Classification of 2-Transitive Groups

Here we want to prove the following well-known theorem.

In fact, it is proved here. This notes come from GroupTheory2024Spring, but I don’t understand the idea of our professor ( )

Burnside

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 (see here);
• elementary abelian, where $G$ is called affine (see here).

Recall that for a quasiprimitive permutation group $G$, we have $\mathrm{soc}(G)=T^n$ where $T$ is a simple group and $n\ge 1$. The proof is as follows.

Transclude of praegerNanScottTheoremFinite1993#^9rlaf2

Let $G$ be a $2$-transitive group. Then $G$ is primitive and quasiprimitive, and so $\mathrm{soc}(G)=T^n$ with simple $T$. Then we have the following lemma.

Lemma

Let $G\le \mathrm{Sym}(\Omega )$ be a $2$-transitive group. Let $N$ be a normal subgroup of $G$ such that $N=T^k$ with $k\ge 2$ and $T$ non-abelian simple. Then:

• $N$ does not have a proper normal subgroup which is regular on $\Omega$. Furthermore, the group $G$ is not HS, HC, SD, CD.
• $N$ is not regular on $\Omega$. Furthermore, $G$ is not TW.

\begin{proof} Assume that there is a regular $M⊲N$. Note that $N_{\omega }\ne 1$ is half-transitive on $\Omega \setminus \{\omega \}$ as $N_{\omega }⊲G_{\omega }$, and define $m$ as the size of an orbits of $N_{\omega }$ on $\Omega \setminus \{1\}$. Then both $|\Omega |-1$ and $|N_{\omega }|$ are divisible by $m$. Furthermore, because $N=M:N_{\omega }$, we have $|\Omega |=|M|=|T{|}^l$ and $|N_{\omega }|=|T{|}^{k-l}$ with $1\le l<k$. So $m$ divides $\gcd (|\Omega |-1,|N_{\omega }|)=(|T{|}^l-1,|T^{k-l}|)=1$ and it yields $m=1$, which is impossible.

Suppose that $N$ is regular on $\Omega$. Then we have $G=N:G_{\omega }$ and so $G_{\omega }$ acting on $\Omega$ can be identified by $G_{\omega }$ acting $N$ by conjugation. By Burnside’s Theorem $|N|$ has at least $3$ prime divisors $r,s,t$. Let $x,y,z$ be elements of $N$ such that $|x|=r$, $|y|=s$ and $|z|=t$. Then $x^{G_{\omega }}$, $y^{G_{\omega }}$ and $z^{G_{\omega }}$ are three orbits and so $\mathrm{rank}(G)\ge 4$. Contradiction. \end{proof}

Corollary

Let $G$ be a quasiprimitive permutation group on $\Omega$, and let $N=\mathrm{soc}(G)$. If $N$ is regular on $\Omega$, then $\mathrm{rank}(G)\ge 4$.

\begin{proof} By the proof of (ii) of ^ou5ao7. \end{proof}

Theorem

Let $G\le \mathrm{Sym}(\Omega )$ be a $2$-transitive group. Then $G$ acts on $\Omega$ in product action, that is, $\Omega ={\Delta }^k$ and $G\le \mathrm{Sym}(\Delta )\wr S_k$ acting on $\Omega$ by

$$({\delta }_1,\cdots ,{\delta }_k{)}^{({\tau }_1,\cdots ,{\tau }_k)\pi }=({\delta }_1^{{\tau }_1},\cdots ,{\delta }_k^{{\tau }_k}{)}^{\pi }=({\delta }_{1^{{\pi }^{-1}}}^{{\tau }_{1^{{\pi }^{-1}}}},\cdots ,{\delta }_{k^{{\pi }^{-1}}}^{{\tau }_{k^{{\pi }^{-1}}}})$$

where ${\delta }_i\in \Delta$ and $\pi \in S_k$. Furthermore, in this case $G$ is PA.

\begin{proof} Let $N{⊲}_{\min }G$ such that $N=T^k$ with $k\ge 2$ and $T$ nonabelian simple. Note that $N$ is transitive as $G$ is $2$-transitive and quasiprimitive. So $N_{\omega }\ne 1$ and $N$ does not have a normal subgroup which is regular on $\Omega$ by ^ou5ao7. Additionally, by the proof of Minimal Normal Subgroups of Permutation Groups, $N$ is the unique minimal normal subgroup. We write $N$ as $N=T_1\times \cdots \times T_k$ where $T_1\cong \cdots \cong T_k$.

Let $H_i$ be the projection of $N_{\omega }$ into $T_i$. Then $H_i\cong N_{\omega }/(N_{\omega }\cap M_i)\cong N_{\omega }{M}_i/M_i$, where $M_i={\Pi }_{j\ne i}{T}_j$ and $N/M_i\cong T_i$.

• Then $H_i\lesssim T_i$ and $H_1\cong \cdots \cong H_k$.

Since $G=N{G}_{\omega }$ is transitive on $\{T_1,\cdots ,T_k\}$ and $N$ acts on $\mathcal{T}:=\{T_1,\cdots ,T_k\}$ trivially by conjugation, we have $G_{\omega }$ is transitive on $\mathcal{T}$. Also, $G_{\omega }$ is transitive on $\{H_1,\cdots ,H_k\}$ by $N_{\omega }⊲G_{\omega }$.

Then $N_{\omega }\le K:=H_1\times \cdots \times H_k<N$ and $K$ is normalized by $G_{\omega }$. Thus $G_{\omega }\le K{G}_{\omega }<G$. Since $G$ is $2$-transitive and primitive, $G_{\omega }$ is maximal and so $G_{\omega }=K{G}_{\omega }$. Note that $K\le G_{\omega }$ and $K<N$ yield $K\le N_{\omega }$. Therefore, we have $K=N_{\omega }$.

As $N=T_1\times \cdots \times T_k$ acting on $\Omega$ transitively, $\Omega$ can be identified with

$$[N:N_{\omega }]=[(T_1\times \cdots \times T_k):(H_1\times \cdots \times H_k)]=[T_1:H_1]\times \cdots \times [T_k:H_k]:={\Delta }_1\times \cdots \times {\Delta }_k,$$

and $N$ acts on $\Omega$ by

$${\omega }^{(t_1,\cdots ,t_k)}=(H_1{x}_1,\cdots ,H_k{x}_k{)}^{(t_1,\cdots ,t_k)}=(H_1{x}_1{t}_1,\cdots ,H_k{x}_k{t}_k).$$

• Then we have $G_{\omega }\le (L_1\times \cdots \times L_k):S_k$ where $H_i\le L_i\le \mathrm{Aut}(T_i)$ and $G\le (\mathrm{Aut}(T_1)\times \cdots \times \mathrm{Aut}(T_k)):S_k$.

It yields that $G$ acts on $\Omega$ in product action and we complete the proof. \end{proof}

Corollary

By Classification of Quasiprimitive Groups, a $2$-transitive permutation group is one of the three types: HA, AS and PA.

• to be continued