Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 19H / Solution

Write a matrix as $(r,b)$, where $r=a^2$ is a nonzero quadratic residue modulo $11$. The square subgroup isMatrix multiplication becomesThus $H$ is the affine semidirect product of cyclic groups of orders eleven and fiveThere are five choices for $r$ and eleven for $b$, so $|H|=55$. It is nonabelian, since

LetConjugation by the complement givesHence the ten nonidentity translations split into the two orbitseach of size five. For $r\ne 1$,Since $1-r\ne 0$, varying $t$ gives every $b\in {\mathbb{F}}_{11}$. Conjugation cannot change $r$ in the abelian quotient $H/N$. Therefore, withthe seven conjugacy classes areof sizes $1,5,5,11,11,11,11$, agreeing with the conjugacy classes in the affine semidirect product of orders eleven and five.

The commutators with $(4,0)$ generate every translation because multiplication by $4-1=3$ is invertible in ${\mathbb{F}}_{11}$. Hence the commutator subgroup is $N$, and the abelianization isPut $\zeta =e^{2\pi i/5}$. The one-dimensional characters factor through the abelianization, giving five characters

For the remaining characters, put $\omega =e^{2\pi i/11}$ and define a character of $N$ byThe complement has two free orbits on the nontrivial ${\theta }_m$, indexed by $m\in Q$ and $m\in 2Q$. By induction from an abelian normal subgroup with a free character orbit, the charactersare irreducible of degree five and vanish outside $N$.

SetThe quadratic periods modulo eleven give the nonsquare sum $\overline{\eta }=(-1-i\sqrt{11})/2$. Multiplication by a square preserves $Q$ and multiplication by a nonsquare exchanges the two square classes, so the complete character table isFinally,and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the irreducible characters of the affine semidirect product of orders eleven and five.

Solved by gpt-5.6-sol high.