrm{st}_{\phi}(p)$. So $H$ is contained in every
left-$\phi$-stabilizer of $p$, and hence contained in the left-stabilizer of
$p$, as claimed. ∎

######  Lemma 6.4.

Suppose $G$ has ucc, and let $H$ be given by Lemma 6.3. Then there is at most
one unrealized generically stable type concentrated on $H$.

###### Proof.

Suppose that $p(x)\in S_{H}(\mathfrak{C})$ and $q(x)\in S_{H}(\mathfrak{C})$
are both unrealized and generically stable, and let $M$ be any small model
over which they are both invariant. Let $a\models p|_{M}$ and $b\models
q|_{(M,a)}$. Since $p$ is concentrated on $H$, we have $a\in H(\mathfrak{C})$,
and $q$ is left-$H(\mathfrak{C})$-invariant by Lemma 6.3, so $a^{-1}b\models
q|_{(M,a)}$. On the other hand, since $p$ and $q$ are generically stable, we
have $p\otimes q=q\otimes p$ (indeed, this holds if just one of $p$ and $q$ is
generically stable), and so in turn $a\models p|_{(M,a^{-1}b)}$. But now
$a^{-1}b\in H(\mathfrak{C})$, and $p$ is right-$H(\mathfrak{C})$-invariant by
Lemma 6.3, so $b=a(a^{-1}b)\models p|_{(M,a^{-1}b)}$. In particular, $b$
realizes both $q|_{M}$ and $p|_{M}$, and since $M$ was arbitrary thus $p=q$. ∎

If $G$ is stable, then it follows from Lemma 6.4 that the subgroup $H$ given
by Lemma 6.3 is minimal, and hence abelian by Reineke’s theorem [20]. However,
if $G$ is merely ucc, we cannot conclude that $H$ is minimal in general, even
assuming that $H$ has an unrealized generically stable type. For example,
consider the case where $G$ is the additive group of the structure
$(\mathbb{R},+,[0,1])$, studied for instance in [17]. Then $G$ is dp-minimal
and has no proper non-trivial definable subgroups, so in this case $H=G$. One
can check that $G$ has a unique global f-generic type, which is generically
stable, but $G$ is not strongly minimal.

So we cannot conclude in the general ucc case that $H$ is minimal. However, an
argument essentially identical to Reineke’s proof nonetheless gives that, if
it admits an unrealized generically stable type, then $H$ is abelian:

######  Lemma 6.5.

Suppose $G$ has ucc and let $H$ be given by Lemma 6.3, and suppose there is a
unrealized generically stable type $q(x)\in S_{H}(\mathfrak{C})$ concentrated
on $H$. Then $H$ is abelian.

###### Proof.

By Fact 2.15, there is a definable normal abelian subgroup
$A\trianglelefteqslant G$ such that $G/A$ has finite exponent. If $A$ is
infinite, then $H\leqslant A$ by construction, so we are done. Hence we may
assume that $A$ is finite, and it then follows that $G$ has finite exponent.

I claim that, if $a\models H$ is any non-central element of $H(\mathfrak{C})$,
then $q(x)$ is concentrated on the $H$-conjugacy class of $a$. To see this,
fix such an $a$. Then $C_{G}(a)$ must be finite; indeed, if it were infinite,
then (since it is a $\mathfrak{C}$-definable subgroup of $G$) it would contain
$H$, in which case $a$ would be central in $H$. Now, let $N$ be any small
model containing $a$ and over which $q$ is invariant, and let $b\models
q|_{N}$. Since $C_{G}(a)$ is finite, $b\in\operatorname{acl}(a,a^{b})$; since
$q$ is unrealized and $a\in G(N)$, it follows that $a^{b}\notin G(N)$. But
$a^{b}$ realizes $a^{q}|_{N}$, where $a^{q}$ is the pushforward of $q$ under
the map $y\mapsto a^{y}$. Definable pushforwards of generically stable types
are generically stable, so $a^{q}$ is generically stable, and it is unrealized
since $a^{b}\notin G(N)$; by Lemma 6.4 it follows that $a^{q}=q$. In
particular, we indeed have $q(y)\vdash y\in a^{H}$.

It follows that any two non-central elements of $H$ lie in the same
$H$-conjugacy class, and hence that $H(\mathfrak{C})/Z(H(\mathfrak{C}))$ has
at most one non-trivial conjugacy class. But $G$ and hence $H/Z(H)$ have
finite exponent; the only group of finite exponent with a unique non-trivial
conjugacy class is the cyclic group of order $2$, and it follows that $H$ is
abelian.∎

######  Corollary 6.6.

Suppose $G$ has ucc, and let $H$ be given by Lemma 6.3. Then $H$ is abelian,
and so in particular $G$ has an infinite normal abelian subgroup.

###### Proof.

If $H$ admits an unrealized generically stable type, then we are done by Lemma
6.5, so we may assume that there is no unrealized generically stable type
concentrated on $H$. It follows that there is no non-constant totally
indiscernible sequence in $H$, as the limit type of such a sequence would be
an unrealized generically stable type concentrated on $H$.

By the construction of $H$, we have $H=\bigcap_{i\in I}H_{i}$ for some
definable subgroups $H_{i}\leqslant G$, and now by compactness there must be
some $i\in I$ such that $H_{i}$ contains no non-constant totally indiscernible
sequence. But then there is no unrealized generically stable type concentrated
on $H_{i}$, as the Morley sequence of such a type would be non-constant and
totally indiscernible. Applying Corollary 2.20 and Theorem 5.21 to the dp-
minimal definable group $H_{i}$, it follows that $H_{i}$ is virtually
nilpotent. In particular, there is an infinite abelian subgroup $A\leqslant
H_{i}(\mathfrak{C})$. But now $C_{G}(C_{G}(A))$ is $\mathfrak{C}$-definable by
Fact 2.10 and infinite; by the properties of $H$, thus $H\leqslant
C_{G}(C_{G}(A))$, so $H$ is abelian and we are done. ∎

######  Theorem 6.7.

If $G$ has ucc, then it is virtually solvable.

###### Proof.

By Fact 2.15 there is a definable abelian normal subgroup
$A\trianglelefteqslant G$ such that $G/A$ has finite exponent. As $G/A$ is
interpretable on the $1$-tuples of $G$, it is still dp-minimal and still has
ucc. Hence we may replace $G$ with $G/A$ and assume without loss that $G$ has
finite exponent.

Now, by Fact 2.10, $G$ is an MC-group of finite exponent, and so by Fact 2.7
and Fact 2.1 there is a definable normal solvable subgroup
$R\trianglelefteqslant G$ such that $G/R$ has no non-trivial normal abelian
subgroup. $G/R$ is again dp-minimal and ucc, so by Corollary 6.6 it must be
finite, and we are done. ∎

## References

  * [1] Ricardo de Aldama. Definable nilpotent and soluble envelopes in groups without the independence property. Mathematical Logic Quarterly, Vol. 59, No. 3 (2013). pp. 201-205.
  * [2] John Baldwin and Jan Saxl. Logical stability in group theory. Journal of the Australian Mathematical Society (Series A), Vol. 21 No. 3 (1976). pp. 267-276.
  * [3] RM Bryant and B Hartley. Periodic locally soluble groups with the minimal condition on centralizers. Journal of Algebra, Vol. 61, Iss. 2 (1979). pp. 328-334.
  * [4] Artem Chernikov, Anand Pillay, Pierre Simon. External definability and groups in NIP theories. Journal of the London Mathematical Society, Vol. 90 Iss. 1 (2014). pp 213-240.
  * [5] Artem Chernikov, Pierre Simon. Definably amenable NIP groups. Journal of the AMS, Vol. 31 Iss. 3 (2018). pp. 609-641.
  * [6] Tim Clausen. Dp-minimal profinite groups and valuations on the integers. Preprint (2020), arXiv:2008.08797.
  * [7] Jan Dobrowolski and Frank Wagner. On $\omega$-categorical groups and rings of finite burden. Israel Journal of Mathematics, Vol. 236 (2020). pp. 801-839.
  * [8] Jan Dobrowolski and John Goodrick. Some remarks on inp-minimal and finite burden groups. Archive for Mathematical Logic, Vol. 58 (2019). pp. 267-274.
  * [9] Alfred Dolich, John Goodrick, David Lippel. Dp-minimality: basic facts and examples. Notre Dame Journal of Formal Logic, Vol. 52 No. 3 (2011). pp. 267-288.
  * [10] Clifton Ealy, Krzysztof Krupiński, Anand Pillay. Superrosy dependent groups having finitely satisfiable generics. Annals of Pure and Applied Logic, Vol. 151 (2008). pp. 1-21.
  * [11] Jakub Gismatullin. Model-theoretic connected components of groups. Israel Journal of Mathematics, Vol. 184 (2011). pp. 251-274.
  * [12] Abderezak Ould Houcine. A remark on the definability of the Fitting subgroup and the soluble radical. Mathematical Logic Quarterly, Vol. 59, No. 1-2 (2013). pp. 62-65.
  * [13] Ehud Hrushovski, Ya’acov Peterzil, Anand Pillay. Groups, measures, and the NIP. Journal of the AMS, Vol. 21 No. 2 (2008). pp. 563-596.
  * [14] Ehud Hrushovski, Anand Pillay. On NIP and invariant measures. Journal of the EMS, Vol. 13 No. 4 (2011). pp. 1005-1061.
  * [15] Ehud Hrushovski, Anand Pillay, Pierre Simon. A note on generically stable measures and fsg groups. Notre Dame Journal of Formal Logic, Vol. 53 No. 4 (2012). pp. 599-605.
  * [16] Itay Kaplan, Elad Levi, Pierre Simon. Some remarks on dp-minimal groups, in: Groups, Modules, and Model Theory, Surveys and Recent Developments. Springer (2017).
  * [17] Krzysztof Krupiński and Adrián Portillo. On stable quotients. Notre Dame Journal of Formal Logic, Vol. 63 No. 3 (2022). pp. 373-394.
  * [18] Alf Onshuus and Alexander Usvyatsov. On dp-minimality, strong dependence, and weight. Journal of Symbols Logic, Vol. 71 No. 1 (2006). pp. 1-21.
  * [19] Anand Pillay and Predrag Tanović. Generic stability, regularity, and quasi-minimality. Preprint (2009). arXiv:0912.1115.
  * [20] Joachim Reineke. Minimale Gruppen. Z. Math. Logik Grundlagen Math, Vol. 21 No. 4 (1975). pp. 357–359.
  * [21] Derek Robinson. A Course in the Theory of Groups. Springer (1996).
  * [22] Saharon Shelah. Classification theory for theories with NIP - a modest beginning, Scientiae Mathematicae Japonicae, Vol. 59 No. 2 (2004). pp. 265-316.
  * [23] Saharon Shelah. Dependent first order theories, continued. Israel Journal of Mathematics, Vol. 173 No. 1 (2009). pp. 1-60.
  * [24] Saharon Shelah. Strongly dependent theories. Israel Journal of Mathematics, Vol. 204 (2014). pp. 1-83.
  * [25] Saharon Shelah. Minimal bounded index subgroup for dependent theories. Proceedings of the AMS, Vol. 136 No. 3, (2008). pp. 1087-1091.
  * [26] Pierre Simon. A guide to NIP theories. Cambridge University Press (2015).
  * [27] Pierre Simon. On dp-minimal ordered structures. The Journal of Symbolic Logic, Vol. 76 No. 2 (2011). pp. 448-460.
  * [28] Pierre Simon. Distal and non-distal NIP theories. Annals of Pure and Applied Logic, Vol. 164 No. 1 (2013). pp. 294-318.
  * [29] Patrick Simonetta. An example of a C-minimal group which is not abelian-by-finite. Proceedings of the AMS, Vol. 131 No. 12 (2003). pp. 3913-3917.
  * [30] Patrick Simonetta. On non-abelian C-minimal groups. Annals of Pure and Applied Logic, Vol. 122 (2003). pp 263-287.
  * [31] Atticus Stonestrom. On f-generic types in NIP groups. Preprint (2023). arXiv:2303.13470.
  * [32] Alexander Usvyatsov. On generically stable types in dependent theories. The Journal of Symbolic Logic, Vol. 74, No. 1 (2009). pp. 216-250.
  * [33] Frank Wagner. Stable Groups. Cambridge University Press (2013).
  * [34] Frank Wagner. Stable groups, mostly of finite exponent. Notre Dame Journal of Formal Logic, Vol. 34, No. 2 (1993). pp. 183-192.