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

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.

In the abelianization, the relation $a^{-1}ba=b^{-1}$ becomes $b=b^{-1}$. Together with $b^3=1$ this forces $b=1$, soConsequently there are $2n$ linear characterswhere $\xi =e^{2\pi i/(2n)}$.

The representations from part (i), denoted ${\rho }_{\epsilon }$, are irreducible: the two eigenspaces of $B$ are its only invariant lines, and $A$ exchanges them. The character of a representation agrees for two parameters exactly when their squares agree, so ${\rho }_{\epsilon }\cong {\rho }_{-\epsilon }$ and there are $n$ pairwise nonisomorphic two-dimensional representations. Their squared dimensions sum toThe sum of squares of irreducible degrees proves that no others exist. Hence the complete list isThis is the classification of the Irreducible representations of G6n.

Every element has a unique form $a^s{b}^t$. Since $a^2$ is central, the conjugacy classes, indexed by $0\le r<n$, areThese have sizes one, two, and three and account for all $6n$ elements.

Take ${\rho }_k={\rho }_{{\xi }^k}$ for $0\le k<n$. On an even power of $a$, its representing matrix is scalar; on an odd power it is off-diagonal. Since $\omega +{\omega }^2=-1$, the full Character table of G6n isfor $0\le j<2n$, $0\le k<n$, and $0\le r<n$.