Distinct components commute

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.