Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 10

Write the symmetry group of a square as $D_4=⟨r,s\mid r^4=s^2=e,srs=r^{-1}⟩$, with $r$ a counterclockwise quarter-turn and $s$ reflection across the horizontal axis. The conjugacy classes of the symmetry group of a square areHere $r^2$ is the half-turn and $r^3$ the clockwise quarter-turn. The two members of the fourth class reflect across the horizontal and vertical axes. Those of the fifth reflect across $y=x$ and $y=-x$. The identity and half-turn are central elements, giving singleton classes. The relations $srs=r^3$, $rs{r}^{-1}=r^{2}s$, and $r(rs)r^{-1}=r^{3}s$ show that each displayed pair consists of conjugate elements. More generally, conjugating $r^{k}s$ by $r$ changes $k$ to $k+2$, and conjugating by $s$ changes $k$ to $-k$; parity is preserved, so the two reflection pairs cannot merge.

In a one-dimensional representation, the relation $srs=r^{-1}$ forces the image of $r$ to equal its inverse. Thus $r$ and $s$ can be assigned independent signs $\alpha ,\beta \in \{1,-1\}$. These four choices give four irreducible characters. The ordinary two-dimensional action on the plane gives one more character: a rotation through $\theta$ has trace $2\cos \theta$, and a reflection has trace zero. Therefore the character table of the symmetry group of a square isFor instance the last row has squared character norm $(4+4)/8=1$, so the planar representation is irreducible; its inner product with each one-dimensional row is zero. All five rows obey character orthogonality. Their squared dimensions sum to $1+1+1+1+4=8$, so they exhaust the irreducible representations of this order-eight group.