Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 19G / iv / Solution

Put $\zeta =e^{2\pi i/7}$ and induce the characters ${\chi }_r(a)={\zeta }^r$ of $H=⟨a⟩$. Since $H$ is normal, the induced value outside $H$ is zero, while at $a^m$ it is ${\zeta }^{rm}+{\zeta }^{2rm}+{\zeta }^{4rm}$. The nonzero $r$ split into the two doubling orbits, so the two distinct induced characters have class valueswhere $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6=\overline{A}$. Summing seventh roots gives $A+B=-1$, and multiplication gives $AB=2$, hence $|A{|}^2=|B{|}^2=2$. Their character norms are $(9+3\cdot 2+3\cdot 2)/21=1$, proving irreducibility. Together with the three degree-one characters, the sum of squared degrees is $3+2\cdot 9=21$. Thus these are all the remaining irreducible characters.