If $G$ acts primitively and $\Gamma \subseteq \Omega$ is not a block, then each pair of points could be separated

Question

Let $G$ act on $\Omega$. A subset $\Delta \subseteq \Omega$ is called a block if for each $x \in G$ either $\Delta^x \cap \Delta = \emptyset$ or $\Delta^x = \Delta$, where $\Delta^x := \{ \delta^x : \delta \in \Delta \}$. The singletons of $\Omega$ and $\Omega$ are the trivial blocks, a group acting on $\Omega$ is called primitive if it has no non-trivial blocks.

Now let $G$ act primitively on $\Omega$. If $\Gamma \subseteq \Omega$ is not a block, and $\alpha, \beta \in \Omega$ are arbitrary, does there exists some $g \in G$ such that either $$ \alpha \in \Gamma^g \mbox{ and } \beta \notin \Gamma^g $$ or $$ \alpha \notin \Gamma^g \mbox{ and } \beta \in \Gamma^g. $$

My proof: We use the fact $$ G \mbox{ acts primitive on } \Omega \Leftrightarrow \mbox{ each point stabilizer is a maximal subgroup }. $$ So let $G$ act primitively on $\Omega$ and let $\Delta \subseteq \Omega$ be a proper subset with at least two elements and $\alpha, \beta \in \Omega$ be distinct. If exactly one of $\alpha$ or $\beta$ lies in $\Delta$, nothing has to be shown. Suppose $\{ \alpha, \beta \}\subseteq \Delta$ and consider $$ H = \langle \{ g \in G : \alpha \in \Delta^g \} \rangle. $$ Then of course $G_{\alpha} \le H$ as $\alpha \in \Delta$. Now by primitivity we have $H = G_{\alpha}$ or $H = G$. If $H = G_{\alpha}$ then $$ \alpha^g = \alpha \Leftrightarrow \alpha \in \Delta^g. $$ Now choose $x \in G$ with $\alpha^x = \beta$ (as primitive implies transitive this is possible), then $\beta \in \Delta^x$ but $\alpha \notin \Delta^x$ (for otherwise we would have $\alpha^x = \alpha$). Now with the case $H = G$ I am stuck, any ideas how to proceed from here?

The case $\alpha$ and $\beta$ both not in $\Delta$ could be reduced to the one above, for if $\gamma \in \Delta$ choose $x \in G$ with $\gamma^x = \alpha$, then $\alpha \in \Delta^x$, if $\beta \notin \Delta^x$ we are done, otherwise $\{\alpha,\beta\} \in \Delta^x$, apply the above reasoning to find some $g \in G$ such that $\Delta^{xg}$ contains exactly one of $\alpha$ or $\beta$.

Remark 1: The question comes from this question for transitive $G$, for which it is wrong. But for primitive $G$ it holds.

Remark 2: My proof attempt uses a connection between primitivity and point stabilizers, as this question comes from an (although wrongly stated, see this post) exercise of Dixon & Mortimers book Permutation Groups, a solution without using this connection would be preferable, as in the book this is stated a few pages behind the exercise, and so maybe there is a solution without using it.

So is there a way to fill the gap in my proof, and if not do you know some other proof?

Answer 1

here's how I went about it. To recap, it turns out this problem is wrong, which i learned thanks to your previous question about it. The correct statement reads `Suppose $G$ is a group acting primitively on a set $\Omega$ and that $\Delta$ is a proper subset of $\Omega$ containing at least two points. Show that for each pair of distinct points $\alpha,\beta\in\Omega$ ...'.

The errata suggests the following hint as well: `Show that the relation $\alpha\approx \beta \iff \forall x\in G, \{\alpha,\beta\}\cap\Delta^x\in\{\emptyset,\{\alpha,\beta\}\}$ is a $G$-congruence'.

So, first let's see why it's an equivalence relation. Clearly $\alpha\approx\alpha$; next, if $\alpha\approx\beta$, then $\beta\approx\alpha$ because $\{\alpha,\beta\}=\{\beta,\alpha\}$. And if $\alpha\approx\beta\approx\gamma$, observe that we cannot separate $\alpha$ and $\gamma$ by some $\Delta^x$ unless we also separate either $\alpha$ and $\beta$ or $\beta$ and $\gamma$, for $\beta$ has to be either inside or outside of $\Delta^x$.

Next, let's see why it's a $G$-congruence. For this we want $\alpha\approx\beta\implies \forall y\in G: \alpha^y\approx\beta^y$. Well, if $\alpha\approx\beta$, then for all $x$ we have $\{\alpha,\beta\}\cap\Delta^x\in\{\emptyset,\{\alpha,\beta\}\}$ and taking images under $y$ we get $\{\alpha^y,\beta^y\}\cap\Delta^{xy}\in\{\emptyset,\{\alpha^y,\beta^y\}\}$. As $x$ takes all values in $G$, so does $xy$, and thus $\alpha^y\approx\beta^y$.

Hence $\approx$ is indeed a $G$-congruence, and its equivalence classes are blocks for the action of $G$ on $\Omega$. But observe that if we have some $\alpha\neq\beta$ such that no $\Delta^x$ separates them, $\alpha\approx\beta$ and thus there is an equivalence class of size more than $1$, contradicting the primitivity assumption. Hence every two $\alpha,\beta$ can be separated.

Next, we need to prove the stronger separation result that we can put $\alpha$ on whichever side of $\Delta$ we want in the case when $G$ is finite. So suppose we want to find $x$ such that $\alpha\in\Delta^x,\beta\notin\Delta^x$, and assume there's no such $x$ for the sake of contradiction. So by the previous part there is some $x$ such that $\beta\in\Delta^x$ but $\alpha\notin\Delta^x$.

Let $\Gamma=\Delta^x$, and consider an element $y$ such that $\alpha^y=\beta$ (one exists by transitivity). Let the order of $y$ be $k$, and let's consider the images $\Gamma,\Gamma^y,\ldots,\Gamma^{y^{k-1}}$. We claim that one of them will contain $\alpha$ and not $\beta$, which suffices, as $\Gamma^{y^i} =\Delta^{xy^i}$.

Again for the sake of contradiction, suppose that $\Gamma^{y^{k-1}}$ fails. Since $\alpha^y=\beta\in\Gamma$, we have $\alpha\in\Gamma^{y^{k-1}}$, so it must be the case that $\beta\in\Gamma^{y^{k-1}}$ as well, i.e. $\alpha^x\in\Gamma^{y^{k-1}}$, so $\alpha\in\Gamma^{y^{k-2}}$. Now if $\Gamma^{y^{k-2}}$ fails, it again follows that $\beta\in\Gamma^{y^{k-2}}$, hence $\alpha\in\Gamma^{y^{k-3}}$. Continuing this way, if we don't find a $\Gamma^{y^i}$ that works, after a finite number of steps we get $\alpha\in\Gamma$, contradiction.

Answer 2

I want to add a solution for finite $G$ which is a little bit different from @amakelov's approach.

So let $G$ be a finite primitve group acting on $\Omega$ and let $\Delta$ be a proper subset of $\Omega$ and $\alpha, \beta \in \Omega$. We can suppose $|\Omega| > 1$, for otherwise by transitivity it would be trivial.

Now set $\Gamma := \bigcap_{\alpha \in \Delta^g} \Delta^g$. Then $\Gamma \ne \Omega$ as $\Delta \ne \Omega$ and it is a block. Let $x \in G$, if $\alpha \in \Gamma^x$ then as $\Gamma^x = \bigcap_{\alpha \in \Delta^g} \Delta^{gx}$ we would have that $$ \alpha \in \Delta^g \Rightarrow \alpha \in \Delta^{gx} $$ i.e. the sets $\{ \Delta^g \mid \alpha \in \Delta^g \}$ are a subset of $\{ \Delta^{gx} \mid \alpha \in \Delta^{gx} \}$ and so $$\Gamma^x = \bigcap_{\alpha\in \Delta^g} \Delta^{gx} \supseteq \bigcap_{\alpha \in\Delta^{gx}} \Delta^{gx} = \Gamma.$$ As $|\Gamma^x| = |\Gamma|$ using finiteness of $\Omega$ we find $\Gamma^x = \Gamma$.

Now suppose $\beta \in \Gamma \cap \Gamma^x$. Choose $y \in G$ with $\alpha^y = \beta$, then we have $\alpha \in \Gamma^{y^{-1}} \cap \Gamma^{xy^{-1}} \cap \Gamma$. Hence by the above $\Gamma = \Gamma^{y^{-1}} = \Gamma^{xy^{-1}}$, and so $\Gamma = \Gamma^x$.

This gives that $\Gamma$ is a block, hence by primitivity $\Gamma = \{\alpha\}$, which gives the claim. $\square$

That $\Gamma$ is a block is essential here, and so this proof could not be generalized for infinite $\Omega$ (or $G$). For set $\Omega = \Gamma = \mathbb Z$ with action by addition. Then for $\Delta := \mathbb N = \{1,2,3,4,\ldots \}$ and $\alpha = 1$ we have $$ \alpha \in \Delta^g \Leftrightarrow g \in \{0,-1,-2,-3,-4,\ldots \} $$ and hence $\bigcap_{\alpha\in\Delta^g} \Delta^g = \mathbb N$, which is not a block as for example $\mathbb N \cap (\mathbb N + 1) = \{2,3,4,\ldots\}$.