Classification of Quasiprimitive Rank 3 Groups

Theorem

If $\mathrm{rank}(G)=3$ and $G$ is quasiprimitive on $\Omega$, then $G$ is HA, AS or $N=T^2$ and $G$ acts on $\Omega =\Delta \times \Delta$ in product action, where $G$ induces a $2$-transitive action on $\Delta$.

\begin{proof} By the following lemmas. \end{proof}

Lemma

Let $M⊲G$ be regular on $\Omega$. Then $\mathrm{rank}(G)⩾\#\text{mbox}fusionclassesofelementsofM$. In particular, if $M$ is nonsolvable, then $\mathrm{rank}(G)⩾4$.

\begin{proof} Identify $\Omega$ with $M$ and let $\omega =1$. Then $G=M:G_{\omega }$ and $G_{\omega }$ acts on $M$ by conjugation. Now the action of $G_{\omega }⩽\mathrm{Aut}(M)$ on $M$ partitions into orbits, whose number $⩾$ the number of fusion classes of $M$. Suppose that $M$ is non-solvable. Then by Burnside’s Theorem, there exist $x$, $y$, $z\in G$ such that $|x|=r$, $|y|=s$ and $|z|=t$. Since $x$, $y$, $z$ are in different fusion classes, we have $\mathrm{rank}(G)⩾4$. \end{proof}

Corollary

With the lemma above, $G$ is not of type HS, HC, TW.

Lemma

Let $G$ be a quasiprimitive group of type SD. Then $\mathrm{rank}(G)⩾4$.

\begin{proof} Let N=T^k\lhd_\min G. Write $N=T_1\times \cdots \times T_n$ where $T_i$ are isomorphic. Then $N_{\omega }=\{(t,\cdots ,t):t\in T\}\cong T$, and $R=T_1\times \cdots \times T_{n-1}\times \{1\}$ is regular on $\Omega$. Note that $\Omega =[N:N_{\omega }]$ and $N_{\omega }(t_1,\cdots ,t_n)\in [N:N_{\omega }]$ with a representation $(t_k^{-1}{t}_1,\cdots ,t_k^{-1}{t}_{k-1},1)\in R$. Observe that $N_{\omega }(s_1,\cdots ,s_{n-1},1)=N_{\omega }(t_1,\cdots ,t_{n-1},1)$ iff $(s_1,\cdots ,s_{n-1},1)=(t_1,\cdots ,t_{n-1},1)$.

Note that $G=N:G_{\omega }=(T_1\times \cdots \times T_k):G_{\omega }$, $G_{\omega }$ normalizes $N$ and induces an action on $\{T_1,\cdots ,T_k\}$ transitively by N\lhd_\min G (need proof). The stabilizer $G_{\omega }⩽\mathrm{Aut}(T)\times S_k$ and $G_{\omega }$ acts on $\{T_1,\cdots ,T_n\}$ by conjugation transitively and faithful ($C_G(N)⊲G$ and $C_G(N)\cap N=1$ yields $C_G(N)=1$). Each element of $G_{\omega }$ has the form $\sigma \pi$ with $\sigma \in \mathrm{Aut}(T)$ and $\pi \in S_k$.

Pick two elements $x:=(s,1,\cdots ,1)$ and $y:=(t,1,\cdots ,1)$ are two nontrivial elements of $R$. Then $x,y$ lie in the same orbits of $G_{\omega }$ iff there exists $\sigma \pi \in G_{\omega }$ such that $(s,1,\cdots ,1)=(t,1,\cdots ,1{)}^{\sigma \pi }=(t^{\sigma },1,\cdots ,1{)}^{\pi }$.

• If $1^{\pi }=1$, then $s=t^{\sigma }$ and so they are conjugate in $\mathrm{Aut}(T)$.
• If $1^{\pi }=k$, the $(s,1,\cdots ,1)=(1,\cdots ,1,t^{\sigma })$ where the latter is conjugate to $\left((t^{\sigma }{)}^{-1},\cdots ,(t^{\sigma }{)}^{-1},1\right)$. Then $(s,1,\cdots ,1)=\left((t^{\sigma }{)}^{-1},\cdots ,(t^{\sigma }{)}^{-1},1\right)$ and so $k=2$, i.e., $(s,1)=(t^{{\sigma }^{-1}},1)$ and so $s,t^{{\sigma }^{-1}}$ are conjugate by $\sigma \in \mathrm{Aut}(T)$.
• If $1^{\pi }\notin \{1,k\}$, then $(s,1,\cdots ,1)=(1,\cdots ,1,t^{\sigma },1,\cdots ,1)$, which is impossible.

Therefore, $(s,1,\cdots ,1)$ and $(t,1,\cdots ,1)$ lie in the same $G_{\omega }$-orbits iff

• $s=t^{\sigma }$, or
• $k=2$ and $s=(t^{-1}{)}^{\sigma }$, and we say $s$ and $k$ are inverse conjugate in $\mathrm{Aut}(T)$.

Thus $\mathrm{rank}(G)⩾\#\text{mbox}orbitsof\mathrm{Aut}(T)\text{mbox}on\left\{\{t,t^{-1}\}:t\in T\right\}⩾4$.

\end{proof}

Example. $G=(A_5\times A_5).(2\times S_2)$ and $G_{\omega }=S_5\times S_2=\mathrm{Aut}(T)\times S_2$. Then $\mathrm{rank}(G)=4$.

Let $T⩽\mathrm{Sym}(\Delta )$, where $\Delta =\{0,1,\cdots ,m-1\}$. For $k⩾2$, define $G⩽T\wr S_k$ and $\Omega ={\Delta }^k$ such that $G$ acts on $\Omega$ naturally. We say $T\wr S_k$ acting on $\Omega$ is a blow-up of $T$ on $\Omega$.

Lemma

Using the notation above, we have

• $\mathrm{rank}(G)⩾k+1$;
• $\mathrm{rank}(G)=3$ iff $k=2$ and $T$ is $2$-transitive on $\Delta$.

\begin{proof} (i) Fix ${\omega }_0=(0,\cdots ,0)$. For a point $\omega =({\delta }_1,\cdots ,{\delta }_k)\in \Omega$, define $\mathrm{weight}(\omega )=\#i\text{mbox}with{\delta }_i\ne 0$. Then if $\mathrm{weight}({\omega }_1)\ne \mathrm{weight}({\omega }_2)$, then ${\omega }_1^g\ne {\omega }_2$ for any $g\in G_{{\omega }_0}$. Since $g$ preserves the weight of each point $\omega$ and the weight function ranges $\{0,1,\cdots ,k\}$, group $G_{{\omega }_0}$ has at least $k+1$ orbits and $\mathrm{rank}(G)⩾k+1$.

(ii) Assume $\mathrm{rank}(G)=3$, then $k=2$ and $\Omega =\Delta \times \Delta$. The set of all points with weight $1$ forms an orbits of $G_{{\omega }_0}$. Thus $T_0$ is transitive on $\Delta \setminus \{0\}$ and so $T$ is $2$-transitive on $\Delta$. Conversely, assume that $k=2$ and $T$ is $2$-transitive on $\Delta$. Since $T_0$ is transitive on $\Delta \setminus \{0\}$, then the set of all points with weight $1$ is an orbit and so for weight $2$. \end{proof}

Corollary

CD is not rank $3$.

\begin{proof} Since CD is a blow up of SD, and SD is not $2$-transitive. So CD is not rank $3$ by the lemma above. \end{proof}