18. Orthogonal Groups We Will Not Give a Full Treatment of the Orthogonal Groups, As We Do Not Have Time, but We’Ll Try and Give a Broad Overview

FINITE PERMUTATION GROUPS AND FINITE CLASSICAL GROUPS 83

18. Orthogonal groups We will not give a full treatment of the orthogonal groups, as we do not have time, but we’ll try and give a broad overview. Throughout this sectionV is ann-dimensional vector space over the ﬁeldk=F q and Q:V F q is a non-degenerate quadratic form. → Recall ﬁrst that we have the following possibilities for (V,Q). (Note that, in all cases, fori=1, . . . , r,(v i, wi) are mutually orthogonal hyperbolic pairs, withQ(v i) =Q(w i) = 0.) (O+ ) with basis v , w , . . . , v , w . 2r { 1 1 r r} (O2r+1) with basis v 1, w1, . . . , vr, wr, u where u is anisotropic and orthogonal to v 1, w2, . . . , vr, wr . We can prescribe,{ moreover, thatQ(u)} = 1 or,� ifq� is odd,Q(u) is 1 or a non-square.� � (O2r− +2) with basis v 1, w1, . . . , vr, wr, u, u� where u, u � is anisotropic and orthogonal to v 1, w2, . . . , vr, wr . { } � � 2 � � We can prescribe, moreover, thatQ(u) = 1,Q(u �) =a andx +x+a is irreducible inF q[x].

Note that, although there are two non-isomorphic spaces O2r+1, the corresponding polar spaces, and hence the corresponding isometry (resp. similarly/ semisimilarity) groups are all isomorphic. This remark allows us to make the following definitions. Note that, throughout, + or , ifn is even; ε is blank−, ifn is odd. � ε ΓO n(q) is the semisimilarity group ofQ; • ε GO n(q) is the similarity group ofQ; • ε O n(q) is the isometry group ofQ; • ε ε SOn(q) is the special isometry group ofQ, i.e. it equals O n(q) SL n(q). • ε � ε ∩ Ω (q)=(O (q))�, a subgroup of SO (q) of index 1 or 2. • n n n For all of the listed groupsX, there is a projective version PX= X/(X K) whereK is the set of scalar matrices.46 ∩ ε The groups we’re primarily interested in are PΩn(q) as these are simple unlessn andq are in a certain small range. Our treatment begins similarly to the other classical groups: Lemma 18.1. ε (1) Letx n be the number of non-trivial singular vectors. Then ε m m 1 x 2m = (q ε1)(q − +ε1); • 2m− x 2m+1 =q 1. • − ε n 2 (2) The number of hyperbolic pairs isx q − . n · + Proof. Clearlyx 1 =x 2− = 0. On the other hand, a space of typeO 2 is a hyperbolic line, thus if (v, w) is a + hyperbolic pair, thenQ(av+bw)=ab and so the singular vectors lie in v w andx 2 = 2(q 1). Now for anyn 3, an orthogonal space admits a basis which is an orthogonal� �∪� � direct sum of a set− of mutually orthgonal hyperbolic≥ lines with one of the spaces already covered. Consider the different cases in turn. + (O2m) withQ( aivi + biwi) = aibi. ThenQ(v) = 0 iff either + + –a 1 = 0,b 1 is anything and the ‘tail’ of the vector in O2r 2 is singular. This givesq(x 2m 2 + 1) 1 possibilities.� � (The ‘+1’� and the ‘ 1’ are there to account− for zero vectors.) − − 1 m − 2m 2 –a 1 = 0 andb 1 =a 1− i=2 biwi. This gives (q 1)q − possibilities. � + 2m 2 + − We conclude thatx 2m = (q 1)q − +q(x 2m 2 + 1) 1 and the result follows by induction. �− − − (O2r+1) Exactly the same reasoning as before implies that 2m 2 + x2−m = (q 1)q − +q(x 2m 21) 1 − − − and the result follows by induction. (O2r− +2) This time we obtain that 2m 2 + x2m+1 = (q 1)q − +q(x 2m 11) 1 − − − and the result follows.

46Some authors label orthogonal groups slightly diﬀerently. I’ve chosen terminology that is consistent with [KL90] but, for ε instance, some people write GOn(q) for the isometry group ofQ, rather than the similarity group. 84 NICK GILL

To calculate the number of hyperbolic pairs (v, w), simply observe that the number of choices for the ﬁrst entryv isx . To ﬁndw we choose any vector in the complement of ker(β ) whereβ :V k,w β(v,w). n v v → �→ Sinceβ v is a non-zero linear functional, its kernel has dimensionn 1 and the number of vectors in the n n 1 − complement of the kernel is, therefore,q q − . Now we must restrict to those elements for whichβ(v,w)=1 1 n n 1 − and we obtain q 1 (q q − ) as required. � − − ε We will use Lemma 18.1 to calculate the size ofO n(q) using induction onn. The base cases are treated in the following exercise. ε (E18.1)O 1(q)= I andO 2(q) = D2(q ε1). {± } ∼ − Lemma 18.2. m 1 ε m(m 1) m − 2i O2m(q) =2q − (q ε1) (q 1). • | | − i=1 − m m2 �2i O2m+1(q) =(2, q 1)q (q 1). • | | − i=1 − Proof. As in previous sections we use� the fact (which follows from Witt’s Lemma) that Isom(Q) acts regularly on the set of orthogonal bases. To count orthogonal bases we choose (x, y) to be a hyperbolic pair and invoke Lemma 18.1, before using induction to count the number of orthogonal bases in x, y ⊥. The result follows, � � using (E18.1) for the base case. � ε ε (E18.2*) Prove that ifg O n(q), then det(g)= 1. Prove that I O n(q). Conclude that ∈ ± − ∈ 1 SOε (q) = PO ε (q) = Oε (q) . | n | | n | (2,q 1) | n | −

Lemma 18.3. (1) Ifq is even, thenΩ 2m+1(q) ∼= Sp2m(q). (2) Ifq is odd, then PΩ 2m+1(q) and PSp2m+1(q) have the same order. If, in additionm>2, then PΩ (q) = PSp (q). 2m+1 � ∼ 2m+1 Proof. LetQ be a non-degenerate quadratic form of type O 2m+1 and assume thatq is even. The polarization ofQ,β Q is alternating and, since the dimension is odd, it must be degenerate. However (E13.14) implies that Rad(βQ) has dimension 1. Let Rad(βQ) = z and choosez so thatQ(z) = 1. Now the spaceV/ z is non-degenerate and symplectic of order 2m. � � � � The action of SO (q) onV induces an action by isometry onV/ z and we obtain a homomorphism 2m+1 � � SO2m+1(q) Sp 2m(q). One can check that the kernel of this homomorphism is trivial, hence we obtain an embedding.→ However checking orders we see that the two groups have the same cardinality and the result follows. Result (2) follows from the following exercise. (E18.3)Letq be odd. Show that PSp (q) has m +1 conjugacy classes of involutions, whilePΩ (q) 2m � 2 � 2m+1 hasm conjugacy classes of involutions. �

Some remarks about Ω2m+1(q): Suppose thatq is even. In light of Lemma 18.3 most authors tend not to study Ω (q), opting instead • 2m+1 to study the isomorphic group Sp2m(q) (see, for instance, [KL90]). The proof of Lemma 18.3 implies that, ifq is even, then Ω 2m+1(q) = SO2m+1(q), and that this group is • simple, except when (m, q)=(1,2) or (2,2). � � In fact this is the only situation whenn<1andΩ n(q) has index 1 in SOn(q). In all other cases, providedn>1, SO � (q):Ω� (q) = 2. | n n | Suppose thatq is odd. (E18.2) implies that SO 2m+1(q) does not contain any scalar matrices and, in • particular, we have that

PSO2m+1(q) = SO2m+1(q) and PΩ2m+1(q)=Ω2m+1(q). FINITE PERMUTATION GROUPS AND FINITE CLASSICAL GROUPS 85

ε The following couple of results show in addition that, whenn 6, PΩ n(q) does not yield a new simple group. Indeed (E18.1) implies that already forn 2. ≤ ≤ Lemma 18.4. Ifq is odd, thenΩ 3(q) ∼= PSL2(q). 47 Proof. Let Ω be the set of homogeneous polynomials overF q in variablesx andy of degree 2, i.e. Ω := rx 2 + sxy+ty 2 r, s, t Fq . { | ∈ } ThenG = GL 2(q) acts on Ω by substitution, i.e. given a b g := G c d ∈ � � we define xg = ax+ by andy g =cx+dy and observe that we have a well-defined action. Indeed (by observing that Ω is a 3-dimensional vector space overF q) we can check that the groupG acts on 2 2 Ω as an object from VectFq : we represent an elementf=rx + sxy+ ty by the vector r s t and observe thatf g = r s t ρ(g) where � � � � a2 2ab b2 a b ρ:GL 2(q) GL 3(q), ac ad+ bc bd . → c d �→  2 2 � � c 2cd d   Clearlyρ is the associated homomorphismG Aut(Ω), where Aut(Ω) = GL (q) is the automorphism group → 3 of Ω as an object from VectFq . 3 Observe that ker(ρ) = I and define a quadratic form onV=(F q) via {± } Q r s t = 4rt s 2. − One can check thatQ is non-degenerate and that,� forg� G, ∈ Q(f ρ(g)) =Q(f)(det(g)) 2.

This implies that SL2(q) acts on (V,Q) as an object from IVectFq and, by restricting the domain ofρ, we obtain ρ:SL (q) Aut(V,Q) = Isom(Q)=O (q). 2 → 3 Now the ﬁrst isomorphism theorem of groups, implies that PSL (q) = SL (q)/ I = Im(ρ) O (q) 2 ∼ 2 �− � ∼ ≤ 3 By checking orders we obtain that Im(ρ) is an index 4 subgroup of O3(q). Ifq> 3, then the simplicity of PSL2(q) can be used to prove that Im(ρ) is a normal subgroup of O3(q); indeed it must be the derived subgroup of O3(q) (since it is perfect and of index 4), and the result follows. Forq = 3 we omit the proof.� The proof of the following lemma is omitted. It is proved in a similar fashion to the last lemma. Lemma 18.5. + (1) PΩ4 (q) = PSL2(q) PSL 2(q). ∼ 2× (2) PΩ4−(q) ∼= PSL2(q ). (3) PΩ5(q) = PSp4(q). + ∼ (4) PΩ6 (q) ∼= PSL4(q). (5) PΩ6−(q) ∼= PSU4(q).

47 2 2 An equivalent formulation is to take Ω to be equal to Sym (V ), the symmetric square ofV=F q. Clearly GL2(q) acts onV naturally via the homomorphismρ deﬁned below. 86 NICK GILL

ε 18.1. Simplicity. We conclude with a statement concerning the simplicity of PΩn(q). The last two lemmas imply the result forn = 5 and 6. Indeed they also imply that PΩ 3(q) and PΩ4−(q) are simple, but we do not include this in the statement. Theorem 18.6. Ifn 5, thenPΩ ε (q) is simple. ≥ n The proof of this theorem is a little different to the previous cases we have studied, and we will not write it down. The following exercise highlights one major difference. ε (E18.4*)SOn(q) contains a transvection if and only ifq is even. ε ε It is worth mentioning a second major hurdle. We have defined Ωn(q) to be the derived subgroup ofO n(q), ε but in practice this definition is not adequate. We will finish by sketching a more explicit definition of Ωn(q). Letv V be a non-singular vector and define the reflection inv as the map ∈ β (v,x) r :V V,x x Q v. v → �→ − Q(v)

(Observe thatr v satisfies the first condition for a map to be a transvection, sincer v I has rank 1, but it does not satisfy the second, since (r I) 2 = 0.) One can check thatr Isom(Q), thatr−2 = 1, and that v − � v ∈ v 1, ifq is odd; det(r ) = v 1−, ifq is even. � Now the following result is [KL90, Prop 2.5.6]. Lemma 18.7. Isom(Q)= r Q(v)= 0 , provided Isom(Q)= O +(2). � v | � � � 4 Now our definition is as follows: + Suppose thatq is even and that Isom(Q)= O 4 (2). We can assume thatn is even by Lemma 18.3 and • ε ε � ε thus, by (E18.2), On(q)=SOn(q) and that, by Lemma 18.7, every element of SOn(q) can be written as a product of reflections. Now the subgroup ofS consisting of products of an even number of reflections ε ε has index 2 in SOn(q) and this is the group Ωn(q). It is not a priori clear that this action yields an index 2 subgroup - the next exercise proves that it does whenε=+. (E18.5*)Prove that this definition yields an index2 subgroup whenε=+ by showing that, in the natural action ofG on , the set of maximal totally singular subspaces, any reflection acts as an odd U r permutation on r. U 2 48 Suppose thatq is odd and thatn 2. Consider the groupF q∗/(Fq∗) which has order 2. Lemma 18.7 • ≥ε implies that every element of SO (q) can be written as an even number of reflectionsg=r r , for n v1 ··· vk some non-singular vectorsv i. Define the spinor norm, k ε 2 2 θ:SO n(q) F q∗/(Fq∗) , g βQ(vi, vi) (mod (Fq∗) ). → �→ i=1 � It turns out thatθ is a well-defined homomorphism, and that it is surjective. In particular ker(θ) is an ε ε index 2 subgroup of SOn(q), and this is the subgroup Ωn(q). ε (E18.6)Calculate the order of Ω n(q) when(n,q,ε)= (4,2,+). +| | � We do not give a definition of Ω4 (2). Those interested should consult [KL90, p. 30].

48 2 We write (Fq∗) for the set of non-zero squares inF q∗. It is an index 2 subgroup ofF q∗.