Proof To simplify notation we assume $n=2$ and write $X,Y$ for $X_1,X_2$ .

Assume (1). Let $p(x,y)$ be a type over ${\mathcal M}$ concentrated on $X\times Y$ . Let $\varphi (x,y)$ be a formula over ${\mathcal M}$ defining a subset of $X\times Y$ . The set Z defined by $\varphi $ is a finite union of boxes $U_i\times V_i$ . It follows that $p_1(x) \cup p_2(y) \vdash p(x,y)$ .

Assume (2). Let $Z\subseteq X \times Y$ be definable. We must show that Z is a finite union of boxes $U_i \times V_i$ . For any type $p(x,y)$ containing the defining formula of Z, we have that $p_1(x) \cup p_2(y) \vdash (x, y) \in Z$ . By compactness, there are $\varphi ^p_1(x)\in p_1(x)$ and $\varphi ^p_2(y) \in p_2(y)$ , such that $\varphi ^p_1(x) \land \varphi ^p_2(y) \vdash (x,y)\in Z$ . Again by compactness, $(x,y)\in Z$ is equivalent to a finite disjunction of formulas of the form $\varphi ^p_1(x) \land \varphi ^p_2(y)$ (if not we reach a contradiction considering a type containing the defining formula of Z and the negation of all the formulas $\varphi ^p_1(x) \land \varphi ^p_2(y)$ ). It follows that Z is a finite union of boxes $U_i\times V_i$ as desired.

Proof Let S be a definable subset of $X^n\times Y\times Z$ . Given $z\in Z$ , consider the fiber $S_z \subseteq X^n\times Y$ consisting of the pairs $(x,y)$ such that $(x,y,z)\in S$ . Let $E_z \subseteq X^n\times X^n$ be the following equivalence relation:

$$\begin{align*}a\,E_z\,b \quad \Longleftrightarrow \quad \forall y\in Y \;\; (a,y)\in S_z \longleftrightarrow (b,y) \in S_z\;\text{.}\end{align*}$$

Then $\{E_z \mid z\in Z\}$ is a family of subsets of $X^n\times X^n$ indexed by Z, so by the hypothesis applied to $X^{2n}\times Z$ , it is finite. By the hypothesis on $X^n\times Y$ , each equivalence relation $E_z$ has finitely many equivalence classes; in fact, $S_z$ is a finite union of $(X^n,Y)$ -boxes and the equivalence classes of $E_z$ are the atoms of the Boolean algebra generated by the projections of these boxes on the component $X^n$ . It follows that $E = \bigcap _z E_z \subseteq X^n\times X^n$ is again an equivalence relation with finitely many classes. On the other hand, for $a,b\in X^n$ ,

$$\begin{align*}a\,E\,b \quad \Longleftrightarrow \quad \forall y\in Y, z\in Z \;\; (a,y,z)\in S \longleftrightarrow (b,y,z)\in S\text{.}\end{align*}$$

We have thus proved that there are finitely many subsets of the form $\pi _{Y\times Z}(\pi _{X^n}^{-1}(x)\cap S)$ with $x\in X^n$ , which is desired result.

Proof If suffices to show that each $X_i$ is orthogonal to the product of the other sets $X_j$ . This follows by Corollary 2.6 and induction on n.

Proof Suppose that all definable subsets of $X\times Y$ are finite unions of $(X,Y)$ -boxes. Let S be a definable subset of $X^n \times Y$ . It suffices to show that S is a finite union of $(X^n,Y)$ -boxes (by Corollary 2.8). We reason by induction on n. The case $n=1$ holds by the assumptions. Assume $n>1$ . For $t\in X$ , consider the set

$$ \begin{align*}S_t = \{ (x,y) \mid (t,x,y) \in S\} \subseteq X^{n-1}\times Y.\end{align*} $$

By induction, $S_t$ is a finite union of $(X^{n-1},Y)$ -boxes. Let $R_t\subseteq Y\times Y$ be the equivalence relation defined by $uR_t v \iff (\forall x\in X^{n-1})\; (x,u)\in S_t \leftrightarrow (x,v)\in S_t$ . Note that $R_t$ has finitely many equivalence classes. By saturation (or elimination of $\exists ^\infty $ ), there is a uniform bound $k\in \mathbb N$ , such that for all $t\in X$ , there are at most k equivalence classes modulo $R_t$ .

We claim that there is a finite subset A of Y such that for all $t\in X$ each equivalence class of $R_t$ intersects A. To this aim we prove, by induction on $j\leq k$ , that there is a finite subset $A_j$ of Y such that for all $t\in Y$ there are at most $k-j$ equivalence classes of $R_t$ which do not intersect $A_j$ . Clearly we can take $A_0$ to be the empty set. Let us construct $A_{j+1}$ . Consider the definable family $(Q_{t})_{t\in X}$ where $Q_{t}\subseteq Y$ is the union of the $R_t$ -equivalence classes which intersect $A_j$ . By the assumptions the family $(Q_{t})_{t\in X}$ contains finitely many distinct subsets $Q_1,\ldots , Q_l$ of Y. We obtain $A_{j+1}$ adding to $A_j$ a point in $Y \setminus Q_i$ for each $i=1,\ldots , l$ such that $Q_i \neq Y$ . The claim is thus proved taking $A = A_k$ .

Now we claim that the equivalence relation $R = \bigcap _{t\in X} R_t$ has finitely many equivalence classes. Indeed, given $a\in A$ consider the family $(P_{t,a})_{t\in X}$ where $P_{t,a}\subseteq Y$ is the $R_t$ -equivalence class of a. For each $a\in A$ , by the assumption there are finitely many sets of the form $P_{t,a}$ for $t\in X$ . Each $R_t$ -equivalence class is of the form $P_{t,a}$ . Hence each R-equivalence class belongs to the finite Boolean algebra generated by the sets $P_{t,a}$ . The claim is thus proved.

Now for each R-equivalence class $B\subseteq Y$ and for each $y_1,y_2\in B$ , we have $\forall t\in X, \forall x \in X^{n-1} \; (t,x,y_1)\in S \iff (t,x,y_2) \in S$ . Thus $S \cap \pi _Y^{-1}B$ is a box where $\pi _Y:X^n\times Y \to Y$ is the projection. Therefore S is a finite union of $(X^n,Y)$ -boxes.

Fact 2.11 [Reference Berarducci and Mamino2, Lemma 2.2].

Let X and Y be orthogonal definable sets and let $(T_s \mid s\in S)$ be a definable family of subsets of a Y-internal set T indexed by an X-internal set S. Then the family contains only finitely many distinct sets $T_s \subseteq T$ .

Fact 2.12 [Reference Berarducci and Mamino2, Proposition 2.3].

Let X and Y be orthogonal definable sets with $|X|\geq 2$ , and S a definable subset of $X\times Y$ . Then S is X-internal if and only if its projection onto Y is finite.

Proof Let $S\subseteq U^{k}\times V^{l}$ be definable. By the assumption there are definable surjective maps $f:X^m\to U^{k}$ and $g:Y^n\to V^l$ . Then, by orthogonality, $(f\times g)^{-1}(S)$ is a finite union $\bigcup _i A_i\times B_i$ of $(X^m, Y^n)$ -boxes. So $S = \bigcup _i f(A_i) \times g(B_i)$ is a finite union of $(U^{k},V^{l})$ -boxes.

Proof (1) Let X be internal to Y. If A and B are non-orthogonal to X, then by Proposition 2.13 they are non-orthogonal to Y. Thus if Y is cohesive, so is X.

We prove (2). Let Z be a cohesive set and suppose that Z is internal to $X\times Y$ where X and Y are orthogonal. By definition there is $m\in \mathbb N$ and a surjective definable map $f: X^m\times Y^m \to Z$ . Since $X^m$ and $Y^m$ are orthogonal, for the sake of our argument we can assume $m=1$ . So we have a surjective definable map $f:X\times Y \to Z$ and we need to show that Z is X-internal or Y-internal. For $x\in X$ and $y\in Y$ , let $f_x(y) = f(x,y) = f^y(x)$ . The image of $f_x$ is Y-internal and the image of $f^y$ is X-internal. These two images are then orthogonal (Proposition 2.13), so they cannot be both infinite by the hypothesis on Z. It follows that either $\operatorname {\mathrm {Im}}(f_x)$ is finite for all $x\in X$ or $\operatorname {\mathrm {Im}}(f^y)$ is finite for all $y\in Y$ . By symmetry let us assume that $\operatorname {\mathrm {Im}}(f^y)\subseteq Z$ is finite for all y. Let $E_y\subseteq X^2$ be the equivalence relation defined by $xE_y x' \iff f(x,y) = f(x',y)$ . Then $E_y$ is a definable equivalence relation on X of finite index. Since $(E_y \subseteq X\times X \mid y\in Y)$ is a definable family of subsets of an X-internal set indexed by a Y-internal set, by orthogonality there are finitely many sets of the form $E_y$ for $y\in Y$ (Fact 2.11). The intersection $E = \bigcap _{y\in Y}E_y$ is then again a definable equivalence relation of finite index on X. Let $x_1, \dots , x_k$ be representatives for the equivalence classes of E. Then Z is the image of the restriction of f to $\bigcup _{i\leq k} \{x_i\} \times Y$ , and therefore it is Y-internal.

Proof Let A and B be non-orthogonal to $X\times Y$ . We need to prove that A and B are non-orthogonal. By Corollary 2.6 either X or Y is non-orthogonal to A. Similarly, either X or Y is non-orthogonal to B. There are four cases to consider, but by symmetry we can consider the following two cases.

Case 1. X is non-orthogonal to both A and B.

Case 2. X is non-orthogonal to A, and Y is non-orthogonal to B.

In the first case by the cohesiveness of X the sets A and B are non-orthogonal. In the second case, since X and Y are non-orthogonal and Y is non-orthogonal to B, by the cohesiveness of Y we conclude that X is non-orthogonal to B, so we have a reduction to the first case.

Proof Let A and B be definable sets non-orthogonal to G. We need to prove that A and B are not orthogonal. Let us first concentrate on A. By Corollary 2.8 there are $n\in \mathbb N$ and a definable relation $R\subseteq A^n \times G$ which is not a finite union of boxes. Let $\mathcal P (G)$ be the power set of G and let $f\colon A^n \to \mathcal P(G)$ be defined by $f(a) = \{g\in G \mid (a,g) \in R\}$ . Since $f(a)$ has dimension $\leq 1$ , its boundary $\delta f(a)$ , closure minus interior, is finite. By the assumption on R, the image of f is infinite; thus also the image of $\delta f\colon A^n\to \mathcal P (G)$ is infinite, since in a group of dimension $1$ there can only be finitely many disjoint definable subsets with a given boundary (because, given a cell decomposition compatible with the boundary any definable subset with the given boundary is a union of some cells of the decomposition). Now recall that in an o-minimal structure a definable family of finite sets is uniformly finite. So there is $k\in \mathbb N$ such that $\delta f(a)$ has at most k elements for every $a\in A^n$ . By ordering the points of $\delta f(a)$ lexicographically, we have a map $h\colon A^n \to G^k$ with infinite image. It follows that there is $i\leq k$ such that $\pi _i \circ h\colon A^n \to G$ has infinite image, where $\pi _i\colon G^k\to G$ is the projection onto the i-th component. We have thus proved that there is a definable map $f_A\colon A^n \to G$ with infinite image. Similarly, there is $l\in \mathbb N$ and a definable map $f_B\colon B^l\to G$ with infinite image. Infinite definable subsets of a one-dimensional group have non-empty interior; thus, composing with a group translation, we can assume that the two images have an infinite intersection. The relation $xQy :\iff f_A(x) = f_B(y)$ then witnesses the non-orthogonality of A and B.

Notation 4.2. If $S \subseteq \prod _{j=1}^k X_{t(j)}$ we define

$$ \begin{align*}\pi_i: S \to X_i^{k_i}\end{align*} $$

as the projection of S onto the $X_i$ -components, where $k_i$ is the number of indexes j with $t(j) =i$ . For instance, if $S \subseteq X_1\times X_2 \times X_1$ , then $\pi _1:S \to X_1^2$ maps $(a,b,c)$ to $(a,c)$ .

Fact 4.4. Let $f:A\to B$ be a function in ${\text {Def}(\Pi _i X_i)}$ . Then there is a finite partition ${\mathcal D}$ of $\operatorname {\mathrm {dom}}(f)$ into definable sets, such that for each $D\in \mathcal D$ , the restriction of f to D splits.

Proof By orthogonality of $X_1,\ldots , X_n$ , up to a permutation of the variables, f is a finite disjoint union $\bigcup _j f_j$ , where $f_j = U_{1,j} \times \dots \times U_{n,j}$ and $U_{i,j} \subseteq \pi _i(A) \times \pi _i(B)$ . We conclude observing that each $f_j$ splits.

Fact 4.5. Let $h, f_1, \dots , f_m$ be maps in ${\text {Def}(\Pi _i X_i)}$ , and suppose that the map $h \circ (f_1\times \dots \times f_m)$ is defined $($ in the sense that the range of $f_1\times \cdots \times f_m$ is contained in the domain of $h)$ . If $h,f_1,\ldots , f_m$ split, so does $h \circ (f_1\times \dots \times f_m)$ .

Proof Straightforward.

Observation 4.7. Let H be a finite index subgroup of G. If H admits an orthogonal decomposition with respect to $X_1,\ldots ,X_n$ , then so does G.

Proof Assume $(1)$ . Consider the projections $H_1,\ldots ,H_n$ of G on $X_1,\ldots ,X_n$ , respectively. The group operation $\mu $ , by the splitting hypothesis, takes the form

$$\begin{align*}\mu((x_1,\ldots,x_n),(y_1,\ldots,y_n)) = (\mu_1(x_1,y_1),\ldots,\mu_n(x_n,y_n)).\end{align*}$$

It is straightforward to check that $\mu _i$ is a group operation on $H_i$ , and $(2)$ is thus established.

Now assume (2). Let $\pi _i: \prod _i X_i \to X_i$ be the projection onto the i-th component. Replacing $H_i$ with $\pi _i(G)$ we can assume, without loss of generality, that $\pi _i(G) = H_i$ for each i. We prove that G has finite index in $H_1\times \cdots \times H_n$ . Let

$$ \begin{align*}p_i:\prod_j X_j \to \prod_{j\ne i} X_j\end{align*} $$

be the projection that omits the i-th coordinate. Fix an index i and consider an element k of $\prod _{j\neq i} H_j$ . Let $H_i(k) = \pi _i(G\cap p_i^{-1}(k))$ . Observe that $L_i:=H_i(p_i(e))$ is a subgroup of $H_i$ . We claim that $L:=L_1\times \cdots \times L_n$ has finite index in $H_1\times \cdots \times H_n$ . Assuming the claim, since $L<G$ , also G must have finite index. It suffices to prove that $L_i$ has finite index in $H_i$ . The coset $h L_i$ , for $h\in H_i$ , coincides with $H_i(p_i(g))$ for any $g\in G\cap \pi _i^{-1}(h)$ . Thus, in particular, the cosets of $L_i$ belong to the family $H_i(-)$ indexed over $\prod _{j\ne i} X_j$ . This family is finite by Fact 2.11, and this concludes the proof of $(3)$ .

It is immediate that $(2)$ , hence also $(3)$ , implies $(1)$ . It remains to show that G admits a decomposition. To this aim, observe that L admits a decomposition: it is, in fact, the product of the subgroups $G\cap p_i^{-1}(p_i(e)) \cong L_i$ . We conclude by Observation 4.7.

Proof Point (1) holds for all groups with fsg [Reference Hrushovski, Peterzil and Pillay9, Proposition 4.2].

To prove point (2) we may assume ${\mathcal M}$ is $\kappa $ -saturated for some sufficiently big cardinal $\kappa $ . We make use of the infinitesimal subgroup $G^{00}$ (see [Reference Simon17] for the definition). Specifically, we need to observe that in a compactly dominated group G, a subset S is generic if and only if some translate of S contains $G^{00}$ [Reference Berarducci1, Proposition 2.1]. If $G = HK$ with $H<G$ and $K\lhd G$ , then $G^{00} = (HK)^{00} = H^{00}K^{00}$ [Reference Conversano4, Theorem 4.2.5] (this holds without the hypothesis that G is compactly dominated). Now suppose $S\subseteq G$ is generic. So there is $g\in G$ such that $gG^{00} \subseteq S$ . Now, $K^{00}$ is a normal subgroup of G (being a definably characteristic subgroup of K), and so, writing $g=kh$ for $h\in H$ and $k\in K$ , we get $gG^{00} = kh K^{00}H^{00} = (kK^{00})(hH^{00}) \subseteq S$ . By $\kappa $ -saturation there are definable sets $U,V$ with $K^{00}\subseteq U \subseteq K$ and $H^{00}\subseteq V \subseteq H$ such that $(kU)(hV) \subseteq S$ . By construction, $kU$ and $hV$ are generic in K and H, respectively.

Point (3) follows from [Reference Starchenko18, Corollary 3.17] (the product of smooth measures is smooth) and the fact that a group is compactly dominated if and only if it has a smooth left-invariant measure [Reference Simon17, Theorem 8.37].

Fact 5.2. Definably compact groups in an o-minimal structure are compactly dominated.

Proof This was first proved for o-minimal expansions of a field in [Reference Hrushovski and Pillay11]. We give some bibliographical pointers to obtain the result for arbitrary o-minimal structures. First one shows that a definably compact group in an o-minimal structures has fsg [Reference Peterzil, Delon, Kohlenbach, Maddy and Stephan12, Theorem 8.6]. From this one deduces that G admits a generically stable left-invariant measure [Reference Simon17, Proposition 8.32]. In an o-minimal structure (and more generally in a distal structure), a generically stable measure is smooth [Reference Simon17, Proposition 9.26]. Finally, a NIP group with a smooth left-invariant measure is compactly dominated [Reference Simon17, Theorem 8.37].

Proof There is a finite subset $I\subseteq G$ such that $A \subseteq \bigcup _{h\in I} hB$ . By Proposition 5.1(1), there is $h\in I$ such that $A \cap hB$ is generic.

Proof Let $P_n: G^n\to G$ be the function sending $(x_1,\ldots ,x_n)$ to $\prod _{i=1}^{n}x_i$ . Then $P_n\in {\text {Def}(\Pi _i X_i)}$ . Since G is compactly dominated, so is $G^n$ (Proposition 5.1(3)). By Fact 4.4 and Proposition 5.1(1), $P_n$ splits on a generic definable set $S\subseteq G^n$ . By Proposition 5.1(2) we find definable generic sets $A_1, \ldots , A_n \subseteq G$ such that $A_1\times \cdots \times A_n \subseteq S$ . By Proposition 5.3 we find $a_1, \ldots , a_n \in G$ such that the set

$$ \begin{align*}U = \bigcap_{i=1}^n a_i A_i\end{align*} $$

is generic in G. Again by Fact 4.4, there is a generic set $D\subseteq U$ such that for every $i=1,\ldots , n$ the function

$$ \begin{align*}f_i: x \in D \mapsto a_i^{-1}x\end{align*} $$

splits on D. Since $D\subseteq a_i A_i$ , the image of $f_i$ is contained in $A_i$ . It follows that the function

$$ \begin{align*}f_1 \times \cdots \times f_n: D^n \to A_1\times \cdots \times A_n\end{align*} $$

splits. By Fact 4.5, since $P_n$ splits on $A_1\times \cdots \times A_n$ , we deduce that the function

$$ \begin{align*}f= P_n \circ (f_1\times \cdots \times f_n)\end{align*} $$

splits on $D^n$ . Since G is abelian,

(6.1) $$ \begin{align} f(x_1, \ldots, x_n) = a \prod_{i=1}^{n}x_i, \end{align} $$

where $a = \prod _{i=1}^n a_i^{-1}$ . By the orthogonality assumption, D is a finite union of sets of the form $U_1\times \cdots \times U_n$ with $U_i \subseteq X_i$ . By Proposition 5.1(1) one of these sets is generic, so by replacing D with a smaller set we can assume that

$$ \begin{align*}D = U_1\times \cdots \times U_n.\end{align*} $$

Let $\pi _i: \prod _i X_i\to X_i$ be the projection onto the i-th component and let

$$ \begin{align*}p_i:\prod_j X_j \to \prod_{j\ne i} X_j\end{align*} $$

be the projection that omits the i-th coordinate. Now fix $k\in D$ and let

$$ \begin{align*}D_i = p_i^{-1}p_i(k) \cap D = \{x\in D \mid \forall j\neq i \; \pi_j(x)= \pi_j(k) \}.\end{align*} $$

Notice that $D_i$ is $U_i$ -internal; hence, a fortiori it is $X_i$ -internal. We claim that

(6.2) $$ \begin{align} k^{n-1}D \subseteq D_1\dots D_n. \end{align} $$

To prove the claim, let $g\in D$ and let $x_i\in G$ be such that

$$ \begin{align*}\pi_i(x_i) = \pi_i(g)\; \& \; \pi_j(x_i) = \pi_j(k) \; \text{for} \; j\neq i.\end{align*} $$

Notice that $x_i\in D_i$ . Since f splits on $D^n$ , the value of $\pi _i f(x_1,\ldots , x_n)$ does not change if we replace $x_i$ with g and $x_j$ with k for $j\neq i$ . By Equation (6.1) we then obtain

$$ \begin{align*}\pi_i f(x_1,\dots, x_n) = \pi_i(ak^{n-1}g).\end{align*} $$

Since this holds for every i, we deduce that

$$ \begin{align*}f(x_1,\dots, x_n) = ak^{n-1}g,\end{align*} $$

and since $f(x_1,\ldots , x_n) = a \prod _{i=1}^n x_i$ we obtain Equation (6.2).

We have thus shown that a translate of D is contained in a product of $X_i$ -internal sets. Since D is generic, the same holds for G, so G is a product of $X_i$ -internal sets.

Proof Suppose $X\subseteq M^m$ and let $\pi _i:M^m \to M$ be the projection onto the i-th coordinate ( $i = 1,\dots ,m$ ). Fix parameters $a_1 < \dots < a_m$ in M, let $Z_i = \{a_i\}\times \pi _i(X)$ and let $Z = \bigcup _i Z_i \subseteq M\times M $ . Then Z has dimension $1$ and is bi-internal to X. Each $\pi _i(X)\subseteq M$ is a finite union of points and intervals with the induced order from ${\mathcal M}$ , and we have an induced order on $Z_i$ via the obvious bijection. We then order Z lexicographically by stipulating that all the elements of $Z_i$ precede all the elements of $Z_{i+1}$ . We thus obtain a definable linear order $<_Z$ on Z, but notice that Z need not be dense, and even if it is, $Z_i$ need not be a finite union of points and intervals in the order $(Z,<_Z)$ (e.g., $Z_i$ may be bounded but with no supremum in Z). We can remedy this by adding and removing from Z finitely many points, thus obtaining a definable set Y with a dense linear order $\prec $ without end-points which agrees with $<_Z$ on $Y\cap Z$ (for example, if ${\mathcal M} = \mathbb R$ and $Z = [1,2) \cup (3,4] \cup [5,6) \subset M$ , we can add the point $2$ and remove $1$ and $4$ ). The set Y can be constructed as a disjoint union $\bigcup _{i=1}^m Y_i$ where, for each i, the symmetric difference $Y_i \triangle Z_i$ is finite and $Y_i$ is a finite union of intervals and points of $(Y,\prec )$ , so in particular it is definable in $(Y,\prec )$ .

To prove (3) it suffices to observe that any definable subset D of Y is the union of its traces $D\cap Y_i$ on the various $Y_i$ , so it is a finite union of points and intervals of $(Y,\prec )$ .

To prove the remaining points we make some preliminary observations (where all the relevant sets are assumed to be definable and n is arbitrary):

1. i) $A_1\times \cdots \times A_n$ can be definably embedded in $(A_1\cup \cdots \cup A_n)^n$ .

2. ii) If A is infinite and F is finite, $A\cup F$ can be embedded in $A\times A$ .

3. iii) $A_1\cup \cdots \cup A_n$ can be definably embedded in $A_1\times \cdots \times A_n\times B$ where B is any infinite set. If, for some i, the set $A_i$ is infinite, we can take $B=A_i$ and embed $A_1\cup \cdots \cup A_n$ in $A_1 \times \cdots \times A_n \times A_i$ , hence a fortiori in $A_1^2\times \cdots \times A_n^2$ .

By i) and iii) $A_1 \times \cdots \times A_n$ is bi-internal to $A_1\cup \cdots \cup A_n$ provided at least one of the sets $A_i$ is infinite.

The set X is included in the Cartesian product of its projections $\pi _i(X)$ , so it can be embedded in the Cartesian product of the sets $Z_i = \{a_i\} \times \pi _i(X)$ . On the other hand, by i), $Z_1\times \cdots \times Z_n$ can be embedded in $(Z_1 \cup \cdots \cup Z_m)^m$ . Moreover, $Z_1 \cup \cdots \cup Z_m$ differs from the o-minimal envelope Y by a finite set, so by ii) it can be embedded into $Y^2$ (note that Y must be infinite as X is assumed to be infinite). It follows that X can be embedded into $Y^{2m}$ , thus proving (2).

By a similar argument, using ii), Y can be embedded into $(Z_1\cup \cdots \cup Z_n)^2$ , which in turn can be embedded, by iii), into the Cartesian product $\prod _i \pi _i(X)^4$ of the projections of X to the $4$ -th power. In particular X and Y are bi-internal and we get (1).

To prove point (4), let $X = X_1\times \cdots \times X_n$ where each $X_i\subseteq M^{n_i}$ is infinite. Let $X_i'$ be the product of the projections of $X_i$ to the $4$ -th power. Then the o-minimal envelope of X can be embedded in $X_1' \times \cdots \times X_n'$ and we get (4).

Proof We can assume that G is infinite as otherwise the result is clear. We can then also assume that each $X_i$ is infinite, because if $X_n$ is finite, say, then G is internal to $X_1\times \cdots \times X_{n-1}$ .

By Lemma 7.2, G is internal to $Y= {[X_1\times \dots \times X_n]}^{\text {o-min}}$ , so it can be considered as an interpretable group in the o-minimal structure $\mathcal Y$ . By [Reference Eleftheriou, Peterzil and Ramakrishnan8] interpretable groups in an o-minimal structure are definably isomorphic to definable groups (in the sense of Definition 7.3). It follows that there is a definable injective map $f:G \to Y^k$ for some $k\in \mathbb N$ . By the construction of the o-minimal envelope, Y can be embedded into a product of sets $Y_i$ , where $Y_i$ is bi-internal to $X_i$ (Lemma 7.2(4)). Now it suffices to take $X^{\prime }_i = Y_i^k$ .

Proof of Theorem 7.1

By Proposition 7.4 there are definable sets $X_1',\dots , X_n'$ with $X_i'$ bi-internal to $X_i$ such that G is definably isomorphic to a group $G'$ contained in $X_1'\times \dots \times X_n'$ . Since G is definably compact, $G'$ also is. By Theorem 6.1 $G'$ admits a decomposition with respect to $X_1', \dots , X_n'$ , hence also with respect to $X_1,\dots , X_n$ . Hence so does G.

Proof By assumption, we can write $H = H_1 \dots H_n$ and $N = N_1 \dots N_n$ , where $H_i$ and $N_i$ are $X_i$ -internal definable sets (not necessarily subgroups). We have

$$ \begin{align*}G = f^{-1}(H_1) \dots f^{-1}(H_n).\end{align*} $$

By [Reference Edmundo7, Theorem 2.5], there is a definable section $\sigma : H\to G$ . Since $f^{-1}(H_i) = \sigma (H_i) N$ , we have

$$ \begin{align*}G = \sigma(H_1)N \dots \sigma(H_n)N.\end{align*} $$

Since $N= N_1\dots N_n$ is contained in the center of G, it follows that $G = U_1\dots U_n$ , where $U_i$ is the $X_i$ -internal set $\sigma (H_i) N_i\dots N_i$ (n occurrences of $N_i$ ).

Proof We reason by induction on dimension. For $\dim (G)=1$ , G is cohesive by Theorem 3.4, so it is indecomposable (Proposition 3.2); hence, it is internal to one of the $X_i$ . Let $\dim (G)>1$ . If G is definably compact, then we conclude by Theorem 7.1. If not, then by [Reference Peterzil and Steinhorn14, Theorem 1.2], G has a one-dimensional torsion-free definable subgroup $H<G$ . Since $\dim (H)=1$ , H admits a decomposition. By induction on dimension, so does $G/H$ . Therefore, since G is abelian, we can apply Lemma 8.1.

Fact 8.3. Let G be a connected group definable in an o-minimal structure. If $Z(G)$ is finite, then $G/Z(G)$ is centerless.

Proof Let $a\in G$ , such that $aZ(G)$ is in the center of $G/Z(G)$ . We want to prove that $a\in Z(G)$ . We have that for all $b\in G$ , $a^{-1}b^{-1}ab\in Z(G)$ . Since G is connected, the image of the map $f:G\to Z(G)$ sending $b\in G$ to $a^{-1}b^{-1}ab \in Z(G)$ is connected. Since $Z(G)$ is finite, f must be constant. Since f maps the identity $e\in G$ to e, we have $a^{-1}b^{-1}ab = e$ . Thus $a\in Z(G)$ , as needed.

Proof We observe that, since the connected component of the identity $G^0$ has finite index in G, by Observation 4.7, it suffices to find a decomposition of $G^0$ . We may thus assume that G is connected.

We prove, now, the theorem, by induction on $\dim (G)$ . For $\dim (G)=0$ , it is obvious. Assume $\dim (G)>0$ .

Case 1. Suppose that the centre $Z(G)$ is finite. Then $Z(G)$ admits an orthogonal decomposition, and by Lemma 8.1 it suffices to prove that $H=G/Z(G)$ has a decomposition. Since $Z(G)$ is finite, H is centerless (Fact 8.3).

By [Reference Eleftheriou, Peterzil and Ramakrishnan8], H is definably isomorphic to a definable group. By [Reference Peterzil, Pillay and Starchenko13, Theorems 3.1 and 3.2], it follows that there are definable real closed fields $R_1, \dots , R_k$ and definable linear groups $H_i<GL(n,R_i)$ such that H is definably isomorphic to $H_1\times \dots \times H_k$ . A definable real closed field in an o-minimal structure has dimension $1$ [Reference Peterzil and Steinhorn14, Theorem 4.1]. We can arrange so that $R_i$ is internal to $X_1\times \cdots \times X_n$ because we can apply the cited results inside the o-minimal structure ${[X_1\times \cdots \times X_n]}^{\text {o-min}}$ (as defined in Lemma 7.2). By Theorem 3.4 $R_i$ is cohesive, so it is internal to some $X_j$ by Proposition 3.2(2). It follows that each $H_i$ is internal to some $X_j$ . We have thus proved that H admits a decomposition with respect to the orthogonal sets $X_1,\ldots ,X_n$ . Therefore we can conclude by Lemma 8.1.

Case 2. Suppose $Z(G)$ is infinite. By the abelian case (Proposition 8.2), $Z(G)$ has a decomposition with respect to $X_1,\ldots ,X_n$ . By induction on the dimension, $G/Z(G)$ has a decomposition too. Therefore we can again conclude by Lemma 8.1.

Proof Let G be definably simple. By the proof of Theorem 8.4 there is a definable real closed field R such that G is definably isomorphic to a definable subgroup of $GL(n,R)$ . Since $\dim (R) = 1$ , by Theorem 3.4 R is cohesive. But $GL(n,R)$ is internal to R; thus, all its definable subsets are cohesive.

Proof By [Reference Eleftheriou, Peterzil and Ramakrishnan8, Theorem 3], there is a definable injective map $f:G \to \prod _{j=1}^k G_j$ where each $G_j$ is a one-dimensional definable group. Define a relation R on $\{1,\dots , k\}$ by $iRj$ if $G_i$ and $G_j$ are not orthogonal. By Theorem 3.4, the groups $G_i$ are cohesive; hence, R is an equivalence relation. Suppose there are n equivalence classes. Let $X_i$ be the product of the groups $G_j$ , with j in the i-th class. Using f, it is easy to define (by a permutation of the coordinates on the image) the injective map $h:G\to \prod _{i=1}^n X_i$ .

The sets $X_i$ are cohesive by Proposition 3.3 and mutually orthogonal by Corollary 2.6.

Proof Let $X_1, \dots , X_n$ be the sets provided by Lemma 8.6. By Theorem 8.4, there are $X_i$ -internal sets $A_i\subseteq G$ , $i=1,\dots , n$ , such that

$$ \begin{align*}G=A_1 \dots A_n.\end{align*} $$

Clearly, the $A_i$ ’s are orthogonal and cohesive, as they inherit those properties from the $X_i$ ’s.

Proof Suppose that G is infinite and fix a cohesive orthogonal decomposition $G = A_1\ldots A_n$ . Then at least one $A_i$ is infinite, say $A_1$ . If some $A_i$ is finite we may replace $A_1$ with $A_1A_i$ and omit $A_i$ obtaining another cohesive orthogonal decomposition. So we may assume that $A_1, \ldots , A_n$ are all infinite. Now consider another cohesive orthogonal decomposition $G = B_1\ldots B_m$ into infinite sets. Fix $B_i$ and observe that $B_i$ is internal to G, which is internal to the Cartesian product $A_1\times \cdots \times A_n$ . Since $B_i$ is indecomposable, it must be internal to some $A_j$ . Moreover, j must be unique, because if $B_i$ is internal to both $A_j$ and $A_h$ , with $j\ne h$ , then it is non-orthogonal to both, so by cohesiveness of $B_i$ , $A_i$ and $A_h$ are non-orthogonal, a contradiction. The argument also shows that $m=n$ and $B_i$ is in fact bi-internal to the corresponding $A_j$ .

Proof The structure $\bigsqcup _i {\mathcal X}_i$ is bi-interpretable with the o-minimal structure ${\mathcal M}$ obtained by concatenating ${\mathcal X}_1, \ldots , {\mathcal X}_n$ in the given order and adding $n-1$ points to separate ${\mathcal X}_i$ from ${\mathcal X}_{i+1}$ for $i<n$ . We can therefore apply to ${\mathcal M} = \bigsqcup _i {\mathcal X}_i$ the various results concerning o-minimal structures. By Theorem 8.4, the group G admits a decomposition $G = A_1\ldots A_n$ with respect to $X_1,\ldots , X_n$ , where $X_i$ is the domain of ${\mathcal X}_i$ . Let $\langle A_i \rangle $ be the locally definable subgroup of G generated by $A_i$ .

We claim that $\langle A_i \rangle $ is locally definably isomorphic to a definably generated group $G_i$ in the structure ${\mathcal X}_i$ . To this aim, let $A_i^{(m)} \subseteq G$ consist of the m-fold products $a_1\ldots a_m$ , where each $a_i$ is either an element of $A_i$ or is the group-inverse of an element of $A_i$ . Without loss of generality, after permuting coordinates, we can choose $k_1,\ldots , k_n\in \mathbb N$ such that G is included in $X_1^{k_1}\times \cdots \times X_n^{k_n}$ . Since $A_i^{(m)}$ is $X_i$ -internal and included in G, by Fact 2.12 it must have a finite projection on the factors different from $X_i$ . Thus we can write $A_i^{(m)} \subseteq L_m \times X_i^{k_i} \times F_m$ where $L_m$ and $F_m$ are finite sets. It follows that the subgroup $\langle A_i \rangle $ of G generated by $A_i$ is included in $L\times X_i^{k_i}\times F$ where $L = \bigcup _m L_m$ and $F= \bigcup _m F_m$ are countable sets. Consider a bijection sending $L\cup F$ to a countable subset of $X_i$ . This induces a locally definable bijection between $\langle A_i \rangle $ and a locally definable subset $G_i \subseteq X_i^{k_i+1}$ . We can endow $G_i$ with a group operation via the bijection. The resulting group $G_i$ will then be locally definable, and in fact definably generated, in the structure $\mathcal {X}_i$ . There is a locally definable group homomorphism $f: G_1\times \cdots \times G_n \to G$ induced by the composition $G_1 \times \cdots \times G_n \cong \langle A_1 \rangle \times \cdots \times \langle A_n \rangle \to G$ , where the last map sends $(x_1,\ldots , x_n)$ to their product in G. Note that f is a homomorphism since G is abelian.

It remains to prove that the kernel $\Gamma $ of the above f is discrete. To this aim fix, for $i=1,\ldots , n$ , a definable subset $U_i$ of $\langle A_i \rangle $ and let S be the set of all tuples $(a_1,\ldots , a_n)\in U_1\times \cdots \times U_n$ such that $a_1 \ldots a_n = 1_G$ . It suffices to show that S is finite. The sets $U_1,\ldots , U_n$ are orthogonal by Proposition 2.13, since they are internal to $A_1,\ldots , A_n$ respectively. It follows that S is a finite union of sets of the form $B_1\times \dots \times B_n$ with $B_i\subseteq U_i$ . However, each $B_i$ can only be a singleton because any choice of $n-1$ elements from $a_1,\ldots , a_n$ determines the last one via the equation $a_1\ldots a_n = 1_G$ .

Proof Let p be an odd prime. The Heisenberg group mod p is the semialgebraic group H of matrices of the form $\begin {bmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end {bmatrix}$ where $c \in \mathbb R/p\mathbb Z$ and $a,b\in \mathbb Z/p\mathbb Z$ . We claim that H is not Lie isomorphic to a semidirect product of a connected real Lie group and a discrete group. Taking $a,b=0$ we obtain the center $Z(H)$ of H, which coincides with $H^0$ and it is isomorphic to the circle group $\mathbb R/p\mathbb Z$ . The quotient $H/H^0$ is isomorphic to $(\mathbb Z/p\mathbb Z)^2$ . Since H is not abelian, H is not isomorphic to the direct product $H^0 \times (\mathbb Z/p\mathbb Z)^2$ . Moreover the direct product is the only possible semidirect product in the Lie category because $(\mathbb Z/p\mathbb Z)^2$ has no non-trivial continuous action on $\mathbb R/p\mathbb Z$ (since the only non-trivial definable automorphism of $\mathbb R/p\mathbb Z$ is the inverse, and p is odd).

Proof Assume that G is the direct product of two definable infinite subgroups $G_1$ and $G_2$ . Then $\dim (G_1) = \dim (G_2) = 1$ . Since definable groups of dimension $1$ are cohesive (Theorem 3.4), we may assume that $G_1$ is $R_1$ -internal and $G_2$ is $R_2$ -internal. Consider the natural (surjective) projections $\pi _i: G \to H_i$ .

We claim that $G_1^0=\pi _2^{-1}({1_G})$ and $G_2^0 = \pi _1^{-1}({1_G})$ where $G_i^0$ is the connected component of the identity of $G_i$ . Consider for instance $G_1$ . Clearly $\pi _2(G_1)$ is finite by orthogonality. Hence $\pi _2^{-1}({1_G}) \cap G_1$ has finite index in $G_1$ , so it is infinite. Moreover $\pi _2^{-1}({1_G})$ is $H_1^0\times \{{1_G}\}$ ; hence, it is connected and, since two definably connected one-dimensional groups having infinite intersection coincide, it must coincide with $G^0_1$ . The claim is thus proved.

Observe that $Z(G) = G^0 = G^0_1 \times G^0_2$ ; thus, $[G_1:G^0_1][G_2:G^0_2]= [G:G^0] = 9$ . So, there are three cases for the possible values of the indexes of $G^0_1$ and $G^0_2$ : $(1,9), (3,3), (9,1)$ .

First case: $[G_1:G_1^0]=1$ and $[G_2:G_2^0]=9$ (observe that the third case is symmetric). In this case $G_1$ is connected; thus, $G_1=G_1^0$ , and $|\pi _1(G_2)| = 9$ because $G^0_2 = \pi _1^{-1}(1_G)$ has index $9$ in $G_2$ . On the other hand,

$$ \begin{align*}H_1 = \pi_1(G) = \pi_1(G_1)\pi_1(G_2) = H_1^0\pi_1(G_2) = Z(H_1)\pi_1(G_2),\end{align*} $$

and since $\pi _1(G_2)$ has nine elements and $Z(H_1)$ has index $9$ in $H_1$ , it follows that $H_1$ must be the direct product of $Z(H_1)$ and $\pi _1(G_2)$ , contradicting the claim in Example 11.1.

Second case: $[G_1:G_1^0]=3 = [G_2:G^0_2]$ . In this case we show that $G_1$ and $G_2$ are abelian and we reach a contradiction since G is not abelian. By symmetry it suffices to show that $G_1$ is abelian. First recall that $G_1^0<Z(G)$ , so in particular $G_1^0$ is central in $G_1$ , and definably isomorphic to $\mathbb R/3\mathbb Z$ . It follows that $G_1$ is definably isomorphic to a central extension of $\mathbb R/3\mathbb Z$ by $\mathbb Z/3\mathbb Z$ . We claim that such a group is necessarily abelian. To this aim we show that there is a copy of $\mathbb Z/3\mathbb Z$ which is a complement of $G_1^0$ . Let $G_1^0, aG_1^0, bG_1^0$ be the three connected components of $G_1$ . Note that the map $x\mapsto x^3$ has image contained in $G_1^0$ and its restriction to $G_1^0$ is onto. Consider the map sending $x\in G_1^0$ to $(ax)^3 \in G_1^0$ . Since $G^0_1$ is central, $(ax)^3 = a^3 x^3$ . Now let $y\in G^0_1$ be such that $y^3 = a^3$ . Then $ay^{-1}$ has order three and generates a complement C of $G^0_1$ in $G_1$ . Since $G^0_1$ is central we conclude that $G_1$ is the direct product of C and $G^0_1$ , hence it is abelian.

Proof We need the following facts:

1. (1) k and $\Gamma $ are internal to K.

2. (2) k and $\Gamma $ are orthogonal.

3. (3) K is internal to any infinite definable subset of K.

4. (4) K is not internal to $k\times \Gamma $ .

Point (1) is obvious. The proof of (2) is in [Reference van den Dries6, Corollary 5.25]. (3) follows from the fact that any definable subset of K is a Boolean combination of valuation balls [Reference van den Dries6, Corollary 3.32], so in particular if it is infinite it contains a valuation ball, and K itself is internal to any valuation ball. Point (4) is clear if k and $\Gamma $ have cardinality less than K and we can reduce to this case going to an elementary equivalent model (for instance, by the theorem of Ax-Kochen and Ershov (see [Reference van den Dries6, Theorem 5.1]) we can take $K = {\mathbb Q}^{ac}((t^{\mathbb Q})), \Gamma = {\mathbb Q}, k = {\mathbb Q}^{ac}$ where ${\mathbb Q}^{ac}$ is the algebraic closure of $\mathbb Q$ ).

Granted (1)–(4) we continue as follows. By (1) and (2) K is not cohesive. We claim that K is indecomposable. So let K be internal to a product $X\times Y$ of orthogonal definable sets. We must show that K is internal to either X or Y. If this is not the case, then by (3) neither X nor Y can have an infinite projection on K. But then both X and Y are internal to $k\times \Gamma $ . So K itself is internal to $k\times \Gamma $ , contradicting (4).