Observation 2.3. If $G\leq H$ are permutation groups, then $f_n(H)\leq f_n(G)$ where f is any of the operators $o,o^i$ , or $o^s$ . In particular every group containing an oligomorphic group is oligomorphic.

Proof (1) A permutation $\sigma $ of $X\times Y$ is contained in $\operatorname {Sym}(X)\wr \operatorname {Sym}(Y)$ if and only if it preserves the equivalence relation $E:=\{((x_1,y_1),(x_2,y_2))\colon y_1=y_2\}$ . This implies immediately that the group $\operatorname {Sym}(X)\wr \operatorname {Sym}(Y)$ is closed. Now let $(\sigma _n)_n$ be a sequence in $G\wr \operatorname {Sym}(Y)$ converging to a permutation $\sigma \in \operatorname {Sym}(X\times Y)$ . Then there exist $\beta _n\in \operatorname {Sym}(Y),\alpha _{y,n}\in G$ such that for all $x\in X$ and $y\in Y$ we have $\sigma _n(x,y)=(\alpha _{y,n}(x),\beta _n(y))$ . Since $\operatorname {Sym}(X)\wr \operatorname {Sym}(Y)$ is closed we know that $\sigma \in \operatorname {Sym}(X)\wr \operatorname {Sym}(Y)$ , and thus there exist $\beta \in \operatorname {Sym}(Y)$ and $\alpha _y\in \operatorname {Sym}(X)$ such that for all $x\in X$ and $y\in Y$ we have $\sigma (x,y)=(\alpha _y(x),\beta (y))$ . We claim that the sequence $(\alpha _{n,y})_n$ converges to $\alpha _y$ for all $y\in Y$ . Indeed let F be an arbitrary finite subset of X. Then since $(\sigma _n)_n\rightarrow \sigma $ it follows that we can choose an index $n\in \omega $ such that for all $i\geq n$ we have $\sigma |_{F\times \{y\}}=(g_i)|_{F\times \{y\}}$ for all $y\in Y$ , and thus $\alpha _{i,y}|_F=\alpha _y|_F$ . We have shown that $(\alpha _{n,y})_n$ converges to $\alpha _y$ , and thus $\alpha _y\in G$ since G is closed. This shows that $\sigma \in G\wr \operatorname {Sym}(Y)$ .

(2) Let $n\in \omega $ , and let Z be any n-element subset of Y. Note that in this case every n-orbit of $G\wr \operatorname {Sym}(Y)$ contains a tuple in $X\times Z$ . Indeed, if $\vec {a}\in (X\times Y)^n$ then $\vec {a}\in (X\times Z')^n$ for some $Z'\subset Y$ with $|Z'|\leq n$ . Let $\beta \in \operatorname {Sym}(Y)$ such that $\beta (Z')\subset Z$ . Then the map $(x,y)\mapsto (x,\beta (y))$ , which is contained in $\{\operatorname {id}(X)\}\wr \operatorname {Sym}(Y)$ , maps $\vec {a}$ into $(X\times Z)^n$ . Let us also observe that $(G\wr \operatorname {Sym}(Y))_{(Z)}=G\wr \operatorname {Sym}(Z)$ . In particular two n-tuples from $X\times Z$ are contained in the same orbit of $G\wr \operatorname {Sym}(Z)$ if and only if they are contained in the same orbit of $G\wr \operatorname {Sym}(Y)$ . From this it follows that $o_n(G\wr \operatorname {Sym}(Y))=o_n(G\wr \operatorname {Sym}(Z))$ . We have already seen that the group $G^{Z}$ is oligomorphic. On the other hand $G\wr \operatorname {Sym}(Z)$ contains $G^Z$ , and thus by Observation 2.3 it follows that $G\wr \operatorname {Sym}(Z)$ is oligomorphic. In particular $o_n(G\wr \operatorname {Sym}(Z))=o_n(G\wr \operatorname {Sym}(Z))$ is finite. Since this holds for all $n\in \omega $ it follows that $G\wr \operatorname {Sym}(Y)$ is oligomorphic.

Notation 2.5. For a structure $\mathfrak {A}$ we denote by $\Sigma (\mathfrak {A})$ the signature of $\mathfrak {A}$ .

Notation 2.6. Let $\mathfrak {A}$ be a relational structure. Then for a relation R on $\mathfrak {A}$ we write $\operatorname {ar}(R)$ for the arity of R. We denote by $m(\mathfrak {A})$ the maximal arity of relations of $\mathfrak {A}$ . $m(\mathfrak {A})$ is defined to be $\infty $ if such a maximum does not exist.

Proof Let us assume that $\mathfrak {A}$ and $\mathfrak {B}$ are bidefinable, and let $\mathfrak {B}'$ and $\mathfrak {A}'$ be structures and let $i\colon A\rightarrow B$ be a bijection witnessing this. Then by definition $\mathfrak {A}$ and $\mathfrak {B}'$ are isomorphic, and we claim that $\mathfrak {B}$ and $\mathfrak {B}'$ are interdefinable. To see this, it suffices to show that $\mathfrak {B}$ is a reduct of $\mathfrak {B}'$ . But this follows from the fact that $\mathfrak {A}'$ is a reduct of $\mathfrak {A}$ , and i defines an isomorphism from $\mathfrak {A}'$ to $\mathfrak {B}$ and from $\mathfrak {A}$ to $\mathfrak {B}'$ . This finishes the proof of the “only if” direction of the lemma.

Now let us assume that $\mathfrak {B}$ and $\mathfrak {B}'$ are interdefinable, and $\mathfrak {B}'$ and $\mathfrak {A}$ are isomorphic. Let i be an isomorphism from $\mathfrak {A}$ to $\mathfrak {B}'$ . Let us define the structure $\mathfrak {A}'$ as follows. The domain set of $\mathfrak {A}'$ is A, $\Sigma (\mathfrak {A}')=\Sigma (\mathfrak {B})$ , and every symbol in $\Sigma (\mathfrak {B})$ is realized in $\mathfrak {A}'$ in such a way that i defines an isomorphism from $\mathfrak {A}'$ to $\mathfrak {B}$ . By definition $\mathfrak {B}'$ is a reduct of $\mathfrak {B}$ . It remains to show that $\mathfrak {A}'$ is also a reduct of $\mathfrak {A}$ . By definition $\mathfrak {B}$ is also a reduct of $\mathfrak {B}'$ . Since $i^{-1}$ defines an isomorphism from $\mathfrak {B}'$ to $\mathfrak {A}$ and from $\mathfrak {B}$ to $\mathfrak {A}'$ , we obtain that $i^{-1}(\mathfrak {B})=\mathfrak {A}'$ is a reduct of $i^{-1}(\mathfrak {B}')=\mathfrak {A}$ . Thus $\mathfrak {A}$ and $\mathfrak {B}$ are bidefinable.

Proof We can assume without loss of generality that $\mathfrak {A}$ is $\omega $ -categorical.

Suppose first that $\mathfrak {A}$ and $\mathfrak {B}$ are bidefinable. Then by Lemma 2.9 we know that $\mathfrak {B}$ is interdefinable with some structure $\mathfrak {B}'$ which is isomorphic to $\mathfrak {A}$ . Then $\operatorname {Aut}(\mathfrak {B})=\operatorname {Aut}(\mathfrak {B}')$ and it is isomorphic to $\operatorname {Aut}(\mathfrak {A})$ .

For the other direction let i be a bijection from A to B such that i induces an isomorphism from $\operatorname {Aut}(\mathfrak {A})$ to $\operatorname {Aut}(\mathfrak {B})$ . Let $\mathfrak {A}':=i^{-1}(\mathfrak {B})$ . Then we have $\operatorname {Aut}(\mathfrak {A}')=\iota (i^{-1})\operatorname {Aut}(\mathfrak {B})=\iota (i^{-1})(\iota (i)(\operatorname {Aut}(\mathfrak {A})))=\operatorname {Aut}(\mathfrak {A})$ . Theorem 2.13 then implies that $\mathfrak {A}$ and $\mathfrak {A}'$ are interdefinable. On the other hand it follows from the definition that $\mathfrak {A}'$ is isomorphic to $\mathfrak {B}$ . Therefore by Lemma 2.9 the structures $\mathfrak {A}$ and $\mathfrak {B}$ are bidefinable.

Fact 2.16. Every homogeneous structure with a finite relational signature is $\omega $ -categorical.

Proof Let $\mathfrak {A}$ be a homogeneous structure with a finite relational signature $\tau $ . We show that $\operatorname {Aut}(\mathfrak {A})$ is oligomorphic. Since $\tau $ is finite, and only contains relational and constant symbols, it follows that every finitely generated substructure is finite. Then using the finiteness of $\tau $ again it follows that for all $n\in \omega $ there exist only finitely many $\tau $ -structures up to isomorphism. In particular $\mathfrak {A}$ has finitely many quantifier-free types. It follows from the homogeneity of $\mathfrak {A}$ that if two n-tuples have the same quantifier-free type then they are in the same orbit of $\operatorname {Aut}(\mathfrak {A})$ . This implies that $o_n(\mathfrak {A})$ is finite. Therefore $\operatorname {Aut}(\mathfrak {A})$ is oligomorphic.

Proof The equivalence of items (1) and (2) is Proposition 6.36 in [Reference Bodirsky and Bodor11]. For the direction (1) $\rightarrow $ (3) we only need to show that item (1) implies that every reduct of $\mathfrak {A}$ is interdefinable with a structure with a finite relational signature. This follows for instance from Lemma 6.35 in [Reference Bodirsky and Bodor11].

In order to show the implication (3) $\rightarrow $ (1) it is enough to show that $\mathfrak {A}$ has finitely many reducts $\mathfrak {B}$ with $m(\mathfrak {B})\leq N$ up to interdefinability. Let R be a relation of $\mathfrak {B}$ with arity at most N. Then R is interdefinable with the relation $\{(x_1,\dots ,x_N)\colon (x_1,\dots ,x_{\operatorname {ar}(R)})\in R\}$ . Therefore we can assume without loss of generality that every relation of $\mathfrak {B}$ is N-ary. Since $\mathfrak {B}$ is a reduct of $\mathfrak {A}$ it follows that every N-ary relation of $\mathfrak {B}$ is a union of some orbits of $\mathfrak {A}$ . Therefore there are at most $2^{o_N(\mathfrak {A})}$ many possible choices for an N-ary relation of $\mathfrak {B}$ , and hence there are at most $2^{2^{o_N(\mathfrak {A})}}$ many choices for reducts $\mathfrak {B}$ of $\mathfrak {A}$ with $m(\mathfrak {B})\leq N$ . Since $\mathfrak {A}$ is $\omega $ -categorical we know that $o_N(\mathfrak {A})$ is finite, and hence $2^{2^{o_N(\mathfrak {A})}}$ is also finite.

The equivalences (1) $\leftrightarrow $ (4) and (2) $\leftrightarrow $ (5) follow from Theorem 2.13 and Lemma 2.14.

Notation 2.19. For a tuple $\vec {a}\in X^n$ we denote by $a_i$ the ith coordinate of $\vec {a}$ . For a function $\pi \colon k\rightarrow n$ we denote by $\vec {a}_{\pi }$ the tuple $(a_{\pi (0)},\dots ,a_{\pi (k-1)})$ .

Notation 2.20. Let $\mathfrak {A}$ be a relational structure, and let $m\in \omega $ . Then we denote by $\Delta _m(\mathfrak {A})$ the structure whose domain is A, and whose relations are all subsets of $A^m$ which are definable in $\mathfrak {A}$ .

Proof The implication (2) $\rightarrow $ (1) is obvious. The converse follows from the fact that if $m'\leq m$ then every $m'$ -ary relation of $\mathfrak {A}'$ is quantifier-free definable from an m-ary relation of $\mathfrak {A}'$ .

The implication (3) $\rightarrow $ (2) is obvious. For the converse implication let us assume that item (2) holds, and let $\mathfrak {A}'$ be a structure as it is dictated in the condition. Then $\Delta _m(\mathfrak {A})=\Delta _m(\mathfrak {A}')$ . Since $\mathfrak {A}$ , and thus also $\mathfrak {A}'$ is $\omega $ -categorical it follows that every definable relation of $\mathfrak {A}'$ of arity m is a finite union of some m-orbits of $\mathfrak {A}'$ . Therefore $\mathfrak {A}'$ is quantifier-free definable in $\Delta _m(\mathfrak {A}')$ . Since $\Delta _m(\mathfrak {A}')$ is reduct of $\mathfrak {A}'$ and $\mathfrak {A}'$ is homogeneous it follows that $\Delta _m(\mathfrak {A}')$ is also quantifier-free definable in $\mathfrak {A}'$ . Therefore $\Delta _m(\mathfrak {A}')$ is interdefinable with $\mathfrak {A}$ , and it is homogeneous.

(3) $\rightarrow $ (4). Let us assume that condition (3) holds. We can assume without loss of generality that $\mathfrak {A}=\Delta _m(\mathfrak {A})$ . We show the contrapositive of the statement in item (4). Let us assume that for all $\pi \colon m\rightarrow n$ we have $\vec {u}_{\pi }\approx _{\mathfrak {A}}\vec {v}_{\pi }$ . Then by definition we have $R(\vec {u}_{\pi })\Leftrightarrow R(\vec {v}_{\pi })$ for all $\pi \colon m\rightarrow n$ and for all relations R of $\mathfrak {A}$ with arity m. Since $\mathfrak {A}=\Delta _m(\mathfrak {A})$ we know that the arity of every relation of $\mathfrak {A}$ is exactly m. This means that the tuples $\vec {u}$ and $\vec {v}$ have the same atomic type over $\mathfrak {A}$ . Since $\mathfrak {A}$ is homogeneous, it has quantifier elimination, and hence $\vec {u}$ and $\vec {v}$ have the same type over $\mathfrak {A}$ .

(4) $\rightarrow $ (3). We show the contrapositive. Let us assume that item (3) does not hold. This means that $\mathfrak {A}$ has a relation R which is not quantifier-free definable over $\Delta _m(\mathfrak {A})$ . Then there exist tuples $\vec {u}\in R, \vec {v}\in \bar {R}$ such that for every relation S of $\Delta _m(\mathfrak {A})$ , and $\pi \colon m\rightarrow n$ we have $\vec {u}_{\pi }\in S$ if and only if $\vec {v}_{\pi }\in S$ . By definition this implies that for all $\pi \colon m\rightarrow n$ the tuples $\vec {u}_{\pi }$ and $\vec {v}_{\pi }$ have the same type over $\mathfrak {A}$ . Thus the tuples $\vec {u}$ and $\vec {v}$ witness that Condition (4) fails.

Notation 2.24. Let $\mathcal {C}$ be a class of finite $\tau $ -structures. Then:

• • We denote by ${\mathcal {C}}^{c}$ the class of those finite $\tau $ -structures which are not in $\mathcal {C}$ .

• • We denote by $\operatorname {Min}(\mathcal {C})$ the class of minimal structures in $\mathcal {C}$ , i.e., the class of those structures in $\mathcal {C}$ which do not have a proper substructure which is also in $\mathcal {C}$ .

Proof The equivalence of conditions (2) and (3) is clear since $\tau $ is finite.

(2) $\rightarrow $ (1). Let $\mathcal {F}$ be a finite subset of $\operatorname {Min}({\operatorname {Age}(\mathfrak {A})}^{c})$ so that every structure in $\operatorname {Min}({\operatorname {Age}(\mathfrak {A})}^{c})$ is represented in $\mathcal {F}$ . We claim that $\operatorname {Age}(\mathfrak {A})=\operatorname {Forb}(\mathcal {F})$ . Let us suppose that $\mathfrak {B}$ is a finite $\tau $ -structure not contained in $\operatorname {Forb}(\mathcal {F})$ . Then $\mathfrak {B}$ has a substructure $\mathfrak {B}'$ which is isomorphic to some structure in $\mathcal {F}$ . In particular $\mathfrak {B}'\not \in \operatorname {Age}(\mathfrak {A})$ . Since $\operatorname {Age}(\mathfrak {A})$ is closed under taking substructures it follows that $\mathfrak {B}\not \in \operatorname {Age}(\mathfrak {A})$ . For the other direction let us assume that $\mathfrak {B}\not \in \operatorname {Age}(\mathfrak {A})$ . Then $\mathfrak {B}$ contains a minimal structure with this property, say $\mathfrak {B}'$ . Then by definition $\mathfrak {B}'$ is isomorphic to some structure in $\mathfrak {C}\in \mathcal {F}$ , and hence $\mathfrak {C}$ embeds into $\mathfrak {B}$ , that is $\mathfrak {B}\not \in \operatorname {Forb}(\mathcal {F})$ .

(1) $\rightarrow $ (3). Let us assume that $\operatorname {Age}(\mathfrak {A})=\operatorname {Forb}(\mathcal {F})$ for some finite set $\mathcal {F}$ of $\tau $ -structures, and let m be the maximum of the sizes of structures in $\mathcal {F}$ . Let $\mathfrak {B}\in \operatorname {Min}({\operatorname {Age}(\mathfrak {A})}^{c})$ . Since $\mathfrak {B}\not \in \operatorname {Age}(\mathfrak {A})=\operatorname {Forb}(\mathcal {F})$ it follows $\mathfrak {B}$ has a substructure $\mathfrak {B}'$ which is isomorphic to some $\mathfrak {C}\in \mathcal {F}$ . But in this case $\mathfrak {B}'$ is also not in $\operatorname {Age}(\mathfrak {A})$ . Then the minimality of $\mathfrak {B}$ implies that in fact $\mathfrak {B}=\mathfrak {B}'$ , and hence $|\mathfrak {B}|\leq m$ .

Notation 2.26. For a structure $\mathfrak {A}$ we denote by $b(\mathfrak {A})$ the maximum of $m(\mathfrak {A})$ (see Notation 2.6) and the maximum of sizes of structures in $\operatorname {Min}({\operatorname {Age}(\mathfrak {A})}^{c})$ . $b(\mathfrak {A})$ is defined to be $\infty $ if such a maximum does not exist.

Notation 2.27. We denote by $\mathcal {FBH}$ the class of those structures which are interdefinable with a finitely bounded homogeneous relational structure.

Proof of Lemma 2.32

Item (1) follows from a straightforward induction using Lemma 3.1 in [Reference Bodirsky8].

As for item (2) let us assume that $\operatorname {Age}(\mathfrak {A})=\operatorname {Forb}(\mathcal {F}_i)$ for some finite sets of structures $\mathcal {F}_i$ for $1\leq i\leq n$ . For a $\Sigma (\mathfrak {A}_i)$ -structure $\mathfrak {B}$ let $(\mathfrak {B};U_i)$ denote the structure $\mathfrak {B}$ expanded by a unary relation denoted by $U_i$ which contains all elements of $B_i$ . Let $\mathcal {G}$ be the set of one-element $\{U_1,\dots ,U_n\}$ -structures $\mathfrak {B}$ such that either $\mathfrak {B}^{U_i}=\emptyset $ for all i or $\mathfrak {B}^{U_i}=\mathfrak {B}^{U_j}=B$ for some $i\neq j$ . Then it follows from the construction that

$$\begin{align*}\operatorname{Age}(\mathfrak{A})=\operatorname{Forb}(\{(\mathfrak{B};U_i)\colon 1\leq i\leq n, \mathfrak{B}\in \mathcal{F}_i\}\cup \mathcal{G}).\end{align*}$$

Therefore $\mathfrak {A}$ is finitely bounded.

Proof of Lemma 2.33

Let us assume that the tuples

$$\begin{align*}\vec{u}=((a_1,m_1),\dots,(a_k,m_k)),\vec{v}=((b_1,n_1),\dots,(b_k,n_k))\end{align*}$$

have the same quantifier-free type over $\mathfrak {A}\wr \omega $ . We have to show that $\vec {u}$ and $\vec {v}$ are in the same orbit of $\operatorname {Aut}(\mathfrak {A}\wr \omega )=\operatorname {Aut}(\mathfrak {A})\wr \operatorname {Sym}(\omega )$ . For an $n\in \omega $ let us denote by $\vec {u}_n$ ( $\vec {v}_n$ ) the subtuple of $\vec {u}$ ( $\vec {v}$ ) containing those elements whose second coordinate is n. By definition it follows that $m_i=m_j$ iff $n_i=n_j$ for all $1\leq i,j\leq k$ . Thus there exists a permutation $\sigma \in \operatorname {Sym}(\omega )$ such that for $\vec {u}_m:=((a_{i_1},m),\dots ,a_{i_k},m))$ we have $n_{i_1}=\dots =n_{i_k}=\sigma (m)$ . Moreover by the definition of $\mathfrak {A}\wr \omega $ it follows that the tuples $(a_{i_1},\dots ,a_{i_k})$ and $(b_{i_1},\dots ,b_{i_k})$ have the same quantifier-free type in $\mathfrak {A}$ . Thus by the homogeneity of $\mathfrak {A}$ it follows that there exists some $\alpha _m\in \operatorname {Aut}(\mathfrak {A})$ such that $\alpha _m(a_{i_j})=b_{i_j}$ for all $j=1,\dots ,k$ . Now let $\gamma \in \operatorname {Sym}(A\times \omega )$ be defined as . Then $\gamma \in \operatorname {Aut}(\mathfrak {A})\wr \operatorname {Sym}(\omega )$ and by our construction $\gamma (a_i,m_i)=(\alpha _{m_i}(a_i),\beta (m_i))=(b_i,n_i)$ for all $1\leq i\leq k$ .

For the second part of the lemma let us assume that $\operatorname {Age}(\mathfrak {A})=\operatorname {Forb}(\mathcal {F})$ for some finite set of structures $\mathcal {F}$ . Let $\mathcal {G}_0$ be a finite set of structures in the signature $\{E\}$ so that $\mathfrak {A}\in \operatorname {Forb}(\mathcal {G}_0)$ if and only if $E^{\mathfrak {A}}$ is an equivalence relation on $\mathfrak {A}$ . We can do this by listing all three-element $\{E\}$ -structures where E does not define an equivalence relation. For a $\Sigma (\mathfrak {A})$ -structure $\mathfrak {C}$ let $\mathfrak {C}^*$ denote the structure $\mathfrak {C}$ expanded by a binary relation, denoted by E which contains every pair in $C^2$ . Let

$$\begin{align*}\mathcal{G}:=\mathcal{G}_0\cup \{\mathfrak{C}^*\colon \mathfrak{C}\in \mathcal{F}\}.\end{align*}$$

Then it is straightforward to check that $\operatorname {Forb}(\mathcal {G})$ is exactly the age of $\mathfrak {A}\wr \omega $ . Therefore $\mathfrak {A}\wr \omega $ is finitely bounded.

Proof Let and . Then is an $\omega $ -partition of G with all components isomorphic to H.

Proof For $i=1,\dots ,k$ let be the $\omega $ -partition of $H_i$ witnessing the fact that $H_i\in \mathcal {H}_n$ , that is, every component of $P_i$ is in $\mathcal {H}_{n-1}$ . Let with and . We claim that P is an $\omega $ -partition of H, and every component of P is a component of one of the $\omega $ -partitions $P_i$ with $i=1,\dots ,k$ . Since these components are all contained in $\mathcal {H}_{n-1}$ by our assumption, this implies the statement of the lemma.

It follows from the definition that H fixes the sets , and $H|_{X_i}=H_i$ . This implies immediately that H fixes the set K and the equivalence relations $\Delta $ and $\nabla $ . It is clear from the definition that K is finite, $\nabla $ has finitely many classes, and every $\nabla $ -class is a union of $\aleph _0$ many $\Delta $ -classes. It remains to show that item (5) of Definition 3.2 holds for the congruences $\nabla $ and $\Delta $ . Let C be a $\nabla $ -class. Then by definition C is also a $\nabla _i$ -class for some $1\leq i\leq k$ . Then we have

$$ \begin{align*}H_{((C))}/\Delta=H_{((C))}/\Delta_i=(H|_{X_i})_{((C))}/\Delta_i=&\\(H_i)_{((C))}/\Delta_i=&\operatorname{Sym}(C/\Delta_i)=\operatorname{Sym}(C/\Delta), \end{align*} $$

where the second-to-last equality above follows from the assumption that $P_i$ is an $\omega $ -partition. This finishes the proof that $P=(K,\nabla ,\Delta )$ is an $\omega $ -partition. Now if X is a $\Delta $ -class, then X is also a $\Delta _i$ -class for some $1\leq i\leq k$ , and thus $H_{((Y))}=(H|_{X_i})_{((Y))}=(H_i)_{((Y))}$ which is by definition a component of $P_i$ . This shows that every component of P is a component of $P_i$ for some $1\leq i \leq k$ .

Proof (i) Let $a\in Y_i$ . Then $\sigma $ maps $u=(a,n_i)$ to for $i'=i(\rho _{\vec {n}}(\sigma )(a))$ . By definition $p(\tau (u'))=\rho _{\psi (\sigma )(\vec {\mathbf {n}})}(\tau )(\rho _{\vec {n}}(\sigma )(a))$ . On the other hand $p(\tau (u'))=p(\tau \sigma (u))=\rho _{\vec {n}}(\tau \sigma )(a)$ . This shows the equality in item (i).

(ii) Using item (i) it follows that

$$\begin{align*}\operatorname{id}(Y)=\rho_{\vec{n}}(\operatorname{id})=\rho_{\vec{n}}(\sigma\sigma^{-1})=\rho_{\psi(\sigma^{-1})(\vec{\mathbf{n}})}(\sigma)\rho_{\vec{n}}(\sigma^{-1}),\end{align*}$$

and the equality (ii) follows.

Proof Let $\sigma ,\tau $ be permutations in $\operatorname {Sym}(\Omega (Y_0,\dots ,Y_k))$ satisfying conditions (a)–(d) in Definition 3.21. We have to show that the permutation $\sigma \tau ^{-1}$ also satisfies conditions (a)–(d). For conditions (a) and (b) this is obvious.

For condition (c) let $\vec {n}\in \{0\}\times \omega ^k$ be arbitrary. Then by Lemma 3.20 we have

$$\begin{align*}\rho_{\vec{n}}(\sigma\tau^{-1})=\rho_{\psi(\tau^{-1})(\vec{\mathbf{n}})}(\sigma)\rho_{\vec{n}}(\tau^{-1})=\rho_{\psi(\tau^{-1})(\vec{\mathbf{n}})}(\sigma)(\rho_{\psi(\tau^{-1})(\vec{\mathbf{n}})}(\tau))^{-1}\in HH^{-1}=H.\end{align*}$$

For condition (d) let $\vec {n},\vec {m}\in \{0\}\times \omega ^k$ . Then we have

$$ \begin{align*} \rho_{\vec{n}}(\sigma\tau^{-1})&= \rho_{\psi(\tau^{-1})(\vec{\mathbf{n}})}(\sigma)(\rho_{\psi(\tau^{-1})(\vec{\mathbf{n}})}(\tau))^{-1}\\ &\equiv\rho_{\psi(\tau^{-1})(\vec{\mathbf{m}})}(\sigma)(\rho_{\psi(\tau^{-1})(\vec{\mathbf{m}})}(\tau))^{-1}=\rho_{\vec{n}}(\sigma\tau^{-1}) \pmod{N}.\\[-34pt] \end{align*} $$

Proof Since K is closed, and $K\subsetneq G$ , we can pick a natural number n such that K has more n-orbits than G. We claim that the n-orbits of K give a G-invariant partition of X. Indeed, suppose that $\vec {u}$ and $\vec {v}$ are in the same n-orbit of K, that is $\vec {v}\in K(\vec {u})$ . Then for any $\alpha \in G$ we have

$$\begin{align*}\alpha(\vec{u})\in \alpha K(\vec{v})=K(\alpha(\vec{v})).\end{align*}$$

Since K is oligomorphic this partition has finitely many parts. Thus, the stabilizer of this partition is a closed finite-index normal subgroup of G. By the choice of n this stabilizer is different from G.

Proof For the ease of notation we identify S with the subgroup $\{\operatorname {id}(\operatorname {Dom}(K))\}\wr S$ .

Let G be a finite-index closed normal subgroup of $K\wr S$ . Then $G\geq S$ by our assumption on S. Let . We have to show that $Q=K^\omega $ . Note that Q is closed normal subgroup of $K^\omega $ , so S acts continuously on $K^\omega /Q$ by conjugation. Moreover, the index $|K^\omega :Q|=|L:G|$ is finite; therefore, the kernel of this action is a closed subgroup of S of finite index which must equal S by our assumption. This implies that for all $g\in S$ and $p\in K^\omega $ the commutator $[g,p]$ is contained in Q.

Now let $k\in K$ be arbitrary, and write $p=(k,\operatorname {id},\operatorname {id},\dots )$ . By the transitivity of S we know that for all $n\in \omega $ there exists some $g_n\in S$ such that $g_n(0)=n$ . Then $q_n=[g_n,p]=(k,\operatorname {id},\dots ,\operatorname {id},k^{-1},\operatorname {id},\dots )\in Q$ where the $k^{-1}$ appears in position n. The limit of the sequence $(q_n)_n$ as $n\rightarrow \infty $ is p. As Q is closed, we obtain $p\in Q$ . The elements $(k,\operatorname {id},\operatorname {id},\dots )$ with $k\in K$ , however, together with S, generate a dense subgroup of $K^\omega $ . Since Q is closed, this means that in fact $Q=K^\omega $ .

Proof Recall that

where $L_0=\{\operatorname {id}(Y_0\times \{0\})\}$ and $L_i=N_i\wr \operatorname {Sym}(\omega )$ for $1\leq i\geq k$ . Let

We have to show that $G_i=L_i$ for all $i\in \{0,\dots ,k\}$ . For $i=0$ there is nothing to prove. Otherwise, $G_i$ is a closed oligomorphic normal subgroup of $L_i$ , and the equality follows from Lemmas 3.23 and 3.24.

Proof Condition (d) says that for a permutation $\sigma \in \mathcal {G}(H;N_0,\dots ,N_k)$ all the permutations of the form $\rho _{\vec {n}}(\sigma )$ are the same modulo N. Then we define $\varphi (\sigma )$ to be the coset of N containing all these elements. Then $\varphi $ defines a homomorphism from $\mathcal {G}(H;N_0,\dots ,N_k)$ to $H/N$ , and the kernel of this homomorphism is $\mathcal {N}(N_0,\dots ,N_k)$ . Note also that $\varphi $ is surjective. Indeed, for an $h\in H$ let us denote by $h^*$ the permutation $(a,n)\mapsto (h(a),n)$ . Then clearly $\varphi (h^*)=h$ . The statement of the lemma then follows from the first isomorphism theorem.

Proof It is enough to show the statement of the lemma for $\vec {m}=\vec {0}=(0,\dots ,0)$ and any $\vec {n}=(n_0,\dots ,n_k)$ . By definition both $\rho _{\vec {0}}$ and $\rho _{\vec {n}}$ map $Y_i$ to $Y_{\phi (\sigma )(i)}$ . Therefore fixes the sets $Y_0,\dots ,Y_k$ setwise.

Now let . We have to show that $\tau _i\in N_i$ . If $n_i=0$ , then there is nothing to prove. Otherwise, let $\nu \in \{\operatorname {id}(\{(0,0)\})\}\times \operatorname {Sym}(\omega )^k$ which flips $(0,i)$ and $(n,i)$ and fixes everything else. Then we have $\hat {\nu }\in \mathcal {N}(N_0,\dots ,N_k)\subseteq G$ . Now let . Then $\mu $ fixes every element outside $Y_i\times \{0,n\}$ , and by Lemma 3.20 we have

$$ \begin{align*} \rho_0^i(\mu)&=\rho_0^i(\hat{\nu}^{-1}\sigma^{-1}\hat{\nu}\sigma)\\ &=(\rho_{\psi_i(\hat{\nu}^{-1}\sigma^{-1}\hat{\nu}\sigma)(0)}^i(\hat{\nu}))^{-1}(\rho_{\psi_i(\sigma^{-1}\hat{\nu}\sigma)(0)}^i(\sigma))^{-1}\rho_{\psi_i(\sigma)(0)}^i(\hat{\nu})\rho_0^i(\sigma)\\ &=(\rho_n^i(\sigma))^{-1}\rho_0^i(\sigma)=\tau_i. \end{align*} $$

Now by conjugating $\sigma '$ by the permutation $\widehat {\nu _{n'}}$ where $\nu _{n'}$ denotes the transposition $((i,n)\,(i,n'))$ we obtain that for all $n'\in \omega $ there exists a permutation $\mu ^{\prime }_{n'}$ in G such that $\rho _0^i(\mu _{n^{\prime \prime }})=\tau _i$ , and $\mu ^{\prime }_{n'}$ fixes every element outside $Y_i\times \{0,n'\}$ . Since G is closed this implies that G contains a permutation $\mu '$ such that $\rho _0^i(\mu ')=\tau $ and $\mu '$ fixes every element outside $Y_i\times \{0\}$ . Then the permutation $(\tau _i,\operatorname {id}(\{0\}))$ is contained in $G_{(((Y_i\times \{0\}))}=N_i\times \{\operatorname {id}(\{0\})\}$ , and thus $\tau _i\in N_i$ , and this is what we wanted to show.

Proof Let $\sigma \in N$ , $\tau \in H$ . We have to show that $\tau ^{-1}\sigma \tau \in N$ . Let $\sigma _0$ be the permutation of $\Omega (Y_0,\dots ,Y_k)$ which maps $(a,0)$ to $(\sigma (a),0)$ for all $a\in Y$ , and fixes every other element. Then by definition $\sigma _0\in \mathcal {N}(N_0,\dots ,N_k)\subset G$ . Let $\tilde {\tau }\in G$ so that $\rho _{\vec {0}}(\tilde {\tau })=\tau $ . Then by permuting the $\Delta $ classes of $\Omega (Y_0,\dots ,Y_k)$ with elements $\{\hat {\nu }: \nu \in \{\operatorname {id}(\{(0,0)\})\}\times \operatorname {Sym}(\omega )^k\}$ we can assume without loss of generality that $\psi (\tilde {\tau })$ fixes all $(0,i)$ with $i=0,\dots ,k$ . In this case the permutation $\sigma _0':=\tilde {\tau }^{-1}\sigma _0\tilde {\tau }\in G$ fixes every element outside $\bigcup _{i=0}^kY_i\times \{0\}$ , and by Lemma 3.20 we have

$$\begin{align*}\rho_{\vec{0}}(\sigma_0')=\rho_{\vec{0}}(\tilde{\tau}^{-1}\sigma_0\tilde{\tau})=(\rho_0(\tilde{\tau}))^{-1}\rho_0(\sigma_0)\rho_0(\tilde{\tau})=\tau^{-1}\sigma\tau. \end{align*}$$

This implies that $\tau ^{-1}\sigma \tau \in N$ .

Proof It follows from our conditions that the groups $G_1$ and $G_2$ both contain the group $\mathcal {N}(N_0,\dots ,N_k)=\{\operatorname {id}(Y_0\times \{0\})\}\times \prod _{i=1}^k(N_i\wr \operatorname {Sym}(\omega ))$ .

Lemma 3.27 implies that for all $\sigma \in G_1$ the coset $\rho _{\vec {n}}(\sigma )N$ does not depend on $\vec {n}$ . Let us denote this coset by $h(\sigma )$ . By Lemma 3.28 we know that N is a normal subgroup of $H_1$ . Then it follows easily from Lemma 3.20 that h defines a homomorphism from $G_1$ to $H_1/N$ , and the kernel of h is exactly $\mathcal {N}(N_0,\dots ,N_k)$ .

Using this if $\sigma \in G_1$ then by definition it follows that $\rho _{\vec {0}}(\sigma ')=\rho _{\vec {0}}(\sigma )$ for some $\sigma '\in G_2$ . This implies $h(\sigma )=h(\sigma ')$ , and thus $h(\sigma (\sigma ')^{-1})=h(\sigma )(h(\sigma '))^{-1}=\operatorname {id}$ , that is $\rho _{\vec {n}}(\sigma (\sigma ')^{-1})$ for all $\vec {n}\in \{0\}\times \omega ^k$ . This means exactly that $\sigma (\sigma ')^{-1}\in \mathcal {N}(N_0,\dots ,N_k)$ . Therefore $\sigma =(\sigma (\sigma ')^{-1})\sigma '\in \mathcal {N}(N_0,\dots ,N_k)G_2=G_2$ . Hence $G_1\subset G_2$ . The other inclusion follows analogously.

Proof Let . By Lemma 3.29 we already know that H normalizes N. Let $G_1=G, G_2=\mathcal {G}(H;N_0,\dots ,N_k)$ , and $H_2=\{\rho _{\vec {0}}(\sigma )\colon \sigma \in G_2\}$ . Then $H_2=H=H_1$ , and thus Lemma 3.29 implies $G=G_1=G_2=\mathcal {G}(H;N_0,\dots ,N_k)$ .

Proof We show that every group in $\mathcal {H}_n'$ is oligomorphic by induction on n. For $n=-1$ there is nothing to show. For the induction step let us assume that $G=\mathcal {G}(H;N_0,\dots ,N_k)$ for some $H,N_0,\dots ,N_k\in \mathcal {H}_{n-1}'$ . Then $G\geq \mathcal {N}(N_0,\dots ,N_k)$ . By item (ii) of Lemma 2.4 we know that the groups $N_i\wr \operatorname {Sym}(\omega ),\ 1\leq i \leq k$ are oligomorphic. By the discussion in Section 2.4 we know that finite direct products of oligomorphic groups are oligomorphic. Therefore, the group $\mathcal {N}(N_0,\dots ,N_k)=\{\operatorname {id}(Y_0\times \{0\})\}\times \prod _{i=1}^k(N_i\wr \operatorname {Sym}(\omega ))$ is oligomorphic. Hence G is also oligomorphic.

Proof We show the statement of the lemma for $\mathcal {H}_n'$ by induction on n. For $n=-1$ there is nothing to show. For the induction step let us assume that $G=\mathcal {G}(H;N_0,\dots ,N_k)$ for some $H,N_0,\dots ,N_k,N=\prod _{i=0}^kN_i\in \mathcal {H}_{n-1}'$ . By Lemma 3.26 it follows that $|G:\mathcal {N}(N_0,\dots ,N_k)|=|H:N|$ . By definition N is a normal subgroup of $H\in \mathcal {H}_{n-1}'$ . Since $N,H\in \mathcal {H}_{n-1}$ this implies that $|H:N|$ is finite by the induction hypothesis. Therefore $|G:\mathcal {N}(N_0,\dots ,N_k)|$ is finite. Now let $G_1\in \mathcal {H}'$ be a closed normal subgroup of G. By Lemma 3.33 we know that $G_1$ is oligomorphic. Let . Then by the second isomorphism theorem we obtain

$$ \begin{align*} G_1/G_2&=G_1/(G_1\cap \mathcal{N}(N_0,\dots,N_k))\\ &\simeq G_1(N_0,\dots,N_k)/\mathcal{N}(N_0,\dots,N_k)\leq G/\mathcal{N}(N_0,\dots,N_k). \end{align*} $$

In particular $G_1/G_2$ is finite. We know that a finite-index subgroup of an oligomorphic group is oligomorphic (see for instance Proposition 6.23 in [Reference Bodirsky and Bodor11]). Therefore $G_2$ is also oligomorphic. $G_2$ is also closed since it is an intersection of two closed groups. Thus by Lemma 3.25 it follows that $\mathcal {N}(N_0,\dots ,N_k)\leq G_2\leq G$ . Since $|G:\mathcal {N}(N_0,\dots ,N_k)|< \infty $ this implies both statements of the lemma.

Proof We show by induction on n that every group in $\mathcal {H}_n'$ is closed. For $n=-1$ there is nothing to prove. For the induction step let us assume that $G=\mathcal {G}(H;N_0,\dots ,N_k)$ for some $H,N_0,\dots ,N_k,N=\prod _{i=0}^kN_i\in \mathcal {H}_{n-1}'$ . By item (i) of Lemma 2.4 we know that the groups $N_i\wr \operatorname {Sym}(\omega )\colon 1\leq i \leq k$ are closed. Thus the direct product $\mathcal {N}(N_0,\dots ,N_k)=\operatorname {id}(Y_0\times \{0\})\times \prod _{i=1}^k(N_i\wr \operatorname {Sym}(\omega ))$ is also closed (see Section 2.4). Since the group $\mathcal {N}(N_0,\dots ,N_k)$ is contained in $\mathcal {H}_n'$ , and it is a normal subgroup of $\mathcal {G}(H;N_0,\dots ,N_k)$ it follows from Lemma 3.34 that $|G:\mathcal {N}(N_0,\dots ,N_k)|$ is finite. This means that G is a finite union of cosets of $\mathcal {N}(N_0,\dots ,N_k)$ , and thus a finite union of closed sets. Hence G is closed.

Proof We prove the lemma by induction on n. For $n=-1$ the statement is trivial. For the induction step let us assume that $G=\mathcal {G}(H;N_0,\dots ,N_k)$ for some $H,N_0,\dots ,N_k,N=\prod _{i=0}^kN_i\in \mathcal {H}_{n-1}'$ . By Corollary 3.18 and Lemma 3.19 it follows that $\prod _{i=1}^k(N_i\wr \operatorname {Sym}(\omega ))\times N_0=\mathcal {N}(N_0,\dots ,N_k)\in \mathcal {H}_n$ . Since $\mathcal {H}_n$ is closed under taking closed supergroups it follows that $\overline {G}\in \mathcal {H}_n$ . Finally we have $\overline {G}=G$ by Lemma 3.35.