The Finite Simple Group Classification Landscape

The image $MZ(Q)\mathrm{/}Z(Q)$ is normal in the simple group $Q\mathrm{/}Z(Q)$, so it is trivial or the whole quotient. In the trivial case $M\le Z(Q)$. In the whole case $Q=MZ(Q)$, and the quotient $Q\mathrm{/}M$ is generated by the image of the abelian group $Z(Q)$; hence $Q\mathrm{/}M$ is abelian. But a quotient of the perfect group $Q=[Q,Q]$ is perfect, and an abelian perfect group is trivial. Thus $M=Q$, contrary to properness. Only the first case remains.

Fix $t\in T$. Since $t$ normalizes $P$, the formula $c_t(p)=tp{t}^{-1}{p}^{-1}$ defines a function $P\to Z(P)$ by hypothesis. Its values are central, so $t(pq)t^{-1}=c_t(p)p{c}_t(q)q=c_t(p)c_t(q)pq$ and therefore $c_t(pq)=c_t(p)c_t(q)$. Thus $c_t$ is a homomorphism to the abelian group $Z(P)$. It kills every commutator of $P$, and $P=[P,P]$, so $c_t$ is trivial. Every $t$ consequently commutes with every $p$, as claimed. The argument also covers a trivial center or a trivial $T$.

Choose a subnormal chain from F1. If $Q\le S_0=S$, the first alternative holds. Otherwise, because $Q\le S_d=G$, there is an index $j<d$ such that $Q\le S_{j+1}$ but $Q\nleqq S_j$. We only select an index from the given finite chain.

Both $Q$ and $S_j$ are normal in $S_{j+1}$, since $Q⊴G$ and $S_j⊴S_{j+1}$. Thus $Q\cap S_j$ is a proper normal subgroup of $Q$, and F2 puts it in $Z(Q)$. For $q\in Q$ and $s\in S_j$, their commutator lies in both $Q$ and $S_j$, hence in $Z(Q)$. Since $S_j$ normalizes the perfect group $Q$, the second clause of F2 gives $[Q,S_j]=1$. As $S\le S_j$, it follows that $[Q,S]=1$. No finiteness of $G$ beyond the finite subnormal chain is needed.

We prove two assertions simultaneously for each finite group $G$: (A) if $Q$ is a component and $N⊴G$, then either $Q\le N$ or $[Q,N]=1$; (B) distinct components commute. Assume both assertions have been proved for every group of order smaller than $\mathrm{\mid }G\mathrm{\mid }$. Each induction call below takes place in a proper normal subgroup $H<G$, so its order is strictly smaller. A subnormal chain for a component in $H$ remains a valid chain there, and conjugation by an element of $G$ preserves the chain inside $H$ because $H⊴G$.

First prove (A). If $Q\le N$ there is nothing to show. Suppose $Q\nleqq N$. If $Q⊴G$, then $Q\cap N$ is a proper normal subgroup of $Q$, so F2 gives $Q\cap N\le Z(Q)$. Since both $Q$ and $N$ are normal in $G$, their commutator lies in $Q\cap N$. The central-action clause of F2, applied to the perfect group $Q$, gives $[Q,N]=1$.

Suppose instead that $Q$ is not normal in $G$. A subnormal chain from $Q$ to $G$ then has a proper term $H⊴G$ containing $Q$; take its penultimate term after deleting repeated terms. Put $M=N\cap H⊴H$. The subgroup $Q$ is a component of $H$ and is not contained in $M$. By the induction hypothesis (A) for $H$, $[Q,M]=1$.

For $n\in N$, conjugation sends $Q$ to another component $Q^n=nQ{n}^{-1}$ of $H$. Indeed $H⊴G$ keeps the conjugated subnormal chain inside $H$. For each $q\in Q$ write $nq{n}^{-1}=c_{q}q$, where $c_q=nq{n}^{-1}{q}^{-1}\in N\cap H=M$; the two normalities give this inclusion. By step 2.1, $c_q$ centralizes $Q$. The induction hypothesis (B) for $H$ says that either $Q^n=Q$ or $Q^n$ centralizes $Q$. In the latter case, for every $q,x\in Q$ the equality $[c_{q}q,x]=[q,x]$ would force $[q,x]=1$. This would make the perfect quasisimple group $Q$ abelian, hence trivial, contrary to its nontrivial simple central quotient. Thus $Q^n=Q$ for every $n\in N$.

It follows that $N$ normalizes $Q$ and $[N,Q]\le N\cap Q$. This is a proper normal subgroup of $Q$, so F2 puts it in $Z(Q)$. The perfect central-action clause of F2 now yields $[N,Q]=1$. This proves (A) for $G$ without invoking (B) for $G$ itself.

Now prove (B), after assertion (A) has been established for this $G$. If $K⊴G$, apply F3 to $K$ and the subnormal subgroup $L$. It gives $[K,L]=1$ unless $K\le L$. In that exceptional case $K⊴L$; because $K\mathrm{\ne }L$, F2 gives $K\le Z(L)$. But $K$ is nontrivial and perfect, so it cannot be abelian or lie in a center. The exceptional case is impossible.

If $K$ is not normal in $G$, take a proper normal subgroup $H⊴G$ containing $K$ from its subnormal chain. Apply assertion (A), already proved for this same $G$ in steps 1.2–4.1, to $L$ and $H$. If $L\nleqq H$, it centralizes $H$ and hence $K$. If $L\le H$, both $K$ and $L$ are distinct components of the smaller group $H$, so the induction hypothesis (B) for $H$ gives $[K,L]=1$. Thus (B) holds for $G$ in every case, completing the strong induction and the stated result.

The result is immediate for $G=1$, since $F^{\ast }(1)=1$.

Assume the result for all finite groups of order smaller than $\mathrm{\mid }G\mathrm{\mid }$. By [L2], $H⊴G$, so $C⊴G$ by [L4].

The subgroup $F(C)$ is characteristic in $C$: an automorphism of $C$ permutes its $p$-cores. Thus $F(C)⊴G$ because $C⊴G$; it is nilpotent by [L1], so $F(C)\le F(G)$. Every component of $C$ is subnormal in $G$, by appending $C⊴G$ to its subnormal chain in $C$. Hence $E(C)\le E(G)$ and $F^{\ast }(C)=F(C)E(C)\le H\cap C\mathrm{.}$

Every element of $C$ centralizes every element of $H$. Consequently $H\cap C\le Z(C)$ and $F^{\ast }(C)\le Z(C)$.

If $C<G$, step 1.2 applied to $C$ gives $C_C(F^{\ast }(C))\le F^{\ast }(C)$. But step 3.1 makes the left side equal to $C$. Therefore $C\le F^{\ast }(C)\le H$, as required.

It remains to consider $C=G$. Then $H\le Z(G)$, in particular $F\le Z(G)$ and $E\le Z(G)$. Suppose $G\mathrm{/}F\mathrm{\ne }1$. Choose a minimal nontrivial normal subgroup $A$ of the finite group $G\mathrm{/}F$, and let $N$ be its inverse image in $G$. Such an $A$ exists by choosing a subgroup of least positive order among the nontrivial normal subgroups of $G\mathrm{/}F$. Thus $F<N⊴G$ and $N\mathrm{/}F=A$.

If $A$ is abelian, then $[N,N]\le F\le Z(G)$. Hence the lower central series of $N$ ends after at most two commutator steps, so $N$ is nilpotent. As $N⊴G$, [L1] gives $N\le F$, contradicting $F<N$.

If $A$ is nonabelian, [L3] writes it as a direct product of isomorphic nonabelian simple groups. Choose a simple direct factor $T⊴A$, and let $N_T$ be its inverse image under $N\to N\mathrm{/}F=A$. Then $N_T⊴N⊴G$, $F\le Z(G)$, and $N_T\mathrm{/}F\cong T$. Put $K=[N_T,N_T]$. The image of $K$ in $T$ is $[T,T]=T$, since a nonabelian simple group is perfect. Thus $N_T=KF$. Because $F$ is central, $K=[N_T,N_T]=[KF,KF]=[K,K],$ so $K$ is perfect. Also $K\mathrm{/}(K\cap F)\cong T$. Since $K\cap F$ is central in $K$ and $T$ has trivial center, $Z(K)=K\cap F$; hence $K\mathrm{/}Z(K)\cong T$ is simple. Finally $K⊴N_T⊴N⊴G$, so $K$ is a component of $G$.

Step 5.2 gives $K\le E\le H\le Z(G)$, whereas $K\mathrm{/}Z(K)\cong T$ is nonabelian. This contradiction rules out the nonabelian case. Together with step 5.1, it follows that $G\mathrm{/}F=1$. Hence $C=G=F\le H$.

Steps 4.1 and 6.1 cover both possibilities for $C$, proving $C_G(F^{\ast }(G))\le F^{\ast }(G)$ for all finite $G$.

Its subgroup of order two is nontrivial and proper, so $C_4$ is not simple.

Thus a finite group need not be simple; CFSG classifies the finite groups that are simple and does not assert otherwise.

Each has a composition series with two factors isomorphic to $C_2$. However, $C_4$ contains an element of order four and $C_2\times C_2$ does not.

The groups are therefore nonisomorphic despite having the same multiset of composition factors.

By L1, $G$ is finite simple and is not cyclic, since every cyclic group is abelian: powers of one generator commute. By L2 the orders of $A_n$ for $0\le n\le 5$ are $1,1,1,3,12,60$. For $n\ge 6$ they are at least $6!\mathrm{/}2=360$, since each further factorial factor is positive and at least one. None equals $168$, so $G$ is not isomorphic to any alternating group.

Thus $G$ is a finite simple group that is neither cyclic nor alternating, refuting the statement with a fully local construction and simplicity proof.

The Oberwolfach report explicitly records an ongoing project, and the current AMS catalog records only the more limited completion accomplished by Number 10.

Thus the cited status evidence supports an ongoing programme, contradicting the claimed completion as of the design check.

It records only the source table's family names and conventions. It does not define root data, Chevalley groups, or the Lie-algebra constructions from which the families arise.

The claim that this page develops the groups through Lie-algebra structure is therefore false.