论文标题
在Burness-Giudici猜想上
On the Burness-Giudici Conjecture
论文作者
论文摘要
令$ g $为$ω$的排列组。如果$ g $ $ g $的$ g $ $ g $的基础是$ω$的子集。由$ b（g）$，我们表示$ g $的最小基数的大小。 $ b（g）= 2 $的每个置换组都包含一些常规的次级孔。 Burness-Giudici在[4]中推测，每个原始排列组$ g $都带有$ b（g）= 2 $的属性，如果$α^g \ not \ inγ$，则$γ\ capγ\ capγ^g \ neq \ neq \ embtyset $，其中$γ$是所有$ g $ g $ a $ g $ a $ a $ g $ a $ g $ a $ a $ a的联合。对于许多零星的简单组和[4]中的一些替代组，已经显示了猜想的肯定答案，但对于简单的lie-type组仍是开放的。我们应该使用的简单的Lie-type群体的第一个候选人可能是$ PSL（2，Q）$，其中$ Q \ geq 5 $。在本手稿中，我们显示了所有具有SOCLE $ PSL（2，Q）$的原始群体的猜想的正确性，请参见定理$ 1.3 $。
Let $G$ be a permutation group on a set $Ω$. A subset of $Ω$ is a base for $G$ if its pointwise stabilizer in $G$ is trivial. By $b(G)$ we denote the size of the smallest base of $G$. Every permutation group with $b(G)=2$ contains some regular suborbits. It is conjectured by Burness-Giudici in [4] that every primitive permutation group $G$ with $b(G)=2$ has the property that if $α^g\not\in Γ$ then $Γ\cap Γ^g\neq \emptyset$, where $Γ$ is the union of all regular suborbits of $G$ relative to $α$. An affirmative answer of the conjecture has been shown for many sporadic simple groups and some alternative groups in [4], but it is still open for simple groups of Lie-type. The first candidate of infinite family of simple groups of Lie-type we should work on might be $PSL(2,q)$, where $q\geq 5$. In this manuscript, we show the correctness of the conjecture for all the primitive groups with socle $PSL(2,q)$, see Theorem $1.3$.
