Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 10

The upper-left block of $U$ is the planar rotation matrix $R(\pi /2)$, and $U$ fixes the third coordinate. ThereforeThe diagonal entries of $V$ square to one, so $V^2=I$. Direct conjugation gives $VUV=U^{-1}$, hence $V{U}^{p}V=U^{-p}$ for every integer $p$. ConsequentlyWhen $p\mathrm{mod}4=0$, the last exponent is four, which is equivalent to zero because $U^4=I$.

The relations in (a) reduce every word in the generators to $U^a{V}^b$, with $0\le a<4$ and $b\in \{0,1\}$. These eight matrices are distinct: the $b=0$ elements have third diagonal entry $1$ and the $b=1$ elements have entry $-1$, while the four planar blocks are distinct in each family. Their multiplication iswith exponents reduced modulo four and two. This proves closure. The identity element is $U^0{V}^0$, the inverse of $U^a$ is $U^{-a}$, and each $U^{a}V$ is its own inverse. Associativity is inherited from matrix multiplication, proving the group axioms. This is the eight-element dihedral group $D_8$.

Set $g_j=U^j$ and $g_{j+4}=U^{j}V$ for $j=0,1,2,3$, consistently with the two labels specified in the question. The row-times-column multiplication table isThe rotation powers commute with one another, and $U^a$ commutes with $U^{c}V$ exactly when $a$ is even. Two elements $U^{a}V,U^{c}V$ commute exactly when $a-c$ is even. These statements also check the symmetric portions of the table.

The character of a representation is the trace of its representing matrix. For the rotation powers,Each $U^{j}V$ has a planar reflection block of trace zero and third diagonal entry $-1$, giving character $-1$. In the chosen element order,Equivalently its values on the conjugacy classes $\{I\}$, $\{U^2\}$, $\{U,U^3\}$, $\{V,U^{2}V\}$, $\{UV,U^{3}V\}$ are $3,-1,1,-1,-1$.

The complete subgroup list isThus there are ten subgroups: one trivial, five of order two, three of order four, and the whole group. The five order-two subgroups are mutually isomorphic. The two noncyclic order-four subgroups are mutually isomorphic Klein four-groups; neither is isomorphic to the cyclic group of order four.

For completeness, Lagrange's theorem permits only orders $1,2,4,8$. An order-four subgroup containing $U$ or $U^3$ must be $⟨U⟩$. Otherwise it can contain only $I,U^2$ and elements $U^{j}V$. The product of two distinct such elements is $U^{j-k}$. To avoid generating $U$ or $U^3$, their exponents must differ by two, giving exactly the two listed Klein four-groups. The order-two subgroups come from the five involutions $U^2$ and $U^{j}V$. This proves completeness.

For any cyclic subgroup $H=⟨s⟩$ of order $n\in \{1,2,4\}$, defineBoth preserve products because scalar phases and planar rotations add their angles. Each sends the generator to an element of exact order $n$, so both have trivial kernel and are faithful representations.

For $⟨U⟩\cong C_4$, take $s=U$: the generator images are $i$ and $R(\pi /2)$. For each order-two subgroup take its nonidentity element $s$; the images are $-1$ and $-I_2$. For the trivial subgroup the unique images are $1$ and $I_2$, which are faithful on a one-element group.

The one-dimensional representation of $C_4$ here must use complex scalars. There is no faithful one-dimensional real representation of $C_4$, since a nonzero real scalar of finite order is $1$ or $-1$, and neither has order four. The two-dimensional representations above are real.