Spreading primitive groups of diagonal type
do not exist

Let $G:=W(T)$, $M:=\rho_{1}(1\times T)$, $\omega_{1}:=1_{T}$, and $D:=G_{\omega_{1}}=\langle\rho_{2}(\operatorname{{\mathrm{Aut}}}(T)),\sigma\rangle\cong\operatorname{{\mathrm{Aut}}}(T)\times C_{2}$. Note that $G=DM$, as $M$ acts transitively on $\Omega$. Let $X:=A\subsetneq\Omega$. Then $\omega_{1}^{1\times B}=B<A=\omega_{1}^{1\times A}$, and $X^{G}=X^{DM}=\{A^{\tau}t\mid\tau\in\operatorname{{\mathrm{Aut}}}(T),t\in T\}$. Next, define $\Delta:={\{Y\in X^{G}\mid Y\cap\omega_{1}^{1\times A}\neq\varnothing\}}$, and observe that $\Delta=\{A^{\tau}a\mid\tau\in\operatorname{{\mathrm{Aut}}}(T),a\in A\}$. Thus the orbits of $1\times A$ on $\Delta$ are $\Delta_{\tau}=\{A^{\tau}a\mid a\in A\}$ for each $\tau\in\operatorname{{\mathrm{Aut}}}(T)$. Additionally, the stabiliser in $1\times A$ of $A^{\tau}\in\Delta_{\tau}$ is $1\times(A\cap A^{\tau})$. As $A=B(A\cap A^{\tau})$, we observe that $1\times B$ is transitive on $\Delta_{\tau}$. The result now follows from Theorem 2.2, applied to the groups $G$, $1\times A$ and $1\times B$. ∎

Let $\Sigma$ be the set of right cosets of $A$ in $R$. Then the point stabilisers in the action of $A$ on $\Sigma$ are precisely the subgroups $A\cap A^{t}$ with $t\in R$. As $B$ is transitive on each orbit in this action, it follows that $A=B(A\cap A^{t})$ for all $t\in R$. Hence we are done if $R=\operatorname{{\mathrm{Aut}}}(T)$. Assume therefore that all $\operatorname{{\mathrm{Aut}}}(T)$-conjugates of $A$ are conjugate in $T$. Then for each $\tau\in\operatorname{{\mathrm{Aut}}}(T)$, there exists $t\in T$ such that $A^{\tau}=A^{t}$, and so $B(A\cap A^{\tau})=B(A\cap A^{t})=A$. ∎

The set $X:=r^{T}$ and the multiset $J:=\Omega+s_{1}^{T}-s_{2}^{T}$ are clearly nontrivial, and (i) implies that $|J|=|\Omega|$. Thus by Definition 2.1, it remains to prove that $\sum_{x\in X^{g}}\mu_{J}(x)$ is constant for all $g\in G:=W(T)$. In fact, we will show that $|X^{g}\cap s_{1}^{T}|=|X^{g}\cap s_{2}^{T}|$ for all $g$, and it will follow immediately that $\sum_{x\in X^{g}}\mu_{J}(x)=|X|$ for all $g$.

As above, $G_{1_{T}}=\langle\rho_{2}(\operatorname{{\mathrm{Aut}}}(T)),\sigma\rangle$, and the transitivity of $M:=\rho_{1}(1\times T)$ on $\Omega$ implies that $G=G_{1_{T}}M$. Hence for each $g\in G$, there exist $\tau\in\operatorname{{\mathrm{Aut}}}(T)$, $\varepsilon=\pm 1$ and $t\in T$ such that $X^{g}=((r^{\tau})^{\varepsilon})^{T}t$. Letting $m:=(r^{\tau})^{\varepsilon}$, $i\in\{1,2\}$ and $h\in s_{i}^{T}$, we observe from [16, Problem 3.9] that the set $\{(x,y)\mid x\in m^{T},y\in t^{T},xy=h\}$ has size

where $\overline{\chi(s_{i})}$ is the complex conjugate of $\chi(s_{i})$. Since $a_{g,i}$ is constant for all $h\in s_{i}^{T}$, and $|m^{T}t^{u}\cap s_{i}^{T}|$ is constant for all $u\in T$, it follows that $|X^{g}\cap s_{i}^{T}|=a_{g,i}|s_{i}^{T}|/|t^{T}|$. As $\chi((r^{\tau})^{-1})=\overline{\chi(r^{\tau})}$ for each character $\chi$ of $T$, (i) and (ii) yield $|X^{g}\cap s_{1}^{T}|=|X^{g}\cap s_{2}^{T}|$ for all $g\in G$, as required. ∎

Let $p$ be the prime dividing $q$, and let $U$ be a Sylow $p$-subgroup of $T$. Then $T$ contains a subgroup $N$ such that $A:=N_{T}(U)$ and $N$ form a $(B,N)$-pair, and the normal subgroup $H:=A\cap N$ of $N$ complements $U$ in $A$ (see [12, Ch. 7.2–9.4]). Since $q\geqslant 3$ (and since $T$ is not the soluble group $A_{1}(3)=\operatorname{{\mathrm{PSL}}}_{2}(3)$), we observe from [12, pp. 121–122] that $|H|>1$, and so $U$ is a proper subgroup of $A$. Furthermore, as $A$ is the normaliser in $T$ of the Sylow subgroup $U$, all $\operatorname{{\mathrm{Aut}}}(T)$-conjugates of $A$ are conjugate in $T$.

By Lemma 2.4, it suffices to show that $A$ and $U$ have the same orbits in the action of $T$ on the set of right cosets of $A$. For each element $w$ of the Weyl group $W:=N/H$, fix a preimage $n_{w}\in N$ of $w$ under the natural homomorphism from $N$ to $W$. In addition, let $x$ and $y$ be elements of $T$ so that $Ax$ and $Ay$ lie in a common $A$-orbit. Then the double cosets $AxA$ and $AyA$ are equal. By [12, Theorem 8.4.3], there exist $a_{x},a_{y}\in A$, $w_{x},w_{y}\in W$ and $u_{x},u_{y}\in U<A$ such that $x=a_{x}n_{w_{x}}u_{x}$ and $y=a_{y}n_{w_{y}}u_{y}$. It follows that $An_{w_{x}}A=An_{w_{y}}A$. As $A$ and $N$ form a $(B,N)$-pair, [12, Proposition 8.2.3] yields $n_{w_{x}}=n_{w_{y}}$. Thus $AxU=An_{w_{x}}U=An_{w_{y}}U=AyU$, i.e. $Ax$ and $Ay$ lie in the same $U$-orbit, as required. ∎

Note that $T\not\cong{}^{2}F_{4}(2)^{\prime}$. As above, let $p$ be the prime dividing $q$, and let $U$ be a Sylow $p$-subgroup of $T$. By [12, Ch. 13–14], $T$ contains subgroups $N$, $H$ and $W$ analogous to those in the proof of Proposition 3.1, so that $A:=U\mkern 3.0mu{:}\mkern 3.0muH$ and $N$ form a $(B,N)$-pair, and $W=N/H$. Furthermore, $H$ is nontrivial for all $q$, as $T$ is not isomorphic to ${}^{2}A_{2}(2)=\operatorname{{\mathrm{PSU}}}_{3}(2)$, ${}^{2}B_{2}(2)=\mathrm{Sz}(2)$ or ${}^{2}F_{4}(2)$. Although [12] directly states only that $A\leqslant N_{T}(U)$, we can show that $A=N_{T}(U)$. Indeed, if this were not the case, then some preimage $n_{w}\in N$ of a fundamental reflection $w\in W$ would normalise $U$ (see [12, Proposition 8.2.2 & Theorem 8.3.2]). However, we deduce from Proposition 13.6.1 and the proof of Theorem 13.5.4 in [12] that no such $w$ exists, and so $A=N_{T}(U)$. We now proceed exactly as in the proof of Proposition 3.1, this time using Proposition 13.5.3 of [12] instead of Theorem 8.4.3. ∎

Since $A_{2}(2)=\operatorname{{\mathrm{PSL}}}_{3}(2)\cong\operatorname{{\mathrm{PSL}}}_{2}(7)$ is addressed by Proposition 3.1, and since the remaining groups $X_{\ell}(2)$ with $\ell\leqslant 2$ are not simple, we shall assume that $\ell\geqslant 3$. Let $\Phi\subseteq\mathbb{R}^{\ell}$ be a root system for $T$, with $\Phi^{+}\subseteq\Phi$ a system of positive roots and $\Pi\subseteq\Phi^{+}$ a system of simple roots (see, for example, [12, Ch. 2]). Then $\Phi$ is the disjoint union of $\Phi^{+}$ and $-\Phi^{+}$, and each node in the Dynkin diagram $D$ corresponding to $T$ is associated with a unique root in $\Pi$ (and vice versa). Hence each diagram automorphism of $D$ induces a permutation of $\Pi$. Let $r$ and $s$ be the simple roots associated with a leaf in $D$ and the adjacent node, respectively. Additionally, let $S$ be the closure of the set $\{s\}$ under the group of diagram automorphisms of $D$, so that $|S|\in\{1,2\}$.

Next, let $U$, $N$ and $W$ be as in the proof of Proposition 3.1. In this case, the group $H$ from that proof is trivial, and so $N_{T}(U)=U$ and $W\cong N$. By [12, Proposition 2.1.8, p. 68 & p. 93], $T$ is generated by a set $\{x_{u}\mid u\in\Phi\}$ of involutions ($x_{u}$ is written as $x_{u}(1)$ in [12]), with $U=\langle x_{u}\mid u\in\Phi^{+}\rangle$, and $N$ is generated by a set $\{n_{v}\mid v\in\Pi\}$ of involutions. Let $A:=\langle U,n_{v}\mid v\in\Pi\setminus S\rangle$ be the parabolic subgroup of $T$ corresponding to the subset $\Pi\setminus S$ of $\Pi$. Then by [12, Ch. 8.5] and the well-known structure of parabolic subgroups of Chevalley groups, $A$ has shape $[2^{m}]:(L_{\{r\}}\times L_{\Pi\setminus(\{r\}\cup S)})$ for some positive integer $m$, where $L_{\{r\}}\cong A_{1}(2)\cong S_{3}$ and $L_{\Pi\setminus(\{r\}\cup S)}$ are Levi subgroups. Since $q=2$, all automorphisms of $T$ are products of inner and graph automorphisms. Moreover, $\Pi\setminus S$ is fixed setwise by all diagram automorphisms of $D$, which correspond to the graph automorphisms of $T$, and so all $\operatorname{{\mathrm{Aut}}}(T)$-conjugates of $A$ are conjugate in $T$. In addition, $A$ contains an index two subgroup $B$ of shape $[2^{m}]:(F\times L_{\Pi\setminus(\{r\}\cup S)})$, where $F:=L_{\{r\}}^{\prime}\cong C_{3}$. We will show that $AtA=AtB$ for all $t\in T$, and the result will follow from Lemma 2.4.

Since $N_{T}(U)=U\leqslant A$, we deduce as in the proof of Proposition 3.1 that, for each $t\in T$, there exists $n\in N$ such that $AtA=AnA$. Thus it suffices to prove that $AnA=AnB$ for all $n\in N$ (it will immediately follow that $AnB=AtB$ for the corresponding $t\in T$). We also observe using [12, Ch. 8.5] (and the fact that $F$ has index two in $L_{\{r\}}$) that $x_{r}\in A\setminus B$ and $x_{-r}x_{r}\in B$. The former containment implies that $AnA=AnB\cup Anx_{r}B$. To complete the proof, we will show that $Anx_{r}B=AnB$.

The Weyl group $W$ acts linearly on $\mathbb{R}^{\ell}$ and fixes $\Phi$. Let $w$ be the element of $W$ corresponding to $n$. By [12, Lemma 7.2.1(i)], $nx_{r}=x_{w(r)}n$ and $x_{-w(r)}n=x_{w(-r)}n=nx_{-r}$. If $w(r)\in\Phi^{+}$, then $x_{w(r)}\in\langle x_{u}\mid u\in\Phi^{+}\rangle=U\leqslant A$, and so $Anx_{r}=An$. Otherwise, $-w(r)\in\Phi^{+}$ and $x_{-w(r)}\in A$, yielding $An=Anx_{-r}$ and hence $Anx_{r}=Anx_{-r}x_{r}$, which lies in $AnB$ by the previous paragraph. In either case, we see that $Anx_{r}B=AnB$, as required. ∎

Since $A_{5}\cong\operatorname{{\mathrm{PSL}}}_{2}(4)=A_{1}(4)$ and $A_{6}\cong\operatorname{{\mathrm{PSL}}}_{2}(9)=A_{1}(9)$ are addressed by Proposition 3.1, we shall assume that $n\geqslant 7$. Let $\Sigma$ be the set of $3$-subsets of a set of size $n$, and let $\alpha\in\Sigma$. Then $A:=T_{\alpha}=(S_{3}\times S_{n-3})\cap T$, and all $\operatorname{{\mathrm{Aut}}}(T)$-conjugates of $A$ are conjugate in $T$. Additionally, $A$ has four orbits on $\Sigma$, namely $\{\alpha\}$ and $\{\beta\in\Sigma\mid|\beta\cap\alpha|=i\}$ for $i\in\{0,1,2\}$. Since the index two subgroup $B:=A_{3}\times A_{n-3}$ of $A$ acts transitively on each of these orbits, and since the action of $T$ on $\Sigma$ is equivalent to its action on the right cosets of $A$, the result follows from Lemma 2.4. ∎

We observe from the Atlas [13] that $T$ has a maximal subgroup $A$, as specified in Table 1, such that either $T=\mathrm{O^{\prime}N}$ or all $\operatorname{{\mathrm{Aut}}}(T)$-conjugates of $T_{\alpha}$ are conjugate in $T$. It is also clear that the group $B$ in the table is a proper normal subgroup of $A$. Note that if $T=\mathrm{M}_{12}$, then $A$ has two normal subgroups isomorphic to $S_{5}$; in what follows, $B$ may be chosen as either.

Now, let $R:=\operatorname{{\mathrm{Aut}}}(T)$ if $T=\mathrm{O^{\prime}N}$, and $R:=T$ otherwise. Additionally, let $\chi$ be the permutation character corresponding to the action of $R$ on the set $\Sigma$ of right cosets of $A$ in $R$. Then for each $t\in R$, the number of points in $\Sigma$ fixed by $t$ is equal to $\chi(t)$. By the Cauchy–Frobenius Lemma, $A$ and $B$ have $c_{A}:=\frac{1}{|A|}\sum_{a\in A}\chi(a)$ and $c_{B}:=\frac{1}{|B|}\sum_{b\in B}\chi(b)$ orbits on $\Sigma$, respectively. It is straightforward to compute $c_{A}$ and $c_{B}$ in GAP [15] using the Character Table Library [9] ($c_{A}$ is most readily calculated as the inner product $[\chi,\chi]$), and in each case we obtain $c_{A}=c_{B}$. Thus Lemma 2.4 yields the result. ∎

Propositions 3.1–3.5 show that (iii) implies (i), which clearly implies (ii). To complete the proof, we will show that if (iii) does not hold, then neither does (ii). If $T\in\{\mathrm{J}_{1},\mathrm{M}_{22},\mathrm{J}_{3},\mathrm{McL}\}$, then we construct $T$ in Magma [7] via the AutomorphismGroupSimpleGroup and Socle functions, and show that (ii) does not hold via fast, direct computations.

Suppose next that $T=\mathrm{Th}$, let $C$ be a proper subgroup of $T$, and let $A$ be a subgroup of $C$. Observe that if $b(T,C)=2$, then (as discussed below the statement of Theorem 1.3) the two-point stabiliser $C\cap C^{t}$ is trivial for some $t$. Hence $A\cap A^{t}=1$, and so $A$ does not satisfy (ii). In general, $b(T,A)\leqslant b(T,C)$. By [11, Theorem 1], $T$ has only two maximal subgroups (up to conjugacy) with corresponding base size greater than two, namely $M_{1}:={}^{3}D_{4}(2)\mkern 3.0mu{:}\mkern 3.0mu3$ and $M_{2}:=2^{5}.\operatorname{{\mathrm{PSL}}}_{5}(2)$, and $b(T,M_{1})=b(T,M_{2})=3$. Hence any proper subgroup $A$ of $T$ satisfying (ii) is a non-simple subgroup of $M_{1}$ or $M_{2}$ with $b(T,A)=3$.

Now, let $\Sigma$ be the set of right cosets in $T$ of a proper subgroup $A$, and let $c$ be the number of orbits of $A$ on $\Sigma$. If $b(T,A)\geqslant 3$, then $|A\cap A^{t}|\geqslant 2$ for all $t\in T$, and it follows from the Orbit-Stabiliser Theorem that $c|A|/2\geqslant|\Sigma|$. If $A$ is a maximal subgroup of either $M_{1}$ or $M_{1}^{\prime}\cong{}^{3}D_{4}(2)$, then the character table of $A$ is included in The GAP Character Table Library. Thus for every such $A\neq M_{1}^{\prime}$, we can compute $c$ as in the proof of Proposition 3.5, and we see that in fact $c|A|/2<|\Sigma|$, and hence $b(T,A)=2$. To show that $b(T,A)=2$ for each maximal subgroup $A$ of $M_{2}$, we construct $T$ and $M_{2}$ in Magma as subgroups of $\operatorname{{\mathrm{GL}}}_{248}(2)$ using the respective generating pairs $\{x,y\}$ and $\{w_{1},w_{2}\}$ given in [19]. Magma calculations (with a runtime of less than 10 minutes and a memory usage of 1.5 GB) then show that $A\cap A^{y}=1$ for all representatives $A$ of the conjugacy classes of maximal subgroups of $M_{2}$ returned by the MaximalSubgroups function. Therefore, $M_{1}$ and $M_{2}$ are the only non-simple subgroups of $T$ with corresponding base size at least three. Further character table computations in GAP show that if $A\in\{M_{1},M_{2}\}$, then for each proper normal subgroup $B$ of $A$, the number of $B$-orbits on $\Sigma$ is greater than the number of $A$-orbits. An additional application of the Orbit-Stabiliser Theorem yields $B(A\cap A^{t})<A$ for some $t\in T$. Therefore, (ii) does not hold.

Finally, suppose that $T=\mathbb{M}$. As above, any proper subgroup $A$ of $T$ satisfying (ii) also satisfies $b(T,A)\geqslant 3$. By [10, Theorem 3.1], the unique (up to conjugacy) proper subgroup $A$ of $T$ with $b(T,A)\geqslant 3$ is the maximal subgroup $K:=2.\mathbb{B}$, with $b(T,K)=3$. Since $K$ is quasisimple, its centre $Z$ of order two is its unique nontrivial proper normal subgroup, and $Z$ lies in each maximal subgroup of $K$. Hence $Z(K\cap K^{t})<K$ for all $t\in T\setminus K$, and so (ii) does not hold. ∎

As each spreading primitive group is synchronising, Theorem 1.1 implies that a spreading primitive group of diagonal type has socle $T\times T$, for some non-abelian finite simple group $T$. For each $T$, let $\Omega:=T$, and let $W(T)$ be the subgroup of $\operatorname{{\mathrm{Sym}}}(\Omega)$ defined in §2. Recall also that each subgroup of a non-spreading group is non-spreading. Thus it suffices to show that $W(T)$ is non-spreading for each $T$. If $T\notin\{\mathrm{J}_{1},\mathrm{M}_{22},\mathrm{J}_{3},\mathrm{McL},\mathrm{Th},\mathbb{M}\}$, then this is an immediate consequence of Theorem 1.3 and Corollary 2.3.

For each of the six remaining groups $T$, let $r$, $s_{1}$ and $s_{2}$ be members of the conjugacy classes of $T$ given in Table 2. By inspecting the character table for $T$ in [13], we observe that these elements satisfy Properties (i) and (ii) of Lemma 2.5. Therefore, that lemma shows that $W(T)$ is non-spreading. ∎