(a) State and prove Burnside's lemma. Deduce that if a finite group $G$ acts 2transitively on a set $X$ then the corresponding permutation character has precisely two (distinct) irreducible summands.

(b) Suppose that $\mathbb{F}_{q}$ is a field with $q$ elements. Write down a list of conjugacy class representatives for $G L_{2}\left(\mathbb{F}_{q}\right)$. Consider the natural action of $G L_{2}\left(\mathbb{F}_{q}\right)$ on the set of lines through the origin in $\mathbb{F}_{q}^{2}$. What values does the corresponding permutation character take on each conjugacy class representative in your list? Decompose this permutation character into irreducible characters.