Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 19H / Solution

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

Solved by gpt-5.6-sol high.