Proof. Let $\Sigma _1$ be a system of imprimitivity for the action of G on $\Omega $ of minimal cardinality. Consider the natural homomorphism

$$ \begin{align*}\bar{\,}:G\to \mathrm{Sym}(\Sigma_1)\end{align*} $$

arising from the action of G on $\Sigma _1$ .

Since the derangement graph of G on $\Omega $ lacks cliques of size $4$ , by Lemma 3.4, the derangement graph $\Gamma _{\bar G,\Sigma _1}$ also lacks such cliques.

Now, suppose the derangement graph $\Gamma _{\bar {G},\Sigma _1}$ has no triangles. As $|\Sigma _1|>1$ , according to [Reference Meagher, Razafimahatratra and Spiga14], we have $|\bar {G}|=|\Sigma _1|=2$ . Consequently, $[G:G_{(\Sigma _1)}]=2$ and $\bar {G}\cong \mathrm {Sym}(2)$ . Notice that each element in $G\setminus G_{(\Sigma _1)}$ acts as a derangement on $\Sigma _1$ and, thus, on $\Omega $ as well.

Consider $g\in G\setminus G_{(\Sigma _1)}$ . If $G_{(\Sigma _1)}$ contains a derangement k in its action on $\Omega $ , then $gk$ and $gkg^{-1}$ are both derangements since $gk\in G\setminus G_{(\Sigma _1)}$ and $gkg^{-1}$ is conjugate to k. Consequently,

$$ \begin{align*}C=\{1,g,k,gk\}\end{align*} $$

forms a clique of size four in $\Gamma _{G,\Omega }$ , which contradicts the assumption that $\Gamma _{G,\Omega }$ has no $4$ -cliques. Hence, $G_{(\Sigma _1)}$ , in its action on $\Omega $ , contains no derangements. Consequently, the derangement graph $\Gamma _{G,\Omega }$ is bipartite with bipartition

$$ \begin{align*}G_{(\Sigma_1)},G\setminus G_{(\Sigma_1)}.\end{align*} $$

However, this contradicts the fact that $\Gamma _{G,\Omega }$ admits a triangle.

This contradiction demonstrates that the derangement graph $\Gamma _{\bar {G},\Sigma _1}$ of $\bar G$ has a triangle but no $4$ -clique. Since $\bar {G}$ is primitive, by [Reference Fusari, Previtali and Spiga8, Theorem 1.2], we conclude that $|\Sigma _1|=3$ and $\mathrm {Alt}(\Sigma _1)\le \bar G\le \mathrm {Sym}(\Sigma _1)$ .

Now, let $\tilde G$ be the preimage under $\bar {\,}$ of $\mathrm {Alt}(\Sigma _1)$ . We have $[G:\tilde G]=1$ when ${\bar G=\mathrm {Alt}(\Sigma _1)}$ and $[G:\tilde G]=2$ when $\bar G=\mathrm {Sym}(\Sigma _1)$ . We claim that $\tilde G$ is transitive on $\Omega $ . This is clear when $G=\tilde G$ . Suppose then $|G:\tilde G|=2$ . If $G_{\omega _1}\le \tilde G$ , then from the bijection between systems of imprimitivity of G on $\Omega $ and the interval lattice of $G/G_{\omega _1}$ , we deduce that G has a system of imprimitivity of size $2$ , because ${|G:\tilde G|=2}$ and $G_{\omega _1}\le \tilde G$ . However, this contradicts the fact that we have shown that the minimal cardinality of a system of imprimitivity for G is $3$ . Therefore, $G_{\omega _1}\nleq \tilde G$ . Since ${|G:\tilde G|=2}$ , we have $G=G_{\omega _1}\tilde G$ . We deduce by the Frattini argument that $\tilde G$ is transitive on $\Omega $ .

By (4-1), we conclude that $G=\tilde G$ , which implies $ \bar G=\mathrm {Alt}(\Sigma _1).$

Suppose $G_{(\Sigma _1)}$ contains a derangement k in its action on $\Omega $ . Now, let $c\in G$ such that $\bar {c}$ acts as a cycle of length $3$ on $\Sigma _1$ , that is, $\bar G=\mathrm {Alt}(\Sigma _1)=\langle \bar {c}\rangle $ . Following the previous argument, we observe that $C=\{1,c,c^2,k,ck,c^2k\}$ forms a clique of size six in $\Gamma _{G,\Omega }$ , which contradicts the assumption that $\Gamma _{G,\Omega }$ has no $4$ -cliques. Therefore, $G_{(\Sigma _1)}$ in its action on $\Omega $ contains no derangement.

Notation 5.1. We establish some necessary notation for the subsequent steps of the proof of Theorem 1.1. From Lemma 4.1, we know that $G_{(\Sigma _1)}$ has no derangements in its action on $\Omega $ , and thus,

(5-1) $$ \begin{align} G_{(\Sigma_1)}=\bigcup_{\omega\in \Omega}G_\omega. \end{align} $$

Let $\Sigma _1=\{\Delta _1,\Delta _2,\Delta _3\}$ and choose $\omega _i\in \Delta _i$ for all $i\in \{1,2,3\}$ .

Since G is imprimitive on $\Omega $ with a system of imprimitivity $\Sigma _1$ , and G induces $\mathrm {Alt}(\Sigma _1)$ on $\Sigma _1$ by Lemma 4.1, we embed

$$ \begin{align*}G\le H \mathop{\mathrm{wr}} \mathrm{Alt}(3),\end{align*} $$

where $H\le \mathrm {Sym}(\Delta )$ , H is transitive, and $\Delta $ is a finite set with $|\Delta |=|\Delta _i|$ . Under this identification, we may write

$$ \begin{align*}\Delta_1=\Delta\times\{1\},\, \Delta_2=\Delta\times\{2\},\, \Delta_3=\Delta\times\{3\},\end{align*} $$

and regard G as endowed with its imprimitive action on $\Omega =\Delta \times \{1,2,3\}$ . Specifically,

$$ \begin{align*}G_{(\Sigma_1)}\le H\times H\times H.\end{align*} $$

For each $i\in \{1,2,3\}$ , let $\pi _i:G_{(\Sigma _1)}\to H$ be the projection onto the $i\,{\mathrm {th}}$ direct factor. Without loss of generality (by potentially replacing H with the permutation group induced by $G_{(\Sigma _1)}$ on its action on $\Delta _i$ ), we may assume that

$$ \begin{align*}H=\pi_1(G_{(\Sigma_1)})=\pi_2(G_{(\Sigma_1)})=\pi_3(G_{(\Sigma_1)}).\end{align*} $$

This allows a concrete representation of the elements of G; indeed, each element of G takes the form $(h_1,h_2,h_3)(1\,2\,3)^i$ for some $h_1,h_2,h_3\in H$ and $i\in \{0,1,2\}$ .

Let $c\in G\setminus G_{(\Sigma _1)}$ with $\Delta _1^c=\Delta _2$ . Then, $c=(h_1,h_2,h_3)(1\,2\,3)$ for some $h_1,h_2,h_3\in H$ and $G=G_{(\Sigma _1)}\langle c\rangle $ . Let $h=h_1h_2h_3$ . Note that

$$ \begin{align*} (1,h_1^{-1},h_2^{-1}h_1^{-1})^{-1}c(1,h_1^{-1},h_2^{-1}h_1^{-1})&= (1,h_1,h_1h_2)(h_1,h_2,h_3)(1\,2\,3)(1,h_1^{-1},h_2^{-1}h_1)\\ &=(h_1,h_1h_2,h_1h_2h_3)(h_1^{-1},h_2^{-1}h_1^{-1},1)(1\,2\,3)\\ &=(1,1,h)(1\,2\,3). \end{align*} $$

This shows that, by replacing G with its conjugate $G^{(1,h_1^{-1},h_2^{-1}h_1^{-1})}$ in $H \mathop {\mathrm {wr}} \mathrm {Alt}(3)$ , we may assume that

$$ \begin{align*} c=(1,1,h)(1\,2\,3). \end{align*} $$

Of course, this substitution does not affect the fact that $\Gamma _{G,\Omega }$ has no $4$ -clique.

Notice that if G is primitive on $\Omega $ , then $\Sigma _1=\Omega $ , which implies that part (1) of Theorem 1.1 holds with $|\Omega |=3$ . Hence, for the remainder of the proof of Theorem 1.1, we assume that G is imprimitive on $\Omega $ , meaning $|\Omega |>3$ .

Proof. From Notation 5.1, we have $\bar G=\mathrm {Alt}(\Sigma _1)$ and, hence, $G_{\{\Delta _i\}}=G_{(\Sigma _1)}$ , because $\mathrm {Alt}(\Sigma _1)$ acts regularly on $\Sigma _1$ . As $G_{\{\Delta _i\}}$ is transitive on $\Delta _i$ , we deduce that

$$ \begin{align*} G_{(\Sigma_1)}\text{ is transitive on }\Delta_i\quad \text{for each }i\in \{1,2,3\}. \end{align*} $$

Since $|\Omega |>|\Sigma _1|=3$ , G admits a system of imprimitivity $\Sigma _2\ne \Sigma _1$ that is a refinement of $\Sigma _1$ . Among all such systems of imprimitivity, choose one so that $|\Sigma _2|$ is as small as possible. Just as in the proof of Lemma 4.1, we may replace G with the permutation group induced by G on $\Sigma _2$ and, hence, we may suppose that $\Sigma _2=\{\{\omega \}\mid \omega \in \Omega \}$ . Therefore, we need to show that the action of G on $\Omega $ itself is permutation isomorphic to the action of $\mathrm {Alt}(4)$ on the six $2$ -subsets of $\{1,2,3,4\}$ .

The minimality of $|\Sigma _2|$ implies that $G_{\{\Delta _1\}}=G_{(\Sigma _1)}$ is primitive in its action on $\Delta _1$ . If $G_{(\Delta _1)}= 1$ , then the hypotheses of Proposition 3.5 are satisfied and, hence, $G_{(\Sigma _1)}$ contains a derangement in its action on $\Omega $ , which contradicts (5-1) (or the conclusion of Lemma 4.1). This contradiction implies that $ G_{(\Delta _1)}\ne 1,$ that is, $G_{\{\Delta _1\}}$ does not act faithfully on $\Delta _1$ . Although we are in a position to apply Proposition 3.6, we choose not to do so here. Instead, we argue using an ad hoc method, as we believe this approach could be applied more generally. However, at present, we do not know how to achieve this broader generalization.

Since $G_{(\Sigma _1)}$ is primitive on $\Delta _i$ , $G_{\omega _i}$ is maximal in $G_{(\Sigma _1)}$ . We claim that

(5-2) $$ \begin{align} G_{(\Delta_1\cup\Delta_2)}=G_{(\Delta_1\cup \Delta_3)}=G_{(\Delta_2\cup\Delta_3)}=1. \end{align} $$

Observe that, since G acts transitively by conjugation on $\Sigma _1=\{\Delta _1,\Delta _2,\Delta _3\}$ , the groups $G_{(\Delta _1\cup \Delta _2)}$ , $G_{(\Delta _1\cup \Delta _3)}$ , and $G_{(\Delta _2\cup \Delta _3)}$ are G-conjugate. Hence, if one of them is trivial, all must be trivial. We argue by contradiction and we suppose that $G_{(\Delta _1\cup \Delta _2)}\ne 1$ . Observe that

$$ \begin{align*}G_{(\Delta_1)}\cap G_{(\Delta_2)}\cap G_{(\Delta_3)}=G_{(\Delta_1\cup\Delta_2\cup\Delta_3)}=G_{(\Omega)}=1.\end{align*} $$

Since $G_{(\Delta _1\cup \Delta _2)}\unlhd G_{(\Sigma _1)}$ and since $G_{\omega _3}$ is a maximal subgroup of $G_{(\Sigma _1)}$ , we deduce that either $G_{(\Sigma _1)}=G_{\omega _3}G_{(\Delta _1\cup \Delta _2)}$ or $G_{(\Delta _1\cup \Delta _2)}\le G_{\omega _3}$ . If $G_{(\Delta _1\cup \Delta _2)}\le G_{\omega _3}$ , then, using the fact that $G_{(\Delta _1\cup \Delta _2)}\unlhd G_{(\Sigma _1)}$ and that $G_{(\Sigma _1)}$ is transitive on $\Delta _3$ , we get

$$ \begin{align*}G_{(\Delta_1\cup\Delta_2)}\le \bigcap_{k\in G_{(\Sigma_1)}}G_{\omega_3}^k=\bigcap_{\omega\in \Delta_3}G_{\omega}=G_{(\Delta_3)}.\end{align*} $$

Thus, $G_{(\Delta _1\cup \Delta _2)}=G_{(\Delta _1\cup \Delta _2\cup \Delta _3)}=1$ , which contradicts the fact that we are assuming $G_{(\Delta _1\cup \Delta _2)}\ne 1$ . Therefore,

$$ \begin{align*}G_{(\Sigma_1)}=G_{\omega_3}G_{(\Delta_1\cup\Delta_2)}.\end{align*} $$

Since $G_{(\Sigma _1)}$ is transitive on $\Delta _3$ , the Frattini argument implies that $G_{(\Delta _1\cup \Delta _2)}$ is transitive on $\Delta _3$ . In particular, from Jordan’s theorem, there exists $k_3\in G_{(\Delta _1\cup \Delta _2)}$ such that $k_3$ is a derangement on $\Delta _3$ and fixes each point in $\Delta _1\cup \Delta _2$ . With a similar argument, we obtain $k_1\in G_{(\Delta _2\cup \Delta _3)}$ acting as a derangement on $\Delta _1$ and $k_2\in G_{(\Delta _1\cup \Delta _3)}$ acting as a derangement on $\Delta _2$ . The support of the permutation $k_i$ is exactly $\Delta _i$ and, hence, $k=k_1k_2k_3\in G_{(\Sigma _1)}$ is a permutation having support exactly $\Delta _1\cup \Delta _2\cup \Delta _3=\Omega $ . Therefore, k is a derangement of $\Omega $ belonging to $G_{(\Sigma _1)}$ , which contradicts (5-1) (or the conclusion of Lemma 4.1). This contradiction has established the veracity of (5-2).

Now, we use the embedding $G\le H \mathop {\mathrm {wr}} \mathrm {Alt}(3)$ ; see Notation 5.1. Let

$$ \begin{align*} A&:=\pi_2(G_{(\Delta_1)})\le \mathrm{Sym}(\Delta), \\ B&:=\pi_3(G_{(\Delta_1)})\le \mathrm{Sym}(\Delta). \end{align*} $$

Thus,

(5-3) $$ \begin{align} G_{(\Delta_1)}\le 1\times A\times B. \end{align} $$

Using the maximality of $G_{\omega _2}$ and $G_{\omega _3}$ in $G_{(\Sigma _1)}$ , we get $G_{(\Sigma _1)}=G_{\omega _2}G_{(\Delta _1)}=G_{\omega _3}G_{(\Delta _1)}$ . Thus, from the Frattini argument, we deduce that $G_{(\Delta _1)}$ is transitive on $\Delta _2$ and on $\Delta _3$ , that is, A and B are transitive subgroups of $\mathop {\mathrm {Sym}}(\Delta )$ .

Conjugating $G_{(\Delta _1)}$ by $c=(1,1,h)(1\,2\,3)$ (see Notation 5.1) we deduce

$$ \begin{align*} G_{(\Delta_2)}&=G_{(\Delta_1)}^c\le (1\times A\times B)^c=(1\times A\times B)^{(1,1,h)(1\,2\,3)}\\ &=(1\times A\times B^h)^{(1\,2\,3)}=B^h\times 1\times A.\nonumber \end{align*} $$

In fact, this computation shows that

(5-4) $$ \begin{align} \pi_1(G_{(\Delta_2)})&=B^h \quad \text{and}\quad \pi_3(G_{(\Delta_2)})=A. \end{align} $$

As $G_{(\Delta _1)},G_{(\Delta _2)}\unlhd G_{(\Sigma _1)}$ , we have

$$ \begin{align*}[G_{(\Delta_1)},G_{(\Delta_2)}]\le G_{(\Delta_1)}\cap G_{(\Delta_2)}=G_{(\Delta_1\cup\Delta_2)}=1,\end{align*} $$

where in the last equality, we have used (5-2). Hence, $G_{(\Delta _1)}$ commutes with $G_{(\Delta _2)}$ . From (5-3) and (5-4), we deduce that $[A,B]=1$ . As A and B are both transitive subgroups of $\mathrm {Sym}(\Delta )$ , from [Reference Cameron4, page 27, 1.5], we get

$$ \begin{align*} B=\mathbf{C}_{\mathrm{Sym}(\Delta)}(A) \quad \text{and}\quad A=\mathbf{C}_{\mathrm{Sym}(\Delta)}(B). \end{align*} $$

We have $A=B$ if and only if A is abelian and, in this case, A acts regularly on $\Delta $ . If $A\ne B$ , then A and B both act regularly on $\Delta $ , and we may identify the action of A on $\Delta $ via its right regular representation and the action of B on $\Delta $ via the left regular representation of A itself.

Conjugating by c again, we deduce

$$ \begin{align*}G_{(\Delta_3)}=G_{(\Delta_2)}^c\le (B^h\times 1\times A)^c=A^h\times B^h\times 1. \end{align*} $$

As above, this computation shows that

(5-5) $$ \begin{align} \pi_1(G_{(\Delta_3)})&=A^h \quad \text{and}\quad \pi_2(G_{(\Delta_2)})=B^h. \end{align} $$

Arguing as above, we see that $G_{(\Delta _3)}$ commutes with $G_{(\Delta _1)}$ and with $G_{(\Delta _2)}$ . From (5-3) and (5-5), we deduce that $[A,B^h]=1$ . As A and $B^h$ are both transitive subgroups of $\mathrm {Sym}(\Delta )$ , from [Reference Cameron4, page 27, 1.5], we get

$$ \begin{align*}B^h=\mathbf{C}_{\mathrm{Sym}(\Delta)}(A)=B.\end{align*} $$

Thus, h normalizes B. As $A=\mathbf {C}_{\mathrm {Sym}(\Delta )}(B)$ , we get

$$ \begin{align*}A^h=\mathbf{C}_{\mathrm{Sym}(\Delta)}(B^h)=\mathbf{C}_{\mathrm{Sym}(\Delta)}(B)=A.\end{align*} $$

Thus, h normalizes A. Now, from (5-3), (5-4), and (5-5), we have

(5-6) $$ \begin{align} G_{(\Delta_1)}&\le 1\times A\times B,\nonumber\\ G_{(\Delta_2)}&\le B\times 1\times A,\\ G_{(\Delta_3)}&\le A\times B\times 1,\nonumber \end{align} $$

where $B=\mathbf {C}_{\mathop {\mathrm {Sym}}(\Delta )}(A)$ and A is a regular subgroup of $\mathrm {Sym}(\Delta )$ .

Now, we use (5-2) combined with (5-6). Indeed, from (5-2) and from the first line in (5-6), we deduce that, for every $a\in A$ , there exists a unique $b\in B$ with $(1,a,b)\in G_{(\Delta _1)}$ . In particular, the mapping $\varphi : A \to B$ , characterized by the property that $(1, a, a^\varphi ) \in G_{(\Delta _1)}$ for every $a \in A$ , defines a group isomorphism from A to B. Similarly, (5-2) and the remaining two lines of (5-6) imply that there exist two more group isomorphisms $\psi ,\eta :A\to B$ with

$$ \begin{align*} G_{(\Delta_1)}&=\{(1,a,a^\varphi)\mid a\in A\},\\ G_{(\Delta_2)}&=\{(a^\psi,1,a)\mid a\in A\},\\ G_{(\Delta_3)}&=\{(a,a^\eta,1)\mid a\in A\}. \end{align*} $$

We are now ready to conclude the proof of this result. Since A and B are commuting regular subgroups of $\operatorname {Sym}(\Delta )$ , we may apply Lemma 3.1. Assume there exist $a \in A \setminus \{1\}$ and $b \in B \setminus \{1\}$ such that $ab$ is a derangement (on $\Delta $ ). Then,

$$ \begin{align*} \underbrace{(a^\psi,1,a)}_{\in G_{(\Delta_2)}} \underbrace{(1,b^{\varphi^{-1}},b)}_{\in G_{(\Delta_1)}} = (a^\psi, b^{\varphi^{-1}}, ab) \in G_{(\Sigma_1)}. \end{align*} $$

The first two coordinates of this element lie in regular subgroups, and are different from the identity since $\varphi $ and $\psi $ are group isomorphisms. Therefore, each coordinate acts as a derangement on its corresponding block and we deduce that $(a^\psi , b^{\varphi ^{-1}}, ab)$ is a derangement on $\Omega $ , which contradicts (5-1). This contradiction shows that only the other case of Lemma 3.1 can occur, namely $|\Delta | = 2$ and $|\Omega | = |\Sigma _1||\Delta | = 6$ .

This has shown that G is a subgroup of $\mathrm {Sym}(2) \mathop {\mathrm {wr}} \mathrm {Alt}(3)$ in its natural imprimitive action of degree $6$ . It can be checked, either by hand or with the aid of a computer, that $\mathrm {Sym}(2) \mathop {\mathrm {wr}} \mathrm {Alt}(3)$ has three conjugacy classes of transitive subgroups, namely: $\mathrm {Sym}(2) \mathop {\mathrm {wr}} \mathrm {Alt}(3)$ ; $\operatorname {Alt}(4)$ in its action on the six $2$ -subsets of $\{1,2,3,4\}$ ; and the cyclic group $C_6$ of order $6$ . Since G is minimally transitive, we have $G \ne \mathrm {Sym}(2) \mathop {\mathrm {wr}} \mathrm {Alt}(3)$ . Moreover, since the derangement graph of $C_6$ has cliques of size $6$ , we also have $G \ne C_6$ . Thus, G is permutation-isomorphic to $\operatorname {Alt}(4)$ in its action on the six $2$ -subsets of $\{1,2,3,4\}$ .

Notation 6.1. Building upon Notation 5.1 and using Lemma 5.2, we introduce additional notation crucial for the subsequent steps of the proof.

We let $\Sigma _2$ be a system of imprimitivity of G on $\Omega $ having cardinality $6$ , which is a refinement of $\Sigma _1$ . We let $\Sigma _2=\{\Delta _{1,1},\Delta _{1,2},\Delta _{2,1},\Delta _{2,2},\Delta _{3,1},\Delta _{3,2}\},$ where $\Delta _i=\Delta _{i,1}\cup \Delta _{i,2}$ for all $i\in \{1,2,3\}.$ Moreover, we choose $\omega _{i,j}\in \Delta _{i,j}$ for all $i\in \{1,2,3\}$ and $j\in \{1,2\}$ .

Using the system of imprimitivity $\Sigma _2$ , we obtain an embedding

$$ \begin{align*}G\le K \mathop{\mathrm{wr}} \mathrm{Alt}(4),\end{align*} $$

where $K\le \mathrm {Sym}(\Delta )$ , $\Delta $ is a finite set with $|\Delta |=|\Delta _{i,j}|$ , and K is the permutation group $G_{\{\Delta _{i,j}\}}^{\Delta _{i,j}}$ . Under this identification, we may write

$$ \begin{align*} \Delta_{1,1}&=\Delta\times\{1\}, \Delta_{1,2}=\Delta\times\{2\},\\ \Delta_{2,1}&=\Delta\times\{3\}, \Delta_{2,2}=\Delta\times\{4\},\\ \Delta_{3,1}&=\Delta\times\{5\}, \Delta_{3,2}=\Delta\times\{6\}, \end{align*} $$

and we may think about G as endowed with its imprimitive action on $\Omega =\Delta \times \{1,2,3,4,5,6\}$ . In particular,

$$ \begin{align*}G_{(\Sigma_2)}\le K\times K\times K\times K\times K\times K.\end{align*} $$

For each $i\in \{1,2,3\}$ and $j\in \{1,2\}$ , let $\pi _{i,j}:G_{(\Sigma _2)}\to K$ be the projection obtained by restricting a permutation of $G_{(\Sigma _2)}$ to $\Delta _{i,j}$ . Since the union of the G-conjugates of $G_{\omega _{1,1}}$ is $G_{(\Sigma _1)}$ by (5-1), we have $G_{\omega _{1,1}}\nleq G_{(\Sigma _2)}$ . As $[G_{\{\Delta _{1,1}\}}:G_{(\Sigma _2)}]=2$ and $G_{\omega _{1,1}}\le G_{\{\Delta _{1,1}\}}$ , we deduce

$$ \begin{align*} G_{\{\Delta_{1,1}\}}=G_{\omega_{1,1}}G_{(\Sigma_2)}. \end{align*} $$

Now, the Frattini argument implies that $G_{(\Sigma _2)}$ is transitive on $\Delta _{1,1}$ . As $G_{(\Sigma _2)}\unlhd G$ , we get that $G_{(\Sigma _2)}$ is transitive on $\Delta _{i,j}$ for every $i\in \{1,2,3\}$ and $j\in \{1,2\}$ .

We let

$$ \begin{align*}L=\pi_{i,j}(G_{(\Sigma_2)}),\end{align*} $$

and observe that L does not depend on $i,j$ and, from the paragraph above, L is a transitive subgroup of K with $[K:L]\le 2$ .

We claim that

$$ \begin{align*} G_{\{\Delta_i\}}=G_{\{\Delta_{i,j}\}}G_\omega\quad \text{for all } i\in \{1,2,3\}, j\in \{1,2\}, \omega\in \Omega\setminus\Delta_i. \end{align*} $$

Indeed, as $[G_{\{\Delta _i\}}:G_{\{\Delta _{i,j}\}}]=2$ , we have either $G_{\{\Delta _i\}}=G_{\{\Delta _{i,j}\}}G_\omega $ or $G_\omega \le G_{\{\Delta _{i,j}\}}$ . In the latter case, if we let $\Delta _{i',j'}$ be the block of $\Sigma _2$ containing $\omega $ , we have $G_\omega \le G_{\{\Delta _{i,j}\}}\cap G_{\{\Delta _{i',j'}\}}=G_{(\Sigma _2)}$ . However, this contradicts the fact that $G_{(\Sigma _2)}$ is transitive on each block of $\Sigma _2$ .

The embedding of G in $K \mathop {\mathrm {wr}} \mathrm {Alt}(4)$ allows a concrete representation of the elements of G; indeed, each element of G is of the form $(k_1,k_2,k_3,k_4,k_5,k_6)\sigma $ for some $k_1,k_2,k_3,k_4,k_5,k_6\in K$ and

$$ \begin{align*} \sigma\in \langle (1\,3\,5)(2\,4\,6),(1\,2)(3\,4)\rangle\cong\mathrm{Alt}(4). \end{align*} $$

In particular, the element c defined in Notation 5.1 can now be written as

$$ \begin{align*} c=(1,1,1,1,k_1,k_2)(1\,3\,5)(2\,4\,6). \end{align*} $$

If $G_{(\Sigma _1)}$ is primitive on $\Delta _i$ , then $\Sigma _2=\Omega $ . Now, to establish part (2) of Theorem 1.1, it is necessary to omit the assumption in (4-1). This is deduced through computation: the derangement graph of every proper overgroup of $\mathrm {Alt}(4)$ in its degree 6 action contains a clique of size $4$ . Therefore, in the rest of the proof of Theorem 1.1, we may assume that $G_{(\Sigma _1)}$ is imprimitive on $\Delta _{i}$ , that is, $|\Omega |>6=|\Sigma _2|$ .

Lemma 5.2 has already shown that $|\Omega |$ is a multiple of $6$ .

Proof. Let $\Gamma = (V, E)$ be an $(a,b)$ -hypergraph that is hyper-transitive for X and suppose that the stabilizer of a point in X fixes some edge of $\Gamma $ . Since X acts transitively on both V and E, it follows that these sets are transitive X-sets.

If $|V| \leq 4$ , then it is clear that $\chi (\Gamma ) \leq 4$ because we may assign to each element of V a different color. Suppose instead that $|V|> 4$ . In particular, the action of X is either permutation isomorphic to the action of $\mathrm {Alt}(4)$ on the $2$ -subsets of $\{1, \ldots , 4\}$ or to the regular action of $\mathrm {Alt}(4)$ on itself by right multiplication.

Assume the first case holds. Then,

$$ \begin{align*} X = \langle (1\,6)(4\,5),\ (1\,6)(2\,3),\ (1\,2\,4)(3\,5\,6) \rangle. \end{align*} $$

Let $e = \{p_1, \ldots , p_b\} \in E$ . By hypothesis, up to replacing e by a suitable conjugate, we may assume $X_e \geq \langle (1\,6)(4\,5) \rangle $ . Thus, we have the following possibilities:

• • $X_e = X$ ;

• • $X_e = \langle (1\,6)(4\,5),\ (1\,6)(2\,3) \rangle $ ; or

• • $X_e = \langle (1\,6)(4\,5) \rangle $ .

If $X_e = X$ , then $E = \{e\}$ because X is transitive on E, and $e = \{\{1,2,3,4,5,6\}\}$ or ${e = \{\{1,2\}, \{3,4\}, \{5,6\}\}}$ , as $X_e$ must be transitive on the b parts of e. In this case, ${\chi (\Gamma ) = 2}$ . A 2-coloring is given by assigning $\chi (x) = \texttt {red}$ when $x \ne 6$ and $\chi (x) = \texttt {blue}$ when $x=6$ .

If $X_e = \langle (1\,6)(4\,5),\ (1\,6)(2\,3) \rangle $ , then

$$ \begin{align*} E = \{\{1,6\}\},\ \{\{4,5\}\},\ \{\{2,3\}\}. \end{align*} $$

A 2-coloring is given by assigning $\chi (x) = \texttt {red}$ for all $x \in \{1,3,5\}$ and $\chi (x) = \texttt {blue}$ for all $x \in \{2,4,6\}$ , so $\chi (\Gamma ) = 2$ .

Finally, if $X_e = \langle (1\,6)(4\,5) \rangle $ , then, up to relabeling the points, we may assume ${e = \{\{1,4\}, \{5,6\}\}}$ and

$$ \begin{align*} E =\{& \{\{1,3\}, \{2,6\}\}, \{\{2,5\}, \{3,4\}\}, \{\{1,5\}, \{4,6\}\}, \{\{1,2\}, \{3,6\}\},\\ &\{\{1,4\}, \{5,6\}\}, \{\{2,4\}, \{3,5\}\} \}. \end{align*} $$

In this case, $\chi (\Gamma ) \leq 3$ , since a proper 3-coloring is given by assigning $\chi (x) = \texttt {red}$ for $x \in \{1,3\}$ , $\chi (x) = \texttt {blue}$ for $x \in \{2,4\}$ , and $\chi (x) = \texttt {black}$ for $x \in \{5,6\}$ .

The case where the action of X is permutation isomorphic to the regular action of $\mathrm {Alt}(4)$ on itself by right multiplication can be treated in a similar manner.

Proof. Since $|\Omega |> |\Sigma _2 | = 6$ , G admits a system of imprimitivity $\Sigma _3\ne \Sigma _2$ that is a refinement of $\Sigma _2$ . Among all such systems of imprimitivity, choose one so that $|\Sigma _3 |$ is as small as possible. Just as in the proof of Lemma 4.1 and of Lemma 5.2, we may replace G with the permutation group induced by G on $\Sigma _3$ and, hence, we may suppose that $\Sigma _3 = \{\{\omega \} \mid \omega \in \Omega \}$ .

If $G_{(\Sigma _2)}=1$ , then G acts faithfully on $\Sigma _2$ and, hence, $G\cong \mathrm {Alt}(4)$ , because the permutation group induced by G on $\Sigma _2$ is $\mathrm {Alt}(4)$ in its degree-six action. Since $12=|G|\ge |\Omega |>|\Sigma _2|=6$ , we deduce $|\Omega |=12$ and G acts regularly on $\Omega $ , which is clearly a contradiction. Therefore,

(6-1) $$ \begin{align}G_{(\Sigma_2)}\ne 1. \end{align} $$

The minimality of $|\Sigma _3|$ implies that $G_{\{\Delta _{1,1}\}}$ is primitive in its action on $\Delta _{1,1}$ . If ${G_{(\Delta _{1,1})}= 1}$ , then the hypotheses of Proposition 3.5 are satisfied and, hence, $G_{(\Sigma _2)}$ contains a derangement in its action on $\Omega $ , which contradicts (5-1). This contradiction implies that

(6-2) $$ \begin{align} G_{(\Delta_{i,j})}\ne 1\quad\text{ for all } i\in \{1,2,3\}, j\in \{1,2\}. \end{align} $$

Suppose that the primitive permutation group $G_{\{\Delta _{i,j}\}}^{\Delta _{i,j}}$ has nonabelian socle. Summing up:

• • $G_{\{\Delta _{i,j}\}}$ acts primitively on $\Delta _{i,j}$ , by Notation 6.1;

• • $G_{(\Sigma _2)} \ne 1$ , by (6-1);

• • $G_{(\Delta _{i,j})} \ne 1$ , by (6-2); and

• • $G_{\{\Delta _{i,j}\}}^{\Delta _{i,j}}$ has nonabelian socle, by the current hypothesis.

Therefore, we are now in a position to apply Proposition 3.6 (with $\Sigma = \Sigma _2$ and ${\Delta = \Delta _{i,j}}$ ) and the notation therein. Since, by (5-1), $G_{(\Sigma _2)}$ has no derangements, we deduce that parts (1)–(5) of Proposition 3.6 are satisfied. In particular, by parts (c) and (d) of Proposition 3.6, there exists an $(a,b)$ -hypergraph $\Gamma $ that is hyper-transitive for $G/_{(\Sigma _2)} \cong \mathrm {Alt}(4)$ and with the property that $G_{\{\Delta _{i,j}\}} / G_{(\Sigma _2)} \cong C_2$ fixes some edge of $\Gamma $ . In particular, the stabilizer of a point in $\mathrm {Alt}(4)$ (in its action of degree $6$ ) fixes some edge of $\Gamma $ . From Lemma 6.2, $\chi (\Gamma ) \leq 4$ .

Observe that, by Burnside’s $p^\alpha q^\beta $ theorem, a nonabelian simple group has order divisible by at least three distinct primes and, hence, the nonabelian simple group T appearing in the statement of Proposition 3.6 has at least $4\, \mathrm {Aut}(T)$ -conjugacy classes. From Proposition 3.6 part (b) and the previous paragraph, we immediately obtain a contradiction.

Assume that $G_{\{\Delta \}}^\Delta $ has abelian socle. Let V be a minimal normal subgroup of G with $V\le G_{(\Sigma _2)}$ . Then, V is an elementary abelian p-group for some prime number p. From (5-1), we have

$$ \begin{align*} V=\bigcup_{\omega\in \Omega}V_\omega. \end{align*} $$

Since $V_\omega $ is normalized by V and by $G_\omega $ , we get $\mathbf {N}_G(V_\omega )\ge G_\omega V=G_{\{\Delta \}}$ and, hence, the union above consists of at most $|\Sigma _2|$ elements. Therefore,

$$ \begin{align*} p^a=|\Delta|=|V:V_{\omega}|<|\Sigma_2|=6. \end{align*} $$

Thus, $|\Omega |=p^a|\Sigma _2|\le 30$ . The rest of the result follows from a computer computation using the database of transitive groups of degree at most $30$ .

Proof. This follows from a computer computation; see also the proof of Lemma 6.2. Given G as above, we constructed all $(a,b)$ -hypergraphs $\Gamma $ such that G is hyper-transitive for $\Gamma $ and such that the stabilizer of a point in G fixes some edge of $\Gamma $ . We give some details of the computation. Let $\Gamma = (V, E)$ be hyper-transitive for G, where V is the set of vertices and E is the set of edges of $\Gamma $ . Since $\Gamma $ is hyper-transitive for G, it follows that V and E are transitive G-sets. In particular, this significantly narrows down the number of possibilities for V and for E. Indeed, V can be any transitive G-set. However, once V is fixed, edge transitivity implies that the set of edges E is uniquely determined by $e^G$ , where $e = \{p_1, \ldots , p_b\}$ is a fixed edge of $\Gamma $ , which, by definition, is a partition of an $ab$ -subset of V into b parts of cardinality a.

This partition e is in turn very limited by the assumption that the stabilizer of a point in G fixes e. Moreover, the choice of e is further restricted because, by definition, the edge stabilizer $G_e$ is transitive on the b parts of e, and either $G_e$ is also transitive on the underlying set of e or $a = 2$ . All of these conditions impose severe constraints on $\Gamma $ .

Then, for each $(a,b)$ -hypergraph $\Gamma = (V, E)$ , we randomly constructed $10^6$ maps $\eta : V \to \{1,2,3,4\}$ and checked whether each map was a coloring of $\Gamma $ . In all cases, we found such a map. Incidentally, these $(a,b)$ -hypergraphs may have a smaller chromatic number. The choice of $10^6$ bears no particular meaning; it was simply large enough to ensure that the computer could find a suitable coloring for each $\Gamma $ by random search within a reasonable amount of time.

Proof. Let $\Phi (X)$ denote the Frattini subgroup of X. If $N \nleq \Phi (X)$ , then there exists a maximal subgroup M with $N \nleq M$ . The maximality of M implies $X = NM$ .

For the final statement, we argue by induction on $|X|$ . If $|X| = 1$ , the result is trivial. If $N \leq \Phi (X)$ , then X is solvable because $X/N$ is solvable and $\Phi (X)$ is nilpotent. Hence, we may take $Y = X$ in this case.

If $N \nleq \Phi (X)$ , then as noted above, $X = NM$ for some maximal subgroup M of X. Since $|M| < |X|$ and $M/(M \cap N) \cong X/N$ is solvable, by the inductive hypothesis, there exists a solvable subgroup Y of M such that $M = Y(M \cap N)$ . It follows that $X = NM = Y(M \cap N)N = YN$ , as required.

Proof Theorem 1.1.

We use the notation established throughout Sections 4, 5, and 6.

Let $\Sigma _4$ be a system of imprimitivity for G that is a refinement of $\Sigma _3$ . Just as in the proof of Lemmas 4.1, 5.2, and 6.3, we may replace G with the permutation group induced by G on $\Sigma _4$ and, hence, we may suppose that $\Sigma _4 =\{\{\omega \} | \omega \in \Omega \}$ .

Suppose first $G_{(\Sigma _3)}=1$ . Then G, as an abstract group, is isomorphic to the permutation group induced by G on $\Sigma _3$ and, hence, it is isomorphic to the transitive group $(18,142)$ or $(30,126)$ in the library of transitive groups in Magma [Reference Bosma, Cannon and Playoust2]. We have checked with a computer that these two permutation groups of degree $18$ and $30$ do not admit a transitive action on a set $|\Omega |>|\Sigma _3|$ such that the derangement graph has no $4$ -cliques. Therefore, $G_{(\Sigma _3)}\ne 1$ . Let $\Delta \in \Sigma _3$ .

Suppose $G_{(\Delta )}=1$ . Then, by Proposition 3.5, $G_{(\Sigma _3)}$ admits a derangement for its action on $\Omega $ , which contradicts (5-1). Therefore, $G_{(\Delta )}\ne 1$ .

Assume that $G_{\{\Delta \}}^\Delta $ has nonabelian socle. We now apply Proposition 3.6 and the notation therein. Since by (5-1), $G_{(\Sigma _3)}$ has no derangements, we deduce that parts (1)–(5) of Proposition 3.6 are satisfied. Observe that, by Burnside’s $p^\alpha q^\beta $ theorem, a nonabelian simple group has order divisible by at least three distinct primes and, hence, the nonabelian simple group T appearing in the statement of Proposition 3.6 has at least $4\, \mathrm {Aut}(T)$ -conjugacy classes. From Proposition 3.6(b) and the previous paragraph, we immediately obtain a contradiction.

Assume that $G_{\{\Delta \}}^\Delta $ has abelian socle. Let V be a minimal normal subgroup of G with $V\le G_{(\Sigma _3)}$ . Then, V is an elementary abelian p-group for some prime number p. From (5-1), we have $V=\bigcup _{\omega \in \Omega }V_\omega .$ Since $V_\omega $ is normalized by V and by $G_\omega $ , we get $\mathbf {N}_G(V_\omega )\ge G_\omega V=G_{\{\Delta \}}$ and, hence, the union above consists of at most $|\Sigma _3|$ elements. Therefore,

$$ \begin{align*}p^a=|\Delta|=|V:V_{\omega}|<|\Sigma_3|. \end{align*} $$

Observe that, since $|G:G_{(\Sigma _1)}|=3$ and $G_{(\Sigma _1)}\unlhd G$ , if $p\ne 3$ , then every p-element of G is contained in $G_{(\Sigma _1)}$ . Hence, G has no derangements of p-power order with $p\ne 3$ by (5-1). Therefore, using Theorem 7.2:

• • $|\Sigma _3|=18$ and $|\Delta |\in \{2,3,4,5,7,8,9,11,16,17\}$ ; or

• • $|\Sigma _3|=30$ and $|\Delta |\in \{2,3,4,7,8,9,11,13,16,17,19,27,29\}$ .

We claim that G is solvable. Observe that $G/G_{(\Sigma _3)}$ is solvable and, hence, by Lemma 7.3, there exists $Y\le G$ with Y solvable and $G=G_{(\Sigma _3)}Y$ . In particular Y is transitive on $\Sigma _3$ and, hence, $VY$ is transitive on $\Omega $ . The minimality of G implies $G=YV$ and, hence, G is solvable.