Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 19F (Representation Theory)

Burnside lemma says that the number of orbits of a finite group $G$ on a finite set $X$ isIndeed, 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. Thenis, 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 $⟨\pi ,\pi ⟩=2$. Transitivity also gives $⟨\pi ,1_G⟩=1$. If $\pi =1_G+\sum m_i{\chi }_i$ is its 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.

Every element of $G{L}_2({\mathbb{F}}_q)$ is conjugate to exactly one representative of one of these types, with the indicated parameter identifications:

The projective line ${\mathbb{P}}^1({\mathbb{F}}_q)$ has $q+1$ points. A line is fixed exactly when it is an eigenline over ${\mathbb{F}}_q$, so the permutation character $\pi$ takes respectively the valuesThe action is two-transitive. Part (a) therefore giveswhere the nontrivial constituent $\mathrm{St}$ is irreducible of degree $q$, the Steinberg representation of GL2 over a finite field. Its values on the four types are