Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 15F (Representation Theory)

If the two-dimensional group representation were reducible, Maschke's theorem would split it into a direct sum of two one-dimensional group representations. In a basis respecting that decomposition, every representing matrix would be diagonal. All would therefore commute, contradicting the specified pair. Hence the group representation is irreducible. Complete reducibility over $\mathbb{C}$ is essential to this argument.

Call the two proposed matrices $A$ and $B$. Direct multiplication gives $A^{2n}=I$, $B^2={ϵ}^{n}I=A^n$, and $B^{-1}AB=A^{-1}$. Both are invertible since $ϵ\ne 0$. They therefore satisfy every defining relation, and the universal property of a group presentation gives the desired group representation.

Put $\zeta =e^{\pi i/n}$. The group representations from the preceding construction with $ϵ={\zeta }^j$, $1\le j\le n-1$, will be denoted ${\rho }_j$. Their diagonal matrices for $a$ have distinct eigenvalues. The only possible invariant lines would be these two eigenspaces, and the matrix for $b$ exchanges them. Thus each ${\rho }_j$ is irreducible. Equivalently their $a,b$ matrices do not commute, so part (a) applies. The choices $j$ and $2n-j$ give equivalent group representations by exchanging basis vectors. The displayed range contains exactly one of each pair, and distinct members have different traces at $a$ because $\cos (\pi j/n)$ is strictly decreasing there.

For a one-dimensional character of a representation, write $a\mapsto \alpha$, $b\mapsto \beta$. The conjugation relation forces ${\alpha }^2=1$, and ${\beta }^2={\alpha }^n$. Hence there are exactly four choices:For even $n$, $\beta =\pm 1$ for either $\alpha$; for odd $n$, $\alpha =1$ has $\beta =\pm 1$ and $\alpha =-1$ has $\beta =\pm i$. Finally,The sum of squares of irreducible representation dimensions proves that these four one-dimensional and $n-1$ two-dimensional group representations exhaust the irreducibles, including $n=1$ where the latter list is empty.

The conjugacy classes are $\{1\}$, $\{a^n\}$, $\{a^r,a^{-r}\}$ for $1\le r\le n-1$, and two classes containing respectively $b{a}^{2s}$ and $b{a}^{2s+1}$, each of size $n$. Conjugation by $b$ inverts powers of $a$, while conjugation by $a$ changes the exponent of $b{a}^r$ by two; these facts prove the list and sizes. The sizes total $4n$.

The complete character table is specified by the following array, with one column for every $r$ and one row for every $j$ in the indicated ranges:Here $1\le r,j\le n-1$; the four pairs $(\alpha ,\beta )$ are those in part (ii). In particular, this formula displays both parity cases without concealing the different values $\pm i$ when $n$ is odd. The zero entries follow because the two-dimensional matrices representing $b{a}^r$ are off-diagonal.