5.4=5.3.2 Construction

Lemma 1

Lemma 1. Let $X$ be almost simple with $L=\mathrm{soc}(X)$. Assume $X=GH$ such that $G$ is metacyclic and $H$ is a $\{2,3\}$-group. Then either

• $L={\mathrm{PSL}}_2(q)$ and $\{G\cap L,H\cap L\}\le \{D_{2(q+1)/d},Z_p^f:Z_{(q-1)/4}\}$ where $q=p^f$ and $d=(2,q-1)$,
• or a few possibilities: $L={\mathrm{PSL}}_2(7),{\mathrm{PSL}}_2(11),{\mathrm{PSL}}_2(23),{\mathrm{PSL}}_3(3),M_{11}$.

Proof. By Kazarin’s classification.

Lemma 2

Lemma 2. Let $\Gamma =\mathrm{Cay}(G,S)$ be connected and of valency 4, where $G$ is metacyclic. Assume $X$ satisfying $G<X\le \mathrm{Aut}\Gamma$ is almost simple and edge-transitive on $\Gamma$, then

$X$	${\mathrm{PGL}}_2(7)$	${\mathrm{PSL}}_2(11)$	${\mathrm{PGL}}_2(11)$	${\mathrm{PGL}}_2(11)$	${\mathrm{PSL}}_2(23)$	${\mathrm{PGL}}_2(23)$
$G$	$7:3$	$11:5$	$11:5$	$11:10$	$23:11$	$23:22$
$X_{\alpha }$	$D_{16}$	$A_4$	$S_4$	$A_4$	$S_4$	$S_4$

Proof. Since $X=G{X}_{\alpha }$ with $G$ metacyclic and $X_{\alpha }\le S_4$, by lemma 1, $\mathrm{soc}(X)$ can be determined. Last time, we have shown that some cases never happen and this proof just concentrates on “construction part”, though all notes of last time I does not understand at all.

As an example, we will construct a graph with $X={\mathrm{PSL}}_2(11)$ and $X_{\alpha }=A_4$.

Let $\Omega =[X:X_{\alpha }]$, where $|\Omega |=55$. Then $X$ is transitive on $\Omega$ by right multiplication.

Suppose $X_{\alpha }$ has an orbit $\Delta$ of size $4$. Then for $\beta \in \Delta$, $|{\beta }^{X_{\alpha }}|=4$ yields $(X_{\alpha }{)}_{\beta }$ has index $4$, i.e. $|X_{\alpha }:X_{\alpha \beta }|=4$ and so $X_{\alpha \beta }=⟨h⟩=Z_3$. Since $X$ is transitive on $\Omega$, there exists a $k\in \Omega$ such that ${\alpha }^k=\beta$ and so $X_{\alpha }^k=X_{{\alpha }^k}=X_{\beta }$. Since $⟨h⟩\le X_{\beta }$ and $⟨h{⟩}^k\le X_{\alpha }^k=X_{\beta }$, $⟨h⟩$ and $⟨h{⟩}^k$ are Sylow subgroups of $X_{\beta }$. By Sylow’s theorem, there is $x\in X_{\beta }$ such that $⟨h⟩=(⟨h{⟩}^k{)}^x=⟨h{⟩}^{kx}$, i.e. $kx\in N_X(X_{\alpha \beta })$.

In $X={\mathrm{PSL}}_2(11)$, $C_X(h)=Z_6\le D_{12}$ by inspecting the “Atlas of finite groups”. Let $g$ be the unique involution of $C_X(h)$. Then $⟨h{⟩}^g=⟨h⟩$ and so $X_{\alpha }\cap X_{\alpha }^g\supset ⟨h⟩=Z_3$. Since $Z_3$ is a maximal subgroup of $A_4$, $X_{\alpha }\cap X_{\alpha }^g=⟨h⟩$.

Let $\Gamma =\mathrm{Cos}(X,X_{\alpha },X_{\alpha }g{X}_{\alpha })$ and $M=⟨X_{\alpha },g⟩\le X$. By Atlas, $M\cong A_5$ or ${\mathrm{PSL}}_2(11)$. Since $gh\in M$ is of order $6$ and $A_5$ does not have elements of order $6$, $M={\mathrm{PSL}}_2(11)=X$. Thus $\Gamma$ is connected. We can also verify $\Gamma$ is undirected and is of valency $4$.

Note that ${\mathrm{PSL}}_2(11)=(11:5)A_4$, $X$ has a subgroup which acting on $\Omega$ is regular and so $\Gamma$ is regular. $■$

HW: Construct 4 graphs left; note that PGL(2,7) has been constructed last time.

Lemma 3

Lemma 3. Let $X$ be non-solvable with $L=\mathrm{soc}(X)$. Assume $X=GH$ such that $G$ is metacyclic and $X_{\alpha }$ is a $\{2,3\}$-group. Then either

• $X$ is almost simple, so $X$ is as above,
• or
  • $X=({\mathrm{PSL}}_2(7)\times Z_l):Z_2$ where $l$ is odd, ${\mathrm{PSL}}_2(7):Z_2={\mathrm{PGL}}_2(7)$ and $Z_l:Z_2=D_{2l}$;
  • $G=(Z_7:Z_3):Z_l$;
  • $X_{\alpha }=D_{16}$. (H=D_8? its order should be 16)

Proof. Suppose $X$ is not almost simple. Let $R$ be the solvable radical of $X$. Then $R$ is intransitive, otherwise $X=R{X}_{\alpha }$ where $X_{\alpha }$ is $\{2,3\}$-group yields $X$ solvable, contradiction. Denote $\overline{X}=X/R$. Then We conclude $(\overline{X},\overline{G},{\overline{X}}_{\alpha })$ is as in lemma 2.

Let $C=C_X(R)$ and $Z=X^{(\infty )}$, where $X^{(\infty )}$ is the smallest normal subgroup of $X$ such that $X/X^{(\infty )}$ is solvable. Then $C\cap Z,CZ⊲X$.

Let $L=\mathrm{soc}(\overline{X})$. Then $L={\mathrm{PSL}}_2(7),{\mathrm{PSL}}_2(11),{\mathrm{PSL}}_2(23)$ by lemma 2.

Case 1: $C$ is non-solvable

Suppose $C$ is non-solvable. Then $Z\le C$.

Case 1.1

If $R\cap X^{(\infty )}\le Z(X^{(\infty )})$, then $R\cap X^{(\infty )}$ is the Schur multiplies of $L$. Thus $R\cap X^{(\infty )}=1$ and $X^{(\infty )}$ is simple. So $X^{(\infty )}=L$ and $R{X}^{(\infty )}=R\times X^{(\infty )}=R\times L$.

If $R\ne 1$, then $L$ is intransitive on $V\Gamma$ and ${\Gamma }_L=C_l$ with $l$ odd. Thus $X$ is not $2$-acr transitive on $\Gamma$ and it follows that $L={\mathrm{PSL}}_2(7)$ and $X=({\mathrm{PSL}}_2(7)\times {\mathbb{Z}}_l){\mathbb{Z}}_2$ due to $X/L$ is arc-transitive on ${\Gamma }_L=C_l$.

Case 1.2

Assume $R\cap X^{(\infty )}\nleqq Z(X^{(\infty )})$. Then $X^{(\infty )}$ acts on $R$ non-transitively. Recall $R$ is metacyclic. Let $M$ be a maximal characteristic subgroup of $R$. Then $M⊲X$ and $R/M=Z_p^d$ for some prime $p$ and $d=1$ or $2$. Thus $X/M\cong Z_p^d:\overline{X}$.

Suppose $X/M\ne Z_p^d\times \overline{X}$. Then $L⊲\overline{X}\le \mathrm{Aut}(Z_p^d)\le {\mathrm{GL}}_2(p)$. By the following fact, $\overline{X}$ is trivial as $L\in \{{\mathrm{PSL}}_2(q):q=7,11,23\}$, which is impossible. So $X/M=Z_p^d\times \overline{X}$.

Fact: ${\mathrm{GL}}_2(p)$ does not contain a nonabelian simple group.

Let $Y$ be the preimage of $\overline{X}$ under $X\to X/M$. Then $Y=M.\overline{X}⊲X$. Now $M<R$. By induction, $Y=M\times \overline{X}$ and then $L⊲X$. So $L\cap R=1$ and $L$ centralizer $R$, a contradiction as $L$ acts non-trivially on $R$.

Case 2: $C$ is solvable.

Suppose $C$ is solvable. Then $X^{(\infty )}$ does not centralize $R$. Let $M\mathrm{char}R$. Then $X/M=R/M.\overline{X}=Z_p^d.\overline{X}$. Use the same argument in 1.2.