Finite Symmetric Graphs with 2-Arc-Transitive Quotients: General Affine Case

Abstract

Let G be a finite group and $\Gamma$ a G-symmetric graph. Suppose that G is imprimitive on $V(\Gamma )$ with B a block of imprimitivity and $\mathcal {B} := \{B^g: g\in G\}$ is a system of imprimitivity of G on $V(\Gamma )$. Define $\Gamma _{\mathcal {B}}$ to be the graph with vertex set $\mathcal {B}$, such that two blocks $B, C \in \mathcal {B}$ are adjacent if and only if there exists at least one edge of $\Gamma$ joining a vertex in B and a vertex in C. Set $v=|B|$ and $k := |\Gamma (C)\cap B|$ where C is adjacent to B in $\Gamma _{\mathcal {B}}$ and $\Gamma (C)$ denotes the set of vertices of $\Gamma$ adjacent to at least one vertex in C. Assume that $k=v-p\ge 1$, where p is an odd prime, and $\Gamma _{\mathcal {B}}$ is (G, 2)-arc-transitive. In this paper , we show that if the group induced on each block is an affine group then $v=6$.

1 Introduction

Let G be a finite group. A graph $\Gamma$ is called G-symmetric if $\Gamma$ admits G as a group of automorphisms acting transitively on the set of vertices and the set of arcs of $\Gamma$, where an arc is an ordered pair of adjacent vertices. Suppose that G is imprimitive on $V(\Gamma )$ with B a block of imprimitivity

$$\begin{aligned} \mathcal {B}:= \{B^g: g\in G\} \end{aligned}$$

is a system of imprimitivity of G on $V(\Gamma )$. Define $\Gamma _{\mathcal {B}}$ to be the graph with vertex set $\mathcal {B}$, such that two blocks $B, C \in \mathcal {B}$ are adjacent if and only if there exists at least one edge of $\Gamma$ joining a vertex in B and a vertex in C. We call $\Gamma _{\mathcal {B}}$ the quotient graph of $\Gamma$ with respect to $\mathcal {B}$. A graph $\Gamma$ is called (G, 2)-arc-transitive if it admits G as a group of automorphisms acting transitively on the set of vertices and the set of 2-arcs of $\Gamma$, where a 2-arc is an oriented path of length two. Denote by $G_B$ the setwise stabilizer of B in G, and define $H:=G_{B}^{\Gamma _{\mathcal {B}}(B)}$ to be the quotient group of $G_B$ relative to the kernel of the induced action of $G_B$ on the set of blocks of $\mathcal {B}$ adjacent to B in $\Gamma _{\mathcal {B}}$. We keep these notation in this paper. In [3], the following question was asked:

Question 1.1

Under the assumption above, when is $\Gamma _{\mathcal {B}}$ a (G, 2)-arc-transitive graph?

This question is studied in [3,4,5,6,7,8] and [9]. In this paper, we investigate the question above when H is an affine group, see [2] for definition of an affine group and other background in permutation groups. To state our main results, we need some definitions and notation.

Throughout this paper, $\Gamma _{\mathcal {B}}$ is a connected (G, 2)-arc-transitive graph with valency b. Let B and C be two vertices in $\Gamma _{\mathcal {B}}$ such that C is adjacent to B. Set $v=|B|$ and $k := |\Gamma (C)\cap B|$, where $\Gamma (C)$ denotes the set of vertices of $\Gamma$ adjacent to at least one vertex in C. In this paper, we assume that $k=v-p\ge 1$, for some odd prime p, and H is the affine group $AGL_n(q)$. In [7], we have shown that if $q=2$, then $v=6$. In this paper, we consider the general case and shall prove theorem below which improves our results in [7]. In fact, in this paper, we show that H cannot be a 2-transitive subgroup of $AGL_n(q)$ for odd prime q.

Theorem 1.2

Assume that $k=v-p\ge 1$, for some odd prime p, and H is an affine group. Then, $q=2$ and $v=6$.

Theorem 1.2 shows that the graph which appears in [9, Theorem 3(e)] is the only graph satisfying the conditions of Theorem 1.2.

2 Proof of the Main Theorem

In this section, we prove Theorem 1.2. Therefore, we keep the assumptions in 1.2. We fix $B \in \mathcal {B}$. Let $\mathcal {U}:= \Gamma _{\mathcal {B}}(B)$ be the set of blocks of $\mathcal {B}$ adjacent to B in $\Gamma _{\mathcal {B}}$. For $\alpha \in B$, let $\Gamma _{\mathcal {B}}(\alpha )$ be the set of blocks in $\mathcal {U}$ containing at least one neighbour of $\alpha$ in $\Gamma$, and let $r:=|\Gamma _{\mathcal {B}}(\alpha )|$. Since $\Gamma$ is G-symmetric and $\mathcal {B}$ is G-invariant, r, v and k are independent of the choice of $\alpha$, B, and C, respectively, for each $C\in \mathcal {U}$. Denote by $G_B$ the setwise stabilizer of B in G, and define $H:=G_{B}^{\Gamma _{\mathcal {B}}(B)}$ to be the quotient group of $G_B$ relative to the kernel of the induced action of $G_B$ on $\mathcal {U}$.

Our strategy to prove Theorem 1.2 is to show that the minimal normal subgroup of H is 2-group. Then, Theorem 1.2 will follow from the main results in [7]. By our assumption, H is a 2-transitive affine group on $\mathcal {U}$. Therefore, we can control the structure of the minimal normal subgroup of H and the parameter b. This has been done in [9]. In fact, in [9] necessary conditions for $\Gamma _{\mathcal {B}}$ to be (G, 2)-arc-transitive were obtained in the case when $k=v-p\ge 1$, where p is an odd prime. In fact, $\Gamma$, $\Gamma _{\mathcal {B}}$, and H satisfy the conditions in the seventh row of Table 1 in [9]. Therefore, with our assumptions, the following theorem follows from the seventh row of Table 1 in [9].

Theorem 2.1

We have $sp+1=b=q^n\ge 2$, where $s\ge 2$, p, and q are primes and $v=q^mp$, where $n-1\ge m\ge 1$.

By Theorem 2.1, we get that $|\mathcal {U}|=sp+1=q^n$ for some prime q and the minimal normal subgroup of H is an elementary abelian group of order $q^n$. We keep the notation in Theorem 2.1.

Now, set

$$\begin{aligned} \mathcal {U}:=\{C,C_1,\ldots ,C_{sp}\},\;\, W := \Gamma (C)\cap B, W_i=\Gamma (C_i)\cap B,\quad {\text {for }}\, i= 1,\ldots ,sp. \end{aligned}$$

Since in this paper, we investigate [9, Theorem 1.1(f)], our parameters are as in the last row [case (f)] of Table 1 in [9]. In fact, $a=q^m$, $v=ap$, $b=|\mathcal {U}|=sp+1=q^n$, $r=q^n(a-1)/a$ and $\lambda =|W\cap W_i|=p(a-2)+(b-a)/as$, for $i=1,2,\ldots ,sp$. We have also that $2\le s\le a-1\le p-2$. We have that $|B {\setminus } W|=p$, by our assumption. Let $H_C$ be the stabilizer of C in H. Then, $H_C$ leaves W and $B {\setminus } W$ invariant. Since $\Gamma _{\mathcal {B}}$ is (G, 2)-arc-transitive, H is 2-transitive on $\mathcal {U}$, and therefore, $H_C$ is transitive on $\mathcal {U}{\setminus } \{C\}$. In fact, $\Gamma$, $\Gamma _{\mathcal {B}}$, and H satisfy the conditions in the seventh row of Table 1 in [9]. Therefore, we have that $H=N\rtimes H_C$ is an affine group (isomorphic to a subgroup of $\mathrm{AGL}(n, q)$), where $N \cong \mathbb {Z}_q^n$ is an elementary abelian group of order $sp+1=q^n$ and is the minimal normal subgroup of H acting regularly on $\mathcal {U}$ with $C_H(N)=N$, and $H_C$ is isomorphic to a subgroup of $\mathrm{GL}_n(q)$ and acts transitively on subgroups of order q in N. We note that since the case $q=2$ is considered in [7], so we may assume that q is an odd prime. Since $sp=q^n-1$, we get that $H_C$ is of even order. Let P(N) be the set of subgroups of order q in N.

Since $\mathcal {U}{\setminus }\{C\}$ has exactly sp element and $H_C$ is transitive on it, sp divides the order of $H_C$. Since p is a prime, $H_C$ contains an element of order p, say, x and an involution say, z. Define

$$\begin{aligned} M := \langle x,z\rangle \le H_C,\;\, P := \langle N,x,z\rangle = N \rtimes M \le H,\;\, X:=\langle x\rangle . \end{aligned}$$

Lemma 2.2

The following hold:

(i):

$C_N(x)=1$ and X has s orbit on $\mathcal {U}{\setminus } \{C\}$.

(ii):

X fixes W and $B {\setminus } W$ setwise and is fixed-point-free on each of them.

(iii):

X is regular on $B {\setminus } W$.

Proof

• (i) By coprime action (see [1, Sect. 24]), we have $N=C_N(X)\oplus [N,x]$. Let I be the set of elements in $\mathcal {U}$ fixed by X. Then, for $y\in C_N(x)$, we have $C^y\in I$. If $C_i\in I$, then there is $y\in N$, such that $C_i=C^y$ which implies that $y\in C_N(x)$. Hence, $|I|=|C_N(x)|$.

  Assume that $|I|\ne 1$. We note that for each $\{C_i, C_j\}\subset I$, $i\ne j$, we have $B{\setminus } W_i$ and $B{\setminus } W_j$ are sets with p element and x acts on each of them. This gives us that x acts trivially on $B{\setminus }( W_i\cap W_j)$. In fact, x acts trivially on $B{\setminus }(\bigcap _{C_i\in I} W_i)$. Let $\alpha \in \bigcap _{C_i\in I} W_i$ and l be the number of orbits of X on $\mathcal {U}{\setminus } I$. Then, we have $r=|\Gamma _{\mathcal {B}}(\alpha )|=|I|+l$. On the other hand, we have $|I|+lp=|\mathcal {U}|=q^n$. These give us that $r-l=q^n-lp$ which implies that $q^n-q^{n-m}-l=q^n-lp$. Now, we have $l(p-1)=q^{n-m}$ which implies that $q=2$ is even, a contradiction to our assumption. This contradiction shows that $\bigcap _{C_i\in I} W_i=\emptyset$, and then, x is trivial on B, a contradiction. Therefore, $|I|=1=|C_N(x)|$ and X has s orbits on $\mathcal {U}{\setminus } \{C\}$. Now i) holds.

• (ii) Since $X \le H_C$, it fixes W and $B {\setminus } W$ setwise. If a vertex $\alpha \in B{\setminus } W$ is fixed by a non-identity element of X, then it is fixed by every non-identity element of X. Since by (a), X has s orbits on $\mathcal {U}{\setminus } \{C\}$, we get that p divides r. However, $r=q^n-q^{n-m}$, and p does not divide r. Therefore, X is fixed-point-free on $B{\setminus } W$. A similar argument shows that X is fixed-point-free on W.

• (iii) Since $|X|=|B {\setminus } W|=p$ is a prime and X acts fixed-point-freely on $B {\setminus } W$, X must be regular on $B {\setminus } W$ and the Lemma holds. $\square$

Lemma 2.3

No nonempty subset of W is N invariant.

Proof

Suppose to the contrary that $\emptyset \ne Y \subseteq W$ is N invariant. Since N is regular on $\mathcal {U}$, for each $C_i$, there exists a unique element $g_i \in N$ such that $C^{g_i} = C_i$. Hence, $W^{g_i} = \Gamma (C_i) \cap B$. Since Y is N invariant, we have $Y = Y^{g_i} \subseteq W^{g_i}$ for $i=1,2, \ldots ,sp$, which implies that $q^n-1=sp=r-1$, a contradiction. $\square$

Lemma 2.4

NX is transitive on B.

Proof

Let $\alpha ^N$ be an N orbit on B, where $\alpha \in B$, and set $A=\cup _{g \in NX} (\alpha ^N)^g$. Since $N \unlhd NX \le H$, $A \subseteq B$ and NX is transitive on A; thus, both A and $B{\setminus } A$ are NX invariant. In particular, both A and $B{\setminus } A$ are N invariant and X invariant. Since $A \ne \emptyset$, by Lemma 2.3, we have $A\cap (B{\setminus } W)\ne \emptyset$. On the other hand, by Lemma 2.2, X is transitive on $B {\setminus } W$. Since A is X invariant and $A\cap (B{\setminus } W)\ne \emptyset$, it follows that $B{\setminus } W \subseteq A$. Now that $B{\setminus } A\subseteq W$ and $B{\setminus } A$ is N invariant, by Lemma 2.3, $B{\setminus } A=\emptyset$, and hence, NX is transitive on $B = A$. $\square$

Let $\alpha \in B$, and set

$$\begin{aligned} \mathcal {F}:=\{(\alpha ^N)^g: g \in X\}. \end{aligned}$$

Since N is normal in P, $\mathcal {F}$ is a system of imprimitivity for P. Then, $|\alpha ^N|=q^m=a$, $|\mathcal {F}|=p$, and $\mathcal {F}$ is the set of all N orbits on B.

Lemma 2.5

We have $|\alpha ^N \cap (B{\setminus } W)|=1$. In fact, each element of $B{\setminus } W$ is in a unique element of $\mathcal {F}$ and each element of $\mathcal {F}$ contains a unique element of $B{\setminus } W$.

Proof

Let $h \in B {\setminus } W$. Since $\mathcal {F}$ is a system of imprimitivity for P, there exists $(\alpha ^N)^g \in \mathcal {F}$, $g \in X$, such that $(\alpha ^N)^g \cap (B{\setminus } W) \ne \emptyset$ and we may assume that $h\in \alpha ^N$. Since X fixes $B{\setminus } W$ setwise and is transitive on $B{\setminus } W$, we have $h^X = B{\setminus } W$. By this and the fact $|\mathcal {F}|=p=|B{\setminus } W|$, we get that $\alpha ^N \cap (B{\setminus } W)=\{h\}$. This shows that each element of $B{\setminus } W$ is in a unique element of $\mathcal {F}$ and each element of $\mathcal {F}$ contains a unique element of $B{\setminus } W$. $\square$

For each $1\ne g\in H$, we define $B_g$ to be the set of all elements in B fixed by g.

Lemma 2.6

For $1\ne y\in N$, we have $|B_y|=a(q^{n-m}-1)/s$. Furthermore, $|B_y\cap (B{\setminus } W)|=(q^{n-m}-1)/s$

Proof

Since N has p orbits on B, by Burnside’s Counting Theorem, we have $p=(ap+sp|B_y|)/q^n$ and then $|B_y|=a(q^{n-m}-1)/s$. Since N acts on $B_y$ and each orbit of N on $B_y$ has a elements, by Lemma 2.4, we get that $|B_y\cap (B{\setminus } W)|=(q^{n-m}-1)/s$ and the lemma is proved. $\square$

Lemma 2.7

$[z,x]\ne 1$.

Proof

Assume $[z,x]=1$. Since $B{\setminus } W$ is of order p and M invariant, we get that z acts trivially on $B{\setminus } W$. Since $[z,N]\ne 1$, we get that there is $y\in N$, such that $y^z=y^{-1}$. Since N is regular on $\mathcal {U}$, we may assume that $C^y=C_1$, and then, $z^y$ acts trivially on $B{\setminus } W_1$. This gives us that $\left\langle z,z^y\right\rangle$ acts trivially on $B{\setminus } (W\cup W_1)$. We note that $y^z=y^{-1}$ which implies that $z^y=y^{-2}z$. This implies that y acts trivially on $B{\setminus } (W\cup W_1)$. Therefore, $|B_y|\ge |B{\setminus } (W\cup W_1)|$. We have $|B{\setminus } (W\cup W_1)|=2p+\lambda -sp$. Now, by this and Lemma 2.6, we get that $sp\ge ap$, which is a contradiction and the Lemma holds. $\square$

Lemma 2.8

If $\left\langle z,x\right\rangle \cong D_{2p}$, then $s=q-1$.

Proof

Assume $s\ne q-1$. Since $sp=q^n-1$, we get that $s=(q-1)f$ where $f\ne 1$ and f divides $(q^n-1)/(q-1)$. Set $N_1=C_N(z)$ and let $|N_1|=q^{n_1}$. By Lemma 2.7, we get that $N_1\ne 1$. Since all p involutions in $\left\langle z,x\right\rangle \cong D_{2p}$ are conjugate and by Lemma 2.2(i) $C_N(X)=1$, we get that $N=N_1\oplus (N_1)^x\oplus \cdots \oplus (N_1)^{x^{p-1}}$. This implies that $pn_1=n$.

Let O be an orbit of X on B. Then $|B_z\cap O|=1$. This tells us that $|B_z|=a=q^m$ and $|B_z\cap (B{\setminus } W)|=1$. Now, by Lemma 2.5, we get that $B_z$ is an N orbit. Furthermore, since z inverts each element in [N, z], we get that [z, N] is trivial on $B_z$. This shows that $n_1=m$, and then, $pm=n$.

Let R be an orbit of X on P(N). Then, $|R\cap P(N_1)|=1$. By this and Lemma 2.2(i), we get that $(q^m-1)/(q-1)=|P(N_1)|\le s/(q-1)$. We note that by Lemma 2.2(i), we conclude that X has $s/(q-1)$ orbits on P(N) and by our assumption $s\le a-1=q^m-1$. Hence, $s=a-1$, and then, $m\ge 2$. Now, we have that $p=(q^{pm}-1)/(q^m-1)=q^{(p-1)m}+\cdots +q+1$ which implies that $q^{2(p-1)}\le q^{(p-1)m}< p$. However, since p is odd, we have $p<2(p-1)$, a contradiction. Now, the Lemma is proved. $\square$

Let y be an element of order q in N and $R_1={C^{\left\langle y\right\rangle }}$ be the orbit of $\left\langle y\right\rangle$ containing C. If $s=q-1$, then N is transitive on $\mathcal {U}$ and X is transitive on P(N). Therefore, $\bigcup _{l\in X} R_1^l=\mathcal {U}$ and $R_1^{x^i}\cap R_1^{x^j}=\emptyset$.

Lemma 2.9

$\left\langle z,x\right\rangle \cong S_p$.

Proof

Assume that $\left\langle z,x\right\rangle \cong D_{2p}$. Then, by Lemma 2.8, we have $s=q-1$ and $x^z=x^{-1}$. By Lemma 2.7, we get that there is $y\in N$ of order q, such that $[z,y]=1$. Let $R_1={C^{\left\langle y\right\rangle }}$ be the orbit of $\left\langle y\right\rangle$ contacting C. Then, z acts trivially on $R_1$. By coprime action, there is subgroup $Y_1\le N$ of order q, such that $Y_1^z=Y_1$ and $y\notin Y_1$. We may assume that $R_2=R_1^x$ is the orbit of $Y_1$ on $\mathcal {U}$ containing C. We note that $(R_1^{x^{-1}})\cap R_2=\emptyset$. However, z acts on $R_2$, and since $x^z=x^{-1}$, we should have $R_2^z=R_1^{x^{-1}}=R_2$, a contradiction. This and Lemma 2.7 show that $\left\langle z,x\right\rangle \cong S_p$ and the Lemma holds. $\square$

Lemma 2.10

We have $|B_y|=a$, $|B_y\cap (B{\setminus } W)|=1$ and $m=n-1$.

Proof

Set $D=\left\langle z,x\right\rangle$. By Lemma 2.9, we have $D\cong S_p$. Therefore, the stabilizer of any element of P(N) in D is isomorphic to $S_{p-1}$. We note that $D\le H_C$. Now, let $Y=\left\langle y\right\rangle \in P(N)$ and $D_1$ be its stabilizer in D. Then, $D_1$ acts on $B_y\cap (B{\setminus } W)$. We note that $D_1$ acts also on W and $B{\setminus } W$. Since $D_1\le D\cong S_p$, we get that $|B_y\cap (B{\setminus } W)|=1$ or $p-1$. This gives us that $|B_y|=(p-1)a$ or a. By this and Lemma 2.6, we get that $|B_y|=a$, and then, $n-1=m$. Therefore, the lemma holds. $\square$

Let $R_1=\{C,C_1,\ldots ,C_{q-1}\}$ and $W_i=\Gamma (C_i)\cap B$, $i=1,2,\ldots ,q-1$. By Lemma 2.10, we may assume that $\{\alpha \}=(B{\setminus } W)\cap B_y$. Then, we have the following lemma.

Lemma 2.11

$\mathcal {U}{\setminus } R_1=\Gamma _{\mathcal {B}}(\alpha )$ .

Proof

Since $\alpha \in B_y$ and $\alpha \notin W$, we get that $R_1\cap \Gamma _{\mathcal {B}}(\alpha )=\emptyset$. We note that by Lemma 2.10, we have $m=n-1$ which implies that $a=q^{n-1}$. Now, for $i=1,\ldots ,q-1$, we have $|W{\setminus }(W\cap W_i)|=k-\lambda = ap-p-(p(a-2)+(q^n-a)/as)=p-(q-1)/s=p-1$. By this and since $\alpha \notin W_i$, we get that $B{\setminus }( W\cup \{\alpha \})\subset W_i$. This gives us that $\mathcal {U}{\setminus } R_1=\Gamma _{\mathcal {B}}(\alpha )$ and the lemma holds. $\square$

Now, we can proof Theorem 1.2.

Proof

By Lemma 2.11, we have $r=q(p-1)=q^n-q^{n-m}$ which is impossible. Therefore, $q=2$ and Theorem 1.2 follows from the main result in [7]. $\square$