Past exam of the algebra course of the University of Oxford / 2011 / 6

The identity column gives the group order as the sum of squared irreducible character degrees:By column orthogonality, the squared norm of a column is the corresponding centralizer order. In source order these orders and the resulting conjugacy class sizes areFor example the fourth column has squared norm $1+1+(|\alpha {|}^2+|\beta {|}^2+|\gamma {|}^2)+1=9$, while the seventh has squared norm $1+1+(|\xi {|}^2+|\eta {|}^2+|\varphi {|}^2)=7$.

An element's order divides its centralizer order. Cauchy's theorem for finite groups provides an involution, and only $C_2$ has an even centralizer order. Hence that entire class consists of involutions. The last three classes consist of elements of order seven. In $C_3,C_4,C_5,C_6$ the possibilities are three and nine. The three real numbers $\alpha ,\beta ,\gamma$ are roots of $t^3-3t+1$, which has no rational root and is irreducible over $\mathbb{Q}$. A real character value at an element of order three belongs to $\mathbb{Q}(e^{2\pi i/3})\cap \mathbb{R}=\mathbb{Q}$. Therefore the irrational real entries in $C_4,C_5,C_6$ rule out order three there. Cubing any of these order-nine elements produces an element of order three, which must belong to the sole remaining class $C_3$. ThusThe three deductions below give nine Sylow subgroups for $2$ with pairwise trivial intersections. Their conjugation action is transitive. Fix one, $S$. If $T\ne S$, put $H=S\cap N_G(T)$. The product $HT$ is a $2$-subgroup, so maximality of $T$ gives $H\le T$. Trivial intersection then gives $H=1$. Thus the orbit of $T$ under $S$ has size eight, exhausting the other eight points. The point stabilizer $N_G(S)$ is therefore transitive on the remaining points: the action is doubly transitive.

For a doubly transitive action, its permutation character has the form $1+\theta$ with $\theta$ irreducible. Here $\theta (1)=8$, and the unique degree-eight row is ${\chi }_6$. Therefore the numbers of fixed points are the entries ofEvery nonidentity element consequently fixes at most two points. This also shows that the action is faithful.

A Sylow subgroup for $2$ has order eight. Every one of its nonidentity elements lies in the only even-order class, so has order two. In a group of exponent two,It is therefore an elementary abelian group and $\boxed{S\cong C_2^3}$.

If Sylow subgroups $S,T$ share $t\ne 1$, they are abelian by the preceding deduction, so $S,T\le C_G(t)$. But $|C_G(t)|=8$ and both subgroups have order eight. Thus $S=C_G(t)=T$. Consequently distinct Sylow $2$-subgroups intersect only in the identity.

All 63 involutions lie in Sylow subgroups for $2$. Each such subgroup has exactly seven nonidentity elements and distinct ones have no nonidentity overlap. Hence

Over ${\mathbb{F}}_q$, the first nonzero column of an invertible $2\times 2$ matrix has $q^2-1$ choices and the second has $q^2-q$ choices. The surjective determinant homomorphism onto ${\mathbb{F}}_q^{\times }$ has kernel ${\mathrm{SL}}_2(q)$. ThusIt acts on the nine one-dimensional subspaces of ${\mathbb{F}}_8^2$, namely the projective line ${\mathbb{F}}_8\cup \{\infty \}$, by fractional linear transformations. Any element acting trivially fixes $0,\infty ,1$, which forces it to be scalar. A scalar matrix in ${\mathrm{SL}}_2(8)$ has $a^2=1$; in characteristic two this implies $a=1$. Thus the action is faithful.

The action is transitive: any nonzero vector can be completed to a determinant-one basis, so any line can be reached. The stabilizer of $\infty =[1:0]$ isOn finite points this is $x\mapsto a^{2}x+ab$. Squaring permutes the seven nonzero field elements, so these are exactly the affine maps $x\mapsto ux+v$. In particular translations act transitively on all eight finite points. Thus the full action is doubly transitive. The point stabilizer has order $8\cdot 7=56$, and all point stabilizers are conjugate.