Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 19I / b / i / Solution

PutThen $A^2={\epsilon }^{2}I$, so $A^{2n}=I$ because ${\epsilon }^{2n}=1$, and $B^3=I$. Moreover, $A$ exchanges the two eigenspaces of $B$, givingThe matrices therefore satisfy every defining relation of $G_{6n}$, and the universal property of a group presentation gives the required complex representation.