Socles and the Onan Scott Landscape

By [A1], the subgroup $[M,N]$ is a normal subgroup of $G$ contained in $M$. Since $M$ is minimal normal, either $[M,N]=1$ or $[M,N]=M$.

The same argument with $N$ shows that either $[M,N]=1$ or $[M,N]=N$. Because $M\mathrm{\ne }N$, the subgroup $[M,N]$ cannot equal both $M$ and $N$.

Therefore $[M,N]=1$. By [L1], every commutator $[m,n]$ is trivial, so $mn=nm$ for all $m\in M$ and $n\in N$.

Let $K$ be a characteristic subgroup of $M$. By [L1], the subgroup $K$ is normal in $G$. Since $K\le M$ and $M$ is minimal normal in $G$, either $K=1$ or $K=M$.

Thus $M$ has no proper nontrivial characteristic subgroup, so [A1] shows that $M$ is characteristically simple.

By [A1], choose a minimal normal subgroup $N⊴G$.

Let $N_1,\dots ,N_r$ be an irredundant family of automorphic images of $N$ that generates the subgroup $H$ generated by all automorphic images. By [A2], the subgroup $H$ is characteristic in $G$, so $H=G$ because $G$ is characteristically simple and $N\mathrm{\ne }1$.

Fix $i$ and put $P_i=⟨N_j:j\mathrm{\ne }i⟩$. The intersection $N_i\cap P_i$ is normal in $G$: both factors are normal, and the intersection is preserved by conjugation. Minimality of $N_i$ makes this intersection either $1$ or $N_i$. The latter would put $N_i$ inside the subgroup generated by the other images, contradicting irredundancy. Hence $N_i\cap P_i=1$ for every $i$. Together with step 2.1, [A2], and [L1], this makes $G=N_1\times \cdots \times N_r$ an internal direct product.

Let $K⊴N_i$. Since the other direct factors centralize $N_i$, conjugation by them fixes $K$, while conjugation by $N_i$ preserves $K$ by normality. Step 3.1 says these factors generate $G$, so $K⊴G$. Minimality of $N_i$ gives $K=1$ or $K=N_i$; thus $N_i$ is simple. All $N_i$ are automorphic images of $N$, so they are pairwise isomorphic to one finite simple group $T$. Therefore $G\cong T^r$.

By [A1], the subgroup generated by all minimal normal subgroups of $G$ is stable under every automorphism of $G$. By definition this subgroup is $\mathrm{soc}(G)$, so $\mathrm{soc}(G)$ is characteristic in $G$.

Let $N_1,\dots ,N_r$ be a maximal family of pairwise distinct minimal normal subgroups of $G$ chosen so that none is contained in the product of the preceding ones; when $G=1$, this family is empty. For $i\ge 1$, [L1] shows that $N_i$ centralizes $N_1\cdots N_{i-1}$, and minimality gives $N_i\cap (N_1\cdots N_{i-1})=1$ because the intersection is a normal subgroup of $G$ contained in $N_i$ but $N_i$ was chosen outside the preceding product. Hence $N_1\cdots N_r$ is an internal direct product, with the case $r=0$ giving the trivial group.

The product $N_1\cdots N_r$ is generated by minimal normal subgroups, so it lies in $\mathrm{soc}(G)$. Conversely, if $M$ is any minimal normal subgroup of $G$ not contained in $N_1\cdots N_r$, then adjoining $M$ would contradict maximality of the chosen family; therefore every minimal normal subgroup of $G$ lies in the displayed product. Thus $\mathrm{soc}(G)=N_1\times \cdots \times N_r$. By [L3], each nontrivial factor $N_i$ is characteristically simple, and [L2] then makes it a direct product of isomorphic simple groups.

By [L1], the normal subgroup $N$ is either transitive or contained in the kernel of the action.

The action is faithful, so [A1] gives trivial kernel. Because $N$ is a minimal normal subgroup, it is nontrivial, so the kernel-contained alternative from step 1.1 is impossible. Hence $N$ is transitive.

By [L2], both $M$ and $N$ are transitive on $\mathrm{\Omega }$. By [L1], every element of $M$ commutes with every element of $N$.

Fix $\alpha \in \mathrm{\Omega }$, and let $m\in M_{\alpha }$. For any $\beta \in \mathrm{\Omega }$, choose $n\in N$ with $n\alpha =\beta$; then $m\beta =mn\alpha =nm\alpha =n\alpha =\beta$. Hence every element of $M_{\alpha }$ fixes every point of $\mathrm{\Omega }$, so faithfulness gives $M_{\alpha }=1$.

The subgroup $M$ is transitive with trivial point stabilizer, so it is regular. By symmetry the same argument applies to $N$.

Suppose that $M_1,M_2,M_3$ are three distinct minimal normal subgroups of $G$. By [L1], each pair among them is regular. In particular, $M_1$ and $M_2$ are both regular.

Fix $\alpha \in \mathrm{\Omega }$. Because $M_1$ is regular, the map $m\mapsto m\alpha$ identifies $M_1$ with the set $\mathrm{\Omega }$, and because $M_2$ is regular there is for each $\beta \in \mathrm{\Omega }$ a unique element $n(\beta )\in M_2$ with $n(\beta )\alpha =\beta$ by [A1].

Applying [L2] to the pair $(M_1,M_3)$ shows that $M_3$ centralizes $M_1$. Hence every element of $M_3$ acts on the identified copy of $M_1$ by right translation. But $M_2$ already has that property by step 2.1, and the right-regular subgroup centralizing the left-regular action of $M_1$ is unique. Therefore $M_3=M_2$, contradicting distinctness. So no third minimal normal subgroup exists.

By [L1], the abelian normal subgroup $V$ is regular on $\mathrm{\Omega }$.

By [L2], the minimal normal subgroup $V$ is characteristically simple; as it is also abelian, [A1] shows that $V$ is elementary abelian. Facts [L3] and [L4] therefore identify $V$ with $({\mathbb{F}}_p{)}^d$ for some prime $p$ and some $d\ge 1$.

Fix $\alpha \in \mathrm{\Omega }$. Because $V$ is regular, every $g\in G_{\alpha }$ acts on $V$ by conjugation and the kernel of this action is $G_{\alpha }\cap C_G(V)$. If a nontrivial element of $G_{\alpha }$ centralized $V$, then it would fix every point $v\alpha$ with $v\in V$, contradicting faithfulness; so the action is faithful. If $W<V$ were a nontrivial proper $G_{\alpha }$-invariant subgroup, then $W$ would be normal in $V{G}_{\alpha }=G$, contradicting minimality of $V$. Thus the action is irreducible, and $G$ is of affine type.

Let $B\subseteq {\mathrm{\Delta }}^{\mathrm{\ell }}$ be a block containing distinct points $x$ and $y$, and choose a coordinate $i$ with $x_i\mathrm{\ne }{y}_i$. By [A2], some $h\in H_{x_i}$ moves $y_i$. Let $a\in H^{\mathrm{\ell }}$ act as $h$ in coordinate $i$ and trivially elsewhere. Then $a$ fixes $x$, so $aB\cap B\mathrm{\ne }\mathrm{\varnothing }$ and the block property gives $aB=B$. Hence $y,ay\in B$ are distinct and differ in exactly coordinate $i$.

Fixing the other coordinates, the set of possible entries in coordinate $i$ among points of $B$ is a block for $H$. It contains the two distinct entries from step 1.1, so primitivity of $H$ makes it all of $\mathrm{\Delta }$. Thus $B$ contains the entire $i$-coordinate fibre through $y$.

The stabilizer in $H\wr K$ of any point of ${\mathrm{\Delta }}^{\mathrm{\ell }}$ induces the transitive group $K$ on the coordinates: a coordinate permutation can be followed by coordinatewise elements of the transitive group $H$ to restore the point. Applying these point-stabilizer elements to the fibre in step 2.1 gives a full fibre in every coordinate. Independent coordinate changes then show that $B={\mathrm{\Delta }}^{\mathrm{\ell }}$. Hence every block is a singleton or the whole set, so the product action is primitive.

Because the action has degree at least $2$, it is nontrivial. The socle analysis [A1] therefore applies to $G$ and reduces the action to the structure of one or two minimal normal subgroups.

The source theorem [A2] says that those socle configurations fall into exactly five families, and [L1] records the names and defining data of those families in the convention used here. Hence $G$ belongs to exactly one of the five listed types.

By [L1], the $2$-transitive action of $G$ is primitive, so the O'Nan-Scott classification applies to it.

The source fact [A1] removes the diagonal, product-action, and twisted-wreath branches from the primitive classification. Therefore the only remaining possibilities are affine type and almost simple type.

In the group $G=T\times T$, each factor $T\times 1$ and $1\times T$ is a minimal normal subgroup, and they are distinct.

By [L2], the socle of $G$ is the direct product of those minimal normal subgroups, so $\mathrm{soc}(G)=T\times T$.

The group $T\times T$ is not simple because each factor is a proper nontrivial normal subgroup. Therefore the claim is false.

The diagonal subgroup is maximal in $T\times T$: if $\mathrm{\Delta }(T)<L$, an element $(a,b)\in L\setminus \mathrm{\Delta }(T)$ yields the nontrivial element $(1,b{a}^{-1})\in L$ after multiplication by $(a^{-1},a^{-1})$; its diagonal conjugates generate $1\times T$ by simplicity, and then $L=T\times T$. Hence the coset action is primitive. Its kernel is the core of $\mathrm{\Delta }(T)$. If $(t,t)$ lies in that core, conjugation by every $(x,1)$ gives $(xt{x}^{-1},t)\in \mathrm{\Delta }(T)$, so $t\in Z(T)=1$ because $T$ is nonabelian simple. Thus the action is faithful. Its two factors $T\times 1$ and $1\times T$ are distinct minimal normal subgroups.

By [L1], those two minimal normal subgroups are regular. So this primitive action has two distinct minimal normal subgroups, contradicting uniqueness. The corollary [L2] shows that this exceptional size is the largest possible.

By [L1], the O'Nan-Scott theorem concerns primitive permutation actions, not the class of all finite simple groups.

The sourced boundary fact [A1] separates the structural reduction from the later theory of finite simple groups. Therefore the two theorems serve different purposes, and the claim is false.

The theorem [L1] is already a completed structural classification of finite primitive permutation groups.

The sourced boundary fact [A1] states that CFSG is used later, not in the theorem itself. So the claim is false.