Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 19I / b

Direct matrix multiplication giveswhile every other pair among $A,B,C,D$ commutes. Also $Z=-I$ is central. Consequently every commutator of words in the generators lies in $⟨Z⟩$, so $G^{\prime }\subseteq ⟨Z⟩$. Since $Z=[A,B]$, the reverse inclusion holds and

Using the commutation relations, every element has a normal formOn squaring, equal generators cancel and every interchange contributes only the central element $Z$, so $g^2\in ⟨Z⟩$. There are at most $2^5=32$ normal forms, and every element has order dividing four. Hence $G$ is a finite p-group with $p=2$, of order at most $32$.

Let $T$ commute with the given representation. Since $A$ and $C$ are diagonal and their four joint pairs of eigenvalues are distinct, commuting with both forces $T$ to be diagonal. Commuting with the permutation matrix $B$, which exchanges coordinates $1\leftrightarrow 2$ and $3\leftrightarrow 4$, equates the first two and last two diagonal entries. Commuting with $D$, which exchanges $1\leftrightarrow 3$ and $2\leftrightarrow 4$, makes all four entries equal. Thus the commutant consists only of scalar matrices. Since the representation is unitary, a proper invariant subspace would have an orthogonal projection in the commutant. No such nonscalar projection exists, so the representation is irreducible.

The subgroups $⟨A,B⟩$ and $⟨C,D⟩$ each have order eight, their elements commute across the two subgroups, and their intersection is exactly $⟨Z⟩$. ThereforeSince $G^{\prime }=⟨Z⟩$, the abelianization $G/G^{\prime }$ has order $16$ and is generated by the four commuting involutions represented by $A,B,C,D$. It is isomorphic to $(C_2{)}^4$, so it has sixteen one-dimensional complex representations, obtained by independently sending $A,B,C,D$ to $\pm 1$ and sending $Z$ to $1$.

These characters contribute $16$ to the sum of squares of irreducible degrees. The given irreducible representation has degree four and contributes another $16$. Part (a) and $|G|=32$ show that no other irreducible representations exist. Thus the complete list is the sixteen linear characters and the given four-dimensional representation.