5.3 generalization of 5.3

Question

Assume $G=⟨a⟩:⟨b⟩$ is a non-abelian group of odd order. Let $S=\{ba,(ba{)}^{-1},b{a}^{-1},(b{a}^{-1}{)}^{-1}\}$ and $\Gamma =Cay(G,S)$ be the Cayley graph.

We have known that $⟨\hat{G},\tau ⟩\le Aut\Gamma$ is vertex-transitive and edge-transitive on $\Gamma$. The question is, is $\Gamma$ arc-transitive?

The key point is to understand $\mathrm{Aut}\Gamma$. Let $X$ be a group satisfying $⟨G,\tau ⟩\le X\le \mathrm{Aut}\Gamma$. Then we will analyse $X$.

Case 1. $X$ is non-solvable

Quotient group of solvable radical in $X$

Let $R$ be the solvable radical of $X$, i.e. the largest solvable normal subgroup. Then $X/R$ is non-solvable. Since $R⊲X$, $R$ is half-transitive (i.e. $R$-orbits have the same length). Let $\mathcal{B}$ be an orbit of $R$ on $V$. Then $\mathcal{B}$ is a block system for $X$ on $V$.

Let $K=X_{(\mathcal{B})}$ be the kernel of $X$ on $\mathcal{B}$. $K\ge R$ is transitive on $B$ and $K=R{K}_{\alpha }$ with $\alpha \in B$. Since $K_{\alpha }⊲X_{\alpha }$ is a $\{2,3\}$-group, $K_{\alpha }$ is solvable by Burnside’s $p^a{q}^b$ theorem. Then $K\ge R$ is solvable. By $R$ solvable radical, $K=R$.

Burnside's Theorem

Burnside’s $p^a{q}^b$ Theorem

Let $p$ and $q$ be prime numbers, and let $a$ and $b$ be non-negative integers with $a+b\ge 2$. If $G$ is a group of order $p^a{q}^b$, then $G$ is not simple.

Link to original

Let $\Sigma ={\Gamma }_K$ be the quotient graph. Since $X$ is egde-transitive and $K⊲X$, the valency of $\Sigma$ divides $\mathrm{val}\Gamma =4$. Thus the valency of $\Sigma$ is 1, 2 or 4.

• If valency is 1, then $|\mathcal{B}|$ is 2. It is contradictory to $|\mathcal{B}|||G|$ and $|G|$ is odd.
• if valency is 2, then $\Sigma$ is a circle whose automorphism group is dihedral. Then $X/K\lesssim Aut\Sigma$ is solvable, which is also impossible.

Therefore, the valency of $\Sigma$ is 4.

Let $\overline{X}=X/K$. Then $\overline{X}=\overline{G}{\overline{X}}_{\alpha }$, where $\overline{G}$ is metacyclic and ${\overline{X}}_{\alpha }$ is a $\{2,3\}$-group.

Centralizer of the minimal normal subgroup of $\overline{X}$ is trivial

Let $M{⊲}_{\min }\overline{X}$. Since $\overline{X}$ is the quotient group of solvable radical, $M$ is non-abelian. Then $M=T^l$, where $T$ is non-abelian simple and $l\ge 1$.

Note that $l\le 2$. Otherwise assume that $l\ge 3$. Since $T$ is non-abelian simple, there exists a prime $p\ge 5$ dividing $T$ by Burnside’s $p^a{q}^b$ theorem. Then $Z_p^3\le M⊲\overline{X}$ and $Z_p^3\le G$ as $X_{\alpha }$ is $\{2,3\}$-group, which is impossible as $Z_p^3$ is not a subgroup of a metacyclic group.

Now assume $l=2$, then $M=T_1\times T_2$ where $T_i$ are isomorphic simple groups. Let $C=C_{\overline{X}}(M)$. Then $C⊲\overline{X}$ and $M\cap C=1$ by $M$ center-free. Thus $M\times C⊲\overline{X}$.

If $C\ne 1$, consider $\overline{X}/C$. If $C$ is transitive on $\mathcal{B}$, then $\overline{X}=C{\overline{X}}_B$ and $M⊲\overline{X}/C=C{\overline{X}}_B/C={\overline{X}}_B/{\overline{X}}_B\cap C$. Thus $M$ is a $\{2,3\}$-group, which is contradictory to $M=T^2$. Thus $C$ is intransitive on $\mathcal{B}$ and ${\Sigma }_C$ is a quotient graph. The argument is similar as above: consider the valency of ${\Sigma }_C$:

• If valency is 1, $|V{\Sigma }_C|=2$, which is impossible as $\mathcal{B}$ is odd.
• If valency is 2, ${\Sigma }_C$ is a circle and so $M\le Aut{\Sigma }_C$ is a subgroup of dihedral group. It derives a contradiction by $M$ is non-solvable.
• If valency is 4, ${\Sigma }_C$ and $\Sigma$ has the same valency and so $C$ acting on $\mathcal{B}$ is semiregular (otherwise valency of ${\Sigma }_C$ is strictly less than 4). Hence $|C|||\mathcal{B}|$. However, it is impossible since $|\mathcal{B}|$ is odd and $|C|$ is even ($C$ is nonsolvable yields $|C|$ is even).

Feit-Thompson Theorem

Every finite simple group (that is not cyclic) has even group order, and the group order of every finite simple non-commutative group is doubly even, i.e., divisible by 4.

Therefore, $C=1$. By N/C lemma, $\overline{X}=\overline{X}/C=N_{\overline{X}}(M)/C\lesssim Aut(M)=Aut(T)\times Aut(T):Z_2$.

$\overline{X}$ is almost simple

Now we have known that

$$T^2\le \overline{X}\le Aut(T{)}^2:Z_2.$$

Let $L=\overline{X}\cap Aut(T{)}^2$. Note that $L⊲\overline{X}$, $|\overline{X}|/|L|\in \{1,2\}$ and $|\mathcal{B}|$ is odd. Thus

$$|B_0^L|\cdot |L_{B_0}|=|B_0^{\overline{X}}|\cdot |{\overline{X}}_{B_0}|,$$

or

$$2\cdot |B_0^L|\cdot |L_{B_0}|=|B_0^{\overline{X}}|\cdot |{\overline{X}}_{B_0}|.$$

Since $|L_{B_0}|||{\overline{X}}_{B_0}|$ and $|B_0^{\overline{X}}|=|\mathcal{B}|$ is odd, $|B_0^L|=|\mathcal{B}|$ and so $L$ is transitive on $\mathcal{B}$.

Consider the quotient graph ${\Sigma }_{T_1}$. Since $T_1$ is non-solvable and $|\mathcal{B}|$ is odd, $T_1$ acting on $\mathcal{B}$ has non-trivial stabilizer. Thus valency of ${\Sigma }_{T_1}$ is 2 and $T_2\le Aut({\Sigma }_{T_1})$ is dihedral, which is impossible.

Therefore, $M=T$ and $\overline{X}$ is almost simple.

Explicit structure of $\overline{X}$ （问题很多）

By Kazarin’s theorem, … see notes of 23/4/13.

Some cases never happen

Case 1. Suppose $M={\mathrm{PSL}}_2(q)$ and ${\overline{X}}_{\alpha }\cap M=D_{q+1}$ where $q=p^f$. If $f\ge 2$, $q^2-1=p^{2f}-1$ has a primitive prime divisor $r$, where primitive prime divisor is defined in the following theorem.

Zsigmondy’s theorem. Let $a,b\in N$ such that $\gcd (a,b)=1$ and $n\in N,n>1$. There exists a prime divisor of $a^n-b^n$ that does not divide $a^k-b^k$ for all $k\in \{1,2,\cdots n-1\}$, which is called primitive prime divisor, except in the following cases:

• $2^6-1^6$,
• $n=2$ and $a+b$ is a power of $2$.

Note that $r\ge 2f+1$: For any primitive prime divisor $s$ of $p^{2f}-1$ with $s\le 2f$, if $s$ divides none of $p^0-1,\cdots ,p^{2f-1}-1$, there exist $k_1\ne k_2$ such that $p^{k_1}-1\equiv p^{k_2}-1\mathrm{mod}s$. Then $p$ is divided by $s$, which is contradictory with $s$ dividing $p^{2f}-1$. So $r\ge 2f+1\ge 5$. However, $r\ge 5$ is a primitive prime divisor of $p^{2f}-1$ yields $p^f+1=q+1$ is divided by $r$. It is impossible as $D_{q+1}$ is a $\{2,3\}$-group.

Therefore, $q=p$ for some prime $p$ and so $L\cap \overline{X}={\mathrm{PSL}}_2(p)$, ${\overline{X}}_{\alpha }\cap L=D_{p+1}$. Then $\overline{X}={\mathrm{PSL}}_2(p)$ or ${\mathrm{PGL}}_2(p)$ and ${\overline{X}}_{\alpha }=D_{p+1}$ or $D_{2(p+1)}$ respectively. 这是什么结论吗？以PSL为socle的群就这些？对应的Xalpha又为什么是这两种？ Let ${\overline{X}}_{\alpha }=⟨a⟩:⟨b⟩$ where $|a|=(p+1)/2$ or $p+1$.

Define ${\overline{X}}_{\alpha \beta }={\overline{X}}_{\alpha }\cap {\overline{X}}_{\beta }$. There is $g\in \overline{X}$ s.t. ${\beta }^g=\alpha$, then ${\overline{X}}_{\beta }^g={\overline{X}}_{\alpha }$.

Suppose $C\le {\overline{X}}_{\alpha \beta }\le {\overline{X}}_{\alpha }$ is cyclic with $|C|\ge 3$. Then $C^g\le {\overline{X}}_{\alpha \beta }^g\le {\overline{X}}_{\alpha }$ is also a cyclic subgroup of a dihedral group. Note that cyclic subgroup with order greater than $2$ is unique. Hence $C=⟨a_0⟩$ for some $a_0\in ⟨a⟩$ and so $C^g=C$. Thus $g\in N_{\overline{X}}(C)=D_{p+1}$ or $D_{2(p+1)}$. (并不知道为什么C的normalizer一定长这样。但g怎么会在${\overline{X}}_{\alpha }$中呢？我觉得这本身就已经矛盾了：因为g没有固定alpha) It is contradictory to $⟨{\overline{X}}_{\alpha },g⟩=\overline{X}$. 既不知道为什么Xalpha和g生成X，也不知道生成了能有什么矛盾－－如果是我上面说的那样，直接从那得到矛盾就可以了啊。 So $|{\overline{X}}_{\alpha \beta }\cap ⟨a⟩|=1$ or $2$.

Since ${\overline{X}}_{\alpha \beta }\subset {\overline{X}}_{\alpha }=⟨a⟩:⟨b⟩$, we have

$$4\ge |{\beta }^{{\overline{X}}_{\alpha }}|=|{\overline{X}}_{\alpha }:{\overline{X}}_{\alpha \beta }|\ge (p+1)/4.$$

Hence $p\le 15$. By prime $p$ satisfying $p\equiv 3\mathrm{mod}4$, $p=7$. Therefore, $\overline{X}={\mathrm{PSL}}_2(7)$ or ${\mathrm{PGL}}_2(7)$ and ${\overline{X}}_{\alpha }=D_8$ or $D_{16}$.

Case 2. Assume $L={\mathrm{PSL}}_2(q)$ and $H:={\overline{X}}_{\alpha }\cap L=S_4$. $|[L:H]|=7$ and $L\curvearrowright [L:H]$ 2-transitive yields the graph is $K_7$, 2trans和K7是怎么来的都不知道。我猜这里是要对一个子图说明valency大于4来出矛盾。 which is impossible as ${\Gamma }_R$ is of valency $4$.

Case 3. Assume $L={\mathrm{PSL}}_3(3)$ and $L_{\alpha }:={\overline{X}}_{\alpha }\cap L=3^2:{\mathrm{\Gamma L}}_1(9)$, where ${\mathrm{\Gamma L}}_1(9)={\mathrm{GL}}_1(9):⟨\varphi ⟩$ and $|\varphi |=2$ is generated by the Frobenius automorphism.

Let $L_{\alpha }^{[1]}$ be the kernel of $L_{\alpha }$ acting on ${\Gamma }_R(\alpha )$. Then $L_{\alpha }/L_{\alpha }^{[1]}\cong L_{\alpha }^{{\Gamma }_R(\alpha )}\le S_4$. … 应该能推出且要推出$L_{\alpha }/L_{\alpha }^{[1]}$是某个阶不是2的幂的群？

Fact. ${\mathrm{A\Gamma L}}_1(9)=Z_3^2:(Z_8:Z_2)$ has unique minimal subgroup $Z_3^2$.

Since $|L_{\alpha }|>|S_4|$, $|L_{\alpha }^{[1]}|$ is not trivial. And $L_{\alpha }^{[1]}⊲L_{\alpha }$ yields $L_{\alpha }^{[1]}\ge Z_2^3$ by fact. Then $L_{\alpha }/L_{\alpha }^{[1]}$ is a $2$-group. Contradiction.

Some cases can be constructed

Case 1. Let $\overline{X}={\mathrm{PSL}}_2(23)$. Then ${\overline{X}}_{\alpha }=S_4$ and ${\overline{X}}_{\alpha \beta }=S_3$. If there exists $g\in \overline{X}$ such that $(\alpha ,\beta )\mapsto (\beta ,\alpha )$, then $g\in N_{\overline{X}}({\overline{X}}_{\alpha \beta })=D_6\times S_2$. 这怎么啪的就出来了，怎么算的（

Fact. $S_4\le {\mathrm{PSL}}_2(23)$ is a maximal subgroup.

By the fact, $\Gamma =Cos(\overline{X},{\overline{X}}_{\alpha },{\overline{X}}_{\alpha }g{\overline{X}}_{\alpha })$ is connected of valency $4$. Let $\overline{G}=Z_{23}:Z_{11}$. Since $\overline{X}=\overline{G}{\overline{X}}_{\alpha }$, $\overline{G}$ is transitive on $[\overline{X}:{\overline{X}}_{\alpha }]$. $|\overline{G}|=|[\overline{X}:{\overline{X}}_{\alpha }]|$ yields $G$ acting on the coset graph is regular and so it is a Cayley graph.

Case 2. Let $\overline{X}={\mathrm{PSL}}_2(7)$. Then ${\overline{X}}_{\alpha }=D_8=⟨a⟩:⟨b⟩$. Let $g\in C_{\overline{X}}(b)\backslash {\overline{X}}_{\alpha }$, where $C_{\overline{X}}(b)\cong D_8$. (why we take g in centralizer?) By Magma, $Y=⟨{\overline{X}}_{\alpha },g⟩\cong S_4$. (so what? does this mean the case of PSL(2,7) is impossible?)

Let $\overline{X}={\mathrm{PGL}}_2(7)$. Then ${\overline{X}}_{\alpha }=D_{16}=⟨a⟩:⟨b⟩{<}_{\max }\overline{X}$. (Why $X_{\alpha }=D_{16}$, instead of $D_8$ or $S_4$?前面提过这个问题了。) Let $g\in N_{\overline{X}}(⟨a^4,b⟩)\backslash {\overline{X}}_{\alpha }$, (similarly, why g is taken from normalizer) where $N_{\overline{X}}(⟨a^4,b⟩)\cong S_4$. Since ${\overline{X}}_{\alpha }$ is a maximal group of $\overline{X}$, $⟨{\overline{X}}_{\alpha },g⟩=\overline{X}$. Using the same argument above, $Cos(\overline{X},{\overline{X}}_{\alpha },{\overline{X}}_{\alpha }g{\overline{X}}_{\alpha })$ is a Cayley graph.

Explicit structure of $X$

Now we have known that $\overline{X}\in \{{\mathrm{PGL}}_2(7),{\mathrm{PSL}}_2(11),{\mathrm{PGL}}_2(11),{\mathrm{PSL}}_2(23)\}$. (i think up to now we only know L is one of groups in the list? 所以Xbar会随之确定？) And $X=R\cdot \overline{X}$ where $R$ is metacyclic and $R⊲X$ is the maximal solvable subgroup of $X$.

Let $M$ be the maximal characteristic group of $R$. Then $M⊲X$ and $R/M$ is characteristically simple, which is $Z_p$ or $Z_p\times Z_p$ by the following lemma.

Def. If $G$ is a group, then $H$ is a characteristic subgroup of $G$ (written $H\mathrm{char}G$) if every automorphism of $G$ maps $H$ to itself.

Prop. If $K\mathrm{char}H$ and $H⊲G$ then $K⊲G$.

Def. A group is said to be characteristically simple if it has no proper nontrivial characteristic subgroups.

Lemma. A group $G$ is characteristically simple iff $G$ is a direct product of isotropic simple groups.

Suppose $R/M=Z_p$. Then $X/M=Z_p\cdot \overline{X}$… 放弃，最后是要证明出$X=\overline{X}$，但是我不仅不知道那几个等号是怎么搞出来的，也不知道这个等式怎么就说明了$X=\overline{X}$。坐等笔记。

… Thus, $X=\overline{X}\in \{{\mathrm{PGL}}_2(7),{\mathrm{PGL}}_2(11),{\mathrm{PSL}}_2(11),{\mathrm{PSL}}_2(23)\}$.