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.

Choose a minimal normal subgroup $N\mathrm{\ne }1$, possible by finiteness. It is transitive by [L1]. If $N$ is regular, identify $\mathrm{\Omega }$ with $N$ by $u\mapsto u\alpha$. Conjugation by $G_{\alpha }$ is transitive on $N\setminus \{1\}$, so all nonidentity elements have the same order, necessarily a prime $p$ by taking powers. By Cauchy's theorem $N$ is a $p$-group; its nontrivial characteristic center equals $N$ by [L3]. Thus $N$ is elementary abelian. A permutation centralizing a regular group is determined by its value at $\alpha$; for abelian $N$ it is the corresponding translation. A second minimal normal subgroup would therefore lie in $N$ by [L2], which is impossible. The affine conclusion follows from [L6].

Assume henceforth that $N$ is nonregular. Since $N_{\alpha }⊴G_{\alpha }$, all its orbits on $\mathrm{\Omega }\setminus \{\alpha \}$ have a common size $m$. Transitivity of $N$ gives the same size at every point. Here $m>1$, for otherwise $N_{\alpha }$ fixes every point and is trivial.

We establish a finite permutation fact: a faithful transitive group $U$ whose nontrivial suborbits all have size $m>1$ is primitive or has trivial two-point stabilizers. Suppose $B$ is a block of size $1<k<n$, with $b\in B$ and $i\mathrm{\notin }B$. Since $U_b$ preserves $B$, $m\mid (k-1)$, so $\gcd (k,m)=1$. The union of the translates of $B$ under $U_i$ is a union both of blocks of size $k$ and of suborbits of size $m$; its size is at most $km$, so equals $km$. Thus the setwise stabilizer $U_{i,B}$ has index $m$ in $U_i$.

For every $b\in B$, $U_{i,b}\le U_{i,B}$ and both have index $m$, so they agree. In particular this group fixes $B$ pointwise. If $b^{\mathrm{\prime }}\in B\setminus \{b\}$, then $U_{i,b}\le U_{(B)}\le U_{b,b^{\mathrm{\prime }}},$ where $U_{(B)}$ is the pointwise stabilizer. All two-point stabilizers have order $\mathrm{\mid }U\mathrm{\mid }\mathrm{/}(nm)$, so both inclusions are equalities. Applying the same argument to the block $C$ containing $i$ gives $U_{(B)}=U_{(C)}$. This holds for every block, so their common pointwise stabilizer fixes all points and is trivial. Both within-block and between-block two-point stabilizers are therefore trivial.

If $N$ were imprimitive, steps 3.1–4.1 would give trivial two-point stabilizers. Put $h=\mathrm{\mid }{N}_{\alpha }\mathrm{\mid }>1$. Counting the elements fixing exactly one point shows that the set $D$ of fixed-point-free elements of $N$ has size $\mathrm{\mid }D\mathrm{\mid }=nh-1-n(h-1)=n-1.$ The set $D$ is conjugation invariant under $G$. The number $a$ of its elements carrying one point to a different specified point is independent of the ordered pair by $2$-transitivity. Counting these incidences gives $n(n-1)=an(n-1)$, hence $a=1$. Conjugating the ordered pair $(\alpha ,d\alpha )$ to $(\alpha ,e\alpha )$ for $d,e\in D$ now shows that $d,e$ are conjugate in $G$.

The action of $N_{\alpha }$ on the other points is free, so $h\mid (n-1)$. For each prime $p\mid n$, Cauchy's theorem gives an element of order $p$ in $N$, necessarily in $D$ since $p\nmid h$. All elements of $D$ have the same order, so $n$ is a power of a single prime $p$. A Sylow $p$-subgroup $P$ of $N$ has order $n$, because $p\nmid h$. All its nonidentity elements are fixed-point-free. Thus $P=D\cup \{1\}$ is a nontrivial proper $G$-normal subgroup of $N$, contradicting minimality. Hence $N$ is primitive.

Suppose $N$ is not simple and choose a nontrivial proper minimal normal subgroup $M$ of $N$. It is not $G$-normal, so it has a distinct $G$-conjugate $M^{\mathrm{\prime }}$. By [L2] for primitive $N$, these are its only two minimal normal subgroups, and they commute and are regular. Thus $M{M}^{\mathrm{\prime }}$ is $G$-normal, whence $N=M{M}^{\mathrm{\prime }}$; also $M\cap M^{\mathrm{\prime }}=1$. The group $M$ is nonabelian: otherwise its permutation centralizer would be $M$, forcing the commuting regular subgroup $M^{\mathrm{\prime }}$ to equal $M$.

Let $H=N_G(M)$. Its index in $G$ is two, it contains $N$, and it is transitive. Consequently $G=H{G}_{\alpha }$ and $[G_{\alpha }:H_{\alpha }]=2$. Its normal subgroup $H_{\alpha }⊴G_{\alpha }$ has either one orbit or two equal-sized orbits on $\mathrm{\Omega }\setminus \{\alpha \}$. Under the regular identification with $M$, these are automorphism orbits on $M\setminus \{1\}$. Thus there are at most two nonidentity element orders in $M$.

If $\mathrm{\mid }M\mathrm{\mid }$ has only one prime divisor, its nontrivial center makes it abelian by [L3] and [L4], contradicting step 7.1. Otherwise Cauchy's theorem shows that there are exactly two prime divisors $p,q$, and the two equal-sized orbits consist of elements of orders $p$ and $q$. No other nonidentity order is possible.

For an element $x$ of order $p$, the centralizer $C_M(x)$ has order a power of $p$: if $q$ divided its order, Cauchy's theorem would give a commuting element $y$ of order $q$, and $xy$ would have order $pq$, an impossibility. By [L5], every $M$-conjugacy class of elements of order $p$ has size divisible by $q$. The total number of such elements, $(\mathrm{\mid }M\mathrm{\mid }-1)\mathrm{/}2$, is therefore divisible by $q$, contrary to $q\mid \mathrm{\mid }M\mathrm{\mid }$. This contradiction proves that $N$ is simple. It is nonabelian, since a faithful transitive abelian group is regular.

Since $N$ is nonregular, [L2] excludes a second minimal normal subgroup of $G$. If its normal centralizer $C_G(N)$ were nontrivial, it would contain a minimal normal subgroup of $G$ and thus contain $N$, contrary to nonabelian simplicity. So conjugation embeds $G$ in $\mathrm{Aut}(N)$. The subgroup $N$ maps to its inner automorphism group, isomorphic to $N$ because $Z(N)=1$. This is the almost simple alternative. Together with step 1.1 it proves the claim.

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.

The natural action of $S_3$ on $\{1,2,3\}$ is $2$-transitive: specifying the images of two distinct points determines a permutation by sending the third point to the remaining point. It is therefore primitive by [L2], and lies in the domain of the O'Nan–Scott classification question.

The subgroup $A=\{1,(123),(132)\}$ is nontrivial and proper in $S_3$. Conjugating either $3$-cycle by a permutation merely relabels its three entries, so it gives one of these same two $3$-cycles. Thus $A$ is normal and $S_3$ is not simple by [L1]. Consequently the two classification questions have different domains: one includes this action of a nonsimple group, whereas the other classifies simple groups up to abstract isomorphism. Their conclusions also ask for different data, action types versus a list of abstract simple groups. They are not the same theorem.

The cited LPS proof invokes Schreier explicitly in two branches. In Case 2(a), the invoked solubility is used to prove $Y=M$; in the simple-socle branch, it supplies the elementary-abelian minimal normal subgroup used in the subsequent argument. These are proof steps, not merely remarks about later refinements.

Schreier is a CFSG consequence. Thus this specific five-type proof does use a CFSG consequence, refuting the stated claim. This establishes neither that CFSG is logically necessary for the theorem nor that another proof could not avoid it.