Action in Groups (transitively)

Question

Let X be a set of order n.

a) If G acts transitively on X then n divides $| G |$.

b) If G acts 2-transitively on X then n(n-1) divides $| G |$

For a) i first prove that if G acts transitively on X, then $\frac{1}{| G |} \sum_{g∈G}χ(g)= 1$ for this i consider the set $ S = \{(g, x) ∈ G × X : gx = x\}$ We have the elements of S in two different ways. First, grouping items according to its second coordinate, we see that $| S |= \sum_{x∈χ}| G_x |$.Now, as the action is transitive, if we fix $x_0∈X$, is $|G_x | = |G_{x_0} |$ for $x∈X$ and then $|S| = |X||G_{x_0} | = |G|$.

Moreover, by grouping the elements of S according to the first coordinate, we see that $| S |= \sum_{g∈G}χ(g)$ , then we got $\frac{1}{| G |} \sum_{g∈G}χ(g)= 1$ using both cases.

The question is how i can conclude a)?.

For b) i do not have idea. Please help me

Answer 1

If $G$ acts 2-transitively on $X$, then it acts transitively on the set of pairs $\{(x,y)|x,y\in X, x\ne y\}$. Now the order of this set is $n(n-1)$, and you can use a).

Answer 2

Hint: since the action is transitive we have, for any $\;x\in X\;$ :

$$n=|X|=|\mathcal Orb(x)|=[G:Stab(x)]\implies |G|=n\cdot|Stab(x)|$$

Answer 3

theorem: Let $G$ be a finite group and $k$-transitive with order $n$, then $ n(n-1)...(n-k+1)||G|.$

proof by induction: let $k=1$, $G$ is transitive then for $x\in\Omega$, $x^{G}=\Omega$, then we have: $|G|=|x^{G}||G_x|=|\Omega||G_x|=n|G_x|‎=> n||G|$.

Suppose the statement is true for any number less than $k$. If $G$ is $k$-transitive, then action $G_x$ on $\Omega-\{x\}$ is $(k-1)$-trannsitive, then for $G_x$ we have: $(n-1)(n-2)...((n-1)-(k-1)+1)||G_x|$ => $(n-1)(n-2)...(n-k+1)||G_x|(*)$, by using $|G|=n|G_X|$ and $(*)$The proof is finished.