Mekler's construction and tree properties

TL;DR

This paper explores how Mekler's construction, which transforms structures into pure groups, preserves various complex model-theoretic properties including NTP$_1$, NSOP$_1$, and NSOP$_2$, expanding understanding of these properties in group theories.

Contribution

The paper demonstrates that Mekler's construction preserves NTP$_1$, NSOP$_1$, and NSOP$_2$, providing new insights into the transfer of model-theoretic properties to pure groups.

Findings

Mekler's construction preserves NTP$_1$ and NSOP$_1$ properties.

The construction also preserves NSOP$_2$, expanding its applicability.

Results imply existence of pure group theories with complex properties derived from non-simple theories.

Abstract

Mekler constructed a way to produce a pure group from any given structure where the construction preserves $\kappa$-stability for any cardinal $\kappa$. Not only the stability, it is known that his construction preserves various model-theoretic properties such as simplicity, NIP, and NTP$_2$. Inspired by the last result, we show that the construction also preserves NTP$_1$(NSOP$_2$) and NSOP$_1$. As a corollary, we obtain that if there is a theory of finite language which is non-simple NSOP$_1$, or which is NSOP$_2$ but has SOP$_1$, then there is a pure group theory with the same properties, respectively.

Full text

The author was supported by Samsung Science Technology Foundation under Project Number SSTF-BA1301-03, and has been supported by an NRF of Korea grant 2018R1D1A1A02085584

Keywords— NTP1, NSOP1, tree indiscerniblility, Mekler’s construction

1. Introduction

Suppose a structure $M$ of finite language has more than one element, then there is a graph $N$ which is bi-interpretable with $M$ [6, Thm 5.5.1]. This implies if $M$ has a model-theoretic property like stability and simplicity, then one can find a graph which has the same properties of $M$. Unlike the graph, it is not easy to see whether there is a group which preserves a model-theoretic property of $M$. A partial answer to this problem was found by Mekler. In [9], he constructed a group $G$ where Th$(G)$ has the same stability spectrum as Th$(M)$. This group is not bi-interpretable with $M$, however, so it does not preserve all the properties of $M$. For example, even though $M$ is $\aleph_{0}$-categorical, the group $G$ may not be.

Later, it is proved that many other properties related with Shelah’s classification program are preserved by Mekler’s construction. Baudisch and Pentzel proved that simplicity is preserved by the construction, and assuming stability, Baudisch proved that CM-triviality is also preserved [1]. Recently, Chernikov and Hempel proved that the construction preserves NIP, $k$-dependence, and NTP2 [3]. Thus, it is natural to expect that the construction preserves NSOP1 and other non-simple theories [3, Conjecture 1].

In this paper we show that the conjecture is true for the following tree properties; NTP1(NSOP2) and NSOP1. To prove them, we analogously follow the argument used in the proof of preservation of NTP2 in [3]. The difference is that parameters witnessing TP2 formula is an array, not a tree. Hence, we find an appropriate generalized indiscernibility for each properties substituting the role of mutual indiscernibility. We use one of the tree indiscernibility, called strong indiscernibility (see Definition 2.4) [8, 12].

SOP1 formula also have parameters of a tree, but the strong indiscernibility is not much helpful. We recall [7, Proposition 2.4], the equivalent conditions of NSOP1, to obtain parameters of array $\omega\times 2$ called SOP1-array (see Definition 3.10).

We preview the corollaries of the main results.

Corollary 1.1**.**

1. (1)

There is a non-simple NSOP1* pure group theory.* 2. (2)

If there is an NSOP2* theory which has SOP1, then there is a pure group theory with the same properties.*

The first one is obtain by the preservation of simplicity and NSOP1. Any example of non-simple NSOP1 theory on finite language can be transformed into a pure group by the construction. Similarly, any NSOP2 theory with SOP1 on finite language can be transformed into a pure group. Thus, NSOP1 and NSOP2 are equivalent if and only if they are equivalent on the pure group theories.

The paper is organized as follows. In section 2, we introduce the notions about strong indiscerniblility on trees from [8] and [12]. In section 3, using strong indiscernibility, we find equivalent conditions of NTP1. And then, we define an SOP1-array and find equivalent conditions of NSOP1. In section 4, we describe and summarize definitions and facts of Mekler’s construction following by [6] and [3]. In section 5, we first observe some combinatorial remarks on trees in [5], then show our main results that Mekler’s construction preserves NTP1 and NSOP1.

2. Tree indiscernibility

Consider a tree ${}^{<\lambda}\kappa$ of height $\lambda$ which has $\kappa$ many branches. Each element in the tree can be regarded as a string. We denote $\langle\rangle$ as an empty string, $0^{\alpha}$ as a string of $\alpha$ many zeros, and $\alpha$ as a string $\langle\alpha\rangle$ of length one.

Definition 2.1**.**

Fix a tree ${}^{<\lambda}\kappa$, and let $\eta,\nu,\xi\in{}^{<\lambda}\kappa$.

1. (1)

(Ordering) $\eta\lhd\nu$ if $\nu\lceil\alpha=\eta$ for some ordinal $\alpha\in$ dom($\nu)$. 2. (2)

(Meet) $\xi=\eta\wedge\nu$ if $\xi$ is the meet of $\eta$ and $\nu$, i.e., $\xi=\eta\lceil\beta$, when $\beta=\bigcup\{\alpha\leq$ dom$(\eta)\cap$dom$(\nu)\hskip 2.84526pt|\hskip 2.84526pt\eta\lceil\alpha=\nu\lceil\alpha\}$. For $\bar{\eta}\in{}^{<\lambda}\kappa$, $\bar{\nu}$ is the meet closure of $\bar{\eta}$ if $\bar{\nu}=\{\eta_{1}\wedge\eta_{2}|\eta_{1},\eta_{2}\in\bar{\eta}\}$. 3. (3)

(Incomparability) $\eta\bot\nu$ if they are $\unlhd$-incomparable, i.e., $\neg(\eta\unlhd\nu)$ and $\neg(\nu\unlhd\eta)$. 4. (4)

(Lexicographic order) $\eta<_{lex}\nu$ if

• (a)

$\eta\lhd\nu$, or

• (b)

$\eta\bot\nu$ and $\exists\alpha(\eta\lceil\alpha=\nu\lceil\alpha\hskip 2.84526pt$and$\hskip 2.84526pt\eta(\alpha)<\nu(\alpha))$.

Definition 2.2**.**

A strong language $L_{0}$ is defined by the collection $\{\lhd,\wedge,<_{lex}\}$

We may view the tree ${}^{<\lambda}\kappa$ as an $L_{0}$-structure.

Definition 2.3**.**

A tree $B={}^{<\lambda^{\prime}}\kappa^{\prime}$ is called a subtree of $A={}^{<\lambda}\kappa$ if $B\subseteq A$ and the inclusion map is an embedding in the language $\{\unlhd,<_{lex}\}$.

Fix a complete first order theory $T$(with language $L$). Let $\mathcal{M}\models T$ be a monster model. From now on, we will work in this $\mathcal{M}$.

Definition 2.4**.**

[8] Fix a structure $\mathcal{I}$ with language $L_{\mathcal{I}}$. For a set $\{b_{i}|i\in\mathcal{I}\}$, we say it is $\mathcal{I}$-indexed indiscernible if for any finite $\bar{i}$ and $\bar{j}$ from $\mathcal{I}$,

qftp($\bar{i})_{\mathcal{I}}$ = qftp($\bar{j})_{\mathcal{I}}$ $\Rightarrow$ $(b_{i})_{i\in\bar{i}}\equiv(b_{j})_{j\in\bar{j}}$.

$\mathcal{I}$ is called the index structure. In particular, we say a set $\{b_{\eta}|\eta\in{}^{<\lambda}\kappa\}$ is strongly indiscernible if it is $\mathcal{I}$-indexed indiscernible for $\mathcal{I}$ the $L_{0}$-structure on ${}^{<\lambda}\kappa$.

Remark 2.5**.**

Let $\{a_{\eta}|\eta\in{}^{<\lambda}\kappa\}$ be a strongly indiscernible tree.

1. (1)

For all $\nu_{1},\nu_{2}\in{}^{\lambda}\kappa$, $(a_{\nu_{1}\lceil\alpha})_{\alpha<\lambda}\equiv(a_{\nu_{2}\lceil\alpha})_{\alpha<\lambda}$ 2. (2)

For all $\bar{\eta_{1}},\bar{\eta_{2}}\in{}^{<\lambda}\kappa$, if $\bar{\nu_{i}}$ is the meet-closure of $\bar{\eta_{i}}$ for each $i=1,2$, then qftp($\bar{\eta_{1}}$) = qftp($\bar{\eta_{2}}$) implies qftp($\bar{\nu_{1}}$) = qftp($\bar{\nu_{2}}$) 3. (3)

For all $\eta\bot\nu\in{}^{<\lambda}\kappa$, and $\xi\in{}^{<\lambda}\kappa$, $\eta<_{lex}\nu\Rightarrow a_{\eta}a_{\nu}\equiv a_{\xi^{\frown}0}a_{\xi^{\frown}0}$ 4. (4)

For any $\eta\in{}^{<\lambda}\kappa$, the tree $(a_{0^{\frown}\eta})_{\eta\in{}^{<\lambda}\kappa}$ is strongly indiscernible over $(a_{\nu_{1}\lceil\alpha})_{\alpha\in dom(\eta)}$

Proof.

See [4] and [8]. ∎

Definition 2.6**.**

[8] Let $\mathcal{I}$ be an index structure.

1. (1)

The EM-type of a set of parameters $A=\{a_{i}\,|\,i\in\mathcal{I}\}$, EM${}_{\mathcal{I}}(A)$, is the collection of formulas $\varphi(x_{i_{1}},\dots,x_{i_{n}})$ in $L$ with variables $\{x_{i}\,|\,i\in\mathcal{I}\}$ such that for all $j_{1},\dots,j_{n}\in\mathcal{I}$, if $j_{1}\dots j_{n}\equiv^{qf}i_{1}\dots i_{n}$, then $\models\varphi(a_{j_{1}},\dots,a_{j_{n}})$. 2. (2)

A set $B=\{b_{\eta}\,|\,\eta\in\mathcal{I}\}$ is based on a set $A=\{a_{\nu}\,|\,\nu\in\mathcal{I}\}$ if for all $\eta_{1},\dots,\eta_{n}\in\mathcal{I}$ and for all $\varphi(x_{\eta_{1}},\dots,x_{\eta_{n}})$ in $L$, there exists some $\nu_{1},\cdots,\nu_{n}\in\mathcal{I}$ such that

• (a)

$\nu_{1}\dots\nu_{n}\equiv^{qf}_{\mathcal{I}}\eta_{1}\dots\eta_{n}$, and

• (b)

$b_{\eta_{1}}\dots b_{\eta_{n}}\equiv_{\varphi}a_{\nu_{1}}\dots a_{\nu_{n}}$

In particular, when $\mathcal{I}$ is $L_{0}$-structure ${}^{<\lambda}\kappa$, we say $B$ is strongly based on $A$ whenever $B$ is based on $A$.

Remark 2.7**.**

Let $B=\{b_{\eta}\,|\,\eta\in\mathcal{I}\}$ and $A=\{a_{\nu}\,|\,\nu\in\mathcal{I}\}$. Then $B$ is based on $A$ if and only if $B\models$EM${}_{\mathcal{I}}(A)$.

Definition 2.8**.**

For an index structure $\mathcal{I}$, we say $\mathcal{I}$-indexed indiscernibles have the modeling property if given any $A=\{a_{\nu}|\nu\in\mathcal{I}\}$, there is an $\mathcal{I}$-indexed indiscernible $B=\{b_{\eta}|\eta\in\mathcal{I}\}$ such that $B$ is based on $A$(or equivalently, $B\models$EM${}_{\mathcal{I}}(A)$).

Fact 2.9**.**

[12] Let ${}^{<\omega}\omega$ be the universe of the index structure. The strong indiscernibles have the modeling property.

As a corollary, we may assume the index structure as ${}^{<\lambda}\kappa$ where both $\lambda$ and $\kappa$ are infinite cardinals.

3. Tree properties

Definition 3.1**.**

We say a subset $\{\eta_{i}\,|\,i<k\}\subseteq{}^{<\lambda}\kappa$ is a collection of $k$ distant siblings if given $i_{1}\neq i_{2}$ and $j_{1}\neq j_{2}$, all of which are less than $k$, $\eta_{i_{1}}\wedge\eta_{i_{2}}=\eta_{j_{1}}\wedge\eta_{j_{2}}$.

We may extend the result in Proposition 2.5(3).

Remark 3.2**.**

Let $\{a_{\eta}\,|\,\eta\in{}^{<\lambda}\kappa\}$ be a strongly indiscernible tree. For any collection of distant siblings $\{\eta_{i}\,|\,i<k\}\subseteq{}^{<\lambda}\kappa$ and for any $\xi\in{}^{<\lambda}\kappa$, if $\eta_{i}<_{lex}\eta_{j}$ for each $i<j<k$ then $a_{\eta_{0}}\cdots a_{\eta_{k-1}}\equiv a_{\xi^{\frown}0}\cdots a_{\xi^{\frown}k-1}$

Definition 3.3**.**

[5, 8, 12] Fix $k\geq 2$.

1. (1)

$\varphi(x;y)$ has SOP1 if there is a $(a_{\eta}\,|\,\eta\in{}^{<\omega}2)$ such that

• (a)

For all $\eta\in{}^{\omega}2$, $\{\varphi(x;a_{\eta\lceil\alpha})\,|\,\alpha<\omega\}$ is consistent,

• (b)

For all $\xi,\nu\in{}^{<\omega}2$, if $\xi^{\frown}0\unlhd\nu$, then $\{\varphi(x;a_{\xi^{\frown}1}),\varphi(x;a_{\nu})\}$ is inconsistent. 2. (2)

$\varphi(x;y)$ has SOP2 if there is a $(a_{\eta}\,|\,\eta\in{}^{<\omega}2)$ such that

• (a)

For all $\eta\in{}^{\omega}2$, $\{\varphi(x;a_{\eta\lceil\alpha})\,|\,\alpha<\omega\}$ is consistent,

• (b)

For all $\xi,\nu\in{}^{<\omega}2$, if $\xi\perp\nu$, then $\{\varphi(x;a_{\xi}),\varphi(x;a_{\nu})\}$ is inconsistent. 3. (3)

$\varphi(x;y)$ has the tree property of the first kind (TP1) if there is $(a_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ such that

• (a)

For all $\eta\in{}^{\omega}\omega$, $\{\varphi(x;a_{\nu\lceil\alpha})|\alpha<\omega\}$ is consistent,

• (b)

For all $\eta\bot\nu\in{}^{<\omega}\omega$, $\{\varphi(x;a_{\eta}),\varphi(x;a_{\nu})\}$ is inconsistent. 4. (4)

$\varphi(x;y)$ has weak $k$-TP1 if there is $(a_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ such that

• (a)

For all $\eta\in{}^{\omega}\omega$, $\{\varphi(x;a_{\nu\lceil\alpha})|\alpha<\omega\}$ is consistent,

• (b)

For any collection of distant siblings $\{\eta_{i}\,|\,i<k\}$, $\{\varphi(x;a_{\eta_{i}})\,|\,i<k\}$ is inconsistent. 5. (5)

We say $T$ has TP1 (resp. SOP1, SOP2) if there is a formula having TP1 (resp. SOP1, SOP2). If not, we say $T$ is NTP1 (resp. NSOP1, NSOP2). We say $T$ has weak-TP1 if there is a formula having $k$-TP1 for some $k$.

Fact 3.4**.**

[4]

1. (1)

$\varphi(x;y)$ has TP1 if and only if there is a strongly indiscernible tree $(a_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ such that

1. (a)

$\{\varphi(x,a_{\eta\lceil\alpha})\,|\,\alpha<\omega\}$ is consistent for some $\eta\in^{\omega}\omega$. 2. (b)

$\{\varphi(x,a_{\nu^{\frown}i})\,|\,i<\omega\}$ is pairwise inconsistent for some $\nu\in{}^{<\omega}\omega$. 2. (2)

$T$ has weak-TP1 if and only if $T$ has TP1.

Remark 3.5**.**

$\varphi(x;y)$ has weak-TP1 if and only if there is a strongly indiscernible tree $(a_{\eta}|\eta\in{}^{<\omega}\omega)$ such that

1. (a)

$\{\varphi(x,a_{0^{i}})|i<\omega\}$ is consistent. 2. (b)

$\{\varphi(x,a_{\nu^{\frown}i})|i<\omega\}$ is $k$-inconsistent for some $\nu\in{}^{<\omega}\omega$ and $k\geq 2$.

To prove the main theorem, we first establish the characterization of given model theoretic property. For example, in [3], Chernikov and Hempel stated the following proposition cited from [2];

Fact 3.6**.**

Let $T$ be a theory and $\mathcal{M}\models T$ a monster model. Let $\kappa:=|T|^{+}$. The following are equivalent:

1. (1)

T is NTP2. 2. (2)

for any array $(a_{i,j}:i\in\kappa,j\in\omega)$ of finite tuples with mutually indiscernible rows and a finite tuple $b$, there is some $\alpha\in\kappa$ satisfying the following:

for any $i>\alpha$, there is some $b^{\prime}$ such that

1. (a)

$(a_{i,j}\,|\,j<\omega)$ is indiscernible over $b^{\prime}$, and 2. (b)

tp$(b/a_{i,0})$=tp$(b^{\prime}/a_{i,0})$.

The basic idea of 3.6 is to find an indiscernible sequence over some $b^{\prime}$ which has the same type with given $b$ over some element in that sequence. The trick in the proof can be seen on [10, Lemma 1.4], too. We apply the same trick on NTP1 with the notions of strong indiscernibility.

Proposition 3.7**.**

Let $\kappa>2^{|T|}$ be some sufficiently large regular cardinal. Then TFAE.

1. (1)

$T$* is NTP*1** 2. (2)

For any strongly indiscernible tree $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ of finite tuples and a finite tuple $b$, there is some $\beta<\kappa$ and $b^{\prime}$ such that

1. (a)

$(a_{{0^{\beta}}^{\frown}i}\,|\,i<\omega)$* is indiscernible over $b^{\prime}$, and* 2. (b)

tp$(b/a_{{0^{\beta}}^{\frown}0})$* = tp*$(b^{\prime}/a_{{0^{\beta}}^{\frown}0})$, 3. (3)

For any strongly indiscernible tree $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ of finite tuples and a finite tuple $b$, there is some $\gamma<\kappa$ satisfying the following:

for any $\beta>\gamma$, there is some $b^{\prime}$ such that

1. (a)

$(a_{{0^{\beta}}^{\frown}i}\;|\;i<\omega)$* is indiscernible over $b^{\prime}$, and* 2. (b)

tp$(b/a_{{0^{\beta}}^{\frown}0})$* = tp*$(b^{\prime}/a_{{0^{\beta}}^{\frown}0})$.

Proof.

(1) $\Rightarrow$ (2). Assume $T$ is NTP1, and let $A=(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ and $b$ be given.

By pigeonhole principle, there is a subsequence $(\alpha_{i}\,|\,i<\omega)$ in the set of successor ordinals smaller than $\kappa$ such that for all $i<j<\omega$, ${\alpha_{i}}^{+}<\alpha_{j}<\kappa$ and tp$(a_{0^{\alpha_{i}}}/b)$ = tp$(a_{0^{\alpha_{j}}}/b)$. We inductively define a subtree $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ in $A$ as follows;

• •

$a^{\prime}_{\langle\rangle}=a_{0^{\alpha_{0}}}$,

• •

for any $\eta\in{}^{<\omega}\omega$, if $a^{\prime}_{\eta}=a_{\rho}$ for some $\rho\in{}^{<\kappa}\kappa$ with Dom$(\rho)=\alpha_{i}$, then for each $j<\omega$, $a^{\prime}_{\eta^{\frown}j}=a_{{\rho^{\prime}}^{\frown}j}$ where $\rho^{\prime}=\rho^{\frown}0^{\delta}$ for some $\delta$ so that Dom$({\rho^{\prime}}^{\frown}j)=\alpha_{i+1}$.

Note that $(a^{\prime}_{0^{i}}\,|\,i<\omega)$ is a subsequence of $(a_{0^{\alpha}}\,|\,\alpha<\kappa)$, and for any $\eta,\nu\in{}^{<\omega}\omega$, if $a^{\prime}_{\eta}=a_{\eta^{\prime}}$, $a^{\prime}_{\nu}=a_{\nu^{\prime}}$, $a^{\prime}_{\eta\wedge\nu}=a_{\xi}$ for $\eta^{\prime},\nu^{\prime}$, and $\xi\in{}^{<\kappa}\kappa$, then $\xi\neq\eta^{\prime}\wedge\nu^{\prime}$. Moreover, the subtree is strongly indiscernible, too.

Let $p(x,a^{\prime}_{0})$ := tp$(b/a^{\prime}_{0})$ and $q(x)$ := $\bigcup_{i<\omega}p(x,a^{\prime}_{i})$. We claim that $q$ is consistent.

Suppose not. Then by compactness and strong indiscernibility, there is a formula $\varphi(x,y)$ such that $\varphi(x,a^{\prime}_{0})\in p(x,a^{\prime}_{0})$ and $\{\varphi(x,a^{\prime}_{i})\,|\,i<\omega\}$ is $k$-inconsistent for some $k<\omega$. On the other hand, $b$ realizes $\bigcup_{i<\omega}p(x,a^{\prime}_{0^{i}})$, so it realizes $\{\varphi(x,a^{\prime}_{0^{i}})\,|\,i<\omega\}$. As a result, $\varphi$ has weak $k$-TP1, and fact 3.4 further says that $T$ has TP1.

By claim, we can find a realization $b^{\prime}\models q(x)$.

We may assume $(a^{\prime}_{i}\,|\,i<\omega)$ is indiscernible over $b^{\prime}$ by Ramsey and compactness. Note that tp$(b/a^{\prime}_{0})$ = tp$(b^{\prime}/a^{\prime}_{0})$ since $b^{\prime}\models p(x,a^{\prime}_{0})$. Thus, if $a^{\prime}_{i}=a_{{0^{\beta}}^{\frown}i}$ for some $\beta$, then this $\beta$ and $b^{\prime}$ is the desired one.

(2) $\Rightarrow$ (3). Assume (2) holds. For a strongly indiscernible tree $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ and a finite tuple $b$, we will say $Q(\beta)$ holds on $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ and $b$ when there is a $b^{\prime}$ such that $(a_{{0^{\beta}}^{\frown}i}\;|\;i<\omega)$ is indiscernible over $b^{\prime}$, and tp$(b/a_{{0^{\beta}}^{\frown}0})$ = tp$(b^{\prime}/a_{{0^{\beta}}^{\frown}0})$. Suppose there is a strongly indiscernible tree $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ and a finite tuple $b$ such that for any $\gamma<\kappa$, there is a $\beta>\gamma$ which does not satisfy $Q$. Then, since cf$(\kappa)=\kappa$, we can find a cofinal map $f:\kappa\rightarrow\kappa$ such that for any $i,j<\kappa$, $1<f(i)^{+}<f(j)$, and $f(i)$ does not satisfy $Q$.

Now, construct the following map $g:{}^{<\kappa}\kappa\rightarrow{}^{<\kappa}\kappa$ . First, if $\eta=\langle\rangle$, then $g(\eta)=\langle\rangle$ For other non-empty string $\eta\in{}^{<\kappa}\kappa$, Dom$(g(\eta))=$ Sup$\{f(l)^{+}\,|\,l\in\text{Dom}(\eta)\}$, and for each $i<\text{Dom}(g(\eta))$,

$(g(\eta))(i)=\begin{cases}\eta(j)&\text{if }i=f(j)^{+}\text{ for some }j<\kappa,\\ 0&\text{otherwise}.\end{cases}$

Then we define $a^{\prime}_{\eta}=a_{g(\eta)}$. The subtree $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ is strongly indiscernible, and for each $\beta<\kappa$, $Q(\beta)$ does not hold on $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ and $b$. This contradicts to (2).

(3) $\Rightarrow$ (1). Assume (3) holds. Suppose $T$ has TP1 witnessed by $\varphi(x;y)$ and $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$. We may assume the tree $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ is strongly indiscernible by the modeling property. Note (a) and (b) in 3.3(3) still hold. Take $b\models\{\varphi(x,a_{0^{\alpha}})\,|\,\alpha<\kappa\}$. By the assumption, we have some ordinals $\gamma<\beta<\kappa$ and $b^{\prime}$ such that

1. (a)

$(a_{{0^{\beta}}^{\frown}i}\;|\;i<\omega)$ is indiscernible over $b^{\prime}$, and 2. (b)

tp$(b/a_{{0^{\beta}}^{\frown}0})$ = tp$(b^{\prime}/a_{{0^{\beta}}^{\frown}0})$.

Choose an automorphism $\sigma$ where it fixes $a_{{0^{\beta}}^{\frown}0}$ and sends $b^{\prime}$ to $b$. Denote $(a^{\prime}_{i}\,|\,i<\omega)$ to $(\sigma(a_{{0^{\beta}}^{\frown}i})\,|\,i<\omega)$, then it is indiscernible over $b$. Since $a^{\prime}_{0}$ is as same as $a_{{0^{\beta}}^{\frown}0}$, $b\models\varphi(x,a^{\prime}_{0})$, and then the indiscernibility implies that for all $i<\omega$, $b\models\varphi(x,a^{\prime}_{i})$. But this is a contradiction because $\{\varphi(x,a^{\prime}_{i})|i<\omega\}$ is 2-inconsistent, as well as $\{\varphi(x,a_{{0^{\beta}}^{\frown}i})\,|\,i<\omega\}$ is.

∎

Analogously, We find a lemma for SOP1. Unlike the case of TP1, we cannot use the strong indiscernibility. For instance, let $A$ be a tree which has the inconsistency condition of SOP1 and let $B$ be a strong indiscernible tree based on $A$. There is no guarantee that $B$ has the inconsistency condition. Hence, we need another indiscernibility which matches up to SOP1.

To find this, we recall results in [4] and [7].

Fact 3.8**.**

[4] Suppose $\varphi(x;y)$ with the tree $(c_{\eta}\,|\,\eta\in{}^{<\kappa}2)$ have SOP1 where $\kappa\geq 2^{|T|}$. Then there is a sequence $(\eta_{i},\nu_{i})_{i<\omega}$ of elements of ${}^{<\kappa}2$ such that

1. (1)

$c_{\eta_{i}}\equiv_{c_{\eta_{<i}}c_{\nu_{<i}}}c_{\nu_{i}}$ for all $i<\omega$, 2. (2)

$\{\varphi(x;c_{\eta_{i}})\,|\,i<\omega\}$ is consistent, and 3. (3)

$\{\varphi(x,c_{\nu_{i}})\,|\,i<\omega\}$ is 2-inconsistent.

Kaplan and Ramsey [7] proved more general result about SOP1.

Fact 3.9**.**

The following are equivalent;

1. (1)

$T$ has SOP1. 2. (2)

There is a formula $\varphi$ and an array $(c_{i,j})_{i<\omega,j<2}$ so that

1. (a)

$c_{i.0}\equiv_{c_{<i,0}c_{<i,1}}c_{i,1}$ for all $i<\omega$ 2. (b)

$\{\varphi(x;c_{i,0})\,|\,i<\omega\}$ is consistent 3. (c)

$\{\varphi(x,c_{i,1})\,|\,i<\omega\}$ is 2-inconsistent. 3. (3)

There is a formula $\varphi$ and an array $(c_{i,j})_{i<\omega,j<2}$ so that

1. (a)

$c_{i.0}\equiv_{c_{<i,0}c_{<i,1}}c_{i,1}$ for all $i<\omega$ 2. (b)

$\{\varphi(x;c_{i,0})\,|\,i<\omega\}$ is consistent 3. (c)

$\{\varphi(x,c_{i,1})\,|\,i<\omega\}$ is $k$-inconsistent for some $k\geq 2$.

From these facts, we derive a new kind of indiscernibility.

Definition 3.10**.**

We say $(c_{i,j})_{i<\omega,j<2}$ is an SOP1-array indiscernible over $A$ if

1. (1)

$(c_{i,0}c_{i,1})_{i<\omega}$ is an (order) indiscernible sequence over $A$, and 2. (2)

$c_{i.0}\equiv_{Ac_{<i,0}c_{<i,1}}c_{i,1}$ for all $i<\omega$.

Note that we may replace the array in 3.9 to an indiscernible SOP1-array .

Remark 3.11**.**

Let an indiscernible SOP1-array $(c_{i,j})_{i<\omega,j<2}$ be given.

1. (1)

For any $\kappa>\omega$, there is a indiscernible SOP1-array $(c^{\prime}_{i,j})_{i<\kappa,j<2}$ such that $c_{i,j}=c^{\prime}_{i,j}$ for all $i<\omega$ and $j<2$. 2. (2)

For any $n<\omega$, $(c_{i,j})_{n\leq i<\omega,j<2}$ is an SOP1-array indiscernible over $\{c_{i,j}\,|\,i<n,j<2\}$.

Proposition 3.12**.**

Let $\kappa>2^{|T|}$ be some sufficiently large regular cardinal. Then TFAE.

1. (1)

$T$* is NSOP*1** 2. (2)

For any indiscernible SOP1*-array $(a_{i,j})_{i<\kappa,j<2}$ of finite tuples and a finite tuple $b$, there is some $\beta<\kappa$ and some $b^{\prime}$ such that*

1. (a)

tp$(b,a_{\beta,0})$* = tp*$(b^{\prime},a_{0,1})$, 2. (b)

$(a_{i,1}\;|\;i<\omega)$* is indiscernible over $b^{\prime}$.* 3. (3)

For any indiscernible SOP1*-array $(a_{i,j})_{i<\kappa,j<2}$ of finite tuples and a finite tuple $b$, there is some $\gamma<\kappa$ satisfying the following:*

for any number $\beta>\gamma$, there is some $b^{\prime}$ such that

1. (a)

tp$(b,a_{\beta,0})$* = tp*$(b^{\prime},a_{0,1})$, 2. (b)

$(a_{i,1}\;|\;i<\omega)$* is indiscernible over $b^{\prime}$.*

Proof.

(1) $\Rightarrow$ (2). Assume $T$ is NSOP1, and let $(a_{i,j})_{i<\kappa,j<2}$ and $b$ in (2) be given. Using the pigeonhole principle, take an indiscernible SOP1-subarray $(a^{\prime}_{i,0}a^{\prime}_{i,1})_{i<\omega}$ in $(a_{i,0}a_{i,1})_{i<\kappa}$ where tp$(a^{\prime}_{i,0}/b)$=tp$(a^{\prime}_{j,0}/b)$ for all $i,j<\omega$.

Now let $p(x,y)=$ tp$(b,a^{\prime}_{0,0})$, and $q(x)=\bigcup_{i<\omega}p(x,a^{\prime}_{i,1})$. We claim that $q$ is consistent. Suppose not. Then by indiscernibility and compactness, there is a formula $\varphi(x,y)$ in $p(x,y)$ such that $\{\varphi(x,a^{\prime}_{i,1})\,|\,i<\omega\}$ is $k$-inconsistent for some natural number $k$. On the other hand, since $\varphi(x,a^{\prime}_{i,0})$ is in tp$(b/a^{\prime}_{i,0})$ for each $i<\omega$, $\{\varphi(x,a^{\prime}_{i,0})\,|\,i<\omega\}$ is consistent. Hence $\varphi$ has SOP1 by Fact 3.9, which is a contradiction.

Let $\bar{y}=(y_{i})_{i<\omega}$ where for each $i<\omega$, $|y_{i}|=|a^{\prime}_{i,1}|$, and let $\Pi(x,\bar{y})$ be the union of $\bigcup_{i<\omega}p(x,y_{i})$, $\Phi(\bar{y})$, and $\Psi(x,\bar{y})$ where $\Phi(\bar{y})=$tp$((a^{\prime}_{i,1})_{i<\omega})$ and $\Psi(x,\bar{y})$ means that $(y_{i})_{i<\omega}$ is indiscernible over $x$. By Ramsey and compactness, $\Pi(x,\bar{y})$ is consistent. Let $(b^{\prime\prime},(a^{\prime\prime}_{i,1})_{i<\omega})$ be a realization of $\Pi(x,\bar{y})$. Since $\Phi(\bar{y})=$tp$((a^{\prime}_{i,1})_{i<\omega})$=tp$((a_{i,1})_{i<\omega})$, we have $(a_{i,1}\;|\;i<\omega)$ is indiscernible over some $b^{\prime}$ where tp$(b^{\prime},a_{0,1})$ = tp$(b^{\prime\prime},a^{\prime\prime}_{0,1})=p(x,y_{0})$ = tp$(b,a^{\prime}_{0,0})$. Note $a^{\prime}_{0,0}=a_{\beta,0}$ for some $\beta<\kappa$.

(2) $\Rightarrow$ (3). For an indiscernible SOP1-array $(a_{i,j})_{i<\kappa,j<2}$ and a finite tuple $b$, we will say $Q(\beta)$ holds on $(a_{i,j})_{i<\kappa,j<2}$ and $b$ when there is a $b^{\prime}$ such that tp$(b/a_{\beta,0})$ = tp$(b^{\prime}/a_{0,1})$, and $(a_{i,1}\;|\;i<\omega)$ is indiscernible over $b^{\prime}$. Assume (2) holds but (3) does not hold, that is, there is an indiscernible SOP1-array $(a_{i,j})_{i<\kappa,j<2}$ and a tuple $b$ such that for any $\gamma<\kappa$, there is some $\beta>\gamma$ which does not satisfy $Q$. From this and cf$(\kappa)=\kappa$, we choose an increasing sequence $(\beta_{i})_{i<\kappa}$ of ordinal numbers such that for each $\beta_{i}$, $Q$ does not hold.

Now take a subarray $(a_{\beta_{i},0}a_{\beta_{i},1})_{i<\kappa}$ in $(a_{i,0}a_{i,1})_{i<\kappa}$. This array is still indiscernible, so by (2), there is some $j<\kappa$ and some $b^{\prime}$ such that tp$(b,a_{\beta_{j},0})$ = tp$(b^{\prime},a_{\beta_{0},1})$, and $(a_{\beta_{i},1}\;|\;i<\omega)$ is indiscernible over $b^{\prime}$. Since tp$((a_{\beta_{i},1})_{i<\omega})$=tp$((a_{i,1})_{i<\omega})$ by indiscernibility, we may assume that there is some $j<\kappa$ and some $b^{\prime}$ such that tp$(b,a_{\beta_{j},0})$ = tp$(b^{\prime},a_{0,1})$, and $(a_{i,1}\;|\;i<\omega)$ is indiscernible over $b^{\prime}$. This contradicts that $Q(\beta_{j})$ does not hold.

(3) $\Rightarrow$ (1). Assume (3) holds. Suppose $T$ has SOP1. By fact3.9(2) and compactness, we have a formula $\varphi(x;y)$ which witnesses SOP1 with an indiscernible SOP1-array $(a_{i,j})_{i<\kappa,j<2}$. Let $b$ be a realization of $\bigwedge_{i<\kappa}\varphi(x,a_{i,0})$. By assumption, there is some ordinals $\gamma<\beta<\kappa$ and some $b^{\prime}$ such that

1. (a)

tp$(b,a_{\beta,0})$ = tp$(b^{\prime},a_{0,1})$, 2. (b)

$(a_{i,1}\;|\;i<\omega)$ is indiscernible over $b^{\prime}$.

From (a), we have $\models\varphi(b^{\prime},a_{0,1})$, and then from (b), we have $\models\varphi(b^{\prime},a_{1,1})$. This contradicts that $\{\varphi(x,a_{i,1})\,|\,i<\kappa\}$ is 2-inconsistent. ∎

4. Mekler’s construction

We recall the definitions and facts from [6].

For a graph $A$ and its vertices $a$ and $b$, we say $R(a,b)$ if $a$ and $b$ are connected by a single edge in $A$.

Definition 4.1**.**

A graph $A$ which has at least two vertices is called nice if

1. (a)

For any two distinct vertices $a$ and $b$, there is some vertex $c\neq a,b$ such that $R(a,c)$ but $\neg R(b,c)$; 2. (b)

There are no triangles nor squares.

Definition 4.2**.**

[1, 9] Fix an odd prime $p$. For a nice graph $A$, let $F(A)$ be the free nilpotent group of class 2 and exponent $p$ generated freely by the vertices of $A$. Assume that $A$ is enumerated with some relation $<$ not in the original language. Then the Mekler group of $A$, denoted by $G(A)$, is defined as follows;

$G(A)=F(A)/\langle\{[a,b]\,|\,a,b\in A,a<b,\text{and }A\models R(a,b)\}\rangle$.

In other words, $G(A)$ is a group defined in the variety of nilpotent groups of class 2 and exponent $p$ such that the generators are the vertices of $A$ and that for any $a$ and $b$ in $G(A)$, $[a,b]=1$ if and only if $a<b$ and $A\models R(a,b)$.

We can see the definition in the point of view of vector spaces. Let $Z(F(A))$ be the center of $F(A)$. Then both $Z(F(A))$ and $F(A)/Z(F(A))$ are all elementary abelian $p$-groups so they can be considered as a $\mathbb{F}_{p}$-vector space with basis $\{[a,b]\,|\,a,b\in A,a<b\}$ and $\{a/Z(F(A))\,|\,a\in A\}$ respectively. The same is true for the Mekler group $G(A)$. If $Z(G(A))$ is the center of $G(A)$, then both $Z(G(A))$ and $G(A)/Z(G(A))$ can be considered as a $\mathbb{F}_{p}$-vector space. The basis of $Z(G(A))$ is $\{[a,b]\,|\,a,b\in A,a<b,\text{ and }A\models\neg R(a,b)\}$, and the basis of $G(A)/Z(G(A))$ is $\{a/Z(F(A))\,|\,a\in A\}$.

Definition 4.3**.**

For any element $g,h$ of $G(A)$, we say

1. (1)

$g\sim h$ if $C(g)=C(h)$, where $C(g)$ is the centraliser of $g$ in $G(A)$, 2. (2)

$g\approx h$ if for some $c$ in the center $Z(G)$ and some $r$ $(0\leq r<p)$, $h=g^{r}\cdot c$, 3. (3)

$g\equiv_{Z}h$ if $g\cdot Z(G)=h\cdot Z(G)$.

Remark 4.4**.**

For any element $g,h$ of $G(A)$, $g\equiv_{Z}h\Rightarrow g\approx h\Rightarrow g\sim h$.

Definition 4.5**.**

Let $g$ be an element in $G(A)$.

1. (1)

$g$ is isolated if every non-central element of $G(A)$ which commutes with $g$ is $\approx$-equivalent to $g$. 2. (2)

We say an element $g$ is of type $q$ if $q$ is the number of $\approx$-classes in the $\sim$-class of $g$. 3. (3)

We say $g$ is of type $q^{\iota}$ (resp. $q^{\nu}$) if $g$ is of type $q$ and isolated (resp. of type $q$ and not isolated).

Definition 4.6**.**

For every element $g$ of type $p$, we say an element $b$ is a handle of $g$ if it is of type $1^{\nu}$ and commutes with $g$.

As a remark, we note that for any $g$ of type $p$, the handle of $g$ exists and is unique up to $\sim$-equivalence.

Fact 4.7**.**

Every non-central element of $G(A)$ is of exactly one of the four type $1^{\nu},1^{\iota},p-1,p$; the classes of elements of each type are 0-definable.

Now, let $G$ be a model of Th$(G(A))$. Let us say an element of $G$ is proper if it is not a product of any elements of type $1^{\nu}$ in $G$.

Definition 4.8**.**

1. (1)

A $1^{\nu}$-transversal of $G$, denoted by $X^{\nu}$, is a set consisting of one representative for each $\sim$-class of elements of type $1^{\nu}$ in $G$. 2. (2)

A $p$-transversal of $G$, denoted by $X^{p}$, is a set of pairwise $\sim$-inequivalent proper elements of type $p$ in $G$ which is maximal with the property that if $Y$ is a finite subset of $X^{p}$ and all elements of $Y$ have the same handle, then $Y$ is a independent modulo the subgroup generated by all elements of type $1^{\nu}$ in $G$ and $Z(G)$. 3. (3)

A $1^{\iota}$-transversal of $G$, denoted by $X^{\iota}$, is a set of representatives of $\sim$-classes of proper elements of type $1^{\iota}$ in $G$ shich is maximal independent modulo the subgroup generated by all elements of types $1^{\iota}$ and $p$ in $G$, together with $Z(G)$. 4. (4)

A subset $X$ of $G$ is called a transversal of $G$ if it is the union of some $1^{\nu}$-transversal $X^{\nu}$, A $p$-transversal $X^{p}$, and $1^{\iota}$-transversal $X^{\iota}$ of $G$.

Note that all the sets in the above definition are definable.

Fact 4.9**.**

Let $A$ be a nice graph. For a model $G\models$ Th$(G(A))$, define an interpretation $\Gamma$ such that $\Gamma(G)$ is a graph where the set of vertices is $\{g\in G\,|\,g$ is a noncentral element of type $1^{\nu}\}/\sim$ and the edge relation is $\{([g]_{\sim},[h]_{\sim})\,|\,[g,h]=1\text{ in }G\}$. Then $\Gamma(G)\models$ Th$(A)$.

From 4.9, we see that if $X^{\nu}$ is a $1^{\nu}$-transversal, then the set can be regarded as a graph which models Th$(A)$.

Fact 4.10**.**

Let $C$ be an infinite nice graph, and $G\models$ Th$(G(C))$. If $X=X^{\nu}\cup X^{p}\cup X^{\iota}$ is a transversal of $G$, then there is a subgroup $H_{X}\leq Z(G)$ such that $G=\langle X\rangle\times H_{X}$ for some $H_{X}\leq Z(G)$. Moreover, if $G$ is saturated and uncountable, then both $\Gamma(G)$ and $H_{X}$ are also saturated.

Since $H_{X}$ is an elementary abelian $p$-group, we sometimes say $G$ is isomorphic to $\langle X\rangle\times\langle H_{X}\rangle$.

Fact 4.11**.**

Let $G$ is a saturated model of Th$(G(C))$ and let $\kappa=|G|$. If $X=X^{\nu}\cup X^{p}\cup X^{\iota}$ is a transversal of $G$, then

1. (a)

for any $x^{\nu}\in X^{\nu}$, the cardinality of $\{x^{p}\in X^{p}\,|\,x^{\nu}\text{ is the handle of }x^{p}\}$ is either zero or $\kappa$, and 2. (b)

$|X^{\iota}|=\kappa$

As we mentioned in 4.9, $X^{\nu}$ can be regarded as a graph where two vertices are joined (connected by a single edge) if they commute in $G$. In this point of view, we can find a supergraph by extending the set of vertices to $X$ and then giving the edge relation with the same rule. Then each $x^{p}\in X^{p}$ is joined to a unique vertex in $X^{\nu}$, which is the handle of $x^{p}$, while each $x^{\iota}\in X^{\iota}$ is joined to no vertex.

This kind of supergraph is called a cover. See [3] for more precise proof.

We give more facts from [3].

Fact 4.12**.**

1. (1)

Let $G$ be a saturated model of Th$(G(C))$, and let $X$ and $H_{X}$ be the sets in 4.10 so that $G=\langle X\rangle\times H_{X}$. If $f$ is a bijection between two small sets $Y=Y^{\nu}\cup Y^{p}\cup Y^{\iota}$ and $Z=Z^{\nu}\cup Z^{p}\cup Z^{\iota}$ of $X$ with the following properties;

1. (a)

$f(Y^{\nu})=Z^{\nu}$, $f(Y^{p})=Z^{p}$, and $f(Y^{\iota})=Z^{\iota}$, 2. (b)

for any $y^{p}\in Y^{p}$, the handle of $y^{p}$ is same as the handle of $f(y^{p})$, 3. (c)

tp$(Y^{\nu})$=tp$(Z^{\nu})$ in $\Gamma$,

then $f$ can be extended to an automorphism $\sigma$ of $G$.

Moreover, for any $\bar{h}$, $\bar{k}\in H_{X}$, if tp$(\bar{h})$=tp$(\bar{k})$ in $H_{X}$, then we may assume $\sigma$ sends $\bar{h}$ to $\bar{k}$. 2. (2)

Let $G$ be a model of Th$(G(C))$, and $\bar{x}=\bar{x}^{\nu}{}^{\frown}\bar{x}^{p}{}^{\frown}\bar{x}^{\iota}$ and $\bar{y}$ be two small tuples of variables. Then there is a partial type $\pi(\bar{x},\bar{y})$ such that $G\models\pi(\bar{a},\bar{b})$ if and only if we can extend $\bar{a}$ to a transversal $X$ of $G$ and find $H$ containing $\bar{b}$ so that $H$ is an independent set in $Z(G)$ and $G=\langle X\rangle\times\langle H\rangle$.

5. Preseervation

In [5], džamonja and Shelah introduced a way to choose a monochromatic subtree when the given tree ${}^{<\kappa}2$ is colored by $\theta$ many colors where $|\theta|<\kappa$ and $\kappa$ is a regular cardinal. The following is the key observation.

Fact 5.1**.**

Let $\kappa$ be a regular cardinal and let $\theta$ be a set of colors such that $|\theta|<\kappa$. For any coloring $f:{}^{<\kappa}2\rightarrow\theta$, there is a color $c\in\theta$ and an element $\nu^{\ast}\in{}^{<\kappa}2$ such that for any $\nu\in{}^{<\kappa}2$ satisfying $\nu^{\ast}\unlhd\nu$, there is $\rho\in{}^{<\kappa}2$ with $\nu\unlhd\rho$, $f(\rho)=c$.

Let us say a subtree $B\subseteq A$ is $f$-monochromatic if there is a color $c$ such that for all $b\in B$, $f(b)=c$.

Lemma 5.2**.**

Let $\kappa$ be an uncountable regular cardinal and $f:{}^{<\kappa}2\rightarrow\omega$ be a coloring.

1. (1)

If $\varphi(x;y)$ and $(a_{\eta}\,|\,\eta\in{}^{<\kappa}2)$ witness SOP2*, then there is a $f$-monochromatic subtree $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}2)$ of $(a_{\eta}\,|\,\eta\in{}^{<\kappa}2)$ such that $\varphi$ and $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}2)$ witness SOP2.* 2. (2)

If $\varphi(x;y)$ and $(a_{\eta}\,|\,\eta\in{}^{<\kappa}2)$ witness SOP1*, then there is a $f$-monochromatic subtree $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}2)$ of $(a_{\eta}\,|\,\eta\in{}^{<\kappa}2)$ such that $\varphi$ and $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}2)$ witness SOP1.* 3. (3)

If $\varphi(x;y)$ and $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ witness TP1*, then there is a $f$-monochromatic subtree $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ of $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ such that $\varphi$ and $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ witness TP1.*

Proof.

(1) Let $c$ and $\nu^{\ast}$ be the element in Fact 5.1. We construct a subtree $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}2)$ in $(a_{\eta}|\eta\in{}^{<\kappa}2)$ inductively as follows;

• •

$a^{\prime}_{\langle\rangle}=a_{\rho}$ where $\rho$ is any element satisfying $\nu^{\ast}\unlhd\rho$ and $f(\rho)=c$,

• •

for any $\eta\in{}^{<\omega}2$, if $a^{\prime}_{\eta}$ = $a_{\rho}$ for some $\rho\in{}^{<\kappa}2$ , then $a^{\prime}_{\eta^{\frown}0}=a_{\rho^{\prime}}$ where $\rho^{\frown}0\unlhd\rho^{\prime}$ and $f(\rho^{\prime})=c$, and $a^{\prime}_{\eta^{\frown}1}=a_{\rho^{\prime}}$ where $\rho^{\frown}1\unlhd\rho^{\prime}$ and $f(\rho^{\prime})=c$.

$\varphi$ and the subtree witness SOP2.

(2) We use a different version of the Fact 5.1 which has one more condition on $\rho$: $\rho$ is of the form $\xi^{\frown}1$ for some $\xi$. This can be proved by the same way used in 5.1, so we omit the proof.

Define $a^{\prime}_{\langle\rangle}$ as same as the previous one, but give a small modification in the induction step. For any $\eta\in{}^{<\omega}2$, let $a^{\prime}_{\eta}$ = $a_{\rho}$ for some $\rho\in{}^{<\kappa}2$. Take $a^{\prime}_{\eta^{\frown}1}=a_{\rho^{\prime}}$ where $\rho\unlhd\rho^{\prime}$, $f(\rho^{\prime})=c$ and $\rho^{\prime}$ is of the form $\xi^{\frown}1$ for some $\xi$. Then take $a^{\prime}_{\eta^{\frown}0}=a_{\rho^{\prime\prime}}$ where $\xi^{\frown}0\unlhd\rho^{\prime\prime}$ and $f(\rho^{\prime\prime})=c$. $\varphi$ and the subtree witness SOP1, too.

(3) In $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$, consider the subtree $(a_{\eta}\,|\,\eta\in{}^{<\kappa}2)$. Find $(a^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}2)$ as in the proof of (1), and then construct $(a^{\prime\prime}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ inductively as follows;

• •

$a^{\prime\prime}_{\langle\rangle}=a^{\prime}_{\langle\rangle}$,

• •

for any $\eta\in{}^{<\omega}\omega$, if $a^{\prime\prime}_{\eta}=a_{\rho}$ for some $\rho\in{}^{<\omega}2$, then $a^{\prime\prime}_{\eta^{\frown}i}=a^{\prime}_{\rho^{\frown}{1^{i}}^{\frown}0}$ for each $i<\omega$.

$\varphi$ and the last subtree witness TP1, too. ∎

Theorem 5.3**.**

For any infinite nice graph C, Th(G(C)) is NTP1 if and only if Th(C) is NTP1.

Proof.

Since $C$ is interpretable in $G(C)$, if Th$(C)$ has TP1, then Th$(G(C))$ also has TP1.

Suppose Th$(C)$ is NTP1 and Th$(G(C))$ has TP1. Let $G$ be a monster model of Th$(G(C))$, $X$ be a transversal of $G$ so that $G=\langle X\rangle\times\langle H\rangle$ for some $H$ as in Fact 4.10. We have a formula $\varphi(x,y)$ and a tree $(a_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ of finite tuples in $G$ for some sufficiently large regular cardinals $\kappa$ so that they witness TP1. Note that for each $\eta\in{}^{<\kappa}\kappa$, $a_{\eta}$ is of the form $t_{\eta}(\bar{x}_{\eta},\bar{h}_{\eta})$ for some terms $t_{\eta}\in L_{G}$, and for some small tuples $\bar{x}_{\eta}=\bar{x}^{\nu}_{\eta}{}^{\frown}\bar{x}^{p}_{\eta}{}^{\frown}\bar{x}^{\iota}_{\eta}\in X$, and $\bar{h}_{\eta}\in H$.

By Lemma 5.2, we may assume $t_{\eta}=t\in L_{G}$, and $|\bar{x}^{\nu}_{\eta}|,|\bar{x}^{p}_{\eta}|,|\bar{x}^{\iota}_{\eta}|,|\bar{h}_{\eta}|$ are constant for all $\eta\in{}^{<\omega}\omega$. To obtain handle correspondence, add handles of elements in the tuple $\bar{x}^{p}_{\eta}$ to the beginning of $\bar{x}^{\nu}_{\eta}$ for all $\eta\in{}^{<\omega}\omega$.

Taking $\varphi^{\prime}(x,y^{\prime}):=\varphi(x,t(y^{\prime}))$ with $|y^{\prime}|=|\bar{x}_{\eta}^{\frown}\bar{h}_{\eta}|$ and $\bar{b}_{\eta}:=\bar{x}_{\eta}^{\frown}\bar{h}_{\eta}$, we have $\varphi^{\prime}\in L_{G}$ and the tree $(\bar{b}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ still satisfy TP1.

By modeling property of strong indiscernibility and compactness, we can find $(\bar{c}_{\eta}\,|\,\eta\in{}^{<\kappa}\kappa)$ with $\bar{c}_{\eta}=\bar{y}_{\eta}^{\frown}\bar{m}_{\eta}=\bar{y}^{\nu}_{\eta}{}^{\frown}\bar{y}^{p}_{\eta}{}^{\frown}\bar{y}^{\iota}_{\eta}{}^{\frown}\bar{m}_{\eta}$ to be a strongly indiscernible tree where $(\bar{c}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ based on $(\bar{b}_{\eta}|\eta\in{}^{<\omega}\omega)$. Note $\varphi^{\prime}$ and the tree still witness TP1. Also, by 4.12(2), we can assume each $\bar{y}_{\eta}$ and $\bar{m}_{\eta}$ are in some $Y$ and $M$ where $Y$ is a transversal of $G$ and $M$ is an independent set in $Z(G)$ and $G=\langle Y\rangle\times\langle M\rangle$. Let $c$ be a realization of $\bigwedge_{\alpha<\kappa}\varphi^{\prime}(x,\bar{c}_{0^{\alpha}})$. Write $c=s(y,m)$ for some terms $s\in L_{G}$, and for some tuples $y=y^{\nu}{}^{\frown}y^{p}{}^{\frown}y^{\iota}\in Y$, and $m\in M$. Again, To obtain handle correspondence, add handles of elements in the tuple $y^{p}$ to the beginning of $y^{\nu}$. Take $\psi(x^{\prime},y^{\prime})=\varphi^{\prime}(s(x^{\prime}),y^{\prime})$, then for all $\eta\bot\nu\in{}^{<\kappa}\kappa$, $\{\psi(x^{\prime},\bar{c}_{\eta}),\psi(x^{\prime},\bar{c}_{\nu})\}$ is inconsistent and $y^{\frown}m$ realizes $\bigwedge_{\alpha<\kappa}\psi(x^{\prime},\bar{c}_{0^{\alpha}})$. Since $y^{\frown}m\cap\bigcup\{\bar{c}_{0^{\alpha}}\,|\,\alpha<\kappa\}$ is finite, we may assume the tree is strongly indiscernible over $y^{\frown}m\cap\bigcup\{\bar{c}_{0^{\alpha}}\,|\,\alpha<\kappa\}$.

Now, consider $y^{\nu}$ and the tree $(\bar{y}_{\eta}^{\nu}\,|\,\eta\in{}^{<\kappa}\kappa)$ in $Y^{\nu}$. Applying 4.9, we can regard the elements as vertices of a graph. This graph satisfies NTP1 theory Th$(C)$, so there is some $\gamma$ satisfying 3.7(2). Then for each $\beta^{+}>\gamma$, we have a tuple ${y^{\prime}}^{\nu}$ such that tp${}_{\Gamma}(y^{\nu}/\bar{y}^{\nu}{}_{{0^{\beta}}^{\frown}0})$=tp${}_{\Gamma}({y^{\prime}}^{\nu}/\bar{y}^{\nu}{}_{{0^{\beta}}^{\frown}0})$ and tp${}_{\Gamma}(\bar{y}^{\nu}{}_{{0^{\beta}}^{\frown}0}/{y^{\prime}}^{\nu})$=tp${}_{\Gamma}(\bar{y}^{\nu}{}_{{0^{\beta}}^{\frown}1}/{y^{\prime}}^{\nu})$.

On the other hand, observe that Th$(\langle M\rangle)$ is a theory of vector spaces, so that the theory is stable and has quantifier elimination. Then for tuple $m$ and the tree $(\bar{m}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ in $\langle M\rangle$, we can apply Corollary 3.7 to have some $\gamma^{\prime}$ satisfying 3.7(2).

Fix a successor ordinal $\beta^{+}$ larger then $\gamma$ and $\gamma^{\prime}$, and let ${y^{\prime}}^{\nu}$ and $m^{\prime}$ be the tuples given by Corollary 3.7. Recall that the tree $(\bar{y}_{\eta}^{\frown}\bar{m}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ is strongly indiscernible over $y^{\frown}m\cap\bigcup\{\bar{c}_{0^{\alpha}}\,|\,\alpha<\omega\}$. So, as in [3, Theorem 5.6], we can find a handle preserving bijection which can be extended by Fact 4.12(1) to have

1. (1)

tp${}_{G}(ym/\bar{y}_{{0^{\beta}}^{\frown}0}\bar{m}_{{0^{\beta}}^{\frown}0})$=tp${}_{G}(y^{\prime}m^{\prime}/\bar{y}_{{0^{\beta}}^{\frown}0}\bar{m}_{{0^{\beta}}^{\frown}0})$, and 2. (2)

tp${}_{G}(\bar{y}_{{0^{\beta}}^{\frown}0}\bar{m}_{{0^{\beta}}^{\frown}0}/y^{\prime}m^{\prime})$=tp${}_{G}(\bar{y}_{{0^{\beta}}^{\frown}1}\bar{m}_{{0^{\beta}}^{\frown}1}/y^{\prime}m^{\prime})$.

From these conditions, we have $G\models\psi(y^{\prime}m^{\prime},\bar{y}_{{0^{\beta}}^{\frown}0}\bar{m}_{{0^{\beta}}^{\frown}0})\wedge\psi(y^{\prime}m^{\prime},\bar{y}_{{0^{\beta}}^{\frown}1}\bar{m}_{{0^{\beta}}^{\frown}1})$, but this contradicts that for any $\eta\bot\nu\in{}^{<\kappa}\kappa$, $\{\psi(x^{\prime},\bar{y}_{\eta}\bar{m}_{\eta}),\psi(x^{\prime},\bar{y}_{\nu}\bar{m}_{\nu})\}$ is inconsistent.

∎

Theorem 5.4**.**

For any infinite nice graph C, Th(G(C)) is NSOP1 if and only if Th(C) is NSOP1.

Proof.

Since $C$ is interpretable in $G(C)$, if Th$(C)$ has SOP1, then Th$(G(C))$ also has SOP1.

Suppose Th$(C)$ is NSOP1 and Th$(G(C))$ has SOP1. Again, we take $G$ to be a monster model of Th$(G(C))$, $X$ to be a transversal of $G$ so that $G=\langle X\rangle\times\langle H\rangle$ for some $H$ as in Fact 4.10. We have a formula $\varphi(x,y)$ and a tree $(a_{\eta}|\eta\in{}^{<\kappa}2)$ of finite tuples in $G$ for some sufficiently large regular cardinals $\kappa$. Note that for each $\eta\in{}^{<\kappa}2$, $a_{\eta}$ is of the form $t_{\eta}(\bar{x}_{\eta},\bar{h}_{\eta})$ for some terms $t_{\eta}\in L_{G}$, and for some small tuples $\bar{x}_{\eta}=\bar{x}^{\nu}_{\eta}{}^{\frown}\bar{x}^{p}_{\eta}{}^{\frown}\bar{x}^{\iota}_{\eta}\in X$, and $\bar{h}_{\eta}\in H$.

By Lemma 5.2, we may assume $t_{\eta}=t\in L_{G}$, and $|\bar{x}^{\nu}_{\eta}|,|\bar{x}^{p}_{\eta}|,|\bar{x}^{\iota}_{\eta}|,|\bar{h}_{\eta}|$ are constant for all $\eta\in{}^{<\kappa}2$. To obtain handle correspondence, add handles of elements in the tuple $\bar{x}^{p}_{\eta}$ to the beginning of $\bar{x}^{\nu}_{\eta}$ for all $\eta\in{}^{<\kappa}2$.

Taking $\varphi^{\prime}(x,y^{\prime}):=\varphi(x,t(y^{\prime}))$ with $|y^{\prime}|=|\bar{x}_{\eta}^{\frown}\bar{h}_{\eta}|$ and $\bar{a}^{\prime}_{\eta}:=\bar{x}_{\eta}^{\frown}\bar{h}_{\eta}$, we have a formula $\varphi^{\prime}\in L_{G}$ and the tree $(\bar{a}^{\prime}_{\eta}|\eta\in{}^{<\kappa}2)$ also witness SOP1.

By Fact 3.8, we can choose an array $(\bar{b}_{i,j})_{i<\omega,j<2}$ in $(\bar{a}^{\prime}_{\eta}\,|\,\eta\in{}^{<\omega}\omega)$ such that

1. (1)

$\bar{b}_{i,0}\equiv_{\bar{b}_{<i,0}\bar{b}_{<i,1}}\bar{b}_{i,1}$ for all $i<\omega$, 2. (2)

$\{\varphi^{\prime}(x;\bar{b}_{i,0})\,|\,i<\omega\}$ is consistent, and 3. (3)

$\{\varphi^{\prime}(x,\bar{b}_{i,1})\,|\,i<\omega\}$ is 2-inconsistent.

Consider an indiscernible sequence $(\bar{c}_{i,0}\bar{c}_{i,1})_{i<\kappa}$ realizing EM$((\bar{b}_{i,0}\bar{b}_{i,1})_{i<\omega})$. The new array $(\bar{c}_{i,j})_{i<\kappa,j<2}$ is an indiscernible SOP1-array and satisfies aforementioned conditions. Also, $\bar{c}_{i,j}$ is of the form $\bar{y}_{i,j}^{\frown}\bar{m}_{i,j}$ and there are some $Y$ and $M$ such that $Y$ is a transversal of $G$ extended from $\{\bar{y}_{i,j}\,|\,i<\kappa,j<2\}$, $M$ is an independent set in $Z(G)$ containing $\{\bar{m}_{i,j}\,|\,i<\kappa,j<2\}$, and $G=\langle Y\rangle\times\langle M\rangle$.

Let $c$ be a realization of $\bigwedge_{i<\kappa}\varphi^{\prime}(x,\bar{c}_{i,0})$. Write $c=s(y,m)$ for some terms $s\in L_{G}$, and for some tuples $y=y^{\nu}{}^{\frown}y^{p}{}^{\frown}y^{\iota}\in Y$, and $m\in M$. Again, To obtain handle correspondence, add handles of elements in the tuple $y^{p}$ to the beginning of $y^{\nu}$. Take $\psi(x^{\prime},y^{\prime})=\varphi^{\prime}(s(x^{\prime}),y^{\prime})$, then for all $i<j<\kappa$, $\{\psi(x^{\prime},\bar{c}_{i,1}),\psi(x^{\prime},\bar{c}_{j,1})\}$ is inconsistent and $y^{\frown}m$ realizes $\bigwedge_{i<\kappa}\psi(x^{\prime},\bar{c}_{i,0})$. Since $y^{\frown}m\cap\bigcup\{\bar{c}_{i,0}\,|\,i<\kappa\}$ is finite, we may assume the array is an SOP1-array indiscernible over $y^{\frown}m\cap\bigcup\{\bar{c}_{i,0}\,|\,i<\kappa\}$.

Now, consider $y^{\nu}$ and the array $(\bar{y}_{i,j}^{\nu}|i<\kappa,j<2)$ in $Y^{\nu}$. Applying 4.9, we can regard the elements as vertices of a graph. This graph is a model of NSOP1 theory Th$(C)$, so there is some $\gamma$ satisfying 3.12(2). Then for each $\beta>\gamma$, we have a tuple ${y^{\prime}}^{\nu}$ such that tp${}_{\Gamma}(y^{\nu}/{\bar{y}^{\nu}}_{i,0})$=tp${}_{\Gamma}({y^{\prime}}^{\nu}/{\bar{y}^{\nu}}_{0,1})$ and tp${}_{\Gamma}({\bar{y}^{\nu}}_{0,1}/{y^{\prime}}^{\nu})$=tp${}_{\Gamma}({\bar{y}^{\nu}}_{1,1}/{y^{\prime}}^{\nu})$.

On the other hand, observe that Th$(\langle M\rangle)$ is a theory of vector spaces, so that the theory is stable and has quantifier elimination. Then for the tuple $m$ and the array $(\bar{m}_{i,j}^{\nu}|i<\omega,j<2)$ in $\langle M\rangle$, we can apply Corollary 3.12 to have some $\gamma^{\prime}$ satisfying 3.12(2).

Fix some ordinal $\beta$ larger then $\gamma$ and $\gamma^{\prime}$, and let ${y^{\prime}}^{\nu}$ and $m^{\prime}$ be the tuples given by 3.12. We will find a handle preserving bijection which can be extended by Fact 4.12(1) to have

1. (1)

tp${}_{G}(ym/\bar{y}_{\beta,0}\bar{m}_{\beta,0})$=tp${}_{G}(y^{\prime}m^{\prime}/\bar{y}_{0,1}\bar{m}_{0,1})$, and 2. (2)

tp${}_{G}(\bar{y}_{0,1}\bar{m}_{0,1}/y^{\prime}m^{\prime})$=tp${}_{G}(\bar{y}_{1,1}\bar{m}_{1,1}/y^{\prime}m^{\prime})$,

for some tuple $y^{\prime}={y^{\prime}}^{\nu}{y^{\prime}}^{p}{y^{\prime}}^{\iota}$.

From the above two type equivalence, we have $G\models\psi(y^{\prime}m^{\prime},\bar{y}_{0,1}\bar{m}_{0,1})\wedge\psi(y^{\prime}m^{\prime},\bar{y}_{1,1}\bar{m}_{1,1})$, and this contradicts that for any $i<j<\kappa$, $\{\psi(x^{\prime},\bar{y}_{i,1}\bar{m}_{i,1}),\psi(x^{\prime},\bar{y}_{j,1}\bar{m}_{j,1})\}$ is inconsistent.

We finish our proof by illustrating the way to find tuple $y^{\prime}={y^{\prime}}^{\nu}{y^{\prime}}^{p}{y^{\prime}}^{\iota}$. For a finite tuple $\bar{z}$ and a natural number $k$, denote $(\bar{z})_{k}$ to be the $k$-th element of $\bar{z}$.

Let ${y}^{\iota}$ be the tuple of length $l$. For each $i<l$, if $(y^{\iota})_{i}$ is $({\bar{y}^{\iota}}_{\beta,0})_{k}$ for some $k$, then choose $y^{\prime\iota}_{i}$ to be $({\bar{y}^{\iota}}_{0,1})_{k}$, and if $(y^{\iota})_{i}$ is not in ${\bar{y}^{\iota}}_{\beta,0}$, then choose $(y^{\prime\iota})_{i}$ to be any element in $Y^{\iota}\setminus{\bar{y}^{\iota}}_{0,1}$.

Let ${y}^{p}$ be the tuple of length $l^{\prime}$. For each $i<l^{\prime}$, either $(y^{p})_{i}$ is in ${\bar{y}^{p}}_{\beta,0}$ or not. Assume first that $(y^{p})_{i}$ is $({\bar{y}^{p}}_{\beta,0})_{k}$ for some $k$. Then choose $(y^{\prime p})_{i}$ to be $({\bar{y}^{p}}_{0,1})_{k}$. Second, suppose $(y^{p})_{i}$ is not in ${\bar{y}^{p}}_{\beta,0}$. Recall that $(y^{\nu})_{i}$ is the handle of $(y^{p})_{i}$. We have either $(y^{\nu})_{i}$ is in ${\bar{y}^{\nu}}_{\beta,0}$ or not. If $(y^{\nu})_{i}$ is $({\bar{y}^{\nu}}_{\beta,0})_{k}$, then choose $(y^{\prime p})_{i}$ to be any element in $Y^{p}\setminus{\bar{y}^{p}}_{0,1}$ whose handle is $({\bar{y}^{\nu}}_{0,1})_{k}$. Otherwise, choose $(y^{\nu})_{i}$ to be any element in $Y^{p}$ whose handle is $(y^{\prime\nu})_{i}$.

Mapping each element in $y{\bar{y}}_{\beta,0}$ to ${y^{\prime}}{\bar{y}}_{0,1}$ by their natural order, we have a bijection which satisfies hypotheses in 4.12(1). As a result, tp${}_{G}(y/\bar{y}_{\beta,0})$=tp${}_{G}(y^{\prime}/\bar{y}_{0,1})$.

It remains to check the bijection from $\bar{y}_{0,1}y^{\prime}$ to $\bar{y}_{1,1}y^{\prime}$ is well-defined. Suppose the $(\bar{y}_{0,1})_{k}$ is equal to the $(y^{\prime})_{k^{\prime}}$ for some $k$ and $k^{\prime}$. Since tp${}_{G}(y/\bar{y}_{\beta,0})$=tp${}_{G}(y^{\prime}/\bar{y}_{0,1})$, $(\bar{y}_{\beta,0})_{k}=(y)_{k^{\prime}}$. This element is in $y^{\frown}m\cap\bigcup\{\bar{c}_{i,0}\,|\,i<\kappa\}$, hence for every $i<\kappa$, $(\bar{y}_{i,0})_{k}$ is $(y)_{k^{\prime}}$. By the definition of indiscernible SOP1-array, the $(\bar{y}_{i,1})_{k}$ is $(y)_{k^{\prime}}$, too. Thus $(y^{\prime})_{k^{\prime}}$, $(\bar{y}_{0,1})_{k}$, $(\bar{y}_{1,1})_{k}$, and $(y)_{k^{\prime}}$ are all the same elements. This comes out again if we assume $k$-th element of $\bar{y}_{1,1}$ is equal to the $k^{\prime}$-th element of $y^{\prime}$. Therefore, we have a well-defined bijection from $\bar{y}_{0,1}y^{\prime}$ to $\bar{y}_{1,1}y^{\prime}$.

∎

Corollary 5.5**.**

1. (1)

There is a non-simple NSOP1* pure group theory.* 2. (2)

If there is an NSOP2* theory which has SOP1, then there is a pure group theory with the same properties.*

Proof.

(1) Fix a structure $M$ of finite language such that Th$(M)$ is non-simple and NSOP1. Then by [6, Theorem 5.5.1, Exercise 5.5.9], there is a nice graph $C$ bi-interpretable with $M$. Since Mekler’s construction preserves simplicity and NSOP1, the theory of Mekler group of $C$ is non-simple and NSOP1, too.

(2) Follow the same argument above. ∎