An exercise from Serre characterizing doubly transitive groups

Question

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:

1. $G$ acts 2-transitively on a set $X.$
2. $G$ acts transitively on the set $(X\times X) \setminus \Delta$, where $\Delta$ is the diagonal of $X\times X.$
3. 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.
4. 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 :

5. $\frac{1}{|G|}\sum_{g\in G}|X^g|^{2} \leq 2$.

6. $\frac{1}{|G|}\sum_{g\in G}|X^g|^{2} = 2$

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.

Answer 1

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 $5 \Rightarrow 6$.