Burnside 2-Transitive

Theorem

A $2$-transitive permutation group $G$ has a unique minimal normal subgroup which is either:

• non-abelian simple group, where $G$ is called almost simple;
• elementary abelian, where $G$ is called affine.

\begin{proof} #todo By ^e3ccf0, $\mathrm{soc}(G)\cong T^n$ with simple group $t$ and some $n\in {\mathbb{N}}_{+}$. If $\mathrm{soc}(G)\cong {\mathbb{F}}_p^n:=V$, then by Frattini’s Argument we have $G=V:G_0$ where $G_0\le \mathrm{GL}(V)$. Therefore, $G$ is affine.

Now we assume that $\mathrm{soc}(G)$ is a direct product of isomorphic non abelian simple groups.

(to be continued)

\end{proof}

Lemma

If $G⩽\mathrm{Sym}(\Omega )$ is a $2$-transitive permutation group, then $G$ has a unique minimal normal subgroup.

\begin{proof} Suppose that $K_1$ and $K_2$ are minimal normal subgroups of $G$. Then $K_1$ and $K_2$ are transitive by Orbits of Normal Subgroups Form a Block System. Since $K_1\cap K_2=\{1\}$, we have $K_1⩽C_G(K_2)$ and $K_2⩽C_G(K_1)$. It yields that $C_G(K_1)=K_2$ and $C_G(K_2)=K_1$ are regular normal subgroups of $G$ by Minimal Normal Subgroups of Permutation Groups. Then we have $G=K_1:G_{\omega }$ and so $G_{\omega }$ acting on $\Omega$ can be identified by $G_{\omega }$ acting $K_1$ by conjugation.

If $K_1$ is solvable, then $K_1$ is elementary abelian. It contradicts with $C_G(K_1)=K_2$ and $K_1\cap K_2=\{1\}$. If $K_1$ is nonsolvable, then $|K_1|$ has at least $3$ prime divisors $r,s,t$ by Burnside’s Theorem. Let $x,y,z$ be elements of $K_1$ such that $|x|=r$, $|y|=s$ and $|z|=t$. Then $x^{G_{\omega }}$, $y^{G_{\omega }}$ and $z^{G_{\omega }}$ are three orbits and $\mathrm{rank}(G)⩾4$, which is impossible. Therefore, the minimal normal subgroup of $G$ is unique. \end{proof}