<Burnside lemma> says that the number of orbits of a finite group $G$ on a finite set $X$ is
$$
|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|.
$$
Indeed, double-count the set $\{(g,x):gx=x\}$. Counting first by $g$ gives the numerator. Counting first by $x$ gives $\sum_x|G_x|$; each orbit $O$ contributes $|O||G_x|=|G|$ by the <orbit-stabilizer theorem>, proving the formula.

Let $\pi(g)=|X^g|$ be the character of the <permutation representation>. Then
$$
\langle\pi,\pi\rangle
=\frac1{|G|}\sum_g|X^g|^2
$$
is, by Burnside's lemma, the number of orbits on $X\times X$. A <two-transitive group action> has exactly two such orbits: the diagonal and the ordered pairs of distinct points. Hence $\langle\pi,\pi\rangle=2$. Transitivity also gives $\langle\pi,1_G\rangle=1$. If $\pi=1_G+\sum m_i\chi_i$ is its <irreducible character>[irreducible character decomposition], then $2=1+\sum m_i^2$, so exactly one nontrivial irreducible character occurs, with multiplicity one. Thus the permutation character has precisely two distinct irreducible summands.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 19F / a / Solution

{wiki=Multiply_transitive_group_action}

= Two-transitive action
{synonym}

= Two-transitive
{synonym}

= Doubly transitive group action
{synonym}

A group action on a set is two-transitive when it acts transitively on ordered pairs of distinct points. Equivalently, a point stabilizer acts transitively on the remaining points.


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 Incoming links: Two-transitive group action