WIS .1006 LIBRARY l Michigan State University This is to certify that the dissertation entitled Subloops of the unit octonions presented by Stephen M. Gagola III has been accepted towards fulfillment of the requirements for the Ph. D. degree in Mathematics GAE/11.32 .UrmiLQ, ’ Major Professor’s Signature "5W. 2: D 2—005” Date MSU is an Affirmative Action/Equal Opportunity Institution PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. MAY BE RECALLED with earlier due date if requested. DATE DUE DATE DUE DATE DUE 2105 chlmtoDueJndd-pjs SUBLOOPS OF THE UNIT OCTONIONS By Stephen M. Gagola III A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 2005 Abstract Subloops of the unit octonions By Stephen M. Gagola III The class of Moufang loops is defined by the identities :r:(y(:z:z)) = ((xy)a:)z, z(x(ya:)) = ((z$)y):r:, and (xy)(z:r:) = (:1:(yz))a:; Non-associative finite simple Moufang loops form the central topic of this work. The emphasis will be on their connections with composition algebras. A composition algebra over some field, not necessarily finite, is either F1 with charF = 2 or an algebra with a non-degenerate quadratic form, 9, that admits composition. If its dimension over the field is eight then we have what we call an octonion algebra. Here we categorize all the subloops of the unit octonions and in particular describe all the finite maximal subloops by using the reflection groups of these Moufang loops. Furthermore, Lagrange’s Theorem for Moufang loops then follows as a corollary. Contents 1 Introduction and Main Results ...................................................... l 2 Notation and Terminology .............................................................. 8 3 General Composition Algebras ...................................................... 10 4 Moufang Loops .............................................................................. 26 5 Structure of Octonion Algebras ...................................................... 32 6 Proof of Theorem 1.3 ...................................................................... 40 7 Proof of Theorem 1.4 ...................................................................... 50 8 Proof of Theorem 1.5 ...................................................................... 56 9 Lagrange’s Theorem for Moufang Loops ........................................ 59 iii Tables 1. A multiplication table for a loop of order five ...................................................... 3 2. The Steiner loop of order ten ............................................................................... 29 iv 1 Introduction and Main Results In this section, we explain the necessary material and introduce the needed notation. After presenting the basic notions, let us briefly summarize the results of this work. Let F be an arbitrary field and A a vector space over F. A is said to be an algebra over F if there is a bilinear multiplication on A satisfying : 1. :r:(y+z) =xy+xz 2. (:r+y)z=:rz+yz 3- (afly = a($y) = $(ay) for all 1:, y, z E A and a E F. Notice that an algebra is not necessarily associative, meaning $(yz) = (a:y)z for all 33,3], 2: E A. Also, an algebra is not always assumed to contain an identity element. A subalgebra of A is a subspace, B, of A that forms an algebra with respect to this multiplication. We will denote this by B g A. A bilinear form on A is a map (-|-) : A x A ——+ F such that 1- (I + yIZ) = (x|z) + (W). 2. (zla: + y) = (zlx) + (zly), and 3- (HI!!!) = a(fly) = (Ilay). for any $,y,z E A and a E F. If B S A then Bi 2 {I E A|(.r|y) = O for all y E B}. A map (1 : A —i F is a quadratic form 011 A if 1. q(a:r:) = a2q(:r:) for all :1: E A and a E F 2. and the map from A x A to F given by (zly) = q(:r: + y) — q(a:) — q(y) is a (symmetric) bilinear form. We say that the quadratic form q is non-degenerate if {:r: E A|(:r|y) = 0 for all y E A} = 0. An algebra A over a field F is a composition algebra if it has a multiplicative norm, that is, a quadratic form, q : A ——> F, with q(ab) = q(a)q(b) for all a,b E A, is nonzero for F 1, and in general is non-degenerate. A composition algebra does not have to contain an identity element but is closely related to one that does contain an identity. Such a connection is an isotopy and every composition algebra is isotopic to an algebra with an identity, see Jacobson [13, p. 418]. The following theorem, dealing with the dimension of composition algebras, was proven by Hurwitz [12]. A proof is given in section three below. Theorem 1.1. Suppose A is a composition algebra over a field F. Then it is of dimension one, two, four, or eight. A composition algebra of dimension four is usually called a quatemion algebra, and such an algebra of dimension eight is usually called an octonian algebra. A quaternion algebra is associative. However, unlike the quaternions, the octonions are nonassociative but do satisfy the Moufang identity, :(.r(y:r)) = ((zx)y);r (see proposition 3.9). One can Show that such an identity, :(.r(y;t')) = ((zat)y):r, is actually equivalent. to: 1. Jib/(Ml) = (($ylxlz and 2. (xy)(z:r) = (z(yz))x (see Lemma 3.1 of [4]). We call a composition algebra a division algebra if it contains no zero-divisors. Otherwise it is said to be split. A quasigroup is a non-empty set S with a closed binary operation, (1:, y) r—> :c - y, such that 1. a - a: = b determines a unique element a: E S given a, b E S and 2. b = y - a determines a unique element 3/ E S given a, b E S. A loop is a quasigroup, L, with an identity element and a subloop of L is a subset of L which, under the binary operation, is a loop. A finite loop L is said to have the Lagrange property if for any subloop K of L the order of K, IKI, divides the order of L, [L]. Lagrange’s Theorem says that finite groups have the Lagrange property, but in general, a finite loop does not satisfy the Lagrange property. For instance, one can easily construct a loop of five elements all of which are of order one or two as one can see in Table 1. 1 a b c d 1 1 a b c d a a 1 d b c b b c 1 d a c c d a 1 b d d b c a 1 Table l: A multiplication table for a loop of order five The reason for this is because without associativity, the coset decomposition breaks down. Definition 1.2. A Moufang loop is a loop, L, that satisfies the Moufang identities. The main results of this thesis are the following four theorems. The first is a version of the Aschbacher-O’Nan—Scott Theorem [2] and is valid for subloops of octonion algebras. Theorem 1.3. Let L be a subloop of an octonion algebra, C, over the field F. Consider the reflection subgroup R = R(L) = (pr ] x E L) where V = F 8. Then there are three possibilities: I. V > [V, R] with one of the following (a) L S S"L for some hexagon line S; { b) L S Q for some nonsplit quatemion subalgebra Q; (c) L S [V, R], where L is totally isotropic and F is a nonperfect field of characteristic two; 2. V = [V, R] is reducible and there is a quaternion subalgebra Q with L g QUQL; 3. R is irreducible on V. Here, [V, R] = span ({—v + or | r E R, v E V}). By a reflection, p1,, of V we are referring to the map PI : V ——+ V sending v to v - (figs. We will let 0(V, q) be the orthogonal group with respect to (1 consisting of all the linear transformations that preserve q. Note that we always have pa, 6 O(V, q) (see [25]). By irreducible, we mean that there does not exist a proper subspace of V that is left invariant under R. Also, for a hexagon line we mean a totally singular 2-space that is contained in 1i such that the multiplication is zero. In the split case, the subalgebras Si and Q are uniquely determined up to the action of Aut(C). See Proposition 5.4 and Theorem 5.6 below. The loops that can occur under Theorem 1.3.1 and 1.3.2 are relatively elementary in structure. The loop (S J*)" is described in detail in Section 5 and is a loop extension of the abelian group F 2 by the group GL2(F). The loops under Theorem 1.3.2 contain subloops of index at most 2 that are contained in the quaternion subalgebra Q. Since quaternion algebras are associative, their subloops are groups. We will only describe the finite subloops that occur in Theorem 1.3.3. Theorem 1.4. Let L be a finite subloop of an octonion algebra C over the field F. Let R 2 (,0... I :r: E L) g O(C, q). IfR is irreducible on C then R is one of the following and unique up to conjugacy in O(C, q): 1. R = +Q,'3‘"(FO) for some finite subfield F0 of F of odd characteristic. We have C split and SLL(FO) S L g F‘ - CLL(FO). 2. R = 0;(F0) for some finite subfield F0 of F of characteristic 2. We have C split and SLL(FO) < L < F‘ - CLL(F0). 3. R = I'V(E8) ”-3 20;”(2) and 2CLL(2) g L g F“ - 2CLL(2) when charF is not 2. Notice that the previous two results give a precise description of all finite subloops of octonian algebras. We have already discussed the structure of such subloops under 1.3.1 and 1.3.2 in general. Since a quaternion algebra always has a quadratic extension that is split, the finite subgroups that may occur are to be found on Dickson’s list [24, Theorem 6.17] of finite subgroups of the matrix algebra M2(F). These two theorems can be thought of as providing an analogue of Dickson’s theorem for octonian algebras. In particular, we find all the maximal subloops of finite octonian algebras. Theorem 1.5. Let C be a finite octonion algebra over a field F and let L be a maximal subloop of B where SLL(F) S B s GLL(F). Then L is one of the following: 1. L = (Si). 0 B for some hexagon line 5'; 2. L = (Q U Qi)’ D B for some quaternion subalgebra Q; 3. L = F’ - CLL(F0) n B for some maximal subfield F0 of F; 4. L = {x E B|q(x) E C} for some maximal subgroup, G, of {q(x)|x E B}; 5. L = F" - 2CLL(2) F) B where F = Fp for some odd prime p. As a corollary to Theorem 1.5, we get the following theorem which has been a conjecture for over forty years. Theorem 1.6. All finite Moufang loops have the Lagrange property. There have been several recent proofs of the Lagrange property for finite Moufang loops. The author of this thesis and his supervisor give a proof in [8]. There is also an independent and earlier proof by A. Grishkov and A. Zavarnitsine [9] and a third proof by E. Moorhouse [18]. All of these proofs make use of the classification of finite simple groups in two ways. First, they use a result of Liebeck [17] to reduce to the case of finite Paige loops P(q). Secondly, they use Kleidman’s list [16] of the maximal subgroups of the triality group P0; (q) : S3 to treat the finite Paige loops P(q). Zavarnitsine has a second paper [27] in which he classifies the maximal subloops of finite octonian algebras, as in Theorem 1.5. Again, he appeals to Kleidman’s list. The proofs of Theorems 1.5 and 1.6 given here are fundamentally different from these others. In particular, Theorem 1.5 is independent of the classification of finite simple groups and Theorem 1.6 is only dependent on Liebeck’s work. Indeed, other than from using Liebeck’s result once, the only place where nonelementary finite group theory is used in this thesis is in the case of characteristic 2 in Theorem 7.2. Also, as noted there, this part of the argument can probably be simplified as well. In this thesis, as well as these other references, the crucial starting point is Doro’s observation [6] that Moufang loops correspond to certain groups with triality. The treatment of groups with triality used here is that of [8] and [10]. The standard reference for general loop theory is Bruck [4] and for octonian algebras Springer and Veldkamp [22]. For group theory see [24], and for geometry see [25]. 2 Notation and Terminology For a field F consider the algebra, C, consisting of all the matrices 0.1) u b where a, b E F and v, u E F 3. Here, addition is defined by a v c a a+c v+a + u b [3 d u + S b + d and multiplication is defined by a v c or ac+v-B aa+du—ux,8 u b B d bfl+cu+vxa bd+u~a where v - u and v x u are the stande dot product and cross product. One can a v obtain an octonion algebra by letting ab — v - u = det be its norm, q. A u b straightforward calculation shows that q is a quadratic form and is multiplicative. Let C be this algebra over the field F with the quadratic form, q : x H det(x). We will denote the associated bilinear form as (xly) = q(x + y) — q(x) - q(y) where x,y E C. For every element, 1), in C we will define the conjugate of v as a a ’27 = —u + (u|1)1. Notice that an element x = E C is invertible if and ,6 b b —a only if det(x) 79 0 and that its inverse is 3% —t3 a 1 0 Clearly det = 0, so C is a split octonion algebra. Indeed, as we will 0 0 see in Proposition 5.4. any split octonion algebra over F is uniquely determined and 8 is therefore isomorphic to C. Indeed if F is finite, then there are always nonzero sin- gular elements. So any octonion algebra over a finite field, F, is split and isomorphic to C. By Proposition 3.9, C satisfies the Moufang identities. In particular, the set of all the invertible elements of C, CLL(F), is a Moufang loop. Let SLL(F) be the set of all the elements in GLL(F) that are of norm one. The center of S LL(F), the set of all elements that commute and associate with the other elements of SLL(F), is Z(SLL(F)) = {i1}. Similarly, Z(GLL(F)) = F‘ ~ 1. Using the natural definition of homomorphisms for loops (see Bruck [4, Chapter IV]) gives rise to what we mean by “normal subloops” and “simple loops”. The subloop N of L is a normal subloop if there exists a homomorphism (p : L ——i L1 such that N = ker(<,o). If the only normal subloops of L are L and {1} then we say that L is simple. It was proven by Paige [20] that if F is a. finite field then PS LL(F ) = SLL(F)/{i1} is a simple Moufang loop and usually denoted by P(F) or P(q) where q = [F]. When F is finite, since dimp(V) is even, there are (up to equivalence) two non- degenerate quadratic forms on V, distinguished by the sign, 6. For the corresponding orthogonal group we will write O§(F). Furthermore, +(2‘8(F) will be used to denote the subgroup of index 2 in O§(F) that is generated by the reflections, p,, with center, x, of square norm. Also, we will use ‘Q§( F) to denote the subgroup of index 2 in O§(F) which is generated by the reflections, p,,, with center, y, of nonsquare norm. In addition, we will let W(E8) represent the Weyl group of the Lie algebra E8. 3 General Composition Algebras We start off with some observations about the composition algebras. Here we will let A be any composition algebra that contains an identity element, 1. By definition, all subalgebras contain the identity element. Using the definition of 17 one can see, from section 1.3 of F. Veldkamp and T. A. Springer [22], that: 1. i7 = —p1(v) for all u E A, 2. i7 2 —-o for all u E 1i, 3. 5 = v for any element '0 E A, and 4. v + u = 17 + a for any elements v,u E A. Lemma 3.1. For any elements x,v, u e A, the following properties hold: 1- (033W?) = (vlu)q(x), 2. (1|u)(x|v) = (vlux) + (xluv), and 3. (xulv) —_— (ulxv) and (uxlv) = (ulux). Proof. By definition of the bilinear form (Wind?) = (1(UI + ua?) - (1(01) - C1(W) = q((v + 10-75) - (1(1’)(1(I) - (100905) = (If!) + u)q(1f) - q(vlqm - (1(lt)(1(r) = l(1(v + U) - (1(0) - (100] (1(1) = (UI‘U)(I(1’)- 10 Since (xlux) + (xluu) + (vlux) + (oluv) = (x + v|u(x + 12)) = (1|?1)q(:c + v) = (llu)[($|v) + q(x) + q(vll = (IIU)($Iv) +(1lu)9($) +(1IU)q(v) = (IIU)($|U) + (Ilux) + (vluv), one can subtract (xlux) + (vluv) from both sides to obtain (1|u)(x|v) = (vlux) + (xluv). We can now use part two to prove part three. (Uliv) = (UK-33 +($l1)1)v) —_- (ul — xv + (x|1)v) = (UI - xv) + (UI(I|1)'U) = -(UI$v) + ($l1)(UIv) = —(u|xu) + (xulu) + (xulu) = — + (xulv) + (ulxv) = (xulv) 11 Also, (Ulvi) = (UM-iv + ($|1)1)) = (u] — vx + (xl1)u) = (Hi - v$)+(UI(-’tl1)v) = -(ulv$)+(UIv)($|1) = —(u|vx) + (uxlu) + (ux|u) = —(u|ux) + (uxlv) + (ulvx) = (uxlv) Lemma 3.2. For any v,u E A 1. 17(22u) = q(v)u = (uv)i7 2. —v=t7fi Proof. Since (MW) - q(v)UIIL‘) = (170111)le - (1(v)(UI$) = (WWII) - (1(1))(141') = (1(v)(‘ltlr) - C1(v)(UIII‘-) =0 for all x E A. by non-degeneracy. 17(vu) = q(u)u. 12 Also, since ((uvl'v - Q(v)UI$) = ((uvlr‘llx) - Q(v)(UII) = (ulev) - q(v)(UI$) = q(UXUIiL‘) - Q(v)(UI$) =0 for all x E A, by non-degeneracy, q(v)u = (uv)t‘i. Likewise, since (1T?! - WISE) = (film) - (WI-1‘) = (1l(uv)=v) - (film) = (fluv) - (fiilv) = (1‘:|uv) - (:r‘Jqu) = 0 for all x E A, by non-degeneracy, an 2 1727.. Lemma 3.3. For all x,v,u E A 1. v(i‘ix) + u(i7x) = (u]u)x 2. ($1011 + (xu)D = (v|u)x 13 Proof. By Lemma 3.2 (u + u)((u + u)x) = q(u + u)x. Therefore, 0 = (u + v)((u—+v)x) — q(u + 22):.- = (u + v)((a + i7)z) — q(u + 22):: = u(ax) + v(i7x) + v(i7.x) + u('Dx) — q(u + 22):: = q(u)x + q(u)x + v(ux) + u(ox) — q(u + v)x = v(ux) + u(i‘)x) — (vlu)x. Hence, v(i‘ix) + u(i7x) = (vlu)x. Likewise, 0 = (x(u + v))u_+v — q(u + u)x = (x(u + v))(a + a) — q(u + v)x = (xu)u + (mow + (xv)i“r + (xu)i‘) — q(u + 21);: = q(u)x + q(v)x + (xv)t’i + (xu)t—) — q(u + v)x = (xv)a + (xu)t7 — (vlu)x. Thus, (xv)ii + (xu)i‘1 = (vlu)x. Lemma 3.4. For all u, u E A, v(uv) = (uu)v 14 Proof. For any 11, u E A, u(uv) = v(uv) + moo) — q(u)fi = (ole-in — q(v)’& = (HIT/)1) - q(vlfi = (uu)u + (017)17. — q(v)i1 = (vu)v. Cl Remark 3.5. Notice that, by 3.3, xy +yx = (xly)1 for any x,y E A. So we obtain that (2337):: + q(x)y = (xly)x. So by Lemma 3.4, xyx + q(x)y = (xly)x. Therefore, if q(x) 75 0 then pe(y) = y - m3: -‘ imam. q(x) — 0(r) Lemma 3.6. Let a E A with q(a) = O and a # 0, then x E aA if and only ifc’ix = 0. Proof. If x E aA then ax E 6(aA) = q(a)A = {0}. Now suppose 6x = 0. Let y E A such that (aly) = 1. By Lemma 3.3, a(yx) + y(c'rx) = (aly)x. Hence, x = a(yx) E aA. C] Theorem 3.7. Let a,b E A with q(a) = 0 = q(b), a 74 0, and b # 0. Then aA and bA coincide if and only if a and b are linearly dependent; their intersection is a line, a subspace of dimension two, if and only if (aIb) = 0 with a and b linearly independent; their intersection is 0 if and only if (alb) # 0. Proof. This is Theorem 4 of van der Blij and Springer [3]. Cl Corollary 3.8. There does not exist a singular plane, a totally singular subspace of dimension three, B, in IL such that xy = 0 for all x,y E B. Proof. If xy = 0 for all x,y E B then fry = 0 for all x,y E B. Thus, by Lemma 3.6, y E 23A for all x,y E B. Therefore B Q xA for all x E B. From Theorem 3.7 we have that dimp(yA fl xA) g 2 for any linearly independent elements x, y E B. Therefore, dimp(B) g 2. [:1 Proposition 3.9. If A is a composition algebra then all of the Moufang identities are satisfied. Proof. Claim: x(xy) = xzy and (yx)x = yx2 for all x,y E A. $(xy) - 12y = (I + i - 5:)(1‘31) - ((1‘ + i - ilxly = (x + flay) - fiery) - ((9: + 5023):! + (Early = (5333)?! — 53(3331) since x + :7: = (x|1)1 = q($)y - (1(x)y by Lemma 3.2 = 0 for all x, y E A. Likewise, (yx)x = yx2 for all x, y E A. 16 Let f1(x1,x2,x3) = (x1x2)x3 — x1(x2x3). Since (x1x2)x3 — x1(x2x3) = (x1x2)x3 — x1(x2x3)+ + (x1 + x2)[(x1 + x2)x3] — (x1 + x2)2x3 = (x1x2)x3 — x1(x2x3)+ + x1(x2x3) + x2(x1x3) — (x1x2)x3 - (x2x1)x3 = 152(331953) — ($2$1)$3 = -fi($2,$1,$3) and (x1x2)x3 — x1(x2x3) = (x1x2)x3 -— x1(x2x3)+ + x1(x2 + x3)2 — [x1(x2 + x3)](x2 + x3) = (x1x2)x3 — x1(x2x3)+ + x1(x2x3) + x1(x3x2) — (x1x2)x3 - (x1x3)x2 = 131(133352) - (5511139132 = ‘f1($1,$3.1‘2l f1($o(i),$o(2),Io(3)) = 59n(0)f1 ($1,532,153). 17 Let f2 : A x A x A x A —-> A be the Kleinfeld function [15], f2(x1,x2,x3,x4) = ((x1x2)x3)x4 — (x1x2)(x3x4)+ — ((x2x3)x4)x1 + ($2$3)($4$1)+ + ((x3x4)x1)x2 — ($35124)($1$2)+ — (($4$1)$2)$3 + ($4131)($2$3). NOLICB that f2($1,$2,$3,$4) = —f2($2,1§3,$4,$1). A130, f2(x1,x2,x3,x4) = ((x1x2)x3)x4 — (x1x2)(x3x4)+ — ((x2x3)x4)x1 + (x2x3)(x4x1)+ + f1(x3x4,x1,x2) — f1(x4x1,x2,x3) = ((x1x2)x3)x4 — (x1x2)(x3x4)+ — ((x2x3)x4)x1 + (x2x3)(x4x1)+ + f1(x2,x3x4,x1) — f1(x2,x3,x4x1) = ((x1x2)x3)x4 — (x1x2)(x3x4)+ — ((x2x3)x4)x1 + (x2x3)(x4x1)+ + (rl'2(1‘34134))171 - ;r;2((x3x4):rl) - (1’21’3)(4U4I1) + 1L‘2(4B3(-L1:~171 )) = f1(.r'1x2,;r:3,.r:.,) -— f1(x2,;r:3,:1:4);r1 — .rgf1(x3,x4,x1). 18 So f2(x1,x2,x3,x3) = 0 for all x,- E A. Thus 0 : f2($lr$2r$3 + $4a$3 + $4) = f2($1, 1:271:31 $4) + f2(x17$2i$4rx3)+ + f2($1,$2, $3,133) + f2($1, $2,334, $4) = f2($1, $2, $3, $4) + f2($1, $2, $4, 533) meaning, f2($1»$2.$4,$3) = "f2($1,$2,$3,$4)- Therefore, since f2($1,$2,$3,$4) = —f2($2.$3,1‘4,1‘1) and f2($1,$2, 334,333) = ’f2($1,$2,$3,$4), f2($a(l)a$a(2)1$a(3)a$a(4)) = sgn(o)f2(x1,x2,x3,x4). So 0 = f2($1,$2,$1,$3) = f1($1$2,$1,$3) — f1($2,$1,$3)$1- $2f1($1,$3,$1) = f1($1$2,$1,$3l - f1($27$1,l‘3)$1 — $2f1($3.$1,$1) :- f1($1$2,$1,I3) _ fl($21$1)x3)$1 which implies that f1(x1x2,x1,x3) = f1(x2,x1,x3)x1. 19 From this we conclude that for any x, y, z E A, ash/(172)) _ ((xy)x)z : x(y(xz)) — (xy)(IZ) + (333/)(132) - (($y)$)z = -f1($.y.$z) - name) = —f1(xz, x,y) — fiat/.1132?) = —f1(z. Lylx - f1(y.$,2)$ : -—fl(z,x,y).’l: + f1(Z,$ayl-T meaning, x(y(xz)) = ((xy)x)z for any x, y, z E A. Hence, A satisfies all the Moufang identities. Another proof can be found in [22] (see Proposition 1.4.1, page 9). 1:] Proposition 3.10. If B is a subalgebra of a composition algebra, A, and v is con- tained in Bi then B+Bv is also a subalgebra with Bu 3 Bi. Moreover, if a = s+tv and b = x + yu then ab = (sx — q(v)yt) + (t5: + ys)v for s,t, x, y E B. Proof. Since B is a subspace containing the identity, B + Bo is a subspace that 20 contains 1. For any a = s + tu,b = x + yo E B + Bo we get that ab =(s+tu)(x + yo) = 3:: + (tv)x + s(yv) + (tu)(yu) = 3:1: + [(le)t - (tx)o] — 3(W) + (—t—u)(yv) by Lemma 3.3 = sx + [0 + (tx)v] — s(og) + (—ot)(yv) by Lemma 3.2 = sx + (tx)v + 5(1):?) + (umyv) = sx + (tx)v + [(ols)g — o(sg)] + v(iy)v by Lemma 3.3 and Moufang’s identity = 3x + (me + [0 + ifs—35] + pimp) = sx + (tap; + (ys)v + (—(gt)o)v by Lemma 3.2 = sx + (tx)v + (ys)u — q(v)yt by Lemma 3.2 = (333 — (1(1))370 + (ta? + y5)v E B + Bo. Therefore, B + Bo is also a subalgebra and since (b1|b2v) = (b-gfhl’v) = 0 for all biEB,B’USB'L. [3 Theorem 3.11. (Dickson’s Theorem) If B is a non-degenerate subalgebra of a com- position algebra, A, with dirnp(B) = n and 'v is contained in Bi with q(u) aé 0 then B + B1) is a non-degenerate subalgebra with dimp(B + Bu) 2 2n. Furthermore, if a = s + to and b = x + go then ab = (sx — q(v)yt) + (ti + ys)u for s, t, x, y E B. Proof. By Proposition 3.10, B + Be is a subalgebra of A. Since q(u) # 0, u is —”—). Thus dim p( B U) = dimpB = 72. Suppose there exists an invertible with v‘1 = W element b1 E B such that b1 u E B. Then, since B is nondegenerate, there exists 21 an element b2 E B such that (bglblv) ¢ 0. Thus (b-lbgl’U) # 0. But b—lbg E B and v E Bi. Hence, by contradiction, B 0 Bo = 0. Therefore, B + Bo is a subalgebra of dimension 2n. Let s + to be some fixed element in B + Bo where t 7t 0. Since B is non-degenerate, there exists an element y E B such that (ylt) aé 0. So (ons + to) = q(s + yv + to) — q(yv) — q(s + to) = (Slyv + tv) + (1(8) + q(yv + tv) - (1(yv) - (8|tv) - (1(8) - (1(tv) = 0 + (1(8) + (yvltv) + (1(yv) + (1(tv) - (1(yv) - 0 - (1(8) - (1(tv) = (yvltv) = (ylt)q(v) by Lemma 3.1 72$ 0. Therefore B + B1) is a non-degenerate subalgebra of dimension 2n. El Theorem 3.12. If B 2 F1 is a proper subalgebra of a composition algebra, A, over the field F of characteristic two and v is contained in A\B with (vll) 79 0 then B + B21 is a non-degenerate subalgebra with dimp(B + Bu) 2 2. Furthermore, if a = s +tu and b 2: 26+ go then ab = (sx— q(v)ty) -l- (sy + (v|1)ty)v for s,t,x, y E B. Proof. First of all notice that such an element, 1), exists since A is non-degenerate and (rill/-31) = q((a + (1)1) — q(nl) — q(BI) =o2+2o,B+BQ—c12—U2 :0 22 for all 01, B1 E F 1. Since B is a subspace containing the identity, B + Bo is a subspace that contains 1. Since 21 E A which is non-degenerate, v2 = v2 + (u|1)v — (vl1)v = -v2 + (v]1)v -(v|1)v = v(—v + (ul1)1) — (u|1)u = m7 + (v|1)v q(v)1 + (vl1)v Foranya=s+to,b=x+quB+vaegetthat ab =(s+to)(x + yo) = sx + (tu)x + s(yu) + (tv)(yo) = sx + (tx)v + (sy)u + (ty)v2 since s,t,x,y 6 F1 = sx + (tx)v + (sy)v + (ty)(q(u)1+ (v|1)v) = (8:8 + q(v)ty) + (ta: + 8y + (ulna/)2) e B + Bo. Therefore, B + B21 is a subalgebra of dimension 2. Let 011 + Bo be some element in B+vaitha,BEF. If6#0then (011+ Bu]1)=(al|1)+((3v|1) = ,t'i('v|1) also. 23 IfB=0anda¢0then ((11 + Bulb) = (ally) = a(1]v) set). Hence B + Bu is a non-degenerate subalgebra of dimension 2. Cl Lemma 3.13. If A is a composition algebra of dimension 8 then A is nonassociative. Proof. Suppose A is of odd characteristic. One can generate A using Dickson’s Theorem 3.11 and letting 221,112,213 E A with q(u) 75 0 for all i, such that v; E 1*, 112 E (1,v1)i, and 113 E (1,v1,u2)i. Then 111(0203) = —UI(U_3772) = -’U1(’U3’Ug) = 273071112) = 03(viv2) : ’(UTUEWa = —(v1v2)vs 75 (U1U2)l’3. Therefore, if A is of odd characteristic then A is nonassociative. Now suppose that A of of characteristic 2. One can generate A using Theorems 3.11 and 3.12 and letting 13,112,213 E A such that ('l'lll) # 0, 02 E (1,1'1)i. and 24 113 E (1,v1,v2)i. Then v1(v2va) = (11073172) = 01(03712) 2 173(le2) = 113 ((111 + (o1|1)1)v2) = v3(v1v2) + (u1|1)v3v2 = (WW—3 + (villlva’vz = (u1v2)v3 + (v1|1)v3u2 7* (11111983- Hence, if A is of characteristic 2 then A is nonassociative. Cl Theorems 3.11 and 3.12 along with Lemma 3.13 can be used to prove Hurwitz’s Theorem 1.1. Proof. Assume there exists a composition algebra of dimension higher than eight, namely A, that contains Bl 2: F 1. By Theorem 3.11 or 3.12, there exists a non- degenerate subalgebra of A, B2, that is of dimension 2. If B2" is a non-degenerate composition algebra of dimension 2" that is properly contained in A then by Theo- rem 3.11, there exists a non-degenerate subalgebra of A, B2n+l, that is of dimension 2"“. B-er-l. Since A is of dimension higher than eight, by induction on n, A contains a non—degenerate subalgebra of dimension sixteen, B + B'L', where B is of dimension eight. By Proposition 3.9, B + Bo satisfies the .\«Ioufang identities. Thus, for every x+yv,s+tvE B+Bv we get that 0 = (x + yv) ((x + yv)(s + tv)) — (x + yv)2(s + to) = [3(838) - (828 — q(v)8(17y) + (1(v)(837)y — q(vfifiy) + (1(v)(58t)y] + [tar2 — ((17):: + yon?) — (yes + q(v)y(it) — q(vltyi)ti v = [0 - q(v)(0) - (1(8) lifiy) - (1005363111 + [0 + M81?) - (118)? + q(v)(0)l '0. Therefore, y(‘x) = (ys):1': for all y,§,x E B. Hence, the octonion algebra B is associative. But, by Lemma 3.13, the octonion algebra B is nonassociative. So by contradiction, there does not exist a composition algebra of dimension higher than eight. [3 4 Moufang Loops The following are some well known properties of Moufang loops that will be used later in this paper. See Bruck [4, Chapter VII] for unexplained material. Lemma 4.1. For a Moufang loop, L, every element x E L has a unique two sided inverse, 13“. Lemma 4.2. If x, y E L where L is a Moufang loop then: 1- I"($y) = y, 2. x(.1:‘1y) = y, 3. y = (y.r).r_1. 26 4- y = (y$"‘)r, and 5. x(yx) = (xy)x. Lemma 4.3. Let x, y E L for some Moufang loop, L. 1. Iflc 2 0 then xky = x(x(......x(xy))) with k copies of x. 2. Ifk < 0 then xky = (x‘1(x‘1(...x‘l(x‘ly)...)) with —lc copies ofx‘l. 3. Ifk Z 0 then yx" = ((...(xy)x...)x)x with k copies of x. 4. Ifk < 0 then yx" = ((...(yx‘1)...)x‘1)x”1 with —k copies ofx‘l. Proof. Part one is easy to see for k = 0 or 1. Suppose it is true for all 0 S k g n. Then x y=(x-x~--x)y with(n+1)copiesofx = (($(ir ' - ° $))$)y = x((x- - - x)(xy)) by the Moufang identity = :r(:r(--‘I((vy)~-)) Thus, x"+1y = x(x(- - --x(xy)--)) with n + 1 copies of x. So, by induction, for all k 2 0. xky = x(x( --x(xy) - )) with k copies of 2:. Similarly, if k 2 0 then yr" 2 ((---(.1:y)x---).r).r with k copies of .r. 27 Part two is easy to see for k = —1. Now suppose that it is true for all n S k S —1. Then x y = (x‘1 ~x“1~-x“1)y with (1— n) copies of x =(($'1($_1°“$-1))$_1)y = x’1((x"1 ~ - - x"1)(x-1y)) by the Moufang identity = s—1(s—1(....—1(.-1y)...)) Thus, xn‘ly = x’1(x‘1(---x‘l(x‘1y)~-)) with 1 — n copies of x‘l. So, by de- duction, for all k < 0, xky = (x‘1(x'1(---x‘1(x‘1y)---)) with —k copies of x‘l. Similarly, if k < 0 then yx" = ((- - - (yx‘l) - - - )x"l)x’1 with —k copies of x‘l. El Lemma 4.4. If x and y are contained in some Moufang loop, L, then x"(xmy) = (UH-my and (yxnhsm = yxn-li-m. Proof. This follows from Lemmas 4.3 and 4.2. C] From this one can show that any Moufang loop is diassociative. That is, any two elements of a Moufang loop generate a group. This was proven by R. Moufang [19] herself back in 1935. However, there are many diassociative loops that are not Moufang loops. Furthermore, as seen in Table 2, there exist diassociative loops that do not satisfy the Lagrange property. In this example we have the Steiner loop of order ten where any two elements generate a group of order four, namely Z2 x Z2. 28 1 be a(bc) b ca b(ca) c ab c(ab) 1 1 a be a(bc) b ca b(ca) c ab c(ab) a a 1 a(bc) bc ab c c(ab) ca b b(ca) bc bc a(bc) 1 a c c(ab) ab b b(ca) ca a( be) a(bc) bc a 1 c(ab) ab c b(ca) ca b b b ab c c(ab) 1 b(ca) ca bc a a(bc) ca ca c c(ab) ab b(ca) l b a a(bc) bc b(ca) b(ca) c(ab) ab 0 ca b 1 a(bc) bc a c c ca b b(ca) bc a a(bc) 1 c(ab) ab ab ab b b(ca) ca a a(bc) bc c(ab) l c c(ab) c(ab) b(ca) ca b a(bc) bc a ab c 1 Table 2: The Steiner loop of order ten A Latin square design is a pair (P, A) of points, P, and lines, A, such that: 1. P is a disjoint union of three parts R, C, and E; 2. each line I E A contains three points and meets each part R, C, and E exactly once; 3. any pair of points that are contained in different parts belongs to exactly one line. For a loop, L, there is a Latin square design of L, DL, that has a point set P = LRULCULE, where L, = {a,la E L} is acopy ofL, and aline set A = {{aR,bc,cE} = [a,b,c] | (ab)c =1 E L}. The automorphism group, Aut(DL), of the Latin square design BL is the set of all bijections f : P —r P that take lines to lines. So [a,b,c] E A if and only if {f(a). f(b), f(c)} E A. For a E R, let To be the map that exchanges the sets C and E such that ra(b) : c and Ta(C) = b if and only if [(1, b, c] E A. 29 Suppose there exist elements f,g E Aut(DL) such that f leug = To = gICUE. Then f g‘1 is trivial on C U E and therefore is trivial on P. Thus, f = 9. Therefore, Ta has at most one extension, and if it exists then it has order two. So we will just call this extension rd. If L is a Moufang loop then such an extension exists, namely: x}; H ax'laR - —1 -1 Ta“ yCHy 0'3 1 l 23 H a‘ 26 This is because if (xy)z = 1 then: xy = 2‘1 => 1 = (s-lz-1)y-1 z. o[(s-1s-1)y-1]o-1 = 1 => [a(x'12_1)] [y-lo-l] :1 => [a(s-1(o(a—1.—1)))] [y-Ia-li =1 => [(ax‘la)(a‘lz_1)] [y‘1a"1] = 1. We will call such an extension a central automorphism of DL with center a. The same holds for central automorphisms Ty and T2 with the centers y E C and z E E. (See [10] and [8] for discussion.) For a Moufang loop, L, we call CL = (rxlx E LR U LC U LE) the triality group of L. In general, a triality group, C, is a group generated by a normal subset, T, of elements that are of order two along with a homomorphism 7r : C ——> 33 such that for any 9, h E T, |gh| = 3 when rr(g) # 7r(h), CL being an example (see [8. 10]). 30 Lemma 4.5. Let L be a subloop of C" where C is an octonion algebra. Suppose L spans C and f,g E CL with f],- = g|,, i E {R,C,E}. Then g(x,) = sf(x,~) and g(xk) = s‘1f(xk) where s E F‘ and {i,j,k} = {R,C,E}. Proof. Without loss of generality we may assume that f I g = gIE. Thus, fg‘l is trivial on E. IL'R H 1130.}; Case 1. If fg’l : 13 H a}; then fg‘l : yc ,_, a'lyc . Assume that a E F1. ZEHZE Since L spans C, there exist elements x, y E L such that (xa)y 7é x(ay). Also, since [ear/.(ayl‘lx‘ll E A, lfg“(:v).fir—Way).f9"‘((ay)“$")l = [$0.11. (aw-122*] e A. But ((xa)y)((ay)‘1x‘1) 75 (x(ay))((ay)‘lx‘l) = 1. Hence, by contradiction, a E F1 meaning g(xR) = sf(xR) and g(xc) = s‘lf(xc) for some 3 E F. 5BR t—r ailic 1 Case 2. Suppose fg" : IR +—> ac. Then fg‘l : yc r—ryafil . Assume that ZEHZE a 5! F1. Since L spans C, there exists an element x E L such that xa 75 ax. Also, since [x‘1,xa,a’l] E A, [fg’1(x"1),f9-l(xa),fg‘1(a'l)] = [x,ax’1,a‘1] E A. But xax‘la‘1 7t xx‘laa‘l : 1. Hence, by contradiction, a E F1 meaning fg'l : (133 H SIEC ye H 8-1313 for some 3 E F. Since L spans C, there exist elements I, y E L\F1 let—’25 such that Iy 79 yet But lfg"(~().f9"(y).fg“(y":(“‘)l = [8"y.s;r,y";c“l (i? A since s“1y(sx)y‘1x"1 = yrfy—lr’l yé.'t‘3/g/_1;1:‘l = 1. Hence fg“(1R) E LC. 31 This completes the proof of Lemma 4.5. C] 5 Structure of Octonion Algebras Here we will let C be an octonion algebra over a field F. Lemma 5.1. If L is a subloop of C then the subspace spanned by L over F, spanp(L), is a subalgebra of C. Proof. For klll + + lc,,l,,,lc’ll'l + + kfinlfin E A where k,,k£ E F and l,-,l£ E L, (klll + + lc,,l,,)(lt:’ll’1 + + kfinlfin) = klk’l (lll’l) + + knkfinanlfin) E A. Hence, A is a subalgebra of C. [:1 Lemma 5.2. Let L1 3 L2 be subloops of C that span C over F. Then R(Ll) = R(Lg) if and only if L2 Q F“ - L1. Proof. Suppose R(Ll) = R(Lg). Then, for every x E L2\L1, p; = pwpy, - - - pyn for some py, E L1. Thus, pI(L1) C F‘ -L1 and {%x|l E L1} C F-Ll. Hence, since L1 spans C over F, {(l]x)]l E L1} 76 {0} and x E F‘ - L1. Therefore, if R(Ll) = R(Lg) then L2 g F“ - L1. Now suppose that L2 Q F“ - L1. Then for any x E L2, x = Icy for some y E L, and k E F‘. Since p,C E O(C, q), ,oI = pky = py. So for all x E L2 we get that pr E R(Ll). Therefore, if L2 g F‘ - L1 then R(Ll) = R(Lg). Cl Lemma 5.3. Let L be a subloop of C‘ such that L spans C over F and R(L) is irreducible on C. Then a copy of either R(L) or R(L)/{i1} is contained in CL/mn with a normal subgroup of inder two. 32 Proof. Let (p : R(L) ——) GL/anl be defined as: (rape) = T138 Claim : (p is well defined. Proof of claim If p31 = p1,, then by Lemma 5.2, x2 = kxl for some k E F. Thus, x1 = x2 E L/LflFl and 3,er2. A Claim : (,0 is a homomorphism. Proof of claim The map (,0 is a homomorphism as long as (a(psmss - - rs.) = regress ~~r,,,E is well defined. Suppose pup$2 - - - p3,, = pwpy, - - -pym where m and n are both odd. Thus, —1?,,("'(TE(IIEI1)'ITg-)"')1?" = —ym("'(m(yl§yl)-y—2)m)ym (1(11)(1(I2)---(1(xn) q(y1)(1(y2)~-(1(ym) for all z E L. Therfore, In('"(I§‘(Iiz“r1):r§‘)-~)r. Q(I1)Q(~F3)"'Q(In) (1(Z)(1(8:2) - ° '(1(a:o_1) yum-(112“(y1:"‘yi)y2“)~- )11... (1(y1)(1(y3)--~(1(ym) = (1(Z)(1(1/2)-~(1(ym_1) 33 for all z E L. Thus, $n("'(172-1($lz_1$1)$2-1)°”)$n = ym(- - - (y§‘(y12“y1)ys‘l) ° ' - )ym EL/LflFI for all z E L/L 0 F1. Hence TI13T123"'TInE(ZE) = ry,Ery,E -~rymE (zE) for all z E L/LflFl. So, by Lemma 4.5, 713137.123 - - - ran = 7111571128 - - - ryms. Likewise, re- guardless of what m and n are, ifpnpx, - - - 102:" = pmpy, - - -pym then TIIEszs - - wring = alarms - - - rymE. Hence, (,0 IS a homomorphism. A Claim : There is a one-to—one correspondance between the reflections, pv, of R(L) and the central automorphisms, T“, for x E L/ L 0 F1. Proof of claim Clearly, by definition of (p, for every x E L/ L 0 F1 there exists an element 11 E L such that tp(p,,) = TIE. If v,u E L such that 99(Po) = 7'21: = <,o(p,,) then u = av 2 for some a E F. Thus, [1,,(2) = —;‘(i“) = —‘;'{:‘:]’ = —:.—,:a—”) = —;]”(—Z;'i) = pv(z) for all z E L. Therefore pv = pa and there is a one-to-one correspondance between the reflections, ,0”, and the central automorphisms, TIE. A Moreover, if tp(p,,,p,,2 - - - p12") = r 11:157- 12;; "'Tx-2..E = 1 then x2,,(- - -x2(x1-lzx1_l)x2 - - - )Ll‘gn = z for all z E L/L 0 F1. Thus, for all z E L there exists an element kz E F such that 34 . . . x2n("'($2(3-1215—5332)“)332n _ q(xl) 9(332n—1) Q($2)ql(174)"1'<2($2n) — kzz' Hence, $2n("'($2($—12$—1)$2)"')xzn = [€22 1021,0222 ' ' ' p$2n(Z) : q(x1)q(x2)-"q(12n) for all z E L. Furthermore, since q(z) = q(pxipxz ° ' ' pxzn (2)) = (10922) = 1831(2). kz E {321} for all z E L. Now suppose 21, 22 E L such that kl, 74 k2,. Then charF aé 2 and without loss of generality we may assume that kz, = 1 and kl, = —1. Therefore, since p3,. E O(C, q), (21]22) = (21] — 22) which implies that (z1|22) = 0. Thus, L = {z E Lllt:z = 1}U{z E L|lcz = —1} g UUUi for some subalgebra, U, of C. Notice that for all x, u E UflL andv E ULflL, p,,(x) = x E UflL and pu(x) = x— %u E UflL. But since R(L) is irreducible on C, either U n L = G or U i H L = 0. Hence, k2, = k2,, for all 21, 22 E L. Therefore ker(tp) 3 {i1}. Hence, there exists a copy of either R(L) or R(L)/{i1} in CL/anl with a normal subgroup, (the rotation subgroup), of index two. 1:] Proposition 5.4. If a composition algebra of a given dimension, C, over a field, F, is split then it is unique up to isomorphism. Moreover, if F is a finite field and C is of dimension greater than one then C is split. Such an algebra of dimension eight is of type 08+ (F). Proof. Let C be a composition algebra over the field F that is split. Suppose there does not exist an element x E C such that q(x) = 0 and (xll) 75 0. So for every x E C with q(x) = 0, x + :7: = 0. Choose a nonzero element y E C such that q(y) = 0. Since q(yx) = q(y)q(x) = 0 for all x E C, yx + W: = 0 for all x E C. So (ylx)1 = ya": + xy = 0 for all x E C. But C is a non-degenerate algebra. By contradiction, there exists an element x E C such that q(x) = 0 and (xll) = 1. Note that x + r: = (x]1)1 = 1. Thus C contains a subalgebra, Fx + F5: 2 F1+ Fx, of dimension two over F that is isomorphic to F (B F. So any two split composition algebras of dimension two over some field, F, are isomorphic. Now suppose that A; and A2 are proper composition subalgebras of 01 and C2, respectively, and are of dimension greater than one with A1 21 A2. Let b,- E A3“ for i E {1,2}. Since C,- is non-degenerate, there exists an element c,- E C,- such that (b,|c,~) 75 0. Let c,- = a,- + a; where a,- E A,- and a; E Af. Thus, (1911+ bi) - (1(02) — (1(1):): (bald?) = (b,|a,—) + (balal) = (b,|a,~ + (1;) yé 0. Hence, there exists an element 5, E A} with q(s,) 75 0. Let a? E A, such that q(ag’) : q(L)‘ Then there exists an element t, = aj’s, E A} such that q(t,) : 1. 36 Therefore, by Dickson’s Theorem 3.11, A1 + Altl g Cl and A2 + Agtg 3 C2 are isomorphic composition algebras of dimension twice that of A1. So by induction on the dimension of A,, we get that any two split composition algebras of the same dimension over some field, F, are isomorphic. El Definition 5.5. A hexagon line of a split octonion algebra C is a totally singular 2-space that is contained in 1i such that the multiplication is zero. A hexagonic structure contains points and lines where any two points can be joined by a path containing at most three lines and there does not exist any 2, 3, 4, or 5—gons. For an octonion algebra, C, we will call a point, a singular 1-space, P, with P2 = 0. We say that a point P is incident to a hexagon line B if P C E. This induces a hexagonic structure which we call the generalized hexagon. Details of this construction can be found in [21]. Theorem 5.6. The automorphism group of a split octonion algebra, C, is of type 02(F) and is transitive on the hexagon lines (see chap. 2 in [22]). Definition 5.7. The stabilizer of a subspace, U, of C is the subspace StabC(U) == {xECIquUforalluEU}. Let H be an additive subgroup of an octonion algebra, C, such that {h— 1|h E H} is contained in a hexagon line of C. Also, let Q be a subalgebra of C that is contained in the stabilizer of {h — 1 I h E H}. The product Q o H is defined to be the algebra with underlying set {(x, h)]x E Q. h E H} and the binary operation (I. .(1)(y. h) =(r11-1+ 17(12 - 1) + (.(1 - 1)y)- 37 Example 5.8. For a field F let C be the split octonion algebra defined in section. 2. a (121,112,113) (7:: larbivhzh 6 P1 (U1, ”2) “3) b 0 (0, 122,0) The subspace S = |v2,u3 E F C C is an example of a (0,0,U3) O hexagon line of C. The stabilizer of such a hexagon line in C is the subspace a (v1, v2, 0) = 5* (u1,0, u;;) b a (v1, 0, 0) 1 (0, v2, 0) g 0 (ul , 0, 0) b . (0,0, u;;) 1 e M2(F) 0 F2. Such a subspace is an algebra that is maximal in C ( see section 6). Now let K be a subloop of CLL(F) such that K g L = (SLY for some hexagon line S. Without loss of generality, by Theorem 5.6, we may assume that S = 0 (0,3,0) |s,t E F (0,0,t) 0 Thus a (v1,v2,0) Si = |a,b,v1,v2,ul,u3 E F (u1,0, 113) b Define a (v1,0,0) 7r : L —-) |a,b,vl,ul E F,ab — v,u] aé O ’35 CL2(F) (U1,0,O) b as 38 a (U1, "(12, 0) a (U1, 0,0) 1r : (u1,0, u3) b (u1,0, 0) b Note that a (v1, v2, 0) c (a1, a2, 0) rr (u1,0, u3) b (,81, 0, B3) d ac + vlfil (aal + dv1,a012 + d‘Ug + ulflg — u3[31, 0) = 7r (bfi1 “‘1‘ C711, 0, 1113;; + 0113 + U102 — 12201) bd + U101 CC + U131 (0.01 + (1111, 0, 0) (bfi1+C'll1,0,0) bd+ulal (a (v1, 0, 0) c ((11, 0, 0) (U1 , 0,0) b (,31, O, 0) d 0' (U1,U2,O) C (0110210) : 1f 7r (U1,0,U3) b (filioifi3) d 1 (0, 112,0) So it is a homomorphism with ker(7r) 3 ]U2, u;; E F (01 01 U3) 1 Note that both the ker(rr|K) and I m(7r| K) are groups. This brings up the ques- tion: “Does there exist a homomorphism d from I m(7r| K) o ker(7r) onto I m(7r| K) o ker(rr|K)?” If so then oIK is a one-to-one homomorphism from K onto I m(7r| K) o ker(7r|K) and K = ¢II<1(1'7'l-(7T|1<))° keTUTlK) EGOH 39 where C S CL2(F) and H 3 F2. Now let L be a subloop of C“ such that L S Q U Ql for some quaternion subalgebra Q. If L g Q then L is associative and therefore a group that is isomorphic to a subgroup of CL2(F). Otherwise if L i Q then C = L it Q is a subgroup of L with [L ; G] = 2 and o e H g GL2(F). So L = GLJGv’for some v e LnQi and by Dickson’s Theorem 3.11 (g -l- hv)(x + yv) = (gx —- q(v)q(y)y’lh) + (q(x)hx’l — yg)v for any g,h,x,y E C. Lemma 5.9. If F0 is a subfield ofF then GLL(FO) S CLL(F). 6 Proof of Theorem 1.3 In this section we will let L be a subloop of an octonion algebra, C, over a field F and R = R(L) will be the reflection group of L. Lemma 6.1. If L is a subloop of C and U is a subspace of C that is left invariant under R(L) then L Q U U U i and U J- is also left invariant under R(L). Proof. Since U is an invariant subspace of C, for every x E L and every u E U we have that px(u) E U. Therefore, u — %u7’x E U for any x E L and u E U. So if x is an element of L then x E U unless (xlu) = 0 for all u E U. Hence, L E U U UL. Moreover, if v E Ui and x E L then either x E Ui or (xlv) = 0. Therefore, pI(v) = v — %x E Ui and UL is left invariant under R(L). [:1 Lemma 6.2. Suppose I E U where U is a subspace invariant under R(L). Then for any y E L. yU is an R(L)-invariant subspace of C. 40 Proof. Let y be some fixed element in L. Note that for any x E L u z u_ (ill/“)3: ps(y ) y q(x) = u_ (17:8IU)$ y q(x) for all u E U. If (yxlu) 7t 0, then 37.1: ¢ Ui. By 6.1, fax E U U UJ'. Therefore, 335x = v for some v E U. Thus, yo = “331523) = x. Hence px(y’u) = yu _. (1:31; yv E yU. Therefore, yU is an invariant subspace of C. Cl Lemma 6.3. Let U be a non-degenerate subalgebra of C with dimp(U) = n. If U is invariant under R(L) and x E L\U then U + xU is an invariant non-degenerate subalgebra of dimension 2n. Proof. By 6.1 x E U i, so by Dickson’s Theorem 3.11 U + xU is a non-degenerate subalgebra of dimension 2n. In Lemma 6.2 we showed that xU is an invariant subspace. Therefore, the non-degenerate subalgebra, U +xU, is also invariant under R(L). [I Now suppose L is a subloop of C‘ such that L does not span C. For any set S g C we will denote the vector space spanned by S over F as spanp(S). The symbol A will always be used to denote the span of L over the field F. Lemma 6.4. If C is not split and A is properly contained in C then A C Q for some quaternion subalgebra, Q, ofC or A is totally isotropic. Proof. Since C is not split, for every 0 7f .17 E C, q(x) # 0. Let B1 2 Fl 3 A. 41 If A = Bl then by Theorem 3.11 or 3.12, there exists a non—degenerate subalgebra of C, B2, that contains A and is of dimension 2. Since q(x) aé 0 for all x E Bf, by Dickson’s Theorem 3.11, there exists a non-degenerate subalgebra of C, Q, that contains A and is of dimension 4. If A 31$ BI and A is not totally isotropic then by Theorem 3.11 or 3.12, there exists a non-degenerate subalgebra of A, .32, that is of dimension 2. If A = 82 then since q(x) 75 0 for all x E 82*, by Dickson’s Theorem 3.11, there exists a non- degenerate subalgebra of C, Q, that contains A and is of dimension 4. Otherwise, if A 76 82 then there exists some element y E A n Bi!“ with q(y) 74 0. Thus, by Dickson’s Theorem 3.11, there exists a non-degenerate subalgebra of C, Q, that is contained in A and is of dimension 4. Assume A # Q. Then there exists some element 2 E A n QJ- with q(z) 51$ 0 and by Dickson’s Theorem 3.11, A contains the non-degenerate subalgebra, C. But A is prOperly contained in C. Hence, A C Q for some quaternion subalgebra, Q, of C. Cl For now on, in this proof of Theorem 1.3, we will assume that C is split. We now want to prove the following proposition. Proposition 6.5. If a subloop L1 ofC does not span C, then we have one of the following: 1. L1 S Si for some hexagon line 5'; 2. L1 3 Q for some nonsplit quaternion subalgebra Q; 3. L1 is totally isotropic and F is a nonperfect field of characteristic two. 42 Definition 6.6. The radical of an algebra, B, which we will denote by Rad(B), is the subspace {b E B | q(b) = O and (xlb) = 0 for all x E B} of B. Lemma 6.7. Rad(A) is an ideal of A. Proof. Here we want to Show that for v E Rad(A) and u E A, both uv and vu are in the radical of A. So suppose v E Rad(A) and u E A. By Lemma 3.1, (1]u)(x|v) = (vlux) + (xluv) for any x E A. Thus, (1|u)0 = O + (xluv) which tells us that (xluv) = 0. Also, q(uv) = q(u)q(v) = q(u)0 = 0. Therefore, uv E Rad(A). A similar argument shows that vu E Rad(A). Lemma 6.8. The stabilizer of a hexagon line of C is a subalgebra of C. Proof. Let U = spanp(v,u) be a hexagon line of C. Certainly, the stabilizer, stabC(U), is a subspace of C that contains 1. Also, for any x E stabC(U) and a E U, (xla) = (llxa) = 0 since {to = [(xl1)1 -— x]a E U. Thus, stabC(U) Q Ui. If x, y E stabC(U) then for any a E U, we get that for some b, c E U. Therefore, stabC(U) is a subalgebra of C. Cl For now, the symbol B will always be used to represent a maximal composition subalgebra of A such that 1 E B. Lemma 6.9. Either B is nondegenerate and B -l- Rad(A) = A or B = F 1 and A is totally isotropic. Proof. If B is non—degenerate then, by definition of Rad(A), B Fl Rad(A) = {0}. Suppose there exists an element v E Bl DA such that q(v) 75 0. Then, by Dickson’s Theorem 3.11, B + Bv is a composition algebra contained in A. But B is a maximal composition subalgebra of A. Thus, q(v) = 0 for all v E Bi 0 A. Since q(v) = O = q(u) for all v, u E BiflA, (vlu) = q(v+u)—q(v)—q(u) = 0. Hence, BiflA Q Rad(A) and therefore Bi 0 A = Rad(A). So if B is non-degenerate then B -l- Rad(A) = A. Now suppose B = F 1 where F is of characteristic 2 and A is not totally isotropic. Then B 0 Rad(A) = {0} since q(x) = 0 for all x E Rad(A). Therefore, if there exists an element v E A such that (vll) yé 0 then, by Theorem 3.12, B + Bv is a composition algebra contained in A. Since B is a maximal composition subalgebra of A, (vll) = 0 for all v E A. Moreover, (vlu) = (vull) = O for any v,u E A. Let y be any element in A and let Or = q(y). There exists a unique element [3 E F such that 62 = (1. Since q(y—ffl) = (,r/,f31)+q(y)+q(fil) = 0+a+o = 0, there exists an element ,61 = y1 E B and an element y — ,61 = y2 E Rad(A) such that y = y] + yg. Hence if B 2 F1 where F is of characteristic 2 then B {L Rad(A) = A. [:1 44 Lemma 6.10. Suppose dimp(B) = 4. If dimF(A) = 6 then A is the stabilizer of a hexagon line, Rad(A). Otherwise, A is either properly contained in the stabilizer of a hexagon line or is a nonsplit quaternion subalgebra. Proof. Suppose there exists an element v E A 0 Bi. We will first show that B -i- Bv is an algebra of dimension 6 over the field F. So suppose that s+tv, x+yv E B +Bv. By Proposition 3.10 we have that (s + tv)(x + yv) = (sx — q(v)(yt)) + (ti: + ys)v =sx+(tx+ys)v E B+Bv. Let u E Bi with q(u) # 0. Since dimp(B) = 4, B is non-degenerate and, by Dickson’s Theorem 3.11, B + Bu = C. Let U’ E C such that q(v’) = 0' and (vlv’) yé 0. Since Cu and Cv’ are totally singular and C = C(v + v’) is non- degenerate, dimp(Cv) = 4 = dimp(Cv’). So 4 = dimp(Cv) = dimp(Bv + (Bu)v) = dimp(Bu fl Cv) + dimp(B fl Cv) since Bv Q Bi = Bu and (Bu)v Q (Bu)Bi Q (Bu)(Bu) Q B by Proposition 3.10. Thus, since Cu is totally singular and B, Bu are non-degenerate subspaces of dimension four, dimp(Bu fl Cv) g 2 and dimp(B fl Cv) S 2. Thus, we get that dimp(Bv) = dirnp(Bu fl Cv) = 2. Since dimp(B + Bo) = 4 + 2 = 6, dirnp(A) 2 6 with BL‘ Q Rad(B + Bo). We now want to show that Bv = Rad(A). For any r E Rad(A) we have that dimp(B+Bv+Br) g dimp(A) < 8. Thus BvflBr aé Q. So dimp(BvflBr) E {1,2}. But dimp((B+Bv)fl(B+Br)) ¢ 5, otherwise B+Bx Q (B+Bv)fl(B+Br) for some x E Rad(A) and dimp(B+Bx) = 6 for all x E Rad(A). Hence, dimp(BvflBr) = 2 and r E B + Bv. Now if tv,yv E Bv = Rad(A) then (tv)(yv) = —q(v)yt = 0. Therefore, A = B + Bv is a maximal subalgebra of C and stabilizes a totally singular 2-space with multiplication zero, Rad(A). By Lemma 6.8, stabc(Rad(A)) is a subalgebra of C. Therefore, either A = stabC(Rad(A)) or C = stabc(Rad(A)). But for any v E Rad(A), dimp(Cv) = 4. Hence, C 7é stabc(Rad(A)) and A = stabC(Rad(A)). Now suppose A = B and is split. Let v be an element in A1 such that q(v) = 0. Then A is properly contained in B + Bv which is the stabilizer of a hexagon line, Rad(B + Bv). [:1 Lemma 6.11. Let H be a hexagon line and let x E H\{0}. Then xC S StabC(H). Proof. Let y E H\Fx and v E C. By Lemma 3.3, y(xv) + f(yv) = (ylx)v = 0. Thus, y(xv) = —x(yv) = x(yv) E xC fl yC. But, by Theorem 3.7, xCfl yC = H. Hence, y(xC) Q H and xC S StabC(H). [:1 Lemma 6.12. If (Il'lllF(B) S 2 then A is not maximal. Moreover, A is properly contained in the stabilizer of some hexagon line. Proof. Case 1. If dimp(Rad(A)) = 0 then A = B and A = B is properly contained in a quaternion sul_)algel)ra. Moreover, by Lemma 6.10. A is contained in the stabi- 46 lizer of a hexagon line. Case 2. If dimp(Rad(A) = 1 then Rad(A) = Fx with x 75 0. Since Ax Q Rad(A), dimp(A/AnnA(x)) = 1. Thus A = Ann/1(1) + F1. Let H be a hexagon line containing x. Therefore, by Lemma 6.11, AnnA(x) = xC g StabC(H). Hence A S StabC(H). Case 3. If dimp(Rad(A)) = 2 then, since Rad(A) is an ideal of A, A is prOp erly contained in the stabilizer of Rad(A). Let Rad(A) = spanp(a,b). Suppose ab = kla + kgb for k1, k2 E F then 0 = a2b = a(ab) = lcla2 + k2(ab) = k2(ab). Thus either k2 = 0 or ab = 0. If k2 = 0 then ab = ha and 0 = ab2 = (ab)b = klab. Hence ab = 0. Therefore, A is properly contained in the stabilizer of a totally singular 2-space with multiplication zero, Rad(A). Case 4. Suppose dimp(Rad(A)) = 3 with Rad(A) = spanp(a1,a2,a3). By corol- lary 3.8 there exists some a 79 b E {a1,a2,a3} such that ab yé 0. Notice that ab E spanp(a, b). Otherwise, if it were then ab = kla + kgb and 0 = a2b = lcla2 + kgab = kgab 7t 0. Thus, Rad(A) = spanp(a, b, ab). Since Rad(A) is an ideal of A, for any x E B we have that xb = k1a+k2b+k3(ab) for some k1, k2, k3 E F. By multiplying both sides of the equation on the right by b we get that 0 = k1(ab) + 0 + 0. Thus, k1 = 0 and BI) Q spanp(b, ab). Now for any x E B we have that x(ab) = k1a+k2b+k3(ab) for some k1, k2, k3 E F. 47 By multiplying both sides of the equation on the right by b we get that k1(ab) + o + o = (x(ab))b = -l(E)ilb = [(abfilb = a(bxb) = —a('lEb) = —a(xbb) =0. Thus, k1 = 0 and B(ab) Q spanp(b,ab). Moreover, b(ab) = —b(bc‘i) = b(ba) = b2a = 0. Hence, A is properly contained in the stabilizer of a totally singular 2-space with multiplication zero, span p(b, ab). C] Thus, if C is split and L is a subloop that does not span C then L is contained in the stabilizer of a hexagon line, S. This completes the proof of proposition 6.5. We will now show that L spans V if and only if V = [V, R]. Since C is non- degenerate, for every x E L there exists some v E V such that (xlv) 75 0. Therefore, 48 we know that: [V, R] =span({—v + v' | r E R,v E V}) =span ({—v+v- (525’s | x e L,v e V}) = span ({ jays 1 s e 1,. e V}) = span(L). Thus V = [V, R] if and only if L spans V. So if V > [V, R] then L does not span V and therefore L is either contained in a quaternion subalgebra or, by Proposition 6.5, L S S i for some hexagon line S. Now let us suppose that V = [V, R] and that R is reducible. Thus, V contains some proper invariant subspace. We showed in Lemma 6.1 that there exists some proper invariant subspace of V that contains the identity, 1. Let U be a minimal invariant subspace of V such that 1 E U. Lemma 6.13. Rad(U) = {0}. Proof. Since ,0 (u) = u— My E U for all u E U, yu = 0 for all y E L\U. y q(y) Therefore Rad(U) C Rad(spanp(L)) = Rad(V) = {0} Cl We now want to show that U is a composition algebra. By Lemma 3.2, for every x E U n L, x‘1 = #5: = (II—$115 + 21831 E U. We showed in Lemma 6.2 that xU is an invariant subspace of V for any x E U. Since both U and xU are invariant subspaces that contain the identity, 1, xU flU is also an invariant subspace containing the identity. By n'iinimality of U, .rU = U. Therefore for all x. y E U flL, 49 x,y E U H L, xy E U. Since U is invariant under R, L C U U UJ'. Also, since C is non-degenerate, dimp(U)+dimp(U‘L) = 8. So if V = [V, R] then dimp(UflUi) = 0 and U = span(U n L). Hence, U is closed under multiplication and is therefore a composition algebra of dimension 1, 2, or 4. If dimp(U) S 2 then, since span(L) = V, by Lemma 6.1, there exists an element x E L n U J: So from Lemma 6.3 one can use Dickson’s Theorem 3.11 and Theorem 3.12 to find an invariant composition algebra of dimension 4, Q. Hence, we showed in Lemma 6.1 that L is contained in Q U Qi. We now have the material to prove Theorem 1.3. Proof. If V = [V, R] and R = R(L) is reducible then, from the material above, L Q Q U Qi for some quaternion subalgebra Q. If V > [V, R] and C is not split then, by Lemma 6.4, L is contained a quaternion subalgebra, Q. If V > [V, R] and C is split then, by Proposition 6.5, L is contained in Si for some hexagon line S. E] 7 Proof of Theorem 1.4 Proposition 7.1. Let G S CLn(F) with n 2 2, be generated by reflections for char(F) 32$ 2 and generated by transvections for char(F) = 2. Suppose that G is irreducible but imprimitive on V = F ". Then there exists an element 9 E CLn(F) with Cg monomial. Indeed, Cg = D : Sn, where S,, is the group of permutation matrices and D is a group of diagonal matrices. Also, there is a subgroup A of F‘ with either D E” A", being the group of all diagonal matrices with entries from A, or D E An‘l, being the subgroup of determinant 1 matrices. Proof. Let T = To be the generating set of reflections (transvections), elements of order 2 with [V, t] of dimension 1, for all t E T. For W a proper block of imprimitivity in V, there is an element t E T such that Wt aé W. Thus, [W, t] has dimension at least that of W. Therefore W and [W, t] both have dimension exactly 1. Let (I = {I’Vill S i S n} be the n blocks of WC, each a 1-space. The group G permutes Q. Therefore, C may be conjugated by some 9 into the group of all monomial matrices. Identify G with this conjugate. Let D be the kernel of the action of C on I), that is, the subgroup of the diagonal matrices that are contained in C. By irreducibility, the group G / D induced on SI is a transitive subgroup of Symm) E S... Every t E T is either in the diagonal subgroup D, and fixes W,- for all i, or exchanges two of the blocks, say W,- and W], and induces the transposition (i, j) on the block set, Q. A transitive subgroup of Sn that contains transpositions is the whole group 5... Therefore, 0/ D E Symm) and every t E T\D acts as a transposition. In the monomial group we get that, for any t,s E T\D with tD yé sD, ts has the same order, 2 or 3, as tD.sD of the corresponding transpositions in C / D ’95 Sn. For 1 S j S n — 1, let in“ E T inducing (j,j + 1) on Q. Thus, 5 = (t1,2,. . .,t,,_1,,,) E I'V(A,,_1) g 8,, with C = D : S. If 0 ¢ 81 E W1 then E = eS : e1,...e,, is a basis with e' E W, which 5' crmutes as it does Q. l J J P We will write our matrices with respect to the basis E, so that S is the group of all permutation matrices. Let t = tin be the transposition (i, j) E T n S. On Fe,- + Fej it acts as 0 1 1 0 and the monomial conjugates of t inducing the same transposition on (I are of the form for a E F“. Therefore there exists a subgroup A S F“ such that tD n T consists exactly of these elements with a E A. Therefore, the elements t(tD D T) E D generate the determinant 1 group iso- morphic to An‘l. If T n D = G then this subgroup is all of D. If T n D 79 Q, then D is the full diagonal group with entries from A. E] For a finite field K, the group O§(K) has a quasisimple derived group Q§(K). (See [25] for this and other facts about orthogonal groups.) If K is of characteristic 2 then Q§(K) has index 2 in O§(K), which is generated by its unique conjugacy class of transvections. If K is not of characteristic 2 then Q§(K) has index 4 in O§(K), which is generated by its two conjugacy classes of reflections. Let +Sl§(K) be the subgroup of index 2 in 0;,(K) that is generated by the reflections, [1,, with center, x, of square norm. Also, let ‘52:,(K) be the subgroup of index 2 in 0;,(K) which is generated by the reflections, p,,, with center, y, of nonsquare norm. Here we also write i(2:3(K) for O§(K). The rest of this section is devoted to a proof of Theorem 1.4. In particular, we let L, C, F, and R be as in the statement of the theorem. Theorem 7.2. R is one of the following and unique up to conjugacy in O(V, q): 1. R = 59§(Fo) for some finite subfield F0 of F of odd characteristic with 6 equal to — or + and 5 equal to —,+, or :i:; 2. R = O§(Fo) for some finite subfield F0 of F of characteristic 2 with 6 equal to — or +,- 3. R = W(E3) E 20; (2) where F = F? for some odd prime p; 4. R = Sg where charF 75 3; 5. R = Sm where charF = 5; 6. R=A":S3where13£ASF’ andn=7or8. Proof. As R is generated by a conjugacy class of reflections (transvections) of O(V, q), it has a derived group R’ of index 2. Also, R is irreducible on V. In characteristic 2, the theorem follows from Proposition 7.1 and Kantor [14, Theorem 11]. (Kantor restricts attention to transvection groups over finite fields, but this immediately implies the results for any finite group generated by transvections; see [5, Theorem 3.4B].) For now we will assume that the characteristic of F is not 2. For R imprimitive, the theorem holds by Proposition 7.1. In the case where R is primitive we may apply 53 the work of Wagner on primitive finite groups generated by reflections [26, Result, Table II, p. 520]. (Note that this table contains some misprints.) In addition to the conclusions of the theorem, Wagner’s results include groups with derived group S L8(Fo) and SU8(FO), for F0 a finite subfield of F. These groups contain transvec- tions and therefore cannot be contained in any orthogonal groups of characteristic other than 2. [:1 Remark 7.3. 1. In the first two cases of the theorem, R’ contains Siegel elements (long root elements) 3. Such elements have [V, s] totally singular, so these cases can only occur when C is split. 2. Wagner’s results are elementary and self-contained. Kantor’s proof makes use of several big classification theorems from the theory of finite groups. It is likely that in the special case used here ( finite groups generated by orthogonal transvections in characteristic 2 ) can be given an elementary proof following the work of Wagner. Lemma 7.4. R i 08" (F0) and R 35 69; (F0) for any subfield F0 S F. Proof. By [6, Corollary 4], there exists a copy of 83 in the outer automorphism group of R. But, by [23, Theorem 30], 53 is not contained in the outer automorphism group of either 0,;(Fo) or 50,;(F0). Thus R x O§(Fo) and R 32 ’Q§(Fo). El Lemma 7.5. R % 39, R 9: Sm, and R 95 A" : SB for n = 7 or 8. Proof. Suppose R 2 S7,, for some at E {9, 10}. Since Sm does not contain a normal subgroup of order two, by Lemma 5.3, R contains a subgroup, A,,,, of index two that 54 is isomorphic to the rotation group of GL/anl. Thus, by [6, Corollary 4], there exists a copy of S3 in the outer automorphism group of Am. But, by [11, Chapter 11.5], |Out(A,,)| = 2 for all n 75 6. Hence, R 2 S9 and R 3.! Sm. Assume R E A" : S3 for n = 7 or 8. By Lemma 5.3, either R contains a subgroup, A" : A3, of index two that is isomorphic to the rotation group of G L / mp1 or R/{ztI} contains a subgroup, H : A8, of index two that is isomorphic to the rotation group of CL /LnF1- Since CI = A" : A8, and C2 = H : A3 have a normal subgroup, N,, with Ci/N, g A3, by [6, Corollary 1] and [6, Corollary 4], there exists a copy of 53 in the outer automorphism group of G,/N,- 3 A3. But, by [11, Chapter II.5], |Out(A3)| = 2. Hence, R a! A" : 58. Cl Lemma 7.6. If L is a maximal subloop of B = {x E GLL(F)|q(x) E G} and R(L)’ —_— aim) then SLL(FO) g L g F‘ . GLL(FO). Proof. By maximality of L, L = F ’ - L 0 B. Since q( 1) = 1, a square, we have that R(L) E {+QB+(FO),0;(FO)}. Thus by 5.2, either F‘ - L = CLL(FO) orF“ - L = F‘ - {v E CLL(FO)]q(v) is a square}. Thus for any element x E SLL(FO) there exist elements a E F' and y E L such that x = ay. Hence, since ay E F“ - L H B, x E L. Therefore SLL(FO) S L S F" - CLL(FO). [3 Therefore, by Theorem 7.2, if R(L) is irreducible then either R(L) 2 +0; (F0) for F0 S F, R = O§(Fo) for F0 S F, or R(L) E 205*(2). Furthermore, by Lemma 5.2, if L S CLL(F) then L "-3 L1 where either SLL(FO) S L1 S F" - SLL(FO) or 2SLL(2) S L1 S F ‘ -2SLL(2). This completes the proof of Theorem 1.4. 8 Proof of Theorem 1.5 In this section we will let C be a finite, and therefore split, octonion algebra over a field F. Also, we will let L be a maximal subloop of loop B where SLL(F) S B S GLL(F). Lemma 8.1. B = {x E CLL(F)|q(x) E C} for some multiplicative group G S F‘. Proof. Note that (p defined in the follow way: toe—st" :8 ._. q(x) is a group homomorphism from B to F “ with ker(<,o) = SLL(F). Thus B = U xkker((p) where q(xk) = k kEIm(gp) = {I E GLL(F)|(1(I) 6 1mm} Cl Proof of Theorem 1.5. If L is a maximal subloop of B = {x E CLL(F)|q(x) E C} that contains SLL(F) then, by 8.1, L = {x E CLL(F)|q(x) E H} for some subgroup H S C. Furthermore, by maximality of L, H is a maximal subgroup of C. Suppose L is a maximal subloop of B that. does not contain SLL(F). If L does not span C then, by Theorem 1.3, L is contained in and therefore equal to 56 (Si),. 0 B for some hexagon line S. If L spans C and R(L) is reducible then by Theorem 1.3, L is contained in and therefore equal to (QU Qt). D B for some quaternion subalgebra, Q, of C. If L spans C and R(L) is irreducible on C then, by Theorem 1.4 and Lemma 7.6, R(L) is one of the following: 1. +12; (F0) for some proper subfield F0 of F; 2. 0; (F0) for some proper subfield F, of F; 3. 20; (2) when F = F,p for some odd prime p. Thus, since R(G’LL(FO)) = O;(Fo), R(2CLL(2)) = 203(3), and the reflection group of {v E GLL(FO)|q(v) is a square} is +(251(F0), by Lemma 5.2, if R(L) is one of these groups then L is one of the following: 1. L = (F‘ ~ GLL(FO)) D B for some maximal subficld F0 of F; 2. L = (F‘ « {v E CLL(FO)|q(v) is a square})flB for some maximal subfield, F0, of F; 3. L = (F‘ - 2CLL(2)) F] B where F = Fp for some odd prime p. But, since {v E CLL(FO)|q(v) is a square } S CLL(FO) and (F‘ - G'LL(FO)) n B S B, L is one of the following: 1. L = (F‘ - CLL(FO)) F) B for some maximal subfield F0 of F; 2. L = (F‘ - 2CLL(2)) O B where F = Fp for some odd prime p. Therefore, if L is a maximal subloop of B then L is one of the following: 57 1. L = (Si). 0 B for some hexagon line S; 2. L : (Q U Qt). 0 B for some quaternion subalgebra Q; 3. L = F‘ ~ CLL(FO) H B for some maximal subfield F0 of F; 4. L = {x E B|q(x) E C} for some maximal subgroup, C, of {q(x)|x E B}; 5. L : F" - 2CLL(2) D B where F = F1,, for some odd prime p. Corollary 8.2. If B = SLL(F) then L is one of the following: 1. L = (Si): O SLL(F) = SL2(F) o F2 for some hexagon line S; 2. L = (QUQLY flSLL(F) = H UxH where H = Q" “=’ SL2(F) for some quaternion subalgebra Q; s. L = F‘ -GLL(F,,) n SLL(F) .—_ SLL(FO) if F is of characteristic two or dimpoF is odd SLL(FO)2 if F is of odd characteristc for some maximal proper subfield F0 of F; 4. L = F“ - 2CLL(2) fl SLL(F) = 2SLL(2) where F is of odd characteristic. Proof. Parts (1) and (2) follow from Theorem 1.5. If L1 : CLL(FO) S CLL(F) then there is agroup homomorphism q : CLL(F) ———r F‘ such that q(Ll) = F; and q(kx) = k2q(x) for all k E F. Let k be an element of F and .r be in 1., such that kx is in SLL(F). So, since k2q(.r) = q(k.r) = 1, we have 58 that k2 = q(x)“1 E F0. Case 1. If F is of characteristic two then k E F0 and F ‘ - CLL(FO) fl SLL(F) = F; ~ CLL(FO) fl SLL(F) = SLL(FO). Case 2. If F is of odd characteristic then k E H S F ‘ where F; S H is of index two. Thus F‘ - CLL(FO) fl SLL(F) = H - CLL(FO) fl SLL(FO) = 2SLL(FO). This concludes part (3). Now suppose L1 = 2CLL(2) S CLL(F) where F is of odd characteristic. Since Ll/{iI} is simple and q : L/{ztI} —> F is a homomorphism, ker(q|L,/{i,}) = Ll/{ztI}. Thus, if k E F and x E L1 such that ha: E SLL(F) then k2 = k2q(x) = q(kx) = 1. Hence, F‘ - 2GLL(2) n SLL(F) = ZGLL(2) n SLL(F) = 2SLL(2). 1:1 9' Lagrange’s Theorem for Moufang Loops Proof. Assume that the theorem is not true and let L be a Moufang loop of minimum order such that the Lagrange property does not hold. It was proven by Bruck [4, p. 92] that if H is a normal subloop of L such that the Lagrange property holds for both H and L / H then Lagrange’s property also holds for L. Thus, by minimality of L, L does not have any normal sublOOps. Since Lagrange’s Theorem is true for all finite groups, L has to be a finite nonassociative simple Moufang loop. Liebeck [l7] classified such loops as Paige loops, P(q) = PSLL(q). By minimality of L = P(q), there exists a maximal subloop M S L = P(q) such that I111] does not divide |L|. By Corollary 8.2, there are five possibilities for AI. 59 Case 1. For M E PSL2(q) 0 F3, - (13((12-1) (13( “-1) _ IMI ‘7 gcd((1+l.2) gcd(?1+l.2) _ [P(q)] Case 2. For M = H UxH where H E PSL2(q), x E Hi, and q(x) = 1, 2_ 4_ lMl_ 2(1((1 1) (13((1 1) =|P(q)| _ ycd((1+l.2) 90d(q+1.2) Case 3. For M = SLL(qo)/{:l:I} where qu S F, ._ 3( 3*11 (13(4-1) _ W l - 92113.12) series) - '1’ W Case 4. For M = 2SLL(qO) / {i1} where qu < F and is of odd characteristic, |M| = 21.11.1111 I M = |p(q)] 9cd((1+1.2) Case 5. For M = 2SLL(2)/{i1} where F is of odd characteristic, _ 2-120 3("-1) __ IMI — 2 I ,1..,:..,., — |P((1)| Hence, by contradiction, there does not exist a Moufang loop of minimal order such that the Lagrange property fails. Cl 60 References [1] M. Aschbacher, On finite groups generated by odd transpositions. 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