Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 19G (Representation Theory)

Every group element has a unique form $a^r{b}^s$, with $r\mathrm{mod}7$ and $s\mathrm{mod}3$. Conjugation by $b$ doubles the exponent of $a$, giving the two orbits $\{1,2,4\}$ and $\{3,6,5\}$. Conjugation by powers of $a$ runs through all seven elements of each nonidentity coset of $⟨a⟩$. The classes are thereforeTheir sizes $1,3,3,7,7$ sum to twenty-one.

In a one-dimensional representation, the images commute, so $ba=a^{2}b$ forces the image of $a$ to be one. With $\omega =e^{2\pi i/3}$ the three possibilities areThey are inequivalent because they have distinct values at $b$.

For a character $\chi$ of a representation of $H\le K$, the induced character isThe sum is zero when no conjugate of $g$ meets $H$.

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.