Let $G$ be a finite group with a permutation action on the set $X$. Describe the corresponding permutation character $\pi_{X}$. Show that the multiplicity in $\pi_{X}$ of the principal character $1_{G}$ equals the number of orbits of $G$ on $X$.

Assume that $G$ is transitive on $X$, with $|X|>1$. Show that $G$ contains an element $g$ which is fixed-point-free on $X$, that is, $g \alpha \neq \alpha$ for all $\alpha$ in $X$.

Assume that $\pi_{X}=1_{G}+m \chi$, with $\chi$ an irreducible character of $G$, for some natural number $m$. Show that $m=1$.

[You may use without proof any facts about algebraic integers, provided you state them correctly.]

Explain how the action of $G$ on $X$ induces an action of $G$ on $X^{2}$. Assume that $G$ has $r$ orbits on $X^{2}$. If now

$$\pi_{X}=1_{G}+m_{2} \chi_{2}+\ldots+m_{k} \chi_{k}$$

with $1_{G}, \chi_{2}, \ldots, \chi_{k}$ distinct irreducible characters of $G$, and $m_{2}, \ldots, m_{k}$ natural numbers, show that $r=1+m_{2}^{2}+\ldots+m_{k}^{2}$. Deduce that, if $r \leqslant 5$, then $k=r$ and $m_{2}=\ldots=m_{k}=1$.