Proof. Let D and B be normal in G and such that $D/B \cong N = T^s$ is a chief factor of G. Since N is a minimal normal subgroup of G, either $D \cap N = 1$ or $D \cap N = N$ . The latter implies that $N \leqslant D$ , and all assertions follow in this case. We may therefore assume that $D \cap N = 1$ and so D centralises N, that is, $D \leqslant C_G(N)$ . Since $|C_G(N)| \leqslant |N|$ (see [Reference Praeger and Schneider28, Theorem 3.2(i)]) we have

$$\begin{align*}|N| = |D/B| \leqslant |D| \leqslant |C_G(N)| \leqslant |N|. \end{align*}$$

In particular, $B=1$ , so D is a minimal normal subgroup of G, and also $D=C_G(N)$ . Hence, [Reference Praeger and Schneider28, Theorem 3.2] shows that D is regular, and thus D itself is a plinth. In particular, D and N are distinct plinths.

Proof. (a) Let $N\in \mathcal {G}(L/K,M)$ and let $N':= N\cap \widehat {K}$ . Then since N and $\widehat {K}$ are both Galois extensions of K, it follows by our comments above that $N'/K$ is a Galois extension. Also $L\not \leqslant N'$ (since $L\not \leqslant N$ ) so in particular $N'\ne \widehat {K}$ . Thus $N'\in \mathcal {G}(L/K,\widehat {K})$ . Further, $ \mathcal {G}(L/K,\widehat {K}) \subseteq \mathcal {G}(L/K,M)$ by definition, and so each element $N\in \mathcal {G}(L/K,\widehat {K})$ lies in $\mathcal {G}(L/K,M)$ and satisfies $N=N\cap \widehat {K}$ . Thus part (a) is proved.

(b) It follows from part (a) that, if $L/K$ is Galois atomic relative to M, then it is also Galois atomic relative to $\widehat {K}$ . Conversely, suppose that $L/K$ is Galois atomic relative to $\widehat {K}$ , and let $N\in \mathcal {G}(L/K,M)$ . By part (a), $N\cap \widehat {K}\in \mathcal {G}(L/K,\widehat {K})$ , and hence, since $L/K$ is Galois atomic relative to $\widehat {K}$ , we have $(N\cap \widehat {K})\cap L=K$ . Since $L\leqslant \widehat {K}$ , this implies that $K=(N\cap \widehat {K})\cap L=N\cap L$ . Thus $L/K$ is Galois atomic relative to M, and part (b) is proved.

(c) First suppose that $L/K$ is innate relative to $\widehat {K}$ , so there exists $N'\in \mathcal {G}(L/K,\widehat {K})$ that is maximal by inclusion and satisfies $N'\cap L=K$ . By part (a), $N'$ is of the form $N'=N\cap \widehat {K}$ for some $N\in \mathcal {G}(L/K,M)$ . Let $R\in \mathcal {G}(L/K,M)$ be maximal by inclusion such that R contains N. Then $R\cap \widehat {K}$ does not contain L since $R\in \mathcal {G}(L/K,M)$ , so in particular $R\cap \widehat {K}\ne \widehat {K}$ . Also $(R\cap \widehat {K})/K$ is a Galois extension since both $R/K$ and $\widehat {K}/K$ are Galois extensions, and hence $R\cap \widehat {K}\in \mathcal {G}(L/K,\widehat {K})$ . Since $N'\in \mathcal {G}(L/K,\widehat {K})$ is maximal by inclusion, and since R contains N, we conclude that $R\cap \widehat {K}=N'=N\cap \widehat {K}$ . Hence $K=N'\cap L= (R\cap \widehat {K})\cap L = R\cap L$ , and it follows that $L/K$ is innate relative to M.

Conversely, suppose that $L/K$ is innate relative to M, so there exists $N\in \mathcal {G}(L/K,M)$ that is maximal by inclusion and satisfies $N\cap L=K$ . By part (a), the intersection $N\cap \widehat {K}\in \mathcal {G}(L/K,\widehat {K})$ and we have $(N\cap \widehat {K})\cap L=N\cap L=K$ . Suppose that $N\cap \widehat {K}$ is not maximal by inclusion in $\mathcal {G}(L/K,\widehat {K})$ , and let $R'\in \mathcal {G}(L/K,\widehat {K})$ such that $R'$ properly contains $N\cap \widehat {K}$ . Then $N\cap \widehat {K}<R'<\widehat {K}$ and $R'/K$ is a Galois extension. Let R be the subfield generated by N and $R'$ , and write $R=R'N$ and note that $R/K$ is a Galois extension. Then, since the lattice of Galois extensions of K contained in M is a modular lattice, we have $R\cap \widehat {K}= (R'N)\cap \widehat {K} = R'(N\cap \widehat {K})$ . Also $R'(N\cap \widehat {K})=R'$ since $N\cap \widehat {K}<R'$ . It follows that $R\cap \widehat {K}=R'$ . In particular $R\ne M$ (since $R'<\widehat {K}$ ) and $R\ne N$ (since $N\cap \widehat {K}<R'$ ), so $N < R < M$ . The maximality of N in $\mathcal {G}(L/K,M)$ now implies that $L\leqslant R$ , and hence also $L\leqslant R\cap \widehat {K}$ . However, $R\cap \widehat {K}=R'$ and since $R'\in \mathcal {G}(L/K,\widehat {K})$ , the field R does not contain L and we have a contradiction. Thus $N\cap \widehat {K}$ is maximal by inclusion in $\mathcal {G}(L/K,\widehat {K})$ . Thus, since $(N\cap \widehat {K})\cap L=K$ , it follows that $L/K$ is innate relative to $\widehat {K}$ , proving part (c).

Finally, suppose that $M'/K$ is a second finite Galois extension such that $K<L\leq M'$ , and let $\widehat {K}'$ be a Galois closure of $L/K$ in $M'$ . Now the Galois closures $\widehat {K}/K$ and $\widehat {K}'/K$ of $L/K$ are isomorphic; see [Reference Stewart32, Theorem 10.3]. Thus either of the properties of $L/K$ being Galois atomic or innate holds relative to $\widehat {K}$ if and only if it holds relative to $\widehat {K}'$ . The final assertion now follows from parts (b) and (c).

Proof. The subgroup U is a point stabiliser in the transitive action of G on $[G:U]$ , and $\overline {G}$ is primitive if and only if U is maximal in G (see, e.g., [Reference Praeger and Schneider28, Corollary 2.15]). By the Galois Correspondence, this holds if and only if the only subfields $L'$ satisfying $K\leqslant L'\leqslant L$ are K and L, that is, the extension $L/K$ is atomic. Thus (a) is proved, and clearly the property of $L/K$ being atomic is independent of the choice of Galois extension M containing L.

The properties in (b) and (c) both involve transitivity conditions for certain normal subgroups of $\overline {G}$ , so we first examine these. We have $\overline {G}=G/X$ , where $X=\cap _{g\in G} U^g$ is the kernel of the action of G on $[G:U]$ . Each normal subgroup H of G corresponds to the normal subgroup $\overline {H}=HX/X$ of $\overline {G}=G/X$ , and $HX/X$ is the trivial group if and only if $H\leqslant X$ , or equivalently, if and only if $H\leqslant U$ . In terms of intermediate subfields, for the subfield $M_H$ corresponding to H, $M_H/K$ is a Galois extension, and $\overline {H}$ is trivial if and only if $L=M_U\leqslant M_H$ . Suppose that $\overline {H}=HX/X\ne 1$ , so $L\not \leqslant M_H$ , and note that the stabiliser of the point U in the group $\overline {G}$ is $\overline {U}=UX/X$ . Then $\overline {H}$ is transitive on $[G:U]$ if and only if $\overline {G}=\overline {H}\,\overline {U}$ (see [Reference Praeger and Schneider28, Lemma 2.11]), that is, $G=(HX)(UX)= HU$ (as $X\leqslant U$ ). As noted above, the intermediate field $M_H\cap L = M_H\cap M_U$ corresponds to the subgroup $\langle H, U\rangle = HU$ (since $H\unlhd G$ ), and so $G=HU$ if and only if $M_H\cap L = M_G = K$ . In summary, a normal subgroup $H\unlhd G$ is nontrivial on $[G:U]$ if and only if $L\not \leqslant M_H$ , and H is transitive on $[G:U]$ if and only if $M_H\cap L = K$ .

To complete the proofs of parts (b) and (c), recall that $\mathcal {G}(L/K,M)$ is the set of all subfields N such that $K\leqslant N < M$ , $L\not \leqslant N$ , and $N/K$ is a Galois extension. Then $K\in \mathcal {G}(L/K,M)$ , so $\mathcal {G}(L/K,M)\ne \emptyset $ . Also the subfields $N\in \mathcal {G}(L/K,M)$ are precisely those of the form $N=M_H$ such that $H\unlhd G$ (since $N/K$ is a Galois extension) and $\overline {H}\ne 1$ (since $L\not \leqslant N$ ). Therefore $\overline {G}$ is quasiprimitive on $[G:U]$ if and only if each such $\overline {H}$ is transitive on $[G:U]$ , equivalently $M_H\cap L=K$ . In other words, the G-action on $[G:U]$ is quasiprimitive if and only if $L/K$ is Galois atomic, proving part (b).

For part (c), $\overline {G}$ is innately transitive on $[G:U]$ if and only if there exists a normal subgroup H of G containing X such that $H/X$ is a minimal normal subgroup of $\overline {G}$ and H is transitive on $[G:U]$ . Such a subgroup H corresponds to a subfield $N=M_H\in \mathcal {G}$ that is maximal by inclusion (as $H/X$ is a minimal normal subgroup of $\overline {G}$ ) and satisfies $N\cap L=K$ (since H is transitive). Thus $\overline {G}$ is innately transitive on $[G:U]$ if and only if $L/K$ is innate, proving part (c).

Proof of Theorem 1.9.

We assume the hypotheses of Theorem 1.9, so we have a global field k, a finite Galois field extension $K/k$ of degree $n>1$ , and a finite extension L of K such that $L\ne K$ , L and K are weakly Kronecker equivalent, and further we assume that the extension $L/K$ is innate. For the definition of an innate extension, we take a finite extension M of L such that $M/k$ is Galois (noting that the property is independent of the choice of M by Lemma 2.6). Let $A= {\mathrm {Aut}}(M/k)$ , and let $A>G> U\geqslant 1$ be the subgroup chain corresponding to the subfield chain $k<K< L\leqslant M$ (that is, $M_A=k, M_G=K, M_U=L,$ and $M_{\{1\}}=M$ , using the notation above for the Galois correspondence). Thus $G\unlhd A$ of index n (as $K/k$ is a Galois extension of degree n), and the action of G by right multiplication on the set $[G:U]$ of right cosets of U is innately transitive by Lemma 2.7(c) (since $L/K$ is innate). Also the A-prime-power-covering sets $P_A(G)$ and $P_A(U)$ are equal (since $K/k$ and $L/k$ are weakly Kronecker equivalent; see [Reference Stern31, Theorem 1.8] or [Reference Lochter19, Theorem 3^′]). Thus the hypotheses of Theorem 1.6 are satisfied. Therefore $|G:U|\leqslant f(n)$ for the increasing integer function f of Theorem 1.6. By the Fundamental Theorem of Galois Theory, $|G:U|$ is the degree of $L/K$ . In addition, by Theorem 1.6, the action of G on $[G:U]$ is primitive, and hence the extension $L/K$ is atomic by Lemma 2.7(a). This completes the proof of Theorem 1.9.

Proof. Let $S = \mathrm {Soc}(G)$ and let X be a minimal normal subgroup of G, so $X\leqslant S$ . Since $X \neq 1$ and G acts faithfully on $\Omega $ , there exists i such that $X^{\Omega _i}\neq 1$ , and so $X^{\Omega _i}$ is a minimal normal subgroup of $G^{\Omega _i}$ . Since $G^{\Omega _i}$ is an affine primitive group, the induced group $X^{\Omega _i} = \mathrm {Soc}(G^{\Omega _i})$ , which is elementary abelian, and we also have $S^{\Omega _i} = \mathrm {Soc}(G^{\Omega _i})$ . Since $X\leqslant S\leqslant \prod _{i=1}^n S^{\Omega _i}$ , it follows from the minimality of X that $X \cong X^{\Omega _i}=S^{\Omega _i}$ . This implies that X and S are elementary abelian groups. In particular, every element of S is a p-element for some prime p. Since A acts transitively on $\{\Omega _1,\ldots ,\Omega _n\}$ , it follows that for any i we have $S^{\Omega _i} = \mathrm {Soc}(G^{\Omega _i})$ and so S acts regularly on each $\Omega _i$ . In particular, $U\cap S=S_\alpha $ fixes $\Omega _1$ pointwise, so $U \cap S = S_{(\Omega _1)}$ and $|S:U\cap S|=|\Omega _1|=|G:U|$ . For each $a\in A$ , the image $\Omega _1^a=\Omega _i$ for some i, and we have $(U \cap S)^a=U^a\cap S = S_{(\Omega _i)}$ . By Lemma 5.2, the equality $P_A(G)=P_A(U)$ implies that $P_A(S)=P_A(U \cap S)$ . Therefore, since every element of S is a p-element,

$$\begin{align*}S= P_A(S)= \bigcup_{a\in A} (U\cap S)^a = \bigcup _{i=1}^n S_{(\Omega_i)}. \end{align*}$$

Hence $|S|\leqslant n\cdot |U\cap S|$ , and so $|G:U|=|S:U\cap S|\leqslant n$ .

Proof. For $1\leqslant i \leqslant n$ , let $P_i \leqslant G^{\Omega _i}$ denote the unique plinth of $G^{\Omega _i}$ and recall that $K_i = G_{(\Omega _i)}$ , the kernel of the action of G on $\Omega _i$ . Thus for any subgroup H of G we have $H^{\Omega _i} = HK_i/K_i$ . Further, let $E_i$ be the full preimage in G of $P_i$ . This means that $K_i < E_i$ , and if $H \leqslant G$ is such that $H^{\Omega _i} = P_i$ , then $H \leqslant E_i$ .

Let $\mathcal X_i = \{ X \unlhd G \mid X^{\Omega _i} = P_i \}$ , the set of normal subgroups of G that project to $P_i$ . In particular, $E_i \in \mathcal X_i$ , so $\mathcal X_i \neq \emptyset $ .

Let $X, Y \in \mathcal X_i$ . Since X and Y are normal in G, the intersection $X \cap Y$ is normal in G, and moreover $[X,Y] \leqslant X \cap Y$ with $[X,Y] \unlhd G$ . Now, $[X,Y]^{\Omega _i} = [X^{\Omega _i}, Y^{\Omega _i}] = [P_i,P_i] =P_i$ (as $P_i=T^s$ is nonabelian). In particular, $P_i \leqslant (X\cap Y)^{\Omega _i} \leqslant X^{\Omega _i} = P_i$ , and so $X\cap Y \in \mathcal X_i$ . Thus $\mathcal X_i$ is indeed closed under intersection, proving Claim 1.

For $1\leqslant i \leqslant n$ , we set:

$$\begin{align*}S_i := \bigcap _{X \in \mathcal X_i} X.\end{align*}$$

Notice that $S_i \unlhd G$ , and by Claim 1, $S_i\in \mathcal {X}_i$ . Thus $S_i$ is the unique minimal element of $\mathcal {X}_i$ (by inclusion), and if $X \unlhd G$ and $X^{\Omega _i} = P_i$ , then $S_i \leqslant X$ . Further, since A acts transitively on $\{\Omega _1, \ldots , \Omega _n\}$ and normalises G, we see that A permutes the set $\{\mathcal {X}_1,\ldots ,\mathcal {X}_n\}$ transitively, and also the set of subgroups $\{S_1,\ldots ,S_n\}$ transitively. In particular, $|S_i| = |S_j|$ for all $i,j$ .

Let X be such a subgroup. Then $X^{\Omega _i}$ is normal in $G^{\Omega _i}$ and $X^{\Omega _i} \leqslant S_i^{\Omega _i}$ . Since $S_i /(S_i \cap K_i) = S_i^{\Omega _i} = P_i$ is a minimal normal subgroup of $G^{\Omega _i}$ , either $X^{\Omega _i}=S_i^{\Omega _i}=P_i$ or $X^{\Omega _i}=1$ . In the former case, $X^{\Omega _i} = P_i$ and hence $X \in \mathcal X_i$ , which implies that $S_i \leqslant X$ and thus $X=S_i$ since $X \leqslant S_i$ by assumption. In the latter case, $X \leqslant K_i$ and so $X\leqslant S_i \cap K_i$ , as required.

Since $S_i \cap K_j$ is normal in G and contained in $S_i$ , it follows from Claim 2 that either (a) $S_i \cap K_j = S_i$ , or, (b) $S_i \cap K_j \leqslant S_i \cap K_i$ . In case (a), we see $S_i=S_i \cap K_j \leqslant K_j$ , so $S_i^{\Omega _j}=1$ . Suppose now that case (b) holds. By the Third Isomorphism Theorem,

$$\begin{align*}P_i = S_i / (S_i \cap K_i) \cong (S_i / (S_i \cap K_j)) /( (S_i \cap K_i) / (S_i \cap K_j)).\end{align*}$$

This means that $S_i^{\Omega _j} \cong S_i/(S_i\cap K_j)$ is a normal subgroup of $G^{\Omega _j}$ that contains a chief factor isomorphic to $P_i \cong T^s \cong P_j$ . Applying Lemma 2.4 to $G^{\Omega _j}$ , we see that $S_i^{\Omega _j}$ contains a plinth of $G^{\Omega _j}$ . Since $G^{\Omega _j}$ has a unique plinth, namely $P_j$ , it follows that there exists M normal in G such that $S_i \cap K_j \leqslant M \leqslant S_i$ and $M^{\Omega _j}= P_j$ . By the minimality of $S_j$ , we have $S_j \leqslant M \leqslant S_i$ , and since $|S_i | = |S_j|$ , we have $S_i = M = S_j$ , and hence also $S_i^{\Omega _j} = S_j^{\Omega _j} = P_j$ , and the claim is proved.

Let X be a minimal normal subgroup of G with $X\leqslant S_i$ . Since X is nontrivial, there exists j such that $X^{\Omega _j} \neq 1$ . Then $1\ne X^{\Omega _j} \leqslant S_i^{\Omega _j}$ , and it follows from Claim 3 that $X^{\Omega _j}=S_i^{\Omega _j} = P_j$ and $S_i=S_j$ . By the minimality of $S_j$ , we have $ S_j \leqslant X \leqslant S_i = S_j$ , and so $S_i=X$ and therefore $S_i$ is a minimal normal subgroup of G.

An immediate consequence of Claim 4 is that $S_i\cap K_i=1$ and $S_i \cong S_i^{\Omega _i} = P_i \cong T^s$ . Together with the uniqueness from Claim 3, this gives part (a) of the lemma.

We define a relation $\sim $ on the set $\{1,\ldots ,n\}$ as follows: $i\sim j$ if and only if $S_i = S_j$ . Clearly this is an equivalence relation, and the relation is preserved by A. Thus, for some positive integer t that divides n, there are exactly t equivalence classes of $\sim $ . After relabelling, we may assume that $S_1$ , …, $S_t$ are representatives of the equivalence classes. Since each $S_i$ is a nonabelian minimal normal subgroup of G, we have

$$\begin{align*}S:= \langle S_1,\ldots, S_n \rangle = \langle S_1,\ldots,S_t \rangle \cong S_1 \times \cdots \times S_t \end{align*}$$

and since each $S_i \cong T^s$ , the group $S \cong T^{st}$ is characteristically simple. Let $1 \leqslant i \leqslant t$ . For any $1\leqslant j \leqslant t$ with $j \neq i$ , we have $S_i \neq S_j$ , so Claim 3 implies that $S_j^{\Omega _i}=1$ . Hence $S^{\Omega _i} = S_1^{\Omega _i} \times \cdots \times S_t^{\Omega _i} = S_i^{\Omega _i}$ . This completes the proof.

Proof. We will assume for a contradiction that $P_A(G)=P_A(U)$ . Let $S=S_1\times \cdots \times S_t$ be the subgroup defined by Lemma 5.6. By that same lemma, $S_1$ is the unique minimal normal subgroup of G contained in S such that $S^{\Omega _1}=S_1^{\Omega _1}$ . In particular, $S_j^{\Omega _1}=1$ for all j such that $2\leqslant j\leqslant t$ . Hence $S \cap K_1 = S_2 \times \cdots \times S_t$ .

By the definition of $\Omega _1$ , the group U contains $K_1$ , and since we are assuming that $\mathrm {Core}_A(U)=1$ , it follows that U does not contain $S_1$ . Hence $U\cap S=(U\cap S_1)\times K_1$ and $U\cap S_1$ is a proper subgroup of $S_1$ . Also, A permutes transitively by conjugation the minimal normal subgroups $S_i$ of G that are contained in S. For $1\leqslant i\leqslant t$ , choose an element $a_i\in A$ such that $S_1^{a_i}=S_i$ , and take $a_1=1$ . Let $A_1 = N_A(S_1)$ . The conjugation action of A on S induces a homomorphism $\varphi : A\to {\mathrm {Aut}}(S)$ with the property that each element of A induces an automorphism that permutes the $S_i$ and hence lies in $ {\mathrm {Aut}}(T^s)\wr {\mathrm {Sym}}(t)$ . We identify each of the $S_i$ with $T^s$ so that we may write

(5.1) $$ \begin{align} \varphi(a_i)=(\alpha_{i1},\dots,\alpha_{it})\sigma_i \text{ with each } \alpha_{ij}\in{\mathrm{Aut}}(T^s) \text{ and }\sigma_i\in {\mathrm{Sym}}(t) \text{ such that }1^{\sigma_i}=i. \end{align} $$

Clearly $P_{A_1}(U \cap S_1) \subseteq P_{A_1}(S_1)$ , so we just need to prove the reverse inclusion. Let $y_1 \in S_1$ be of prime-power order. Then, for $i=1,\dots ,t$ , the element $y_i:= y_1^{a_i}\in S_i$ , and from the form of $\varphi (a_i)$ given in (5.1) we see that $y_i$ is the element $y_1^{\alpha _{i1}}$ of $T^s$ . In particular, $|y_i|=|y_1|$ , so $y:=(y_1,y_2,\dots ,y_t)\in S$ has prime-power order $|y|=|y_1|$ . Since $P_A(G)=P_A(U)$ , there exists $a\in A$ such that $y^a\in U$ , and since S is A-invariant, $y^a\in U\cap S$ . As we saw above, $U\cap S$ contains $K_1=S_2\times \dots \times S_t$ , and hence $y^a\in U\cap S$ if and only if its first entry lies in $U\cap S_1$ . Now $\varphi (a)=(\beta _1,\dots ,\beta _t)\sigma $ for some $\beta _i\in {\mathrm {Aut}}(T^s)$ and $\sigma \in {\mathrm {Sym}}(t)$ . Suppose that $i^\sigma =1$ , that is, $S_i^a=S_1$ . Then the first entry of $y^a$ is

$$\begin{align*}y_{1\sigma^{-1}}^{\beta_{1\sigma^{-1}}} = y_i^{\beta_i},\ \text{which is the element } y_1^{\alpha_{i1}\beta_i} \text{ of } T^s, \end{align*}$$

and hence $y_1^{\alpha _{i1}\beta _i}\in U\cap S_1$ . Note that the element $a_ia\in A$ satisfies $S_1^{a_ia}=(S_1^{a_i})^a=S_i^a=S_1$ , that is to say, $a_ia\in A_1$ , and

$$\begin{align*}\varphi(a_ia)=(\alpha_{i1},\dots,\alpha_{it})\sigma_i \cdot (\beta_1,\dots,\beta_t)\sigma = (\alpha_{i1}\beta_{1\sigma_i},\dots,\alpha_{it}\beta_{t\sigma_i})\sigma_i\sigma. \end{align*}$$

Since $1^{\sigma _i}=i$ , this implies that $(y_1, 1,\dots ,1)^{a_ia}=(y_1^{\alpha _{i1}\beta _i},1,\dots ,1)$ and lies in $U \cap S$ , with $a_ia\in A_1$ . Removing the t-tuple notation this says that $y_1^{a_ia}\in U\cap S_1$ . Thus $P_{A_1}(S_1)=P_{A_1}(U\cap S_1)$ as claimed.

Now we write $S_1=T_1\times \dots \times T_s=T^s$ . Since $S_1$ is transitive on $\Omega _1$ , we have $G=US_1$ , and since $S_1$ is a minimal normal subgroup of G, it follows that U acts transitively on $\{T_1,\dots , T_s\}$ by conjugation. Since $U \cap S_1$ is a proper subgroup of $S_1$ , after relabelling the $T_i$ if necessary, we may assume that $U\cap T_1$ is a proper subgroup of $T_1$ .

By definition, $P_{A_{11}}(U\cap T_1)\subseteq P_{A_{11}}(T_1)$ , and to prove the reverse inclusion let $y\in T_1$ be an element of prime-power order, and identify y with the s-tuple $(y,1,\dots ,1)\in S_1$ . Since $P_{A_1}(S_1)=P_{A_1}(U\cap S_1)$ by Claim 1, there exists $a\in A_1$ such that $(y,1,\dots ,1)^a\in U\cap S_1$ . Now $T_1^a=T_i$ for some $i\leqslant s$ , and there exists $u\in U$ such that $T_i^u=T_1$ since $G=US_1$ . Thus $T_1^{au}=T_1$ , that is, $au\in A_{11}$ , and therefore $(y,1,\dots ,1)^{au}\in T_1$ . Also $(y,1,\dots ,1)^{au}\in (U\cap S_1)^u=U\cap S_1$ , so $(y,1,\dots ,1)^{au}\in U\cap T_1$ . Thus $P_{A_{11}}(T_1)=P_{A_{11}}(U\cap T_1)$ and Claim 2 is proved.

As noted above, $U\cap T_1$ is a proper subgroup of $T_1$ , and $T_1$ is a finite nonabelian simple group. Let $\rho : A_{11}\rightarrow {\mathrm {Aut}}(T_1)$ be the group homomorphism induced by the conjugation action on $T_1$ . Since $\rho $ restricted to $T_1$ is a monomorphism, we have

$$\begin{align*}\rho(U \cap T_1) < \rho (T_1) \unlhd \rho(A_1) \leqslant {\mathrm{Aut}}(T_1).\end{align*}$$

The equality $P_{A_{11}}(T_1) = P_{A_{11}}(U \cap T_1)$ implies that $P_{\rho (A_{11})}(\rho (T_1)) = P_{\rho (A_{11})}(\rho (U \cap T_1))$ , which is a contradiction to the Guralnick–Saxl Theorem (Theorem 2.1). This final contradiction completes the proof of Theorem 5.5.

Proof. Recall that $K_i$ is the kernel of the action of G on $\Omega _i$ . Thus each $K_i$ is normal in G and A acts transitively on the set $\{K_1,\ldots ,K_n\}$ . Since $G^{\Omega _1}$ contains exactly two minimal normal subgroups, by [Reference Praeger and Schneider28, Corollary 3.11], we may label these two minimal normal subgroups as $P_{1,L}$ and $P_{1,R}$ and we have $P_{1,L}\cong P_{1,R} \cong T^s$ , for some finite nonabelian simple group T and positive integer s. Moreover, $P_{1,L}$ and $P_{1,R}$ are the left and right regular actions of $T^s$ , respectively, and the double-centraliser property holds: $C_{G^{\Omega _1}}(P_{1,L}) = P_{1,R}$ and $C_{G^{\Omega _1}}(P_{1,R}) = P_{1,L}$ (see [Reference Praeger and Schneider28, Theorem 3.6(iii)]). We define $E_{1,L}$ and $E_{1,R}$ to be the full preimages in G of $P_{1,L}$ and $P_{1,R}$ , respectively. In particular, each of $E_{1,L}$ and $E_{1,R}$ is normal in G and for any subgroup H of G with $H^{\Omega _i} = P_{1,L}$ , we have $H \leqslant E_{1,L}$ . Similarly for $E_{1,R}$ . Let

(5.2) $$ \begin{align} \mathcal T = \{a_1,\ldots,a_n\} \end{align} $$

be a transversal to G in A labelled so that $\Omega _1^{a_i} = \Omega _i$ for each i. In particular, we have $a_1 \in G$ . For each $1\leqslant i \leqslant n$ , we define

$$\begin{align*}E_{i,L}:=(E_{1,L})^{a_i} \quad \text{and} \quad E_{i,R}:=(E_{1,R})^{a_i}.\end{align*}$$

Suppose that $\mathcal T' = \{a_1',\ldots ,a_n'\}$ is another transversal to G in A, again labelled so that $\Omega _1^{a_i'}=\Omega _i$ . Since $\Omega _i=\Omega _1^{a_i}$ and A acts regularly on $\{\Omega _1,\ldots ,\Omega _n\}$ with stabiliser G, we have $a_i' = g_i a_i$ for some $g_i\in G$ . Hence $(E_{1,L})^{a_i'} = ((E_{1,L})^{g_i})^{a_i} = (E_{1,L})^{a_i} = E_{i,L}$ , and similarly $(E_{1,R})^{a_i'} = E_{i,R}$ . Thus the definitions of $E_{i,L}$ and $E_{i,R}$ are indeed independent of the choice of transversal. To prove the second assertion, let $a\in A$ and i be arbitrary and consider $(E_{i,L})^a$ . Let j be such that $\Omega _i^a = \Omega _j$ . Now $\Omega _i = \Omega _1^{a_i}$ and $\Omega _j=\Omega _1^{a_j}$ . Thus $\Omega _1^{a_i a a_j^{-1} } = \Omega _1$ . Since A acts regularly on $\{\Omega _1,\ldots ,\Omega _n\}$ with stabiliser G, there exists $g\in G$ such that $a_i a a_j^{-1} = g$ , whence $a = a_i^{-1} g a_j$ . Then, since $E_{1,L}$ is normal in G,

$$\begin{align*}(E_{i,L})^a = (E_{i,L})^{a_i^{-1}g a_j} = ((E_{1,L})^g)^{a_j} = (E_{1,L})^{a_j} = E_{j,L}. \end{align*}$$

It follows that A acts transitively on the set $\{E_{1,L},\ldots ,E_{n,L}\}$ by conjugation. Exactly the same argument shows that A acts transitively on the set $\{E_{1,R},\ldots ,E_{n,R}\}$ by conjugation.

The previous claim allows us to define

$$\begin{align*}\mathcal L _i : = \{ X \unlhd G \mid X^{\Omega_i} = (E_{i,L})^{\Omega_i} = P_{i,L} \} \quad \text{and}\quad \mathcal R_i := \{ X \unlhd G \mid X^{\Omega_i} = (E_{i,R})^{\Omega_i} = P_{i,R} \}\end{align*}$$

and we note that each of the sets $\{\mathcal {L}_1,\ldots ,\mathcal {L}_n\}$ and $\{\mathcal {R}_1,\ldots ,\mathcal {R}_n\}$ is permuted transitively by A. Furthermore, $E_{i,L} \in \mathcal L_i$ and $E_{i,R} \in \mathcal R_i$ , so that $\mathcal {L}_i$ and $\mathcal {R}_i$ are both nonempty.

The proof is the same as the proof of Claim 1 in the proof of Lemma 5.6 and so we omit it here.

For each i we set $L_i := \bigcap _{X \in \mathcal {L}_i} X$ and $R_i := \bigcap _{X\in \mathcal {R}_i} X$ , and note that A permutes each of the sets $\{L_1,\ldots ,L_n\}$ and $\{R_1,\ldots ,R_n\}$ transitively. Hence $|L_i|=|L_j|$ for all $i,j$ and $|R_i|=|R_j|$ for all $i,j$ . Furthermore, since $L_i\in \mathcal {L}_i$ and $R_i\in \mathcal {R}_i$ , we have $L_i^{\Omega _i}=P_{i,L}$ and $R_i^{\Omega _i}=P_{i,R}$ so, in particular, $L_i \neq R_i$ , and both $L_i$ and $R_i$ are nontrivial, for all i.

Let X be a minimal normal subgroup of G. Since G acts faithfully on $\Omega $ , X must act nontrivially on some $\Omega _i$ , that is, $X^{\Omega _i}\neq 1$ and hence $X^{\Omega _i}$ is a minimal normal subgroup of $G^{\Omega _i}$ . Thus $X^{\Omega _i} = P_{i,L}$ or $P_{i,R}$ . In the first case, we have $X \in \mathcal {L}_i$ , and so $L_i \leqslant X$ , which implies $L_i=X$ . In this case, for each $j\leqslant n$ , there is an element $a\in A$ such that $\Omega _i^a=\Omega _j$ , and the conjugate $X^a$ is a minimal normal subgroup of G lying in $\mathcal {L}_i^a=\mathcal {L}_j$ , which implies that $L_j\leqslant X^a$ and hence $L_j=X^a$ . In the second case, a similar argument yields that, for each j, $R_j$ is some A-conjugate of X. This proves the claim, after interchanging the labels L and R in the second case.

Claim 4. For all $i,j$ , one of the following holds:

$$ \begin{align*} \text{(a) } R_i^{\Omega_j}=1,\quad \text{or (b) } R_i^{\Omega_j}=P_{j,R} \text{ and } R_i=R_j,\quad \text{or (c) } R_i^{\Omega_j}=P_{j,L}. \end{align*} $$

To prove the claim, we may assume $R_i^{\Omega _j} \neq 1$ (i.e., part (a) does not hold). Thus $R_i^{\Omega _j}$ contains one of the two minimal normal subgroups of $G^{\Omega _j}$ . Suppose first that $R_i^{\Omega _j} \geqslant P_{j,R}$ . By the Correspondence Theorem, we may choose $R_0 \leqslant R_i$ with $R_0$ normal in G such that $R_0^{\Omega _j}=P_{j,R}$ . Then $R_0 \in \mathcal R_j$ and so by the minimality of $R_j$ , we have $R_j \leqslant R_0 \leqslant R_i$ . The equality $|R_i|=|R_j|$ implies that $R_i=R_j$ and $R_i^{\Omega _j}=R_j^{\Omega _j} = P_{j,R}$ , as in part (b). Now we may assume that $R_i^{\Omega _j} \ngeqslant P_{j,R}$ . The fact that $P_{j,R}$ is a minimal normal subgroup of $G^{\Omega _j}$ , combined with the fact that $R_i^{\Omega _j}$ is normal in $G^{\Omega _j}$ , implies that $R_i^{\Omega _j}\cap P_{j,R} = 1$ , and hence $R_i^{\Omega _j} \leqslant C_{G^{\Omega _j}}(P_{j,R})=P_{j,L}$ . This implies that $R_i^{\Omega _j} = P_{j,L}$ , as in part (c).

Let X be a minimal normal subgroup of G contained in $R_i$ . Now X must act nontrivially on $\Omega _j$ for some j, so we have $1\neq X^{\Omega _j} \leqslant R_i^{\Omega _j}$ . Then by the assumptions in this claim, $X^{\Omega _j}=R_i^{\Omega _j}=P_{j,R}$ . The minimality of $R_j$ means that $R_j \leqslant X \leqslant R_i$ , and since $|R_j|=|R_i|$ , we have that $R_i=X=R_j$ . In particular, $R_i$ is a minimal normal subgroup of G.

Suppose that $R_i^{\Omega _j}=P_{j,L}$ . By the minimality of $L_j$ , we have $L_j \leqslant R_i$ , and since $L_j$ is a minimal normal subgroup of G (by Claim 3) and $L_j$ acts faithfully on $\Omega _j$ , we obtain

$$\begin{align*}R_i = L_j \times (R_i \cap K_j).\end{align*}$$

Now consider $L_j^{\Omega _i}$ . Since $L_j$ is normal in G and $L_j \leqslant R_i$ , we have $L_j^{\Omega _i} = 1$ or $L_j^{\Omega _i}=R_i^{\Omega _i}=P_{i,R}$ . In the latter case, this means that $L_j \in \mathcal {R}_i$ , and so by the minimality of $R_i$ , we obtain $R_i \leqslant L_j$ , whence $R_i=L_j$ , as required. So we may suppose that $L_j^{\Omega _i} = 1$ . This implies that

$$\begin{align*}(R_i \cap K_j)^{\Omega_i}=R_i^{\Omega_i}=P_{i,R} \end{align*}$$

and by minimality of $R_i$ , we have $R_i = R_i \cap K_j$ , which means that $L_j=1$ , a contradiction.

By Claim 4, for each i, the assumptions of one of Claim 5 or Claim 6 hold. We therefore conclude that $R_i$ is a minimal normal subgroup of G for each $i \leqslant n$ . Since Claim 3 shows that each $L_i$ is a minimal normal subgroup of G, and $R_i\neq L_i$ , we have $L_i \times R_i \leqslant \mathrm {Soc}(G)$ . Moreover, $R_i^{\Omega _i} = P_{i,R} \neq P_{i,L} = L_i^{\Omega _i}$ so $\mathrm {Soc}(G^{\Omega _i}) = L_i^{\Omega _i} \times R_i^{\Omega _i} = (L_i \times R_i)^{\Omega _i} \leqslant {\mathrm {Soc}}(G)^{\Omega _i}$ . On the other hand, $\mathrm {Soc}(G)^{\Omega _i} \leqslant {\mathrm {Soc}}(G^{\Omega _i})$ , and so $\mathrm {Soc}(G^{\Omega _i}) = {\mathrm {Soc}}(G)^{\Omega _i}$ , and the proof is complete.

Proof. Let p be a prime divisor of $|T|$ such that the number of $ {\mathrm {Aut}}(T)$ -classes of p-elements in T is equal to $m(T)$ , and recall that the identity $1_T$ is a p-element. Thus there are at least two distinct $ {\mathrm {Aut}}(T)$ -classes of p-elements in T, so $m(T)\geqslant 2$ . Hence

(5.5) $$ \begin{align} \max\{ s, m(T)-1\} \leqslant s\cdot (m(T)-1), \end{align} $$

so to establish the lemma, we need to bound the right-hand side of the above inequality by n. Recall that $S = S_1 \times \cdots \times S_t$ , and each $S_i \cong T^s$ .

As usual, we write elements of S as t-tuples $x=(x_1,\dots , x_t)$ with $x_i\in S_i$ , for each i. Let $\mathcal {T}$ be the set of all $ {\mathrm {Aut}}(T^s)$ -conjugacy classes of p-elements in $T^s$ .

This is a very weak lower bound but sufficient for our purposes. To see that it holds, choose representatives $y_1,\dots , y_{m(T)-1}$ for the $m(T)$ distinct $ {\mathrm {Aut}}(T)$ -classes of nontrivial p-elements in T, and for $1\leqslant \ell \leqslant m(T)-1$ and $1\leqslant j\leqslant s$ , let $x_{\ell ,j}\in T^s$ be the element with entries $1,\dots ,j$ equal to $y_\ell $ and entries $j+1,\dots , s$ equal to the identity. Then the $x_{\ell ,j}$ lie in pairwise distinct $ {\mathrm {Aut}}(T^s)$ -classes. This proves Claim 1.