I'm working on an exercise from Serre's Finite Groups: An Introduction where I am required to prove the equivalence of the following statements:
A group $G$ is said to act 2-transitively on a set $X$ if for every $x_1, x_2, y_1, y_2$ with $x_i\neq y_i$ for $i = 1,2,$ there exists $g \in G$ such that $g\cdot x_i= y_i$ for $i = 1,2$
Theorem: (Exercise 11, Ch.1)
The following are equivalent:
- $G$ acts 2-transitively on a set $X.$
- $G$ acts transitively on the set $(X\times X) \setminus \Delta$, where $\Delta$ is the diagonal of $X\times X.$
- The action of $G$ on $X$ is transitive and for any $x\in X$ the action of $Stab(x)$ on $X \setminus \{x\}$ is transitive.
- The number of orbits of $G$ acting on $X\times X$ is 2.
Suppose that $G$ and $X$ are finite. These properties are also equivalent to :
-
$\frac{1}{|G|}\sum_{g\in G}|X^g|^{2} \leq 2$.
-
$\frac{1}{|G|}\sum_{g\in G}|X^g|^{2} = 2$
My attempt: $1\iff 2$ is immediate by definition. Since $G$ also acts transitively on $\Delta, 2\iff 4$ is also immediate. For $1\implies 3,$ there always exists an element taking $(x, y_{1})$ to $(x, y_2).$ For $3\implies 1$, fix two pairs of distinct elements $(x_1,x_2), (y_1,y_2).$ There exists $g\in Stab(x_1)$ mapping $x_2$ to $y_2$ and similarly $g'\in Stab(y_2)$ mapping $x_1$ to $y_1. g'g$ is then the required group element taking $(x_1,x_2)$ to $(y_1,y_2).$
For $4\implies 6,$ Burnside's lemma applied to $X\times X$ reads:
$$2=\frac{1}{|G|}\sum_{g\in G}|(X\times X)^g|= \frac{1}{|G|}\sum_{g\in G}|X^g|^2$$
I've shown the equivalence of 1-4, and that they imply 6, but don't see how to proceed with the remaining implications. I'd also appreciate any help improving or correcting the attempted proofs here.
Best Answer
Your proof of $4 \Rightarrow 6$ works in both directions, so you have shown that they are equivalent.
In fact there is a mistake in your definition of a $2$-transitive action: it should also include the condition that $|X| \ge 2$. (Your proof of $2 \Leftrightarrow 4$ tacitly assumes that $|X| \ge 2$.)
Given that, the number of orbits of $G$ on $X \times X$ must be at least $2$, because $(x,x)$ and $(x,y)$ are in different orbits for $x,y \in X$ with $x \ne y$. So the expression in Condition 5, which is equal to this number of fixed points, must be at least $2$, and we get $5 \Rightarrow 6$.