Proof. Suppose $(X,G)$ is micro-supported, and let $W \subset X$ be an open set. Note that since a micro-supported action is effective by definition, then $X$ does not have isolated points, and so $W$ is not a singleton. Choose a non-empty open subset $V \subset W$ such that the topological closure $\overline V$ is properly contained in $W$. Then $W \backslash \overline{V}$ is open, and by the definition of a micro-supported action there exists $g \ne 1_G$ such that $g|X \backslash (W \backslash \overline{V}) = {\rm id}$. Then $g|V = {\rm id}$ and $g|W \ne {\rm id}$, since $g \ne 1_G$ and so it must act non-trivially on $W \backslash \overline{V}$.

Proof. Since ${\rm Conj}(H)$ is finite, there exists $g\in G$ such that $x\in \overline{X_H(g)}$. Since $G$ acts on $X$ by homeomorphisms, the continuity of the action implies that $gHg^{-1}\subseteq G_x$.

Proof. If there exists $s\in G$ such that $sHs^{-1} \subsetneq H$, a proper subgroup of $H$, then for $k\geq 1$,

\begin{equation*}s^kHs^{-k} \subsetneq s^{k-1}Hs^{-(k-1)},\end{equation*}

and we obtain a descending infinite sequence of proper subgroups conjugate to $H$. But this is not possible since $H$ is almost normal, and so has finitely many distinct conjugate subgroups.

Proof. By definition, $g\in lN_G(H)$ if and only if $X_H(g)=X_H(l)$. Thus, if $g\in lN_G(H)$ then $\overline{X_H(g)}=\overline{X_H(l)}$. This implies that $\{\overline{X_{H}(g_i)}: 1\leq i\leq m\}$ is a closed covering of $\overline{X_H}$. Since for every $g\in G$ we have $gX_H(1_G)=X_H(g)$, the continuity of the action implies $g\overline{X_{H}(1_G)} = \overline{X_H(g)}$, and (1) follows.

Statement (2) will be a consequence of the following claim:

Proof of claim 1

Let $x\in X$ be such that $\overline{O(x)}\cap X_H\neq \emptyset$. If $O(x) \cap \overline{X_H} \ne \emptyset$, then we are done, so assume that $\overline{X_H}$ contains an accumulation point of $O(x)$. This does not imply that $O(x)$ intersects $\overline{X_H}$, so we cannot use lemma 3.1. Instead, we are going to find a point $y \in \overline{O(x)}\cap X_H$ such that $G_y$ contains $gHg^{-1} \cup lHl^{-1}$, for some $g,l \in G$. Then, since $y \in X_H$, there is $m \in G$ such that $(gHg^{-1}\cup lHl^{-1} )\subset m H m^{-1}$, and lemma 3.2 implies that $lHl^{-1} = mHm^{-1} = gHg^{-1}$.

Since $N_G(H)$ is a finite-index subgroup of $G$, its core $\Gamma=\bigcap_{g\in G}gN_G(H)g^{-1}$ is a normal finite-index subgroup of $G$. Since $\Gamma$ is a finite-index subgroup of $G$, for $y\in \overline{O(x)}\cap X_H$ there exists $d\in G$ such that for every open neighbourhood $V$ of $y$, there exists $\gamma_V\in\Gamma$ such that $d\gamma_Vx\in V$. By continuity of the action, this implies that for every neighbourhood $U$ of $d^{-1}y$, there exists $\gamma_U\in\Gamma$ such that

(3.1)\begin{equation} \gamma_Ux\in U. \end{equation}

Let $g, l \in G$ be such that $gHg^{-1} \cup l H l^{-1}\subseteq G_x$. For every $h\in H$ and every open neighbourhood $W$ of $ghg^{-1}d^{-1}y$, there exists a neighbourhood $U_W$ of $d^{-1}y$ and $\gamma_{U_W} \in \Gamma$ such that

(3.2)\begin{equation} ghg^{-1}\gamma_{U_W}x\in W. \end{equation}

Observe that if $d^{-1}y\neq ghg^{-1}d^{-1}y$, then we can take $W\cap U_{W}=\emptyset$.

Due to $\Gamma \subseteq N_G(gHg^{-1})$, for every $W$ we have

\begin{equation*} \gamma_{U_W}^{-1}ghg^{-1}\gamma_{U_W}x=x, \end{equation*}

since $g H g^{-1} \subset G_x$ by assumption. From this we get

\begin{equation*} ghg^{-1}\gamma_{U_W}x=\gamma_{U_W}(\gamma_{U_W}^{-1}ghg^{-1}\gamma_{U_W})x=\gamma_{U_W}x. \end{equation*}

Thus, from Eqs (3.1) and (3.2), we obtain

\begin{equation*} ghg^{-1}d^{-1}y=d^{-1}y, \end{equation*}

so $gHg^{-1} \subset G_{d^{-1}y}$. Repeating the above argument for $l H l^{-1}$ and using the fact that $\Gamma$ also normalizes $lHl^{-1}$, we obtain $lHl^{-1} \subseteq G_{d^{-1}y}$. Thus $d^{-1}g Hg^{-1}d \cup d^{-1}l H l^{-1}d \subset G_y$, where $y \in X_H$, and the argument at the beginning of the proof shows claim 1.

Since by lemma 3.1 $y\in \overline{X_{H}(g_i)}$ implies $g_iHg_i^{-1}\subseteq G_y$, by claim 1 we deduce that if $y\in \overline{X_{H}(g_i)}\cap \overline{X_{H}(g_j)}$, for some $1\leq i\neq j \leq m$, then $\overline{O(y)}\cap X_H=\emptyset$, and so $E \subset X_H^c$, the complement of $X_H$ in $X$. The set $E$ is closed since it is a finite intersection of closed sets, and it is invariant since the stabilizers of points in the same orbit are conjugate. This shows statement (2).

For (3), let $x\in X_H$, and let $1\leq j\leq m$ be such that $G_x=g_jHg_j^{-1}$. Choosing an open neighbourhood $V$ of $x$ small enough, we can arrange that for every $y\in V$, the group $g_jHg_j^{-1}$ is contained in $G_y$. Indeed, if such $V$ does not exist, then by lemma 3.1 we should have $G_x = g_jHg_j^{-1} \supset g_i H g_i^{-1}$ for $g_i \ne g_j$, which is not possible by lemma 3.2. Thus the action of every $h \in g_jHg_j^{-1}$ fixes $V$, which shows that $x$ has trivial holonomy.

Now, suppose that $x\in X$ has trivial holonomy. Since $X=\overline{X_H}$, there exists $1\leq i \leq m$ such that $x\in \overline{X_H(g_i)}$. This implies that $g_iHg_i^{-1}\subseteq G_x$, and we must show the reverse inclusion. For that, by an argument similar to the one in the previous paragraph, we can find an open neighbourhood $V$ of $x$ such that $X_H \cap V = X_H(g_i) \cap V$, i.e. for any $y \in V$ we have $G_y = g_iHg_i^{-1}$. Let $g \in G_x$. Since $x$ has trivial holonomy, there is an open neighbourhood $W_g \subset V$ such that $g$ fixes every point in $W_g$, and in particular the points in $W_g \cap X_H(g_i)$. This shows that $G_x \subset g_i H g_i^{-1}$, and $x \in X_H$.

To show (4), note that if $(X,G)$ is minimal, then $X = \overline{X_H}$, and by lemma 3.1 for any $y \in X$ the stabilizer $G_y$ contains at most one distinct conjugate of $H$. It follows that $E$ must be empty.

Next, by (3) $X_H$ coincides with the set of points with trivial holonomy of $(X,G)$, and then (4) follows by the result of Glasner and Weiss [Reference Glasner and Weiss13], see also [Reference Boudec and Bon6, Lemma 2.2, Proposition 2.5], since $(X,G)$ is minimal.

Proof. Since $X_H$ is invariant, the minimality of $(X,G)$ implies that $X=\overline{X_H}$. Theorem 3.3, (2), implies that $E=\emptyset$, and the collection ${\mathcal C}$ is a clopen partition of $X$. Theorem 3.3, (1), implies that the map $\phi:X\to G/N_G(H)$ is well defined, continuous, surjective, and commutes with the respective actions.

Proof. Suppose there exists a finite subgroup $K\in {\rm Stab}(X)$ (see §2.3 for the definitions of the space of subgroups $S(G)$ and the map ${\rm Stab}:X\to S(G)$). Then there exists a finite group $H\in \overline{{\rm Stab}(X)}$ with minimal cardinality (that is, $|H|\leq |H'|$ for every $H'\in \overline{{\rm Stab}(X)}$). Consider the set $A=\overline{\{gHg^{-1}: g\in G\}}\subseteq \overline{{\rm Stab}(X)}$. The definition of the Chabauty–Fell topology for countable groups implies that if $L\in A$, then for every finite set $F\subseteq L$ there exists $g\in G$ such that $gHg^{-1}\cap F=L\cap F=F$, which implies that $|L|\leq |H|$. Because the cardinality of $H$ is minimal, it follows that if $L\in A$ then $|L|=|H|$, and hence $L=gHg^{-1}$ for some $g\in G$. Thus $A$ is the collection of conjugates of $H$, that is, a single orbit in $S(G)$. Since $H$ is finite and $A$ is compact, $A$ must be finite. Indeed, if $A$ were infinite, there would exist $L\in A$ which is an accumulation point of a sequence $(g_iHg_i^{-1})_{i\in\mathbb{N}}$ with $g_iHg_i^{-1}\neq g_jHg_j^{-1}$ for every $i\neq j$. Since $|L|=|H|$, there exists $i_0$ such that $g_iHg_i^{-1}=L$ for every $i\geq i_0$, which is a contradiction. Since $A$, the orbit of $H$ in $S(G)$, is finite, it follows that $H$ is almost normal, and its orbit is a minimal subset of $\overline{{\rm Stab}(X)}$. By uniqueness of the stabilizer URS for minimal systems, it follows that $A={\rm Conj}(H)$ is the stabilizer URS of $(X,G)$. Thus (1) holds if and only if (2) holds. Statement (3) is equivalent to (2) by the definition of a URS, and (4) follows from (3) since the set of points without holonomy is residual (and therefore non-empty) in $X$.

Proof. We show that (1) holds if and only if (3) holds. By theorem 3.3, the space $X$ is partitioned into a finite number of clopen sets $X_i:=\overline{X_H(g_i)}$, where $g_1 = 1_G,g_2,\ldots, g_n$ are representatives of the cosets of $N_G(H)$. Fix $\epsilon \lt {\rm min}\{{\rm dist}(X_i,X_j): 1 \leq i \ne j \leq n\}$. Then an open set $U$ of diameter less than $\epsilon$ is contained in $X_i$ for some $1 \leq i \leq n$. Let $g \in G$ and $V \subset U$ be such that $g|V = {\rm id}$, then for any $x \in V$ we have $g \in G_x$. Since the points in $X_H(g_i)$ have equal stabilizers, we get $g \in G_x$ for $x \in U \cap X_H(g_i)$. Since the latter is dense in $U$, we obtain $g \in G_x$ for all $x \in U$, and so $g|U = {\rm id}$. This shows that the action $(X,G)$ is LQA.

For the converse, suppose that $(X,G)$ is LQA, and let $\epsilon \gt 0$ be such that for any open $U \subset X$ with ${\rm diam}(U) \lt \epsilon$, and any $g \in G$, if $g|V = {\rm id}$ for some open $V \subset U$, then $g|U = {\rm id}$. Fix such $U$. Recall that if $x \in U$ is a point without holonomy, and $g \in G_x$, then $g$ fixes an open neighbourhood of $x$ in $U$, and therefore must fix every point in $U$. It follows that all points without holonomy in $U$ have equal stabilizers. Fix one such point without holonomy $x \in U$ and set $H: = G_x$. Next, consider a covering $\mathcal U = \{gU: g \in G\}$ of $X$ by open sets, and choose a finite subcover $\mathcal U' = \{g_iU: 1 \leq i \leq n\}$. Since the orbit of $x$ is dense in $X$, all points without holonomy in $g_i U$ have stabilizers equal to $g_i H g_i^{-1}$, and if $g_iU \cap g_jU \ne \emptyset$, then $g_i H g_i^{-1} = g_jH g_j^{-1}$ by lemma 3.2. Since $\mathcal U'$ is finite, $H$ is almost normal.

The equivalence between (2) and (1) follows from the definition of the stabilizer URS.

Proof of theorem 1.4

The equivalence of statements (1)–(3) is given by proposition 3.6; the equivalence of statements (1)–(3) to (4) is given by theorem 3.7 plus the assumption that there is a point with finite stabilizer.

Proof. Let ${\rm Stab}: X \to S(G)$ be the map introduced in §2.3, and let $\nu={\rm Stab}_*\mu$. If $D\subseteq G$ is a set containing exactly one representing element of each class in $\{gN_G(H): g\in G\}$, then $\{G_x: x\in X_H\}=\bigcup_{g\in D}gHg^{-1}$. Since this union is disjoint, we have

\begin{equation*} 0 \lt \mu(X_H)\leq \nu(\{G_x: x\in X_H\})=\nu(\{gHg^{-1}: g\in D\})=|D|\nu(\{H\}), \end{equation*}

which implies that $|D|=[G:N_G(H)] \lt \infty.$

Proof. Assume (1), and observe that for every $n\geq 1$, the set $\{x\in Y: |G_x|=n\}$ is Borel and invariant. Thus, since

\begin{equation*}\{x\in X: |G_x| \lt \infty \}=\bigcup_{n\geq 1}\{x\in X: |G_x|=n\},\end{equation*}

the ergodicity of $\mu$ implies there exists a unique $n\geq 1$ such that

\begin{equation*}\mu(\{x\in X: |G_x|=n\})=1.\end{equation*}

Since the collection of finite subsets of a countable set is countable (we can think of the subsets of $G$ with cardinality $k$ as elements of $G^k$), we can write the set of all $x\in X$ such that $|G_x|\leq n$, as a disjoint union of sets $\{x\in X: G_x\in {\rm Conj}(H_i)\}$, where $\{H_i: i\in I\}$ is a countable (or finite) collection of subgroups of $G$ with cardinality less than or equal to $n$. Furthermore, since for every $i\in I$ the set $ \{x\in X: G_x\in {\rm Conj}(H_i)\}$ is invariant and Borel, the ergodicity of $\mu$ implies there exists $i\in I$ such that $\mu(X_{H_i})=1$, and by lemma 3.9 $H_i$ is almost normal. Thus (2) holds if and only if (1) holds.

Assume (2), so there exists a finite subgroup $H$ of $G$ such that $\mu(X_H)=1$. Then lemma 3.9 implies that $H$ is almost normal. Thus, the orbit of $H$ in ${\rm Stab}(X)$ is finite, and because ${\rm Stab}^{-1}\{{\rm Conj}(H)\}=X_H$, we deduce that ${\rm Stab}_*\mu({\rm Conj}(H))=1$, which implies that ${\rm Stab}_*\mu(\{H\}) \gt 0$, i.e. (3).

Conversely, if there exists a finite group $H$ of $G$ such that ${\rm Stab}_*\mu(\{H\}) \gt 0$, then $H$ is almost normal, because ${\rm Stab}_*\mu$ is invariant by the action by conjugation. This also implies that $\mu(X_H) \gt 0$. Since $\mu$ is ergodic and $X_H$ is invariant, we get $\mu(X_H)=1$.

Proof. We will prove that (5) in theorem 3.11 implies (1). For that, consider the action of $N_G(H)/H$ on $\overline{X_H(1_G)}$, i.e. on the closure of the set of points with stabilizer equal to $H$, given by $(gH)x=gx$, for every $x\in \overline{X_H(1_G)}$ and $g\in N_G(H)$. This action is well-defined and minimal, and its set of points with trivial stabilizer is $X_H(1_G)$. Furthermore, the measure $[G:N_G(H)]\mu$ is an ergodic measure of $(\overline{X_H(1_G)}, N_G(H)/H)$. On the other hand, since $N_G(H)$ has finite index in $G$, if $G$ is not allosteric, then [Reference Joseph19, Theorem 2.9] implies that $N_G(H)$ is not allosteric. By assumption $N_G(H)/H$ is commensurable to $N_G(H)$, so again by [Reference Joseph19, Theorem 2.9], the group $N_G(H)/H$ is not allosteric. This implies that its measure is supported on the set of points with trivial stabilizers, i.e.  $\mu(X_H(1_G))=\frac{1}{[G:N_G(H)]}$. Then $\mu(X_H)=1$. The rest follows from theorem 3.11.

Proof. Let $H$ be an almost normal finite group of $G$. Since $G$ is residually finite, there exists a finite-index subgroup $\Gamma$ of $G$ such that $H\cap \Gamma=\{1_G\}$. This implies that $(\Gamma\cap N_G(H))/H$ and $\Gamma\cap N_G(H)$ are isomorphic. Since $(\Gamma\cap N_G(H))/H$ and $\Gamma\cap N_G(H)$ are finite-index subgroups of $N_G(H)/H$ and $N_G(H)$, respectively, then the groups are commensurable.

Proof. Let $y\in Y$ and $\pi^{-1}\{y\}=\{x_1,\ldots, x_m\}$. We identify $\pi^{-1}\{y\}$ with the set $\Sigma_m$. Then each $g \in G_y$ defines a permutation $\sigma_g: \Sigma_m \to \Sigma_m $, where $\sigma_g(i) \in \Sigma_m$ is such that $g x_i=x_{\sigma_g(i)}$, and we get a group homomorphism $\phi: G_y\to S_m$ given by $\phi(g)=\sigma_g$, for every $g\in G_y$. If $\sigma_g=id$, then $gx_i=x_i$ for every $1\leq i\leq m$, which implies that $g\in \bigcap_{j=1}^m G_{x_j}$. Thus, $\ker(\phi)$ (the kernel of $\phi$) is a subgroup of $G_x$, for every $x\in \pi^{-1}\{y\}$. Since $G_y/\ker(\phi)$ is isomorphic to a subgroup of $S_m$, we have that $\ker(\phi)$ is a subgroup of $G_y$ of index at most $m!$. Finally, due to $\ker(\phi)\subseteq G_x\subseteq G_y$, we deduce that $[G_y:G_x]\leq m!$, for every $x\in \pi^{-1}\{y\}$.

Proof. Since there exists $y\in Y$ with a finite fibre that contains a point with trivial stabilizer, lemma 4.1 implies that $G_y$ is finite. So, by theorem 1.4, there exists an almost normal finite group $H$ of $G$ such that $Y_H\neq\emptyset$. Suppose that for some $z\in Y_H$ there is some $x\in \pi^{-1}\{z\}$ such that $G_x\neq \{1_G\}$. Since $G_x$ must be finite, $G_x$ cannot be almost normal; otherwise, the orbit of $G_x$ in $\overline{{\rm Stab}(X)}$ would be a minimal set, different from $\{1_G\}$, the unique URS of $(X,G)$. Furthermore, we have $gG_xg^{-1}\subseteq {\rm Stab}(g\pi(x))=gHg^{-1}$, for every $g\in G$. This implies $gG_xg^{-1}\subseteq H$, for every $g\in N_G(H)$. From this we get that there are infinitely many different conjugates $g_1G_xg_1^{-1}, g_2G_xg_2^{-1},\ldots$ in $H$, which contradicts that $H$ is finite. This shows that $\pi^{-1}Y_H\subseteq \{x\in X: G_x=\{1_G\}\}.$

Now suppose $(X,G,\mu)$ is essentially free. From lemma 4.3, there exists a finite $n\geq 1$ such that for $\pi_*\mu$-a.e.  $y\in Y$ we have $|\pi^{-1}\{y\}|=n$. Then, applying lemma 4.1, we get $|G_y|\leq n$, for $\pi_*\mu$-a.e.  $y\in Y$. The rest of the proof follows from proposition 3.10.

Proof. Let $g_1,\ldots, g_k\in G$ be representing elements of each class in $G/N_G(H)$. For every $1\leq i\leq k$, we set $X_i=\phi^{-1}\{g_iN_G(H)\}$. Let us define the following relation on $X$: $x_1\sim x_2$ if and only if there exist $1\leq i\leq k$ and $h\in H$ such that $x_1,x_2\in X_i$ and $x_1=g_ihg_i^{-1}x_2$. This is an equivalence relation, with equivalence classes contained in $X_i$, $1 \leq i \leq k$.

Proof of claim 2

Indeed, if $(x_1,x_2)\in R$ then there exist $1\leq i\leq k$ and $h\in H$ such that $x_1=g_ihg_i^{-1}x_2$. Let $p\in G$, $1\leq j\leq k$ and $l\in N_G(H)$ such that $pg_i=g_jl$. Then

\begin{equation*}px_1= pg_ihg_i^{-1}x_2=g_jlhg_i^{-1}x_2=g_j(lhl^{-1})lg_i^{-1}x_2=g_jh'g_j^{-1}px_2,\end{equation*}

where $lhl^{-1}=h'\in H$. This shows that $(px_1,px_2)\in R$. On the other hand, suppose that $(x,y)\in (X\times X) \setminus R$. If $x\in X_i$ and $y\in X_j$ with $i\neq j$, since $X_i$ and $X_j$ are clopen, there exist neighbourhoods $V_x\subseteq X_i$ and $V_y\subseteq X_j$ of $x$ and $y$, respectively. We have that $V_x\times V_y$ is a neighbourhood of $(x,y)$ contained in $(X\times X)\setminus R$. If $x$ and $y$ are in the same $X_i$, since $X$ is Hausdorff, there exist an open neighbourhood $V$ of $x$, and open neighbourhoods $V_h$ of $g_ihg_i^{-1}y$, such that $V\cap V_h=\emptyset$, for every $h\in H$. Since $H$ is finite, the continuity of the action implies there exists an open neighbourhood $U$ of $y$ such that if $z\in U$, then $g_ihg_i^{-1}z\in V_h$, for every $h\in H$. The set $V\times U$ is an open neighbourhood of $(x,y)$ contained in $(X\times X)\setminus R$. This shows that $R$ is closed.

Since $R$ is a closed $G$-invariant equivalence relation, from [Reference Auslander3, p. 23] we get that $Y=X/\sim$ is compact and Hausdorff, and the projection $\pi: X\to Y$ is a factor map, where the action of $G$ on $Y$ is given by $g\pi(x)=\pi(gx)$, for every $x\in X$. The definition of $\sim$ implies that $\tilde{\phi}: Y\to G/N_G(H)$ given by $\tilde{\phi}(y)=\phi(x)$, where $x$ is any element in $\pi^{-1}\{y\}$, is a well-defined factor map that verifies (1).

Let $y\in Y$ and $g\in G$ be such that $y\in \tilde{\phi}^{-1}\{gN_G(H)\}$. This yields $\pi^{-1}\{y\}\subseteq \phi^{-1}\{gN_G(H)\}$. From this, we get that $G_x$ is a subgroup of $gN_G(H)g^{-1}$, for every $x\in \pi^{-1}\{y\}$. Since $gHg^{-1}$ is normal in $gN_G(H)g^{-1}$, we obtain that $gHg^{-1}G_x$ is a subgroup of $gN_G(H)g^{-1}$, and so of $G$, for every $x\in \pi^{-1}\{y\}$. If $l\in G_y$, then for $x\in \pi^{-1}\{y\}$, we have $\pi(lx)=\pi(x)$. This implies that $lx\sim x$, from which we deduce that there exists $h\in H$ such that $ghg^{-1}lx=x$. From this follows that $ghg^{-1}l\in G_x$, and then $l \in gHg^{-1}G_x$. This shows that $G_y\subseteq gHg^{-1}G_x$. Conversely, if $h\in H$ and $l\in G_x$, then $ghg^{-1}ly=ghg^{-1}l\pi(x)=ghg^{-1}\pi(lx)=ghg^{-1}\pi(x)=\pi(ghg^{-1}x)=\pi(x)=y$. This shows that $G_{y}=gHg^{-1}G_x$, and so

\begin{equation*}gHg^{-1}G_x=G_y =gHg^{-1}G_{x'}, \mbox{for every } x,x'\in \pi^{-1}\{y\}.\end{equation*}

From the definition of $\sim$, the cardinality of $\pi^{-1}\{y\}$, is equal to

\begin{equation*}|gHg^{-1}/(gHg^{-1}\cap G_x)|=\frac{|H|}{|gHg^{-1}\cap G_x|}, \mbox{for every } x\in \pi^{-1}\{y\}.\end{equation*}

Thus, statements (2) and (3) have been established.

Let $(Y',G)$ be a dynamical system such that there exists a factor map $\pi':X\to Y'$, such that (i) and (ii) in the hypothesis of the proposition are satisfied. If $(x_1,x_2)\in R\subseteq X\times X$, then there exists $g\in G$ such that $x_1$ and $x_2$ are in $\phi^{-1}\{gN_G(H)\}$, and $x_1=ghg^{-1}x_2$, for some $h\in H$. Then by (i) we have $\pi'(x_1), \pi'(x_2)\in \tilde{\phi}'^{-1}\{gN_G(H)\}$, and $\pi'(x_1)=ghg^{-1}\pi'(x_2)=\pi'(x_2)$, thanks to point (ii). This shows that

\begin{equation*}R \subseteq R'= \{(x_1,x_2)\in X\times X: \pi'(x_1)=\pi'(x_2)\},\end{equation*}

which implies there exists a factor map $\varphi:Y\to Y'$ such that $\pi'=\varphi\circ\pi$.

Proof. Suppose there exist $x\in X$ and $g\in G$, such that there exists $l\in (G_x\cap gHg^{-1})\setminus\{1_G\}$. This implies that the stabilizer of every point in the orbit of $x$ contains a conjugate of $l$, i.e. a non-trivial element in $C=\bigcup_{g\in G}gHg^{-1}$. Since $(X,G)$ is topologically free, there exists $y\in X$ with trivial stabilizer, and the minimality of $(X,G)$ implies that for every neighbourhood $U$ of $y$ there is some $g_U\in G$ such that $g_Ux\in U$. Since $C$ is finite, this implies there exists $g\in G$ such that $glg^{-1}\in G_{g_Ux}$, for every neighbourhood $U$ of $x$. From the continuity of the action follows that $glg^{-1}y=y$, which is a contradiction.

Proof of theorem 4.9

Suppose that $(X,G)$ is minimal and topologically free, and $H$ is almost normal. By lemma 4.8, we have $G_x\cap gHg^{-1}=\{1_G\}$, for every $x\in X$. Then, if we assume (1.1), proposition 4.6 implies there exists a minimal dynamical system $(Y,G)$, and a finite-to-one factor map $\pi:X\to Y$ such that for every $g\in G$,

\begin{eqnarray*} \pi^{-1}\{y\in Y: G_y=gHg^{-1}\} & = &\{x\in X: G_x\subseteq gHg^{-1}\}\\ & = &\{x\in X: G_x=\{1_G\}\}. \end{eqnarray*}

Thus, we have

\begin{equation*} \pi^{-1}Y_H=\{x\in X: G_x=\{1_G\}\}. \end{equation*}

We claim that $\pi$ is $|H|$-to- $1$. Indeed, by proposition 4.6 $|\pi^{-1}(\pi(x))| = \frac{|H|}{|gHg^{-1} \cap G_{\pi(x)}|}$, for an appropriate $g \in G$, so this is clear for $x \in X$ with $G_x = \{1_G\}$. Suppose $x \in X$ is such that $G_x \ne \{1_G\}$, and let $K = gHg^{-1} \cap G_x$ for some $g \in G$. Since $\bigcup_{g \in G}gHg^{-1}$ is a finite set, $\bigcup_{g \in G} gKg^{-1}$ is a finite set, and so $K$ has a finite number of conjugates. The orbit of $x$ is dense in $X$, and so accumulates at a point $y$ with $G_y = \{1_G\}$. Since $K$ has a finite number of conjugates, we may choose a sequence of points $\{x_i\}_{i\geq 1} \subset O(x)$, converging to $y$, such that $G_{x_i} \supset gKg^{-1}$ for a fixed $g \in G$. Then by continuity of the action $gKg^{-1} \subset G_y =\{1_G\}$, which implies that $K = \{1_G\}$, and $\pi$ is $|H|$-to- $1$.

Assume that we have (1.2). Since $(Y,G)$ is minimal, $H$ is almost normal, and $Y_H\neq \emptyset$, corollary 3.5 implies there exists a factor map $\varphi: Y\to G/N_G(H)$. The composition $\phi=\varphi\circ \pi$ is the map described in (1.1).

Suppose that $(X,G,\mu)$ is minimal and essentially free with respect to an ergodic measure $\mu$. If (2.1) holds, and since essentially free implies topologically free, we get (2.2). If (2.2) holds, then it is direct that (2.3) holds. If we assume (2.3), then lemma 3.9 implies that $H$ is almost normal. Then, by corollary 3.5, there exists a factor map $\varphi: Y\to G/N_G(H)$. The composition $\phi=\varphi\circ \pi$ is the map in (2.1).

Proof. Observe that (2) implies (1) because it is enough to take $H=\{1_G\}$.

Suppose that $G$ is allosteric. Then there exists a minimal system $(X,G,\mu)$ equipped with an ergodic measure $\mu$, such that $(X,G,\mu)$ is topologically free but not essentially free with respect to $\mu$. By remark 4.7, we can apply theorem 1.8 to $(X,G,\mu)$ and $H$. Thus, there exists a minimal system $(Y,G)$ and a $|H|$-to-one factor map $\pi:X\to Y$ such that

\begin{equation*} \pi^{-1}Y_H=\{x\in X: G_x=\{1_G\}\}, \end{equation*}

which implies

\begin{equation*} \pi_*\mu(Y_H)=\mu(\{x\in X: G_x=\{1_G\}\})=0. \end{equation*}

Note that $\pi_*\mu$ is an ergodic measure for the minimal dynamical system $(Y, G/H)$ given by

\begin{equation*} gHy=gy \mbox{ for every } y\in Y, g\in G. \end{equation*}

Since

\begin{equation*} \{y\in Y: (G/H)_{y}=\{1_{G/H}\}\}=\{y\in Y: G_y=H\}, \end{equation*}

we deduce that $(Y,G/H,\pi_*\mu)$ is topologically free but not essentially free w.r.t. $\pi_*\mu$.

Proof. If $(Y,G)$ is a system such that there exists a factor map $\pi:X\to Y$, then $(Y,G)$ is minimal. If in addition, we have $|\pi^{-1}\{y\}| \lt \infty$ for $\pi_*\mu$-a.e.  $y\in Y$, then proposition 4.4 implies there exists an almost normal finite subgroup $H$ of $G$ such that $\pi_*\mu(Y_H)=1$. From theorem 1.8 we get that the equicontinuous system given by the left multiplication action of $G$ on $G/N_G(H)$ is a factor of $(X,G)$. If $(X,G,\mu)$ is weakly mixing, then $G/N_G(H)$ is trivial (see [Reference Auslander3, Theorem 13, p. 133]), which implies that $H$ is normal and then it is in the stabilizer of every element of $Y$. Thus, if the action of $G$ on $Y$ is effective, then $H=\{1_G\}$. This shows that $(Y,G,\pi_*\mu)$ is essentially free.