Blocks Primitivity and Multiple Transitivity

If $g\cdot B$ meets $h\cdot B$, choose $x\in (g\cdot B)\cap (h\cdot B)$. Then $h^{-1}\cdot x\in (h^{-1}g)\cdot B\cap B$, so [L1] gives $(h^{-1}g)\cdot B=B$.

By definition every point of $\bigcup {\mathcal{T}}_B$ has the form $g\cdot b$ with $g\in G$ and $b\in B$, and every such point lies in the translate $g\cdot B$. So $\bigcup {\mathcal{T}}_B=G\cdot B$.

From step 1.1, $g\cdot B=h\cdot B$ whenever the two translates meet. Thus distinct translates are disjoint.

For $k\in G$ one has $k\cdot (g\cdot B)=(kg)\cdot B$, which is again in ${\mathcal{T}}_B$. Hence $G$ permutes the members of ${\mathcal{T}}_B$, and steps 2.1 and 1.2 make it a $G$-invariant partition of $G\cdot B$.

For the forward direction, let $\mathcal{B}$ be a $G$-invariant partition into blocks. The relation ${\sim }_{\mathcal{B}}$ is reflexive because every point lies in its own part, symmetric because “lying in the same part” is symmetric, and transitive because two parts that meet are equal.

For the converse direction, let $\sim$ be a $G$-invariant equivalence relation. Its equivalence classes partition $\mathrm{\Omega }$: every point lies in its own class, and two classes that meet are equal because symmetry and transitivity identify every element of one with every element of the other.

For the converse direction, fix an equivalence class $C=[x]$. Invariance gives $g\cdot C=[g\cdot x]$ for every $g\in G$, so $G$ permutes the equivalence classes. In particular, if $(g\cdot C)\cap C\mathrm{\ne }\mathrm{\varnothing }$, the two equivalence classes are equal; otherwise they are disjoint. Thus every class is a block and the class partition is $G$-invariant.

For the forward direction, if $x{\sim }_{\mathcal{B}}y$ and both lie in a part $B$, then $g\cdot x$ and $g\cdot y$ lie in the part $g\cdot B$ of the same partition by its $G$-invariance. Thus ${\sim }_{\mathcal{B}}$ is $G$-invariant.

Step 1.3 shows that the equivalence classes form a $G$-invariant block system, and steps 1.1 and 2.1 give the converse construction.

Let $B,C\in \mathcal{B}$. Choose $b\in B$ and $c\in C$. By transitivity there is $g\in G$ with $g\cdot b=c$.

Since $g\cdot b\in (g\cdot B)\cap C$, the two blocks $g\cdot B$ and $C$ meet. Because $\mathcal{B}$ is a partition into blocks, they are equal.

The map $B\to C$, $x\mapsto g\cdot x$, is a bijection because every group element acts bijectively on $\mathrm{\Omega }$. Hence $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }C\mathrm{\mid }$.

As $B$ and $C$ were arbitrary, all blocks in $\mathcal{B}$ have the same size. Since the blocks are pairwise disjoint and cover the finite set $\mathrm{\Omega }$, the cardinality of any block divides $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$.

For the forward direction, let $H$ satisfy $G_{\alpha }\le H\le G$ and put $B_H:=H\cdot \alpha$. If $(g\cdot B_H)\cap B_H\mathrm{\ne }\mathrm{\varnothing }$, choose $g{h}_1\cdot \alpha =h_2\cdot \alpha$ with $h_1,h_2\in H$. Then $h_2^{-1}g{h}_1\in G_{\alpha }\le H$, so $g\in H$ and therefore $g\cdot B_H=B_H$. Thus $B_H$ is a block, and clearly $\alpha \in B_H$.

For the converse direction, let $B$ be a block containing $\alpha$, and let $G_B:=\{g\in G:g\cdot B=B\}$. If $s\in G_{\alpha }$, then $s\cdot \alpha =\alpha \in B$, so $(s\cdot B)\cap B\mathrm{\ne }\mathrm{\varnothing }$; [L1] gives $s\cdot B=B$, hence $G_{\alpha }\le G_B\le G$.

For the converse direction, every $g\in G_B$ sends $\alpha \in B$ back into $B$, so $G_B\cdot \alpha \subseteq B$. Conversely, if $\beta \in B$, choose $g\in G$ with $g\cdot \alpha =\beta$ by [L2]. Then $\beta \in (g\cdot B)\cap B$, so [L1] gives $g\cdot B=B$ and therefore $g\in G_B$. Hence $\beta =g\cdot \alpha \in G_B\cdot \alpha$, so $B=G_B\cdot \alpha$.

Step 1.3 shows that $B\mapsto G_B\mapsto G_B\cdot \alpha$ returns $B$. Step 1.1 gives $H\cdot \alpha$ as a block containing $\alpha$, and if $g\in G_{H\cdot \alpha }$ then $g\cdot \alpha \in H\cdot \alpha$, so $g\cdot \alpha =h\cdot \alpha$ for some $h\in H$; thus $h^{-1}g\in G_{\alpha }\le H$, and hence $g\in H$. Therefore $G_{H\cdot \alpha }=H$. The two assignments are mutually inverse.

Because $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }>1$, choose $\beta \in \mathrm{\Omega }$ with $\beta \mathrm{\ne }\alpha$. Transitivity gives $g\in G$ with $g\cdot \alpha =\beta$, so $g\mathrm{\notin }{G}_{\alpha }$. Hence $G_{\alpha }$ is a proper subgroup of $G$.

For the converse direction, suppose $G_{\alpha }$ is maximal proper. Let $B$ be a block containing $\alpha$. By [L2], the corresponding subgroup $G_B$ satisfies $G_{\alpha }\le G_B\le G$, so maximality gives $G_B=G_{\alpha }$ or $G_B=G$. The first case gives $B=\{\alpha \}$ and the second gives $B=\mathrm{\Omega }$ by [L2]. Hence every block containing $\alpha$ is trivial.

Let $C$ be any block and choose $c\in C$. By transitivity, choose $g\in G$ with $g\cdot c=\alpha$. Then $g\cdot C$ is a block containing $\alpha$, so step 1.2 makes it either $\{\alpha \}$ or $\mathrm{\Omega }$. Applying $g^{-1}$ shows that $C$ is respectively a singleton or $\mathrm{\Omega }$. Hence the action is primitive by [L1].

For the forward direction, suppose the action is primitive. By [L2], any subgroup $H$ with $G_{\alpha }\le H\le G$ corresponds to a block containing $\alpha$. By [L1], that block is either $\{\alpha \}$ or $\mathrm{\Omega }$, so [L2] forces $H=G_{\alpha }$ or $H=G$. Together with step 1.1, this makes $G_{\alpha }$ maximal proper.

Step 2.2 proves that primitivity implies maximality, while steps 1.2 and 2.1 prove the converse.

Let $\mathcal{B}$ be a block system and let $B\in \mathcal{B}$. By [L2], the positive integer $\mathrm{\mid }B\mathrm{\mid }$ divides the prime $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$. So [L3] gives $\mathrm{\mid }B\mathrm{\mid }=1$ or $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$.

If $\mathrm{\mid }B\mathrm{\mid }=1$, then every block has size $1$ by [L2], so $\mathcal{B}$ is the singleton partition. If $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$, then $B=\mathrm{\Omega }$ and $\mathcal{B}=\{\mathrm{\Omega }\}$. Thus every block system is trivial, and [L1] makes the action primitive.

Define $x{\sim }_{N}y$ when $y=n\cdot x$ for some $n\in N$. This is an equivalence relation because $1\in N$, inverses in $N$ reverse the relation, and products in $N$ compose it.

The relation ${\sim }_N$ is $G$-invariant: if $y=n\cdot x$ with $n\in N$, then for every $g\in G$ one has $g\cdot y=(gn{g}^{-1})\cdot (g\cdot x)$, and [L3] puts $gn{g}^{-1}$ back in $N$.

By [L2], the ${\sim }_N$-classes form a block system. Since the action is primitive, [L1] makes that block system either the one-block partition or the singleton partition.

In the one-block case, every point lies in the $N$-orbit of every other point, so $N$ is transitive. In the singleton case, every $N$-orbit has one point, so each $n\in N$ fixes every point of $\mathrm{\Omega }$.

If the action is faithful and $N\mathrm{\ne }1$, the second case of step 4.1 is impossible. Hence a nontrivial normal subgroup of a faithful primitive action is transitive.

By [L1], the action of $N$ on $\mathrm{\Omega }$ is transitive.

Fix $\alpha \in \mathrm{\Omega }$, and suppose $n\in N$ fixes $\alpha$. For any $\beta \in \mathrm{\Omega }$, step 1.1 gives $m\in N$ with $\beta =m\cdot \alpha$. Since $N$ is abelian, $n\cdot \beta =n\cdot (m\cdot \alpha )=(nm)\cdot \alpha =(mn)\cdot \alpha =m\cdot (n\cdot \alpha )=m\cdot \alpha =\beta \mathrm{.}$

Step 2.1 shows that any element of $N$ fixing one point fixes every point. Faithfulness of the ambient action therefore forces that element to be the identity. So the action of $N$ is free.

Steps 1.1 and 3.1 make the action of $N$ transitive and free, hence regular by [L2].

To prove $k$-homogeneity, let $A,B\subseteq \mathrm{\Omega }$ be $k$-element subsets. Choose orderings $A=\{{\alpha }_1,\dots ,{\alpha }_k\}$ and $B=\{{\beta }_1,\dots ,{\beta }_k\}$. By [L1], some $g\in G$ sends each ${\alpha }_i$ to ${\beta }_i$, so $g\cdot A=B$.

To prove $j$-transitivity, start with ordered $j$-tuples of distinct points $({\alpha }_1,\dots ,{\alpha }_j)$ and $({\beta }_1,\dots ,{\beta }_j)$. Because $\mathrm{\Omega }$ has at least $k$ distinct points, extend them to ordered $k$-tuples of distinct points $({\alpha }_1,\dots ,{\alpha }_k)$ and $({\beta }_1,\dots ,{\beta }_k)$. Then [L1] gives $g\in G$ with $g\cdot {\alpha }_i={\beta }_i$ for all $1\le i\le k$, in particular for $1\le i\le j$.

Step 1.1 gives $k$-homogeneity and step 1.2 gives $j$-transitivity.

Let $B$ be a block containing some $\alpha \in \mathrm{\Omega }$. If $B=\{\alpha \}$ there is nothing to prove, so suppose $B$ also contains $\beta \mathrm{\ne }\alpha$.

For any $\gamma \mathrm{\ne }\alpha$, [L2] gives an element $g\in G$ with $g\cdot \alpha =\alpha$ and $g\cdot \beta =\gamma$. Then $\gamma \in g\cdot B$, while $\alpha \in g\cdot B\cap B$ because $g$ fixes $\alpha$. So [L1] gives $g\cdot B=B$, and therefore $\gamma \in B$.

Step 1.2 shows that every $\gamma \mathrm{\ne }\alpha$ lies in $B$, so $B=\mathrm{\Omega }$. Thus any block containing more than one point is all of $\mathrm{\Omega }$, and the only block systems are the trivial ones. By [L3], the action is primitive.

Every $G$-orbit on $\mathrm{\Omega }\times \mathrm{\Omega }$ contains some pair $(\alpha ,\beta )$: for $(x,y)$ choose $g\in G$ with $g\cdot x=\alpha$ by [L2], and then $(g\cdot x,g\cdot y)=(\alpha ,g\cdot y)$.

The assignment is well defined. If $(\alpha ,{\beta }_1)$ and $(\alpha ,{\beta }_2)$ lie in the same $G$-orbit, choose $g\in G$ with $g\cdot (\alpha ,{\beta }_1)=(\alpha ,{\beta }_2)$. Then $g\in G_{\alpha }$, so ${\beta }_2=g\cdot {\beta }_1$ and the two second coordinates lie in the same suborbit.

The assignment is surjective because every suborbit has the form $G_{\alpha }\cdot \beta$, and it is the image of the orbital $G\cdot (\alpha ,\beta )$.

The assignment is injective. If $G_{\alpha }\cdot {\beta }_1=G_{\alpha }\cdot {\beta }_2$, choose $g\in G_{\alpha }$ with $g\cdot {\beta }_1={\beta }_2$. Then $g\cdot (\alpha ,{\beta }_1)=(\alpha ,{\beta }_2)$, so the two pairs lie in the same $G$-orbit.

Steps 1.2, 1.3, and 1.4 show that orbital classes on ordered pairs correspond bijectively to suborbits at $\alpha$.

For the forward direction, suppose the action is doubly transitive. Then every $\beta \mathrm{\ne }\alpha$ can be sent to every other $\gamma \mathrm{\ne }\alpha$ by some element fixing $\alpha$, because [L2] applies to the ordered pairs $(\alpha ,\beta )$ and $(\alpha ,\gamma )$. Since $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }>1$, the complement $\mathrm{\Omega }\setminus \{\alpha \}$ is nonempty, so the two $G_{\alpha }$-orbits are exactly $\{\alpha \}$ and $\mathrm{\Omega }\setminus \{\alpha \}$. Hence the rank is two by [L1].

For the converse direction, suppose the rank is two. Then [L1] says the only $G_{\alpha }$-orbits are $\{\alpha \}$ and $\mathrm{\Omega }\setminus \{\alpha \}$, so $G_{\alpha }$ is transitive on the complement of $\alpha$. Given ordered pairs $(x,y)$ and $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$ with $x\mathrm{\ne }y$ and $x^{\mathrm{\prime }}\mathrm{\ne }{y}^{\mathrm{\prime }}$, choose $g\in G$ with $g\cdot x=x^{\mathrm{\prime }}$ by transitivity. The stabilizer satisfies $G_{x^{\mathrm{\prime }}}=g{G}_x{g}^{-1}$, and because the action is transitive the rank-two hypothesis at $\alpha$ implies the same two-suborbit description at $x$. Hence $G_{x^{\mathrm{\prime }}}$ is transitive on $\mathrm{\Omega }\setminus \{x^{\mathrm{\prime }}\}$, so some $h\in G_{x^{\mathrm{\prime }}}$ sends $g\cdot y$ to $y^{\mathrm{\prime }}$. Then $hg$ sends $(x,y)$ to $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$. Therefore the action is doubly transitive by [L2].

The two directions of steps 1.1 and 1.2 prove the equivalence.

Fix one ordered $k$-tuple of distinct points $({\alpha }_1,\dots ,{\alpha }_k)$. Define $\mathrm{\Phi }:G\to T_k(\mathrm{\Omega })$ by $\mathrm{\Phi }(g):=(g\cdot {\alpha }_1,\dots ,g\cdot {\alpha }_k)$, where $T_k(\mathrm{\Omega })$ is the set of ordered $k$-tuples of distinct points of $\mathrm{\Omega }$.

The map $\mathrm{\Phi }$ is bijective: existence in [L1] makes it surjective, and uniqueness in [L1] makes it injective.

The set $T_k(\mathrm{\Omega })$ has $n$ choices for the first entry, then $n-1$ for the second, and so on down to $n-k+1$ for the last. Hence $\mathrm{\mid }{T}_k(\mathrm{\Omega })\mathrm{\mid }=n(n-1)\cdots (n-k+1)$, and step 2.1 gives the same value for $\mathrm{\mid }G\mathrm{\mid }$.

For each $g\in G$, let $k_g$ be the permutation of $\mathrm{\Sigma }$ induced by $g$, so $k_g(C)=g\cdot C$. For each $C\in \mathrm{\Sigma }$, the element $t_C^{-1}g{t}_{g^{-1}\cdot C}$ stabilizes $B$ setwise because $t_C^{-1}g{t}_{g^{-1}\cdot C}\cdot B=t_C^{-1}g\cdot (g^{-1}\cdot C)=t_C^{-1}\cdot C=B\mathrm{.}$ Let $f_g(C)\in H$ be the induced permutation of $B$ defined by this element.

Define $\mathrm{\Phi }(g):=(f_g,k_g)$. For $C\in \mathrm{\Sigma }$, the function component of $\mathrm{\Phi }(g)\mathrm{\Phi }(h)$ at $C$ is $f_g(C)f_h(g^{-1}\cdot C)$, while $t_C^{-1}gh{t}_{(gh{)}^{-1}\cdot C}=\left(t_C^{-1}g{t}_{g^{-1}\cdot C}\right)\left(t_{g^{-1}\cdot C}^{-1}h{t}_{h^{-1}{g}^{-1}\cdot C}\right),$ so it induces the same permutation of $B$ as $f_{gh}(C)$. Also $k_{gh}=k_g{k}_h$. Hence $\mathrm{\Phi }(gh)=\mathrm{\Phi }(g)\mathrm{\Phi }(h)$.

Identify $\mathrm{\Omega }$ with $B\times \mathrm{\Sigma }$ by $\mathrm{\Psi }(b,C):=t_C\cdot b$. Then for every $g\in G$ one has $\mathrm{\Psi }(\mathrm{\Phi }(g)\cdot (b,C))=t_{g\cdot C}\cdot (f_g(g\cdot C)\cdot b)=g\cdot (t_C\cdot b)=g\cdot \mathrm{\Psi }(b,C)\mathrm{.}$ So $\mathrm{\Phi }(g)=1$ exactly when $g$ fixes every point of $\mathrm{\Omega }$.

Step 3.1 shows that $\ker \mathrm{\Phi }$ is the kernel of the given action. Therefore a faithful action makes $\ker \mathrm{\Phi }=1$, so in that case $\mathrm{\Phi }$ is an embedding into $H{\wr }_{\mathrm{\Sigma }}K$.

Let $N⊴G$ be nontrivial. By [L1], $N$ is transitive on $\mathrm{\Omega }$. Hence for every $g\in G$ there is $n\in N$ with $n\cdot \alpha =g\cdot \alpha$, and then $g^{-1}n\in G_{\alpha }$. So $G=N{G}_{\alpha }$.

Fix $g\in G$, and write $g=nh$ with $n\in N$ and $h\in G_{\alpha }$ as in step 1.1. Because $A⊴G_{\alpha }$, one has $hA{h}^{-1}=A$. For $a\in A$, the element $na{n}^{-1}{a}^{-1}$ lies in $N$ by [L3], so $na{n}^{-1}=(na{n}^{-1}{a}^{-1})a\in NA$. Therefore $gA{g}^{-1}=n(hA{h}^{-1})n^{-1}=nA{n}^{-1}\subseteq NA\mathrm{.}$

The conjugates of $A$ generate $G$ by hypothesis, and step 2.1 puts each of them inside $NA$. Hence $G=NA$. Modulo $N$, this says $G\mathrm{/}N$ is generated by the image of $A$; since $A$ is abelian, $G\mathrm{/}N$ is abelian.

Because $G\mathrm{/}N$ is abelian, every commutator of $G$ lies in $N$. By [L2], the subgroup they generate is $[G,G]$, so $[G,G]\le N$.

Step 4.1 holds for every nontrivial normal subgroup $N⊴G$. Therefore if $G=[G,G]$, every nontrivial normal subgroup contains all of $G$ and is equal to $G$. So $G$ is simple.