By STEPHEN D. COHEN and MICHAEL J. GANLEY. [Received 26 November 1982; in revised form 28 June 1983]. 1. Introduction. SEVERAL authors (e.g. [1], [5] ...
S O M E CLASSES OF T R A N S L A T I O N P L A N E S By STEPHEN D. COHEN and MICHAEL J. GANLEY [Received 26 November 1982; in revised form 28 June 1983]
1. Introduction SEVERAL authors (e.g. [1], [5], [6], [7], [8], [9], [10], [11]) have considered problems of the following type. Suppose TT is a translation plane, 770 a subplane of TT and suppose that TT admits a group G of collineations which fixes TT0 (as a set) and acts transitively on the points of L\(LnTr 0 ). What then can be said about v, TT0 or G? Particular questions that have been settled are when G fixes TT0 pointwise (then TT is a Generalized Hall Plane) or when TT0 is the kern subplane and G fixes a quadrangle of reference inside TT0 (then TT is a Hall plane). Following on from this it is a natural question to ask what happens if G fixes a triangle of reference in TT0, SO that G acts as an autotopism group of TT. We consider the particular case when TT0 is the kern subplane. The main object of this paper is to exhibit two infinite families of planes with the above properties, which are not Hall planes and which appear to be previously unknown. We also exhibit a family of planes, of the above type, which are derived from semifield planes. Jha [7] has recently considered a problem similar to the one mentioned above. He supposed that TT was a translation plane of order p r (p prime, r 5= 2) which admitted an autotopism group G with the properties (i) G leaves invariant a set A of p +1 points of L, (ii) G acts transitively on the points of L\A and (iii) G has at most one Sylow p-subgroup. He called such a plane A-transitive and showed that such a plane had order p 2 . Hall planes of order p 2 are A-transitive and in his paper Jha suggested that there were probably no other A-transitive planes. In the light of [7] we shall assume, for most of this paper, that we are dealing with a translation plane TT of order p 2 (p an odd prime) and that TT admits an autotopism group G having an orbit of length p2-p on L. Theorem 2.1 then shows that we may assume that G fixes the kern subplane, so that we then have the situation discussed earlier in the introduction. After Theorem 2.1 we make one further assumption, namely that G acts faithfully on the points of /„. Examples of A-transitive planes are given in section 6 and in section 7 we show how to extend the ideas to planes of order p 2 ', r 5* 2. We would like to thank Vikram Jha for several helpful discussions during the preparation of this paper. Quart. J. Math. Oxford (2), 35 (1984), 101-113
102
STEPHEN D. COHEN AND MICHAEL J. GANLEY
2. The group G 2 THEOREM 2.1. Let -n be a translation plane of order p (p an odd prime) and suppose that TT admits an autotopism group G, fixing 2 points on !„ and having an orbit of length p2—p on („. Then either TT is a Hall plane or else G leaves invariant a subplane of order p. Proof. Since p2—p ||G|, G contains p-elements. Obviously p-elements in G are planar and so must be Baer p-elements. In particular Sylow p-subgroups of G leave invariant Baer subplanes. So, either there is a single such subplane, left invariant by G, or else there are subplanes Ti, 1J2 (of order p), having TTJ ^ n2, each of which is fixed by Baer p-elements. We consider the second possibility. Clearly, we may assume pj= 3, since the only translation planes of order 9 are either Hall or desarguesian (and desarguesian planes do not admit Baer p-elements). So, by Foulser [2, Corollary 4.2], G must contain SL(2, p). But then we have that we can apply results of Walker [12] concerning translation planes of order q2, having GF(q) inside the kern, and admitting SL(2, q) in the translation complement. In the terminology of [12], the p-elements in SL(2, p) are quadratic (this can be easily seen from Lemmas 2.2 and 2.7 of [2D and so by [12, Theorem 2.5] IT must be either a desarguesian plane or a Hall plane. That v cannot be desarguesian has already been observed. This completes the proof of the theorem. As we are not particularly interested in Hall planes we shall make the assumption that G fixes a subplane, v0, of order p. By choosing our quadrangle of reference suitably, we may assume that IT is coordinatised by a quasifield Q and that TT0 is coordinatised by the kern K. Furthermore, G acts transitively on the points of LA(Ln7r 0 ). We make one other assumption, namely that G acts faithfully on L, so that G does not contain any homologies with centre (0,0) and axis L. Remark. We are assuming that Q is a right quasifield, so that Q may be regarded as a 2-dimensional left vector space over K. (Recall that K = {keQ: k(xy) = (kx)y and k(x + y) = kx + ky, Vx, y eQ}). We now give some elementary results about the group G. A typical element g E G may be represented by a triple of mappings g = (A, B, C), where A, B, C are bijections from Q to Q, each fixing 0 e Q. A represents the action of g on the x-axis, B the action on L and C the action on the y-axis. Note that each of A, B, C fixes K (as a set). The following result is standard. 2.2. Let g = (A, B, C) e G, then (i) A, B, C are additive, (ii) (xy)C = (xA)(yB) V x , y e Q ,
LEMMA
SOME CLASSES OF TRANSLATION PLANES
103
(iii) (fcx)B = fc(xB) \/xeQ,keK, (iv) (ikx)C = fc(xC) V x € Q,fc€ K. Note that (iii) and (iv) are true because \K\ = p. In the more general case we would have (fcx)B = k"(xB) and (kx)C = kT(xC), where a, re AutK. DEFINITION. Suppose g = ( A , B , C)eG. Define a,b,ceK l B = b, l C = c. LEMMA
by 1A = a,
2.3. Let g = (A, B, C) e G. Then
(i) kB = kb
)
(ii)fcC= fcc [ VfceK (iii) kA =fca= kcfe-lJ (iv) xC = (b-1c)(xB)
VxeQ.
Proof. The proof follows immediately from Lemma 2.2. As we have already observed, Q is a 2-dimensional left vector space over K. It teQ\K,.then elements of Q can be represented (uniquely) in the form k1 + k2t for some ku k2sK. LEMMA 2.4. Let g = (A, B, C)€ G. Suppose B:t^fi+at, K, a ^= 0. Then (i) B :fcT+ fcjf^^b + (ii) C: k1 + k 2 t ^ ( k 1 c +
where a, & e
Proof. Using Lemmas 2.2 and 2.3 we have (fci + k2t)B = (^B) + (k2t)B = ktb + k2(tB) = ktb + k2(|3 + at) and (kx +fcaOC= (kiC) + (k2r)C = klC + k2(tC) = klC + fc2((b"1c)O + at)) =fcjc+ (fczb-V)^ + at) = (krc + k2b~lcP) + (k2b~1ca)t. LEMMA 2.5.
(i) G contains elements of order p. (ii) Each element of G, of order p, fixes v0 pointwise. (iii) / / g = (A,B, C)eG has order p, then B: t-+@ + t for some /3e K\{0}. (iv) G contains precisely p — 1 elements of order p. Proof, (i) and (ii) have already been noted, in the proof of Theorem 2.1. (iii) From (ii), b = 1. By repeated application of Lemma 2.4(i) we have
104
STEPHEN D. COHEN AND MICHAEL J. GANLEY
B": f->/3(l + a + a 2 + - • • + ap~1) + apt (where B: t-»/3 + at). But B" fixes t, so a p = 1 and hence a = 1. Note that 0 ^ 0 since g has order p. (iv) Since p | |G| there exist 5*p-l elements of order p. If there exist >p — 1 elements of order p, then there exist distinct glt g2 e G, having order p, where g t = (A^ B 1( Q ) , g2 = (A2, B 2 . C2X such that IB, = 1 and tB, = /3 +1 (i = 1, 2) for some /3 e K\{0}. But then BXB^ fixes I. pointwise and this contradicts our assumption that G acts faithfully on L, unless gj = g2. 3 . Hie centralHy of K
In this section we consider whether K is central in Q and although we do not completely prove that K is central in Q until later in the paper, we develop material here which will be required in the next section. The condition of K central in Q is equivalent to saying that v is derivable with respect to the set 7r0nL,. LEMMA 3.1. Let keK then tk=f(k) + kt for some function f: K-+K such thatf(O) =
Proof. Start by writing tk = f(k) + g(k)t, where /, g: K-*K and clearly Let h = (A,B,C)eG be a p-element. By Lemma 2.5(ii), hfixesTT0 pointwise and so a = b = c = l. But then by Lemma 2.2(ii), A=B = C. Thus h = (B, B, B) where IB = 1 and tB = 0 +1 for some 0 € K\{0}. Act on the equation tk =/(fc) + g(fc)f with h, to obtain i.e. pfc So, /3k = g(k)/3. This is true for all 0 e X\{0}, so we must have g(k) = k and the lemma is proved. LEMMA 3.2. Let g = (A, B,C)eG defined in Lemma 3.1, then
and suppose tB =fi+ at. If f is as
f{k) = kbfib'1) +b~2af(kb) V k e K. Proof. We have (tk)C = (f(k) + kt)C = (f(k))C + (kt)C i.e. i.e.
(tA)(kb) = f{k)c + k({b-lc)(tB))
so ((A)(kb)=/(k)c + (W>-1cp) + (kb-1ca)t.
(3.1)
SOME CLASSES OF TRANSLATION PLANES
105
From Lemma 2.2(ii) tA = (tb~l)C =
Hence (tA)(kb) = (fib-^c + b-2cp)(kb) + (ir2ca)(/(W>) + (kb)t) = (fib-^cbk + b~lckp + b~2caf(kb)) + (b-lcka)t.
(3.2)
From (3.1) and (3.2) we have f(b-1)bk + b-2af(kb), as required. We wish to show that in fact K is central in Q. To do this it is clearly sufficient to show that tk = kt (VkeK) or equivalently that f(k) = 0, VIceK We prove this for one particular case here—the remaining cases will be dealt with later. PROPOSITION
3.3. If G does not act regularly on / 0O \(Ln7r 0 ) then K is
central in Q. Proof. If \G\ > p(p — 1) then there are more than p elements of G which fix the point (1) e L= n TT0, (i.e. have b = 1). One of these is the identity and p - 1 of them are p-elements (see Lemma 2.5). Hence there is an element g = (A,B,C)sG such that IB = 1 and tB = /3 + at for some a,PeK with a ^ O and a ^ l . Put this particular value of a (and b = 1) into the equation obtained in Lemma 3.2. This gives f(k)=f(l)k + af(k),
Vfc€K
But /(I) = 0 and a f 1, so f(k) = 0 V k e K. This proves that K is central in Q. We shall now forget the case when G does not act regularly on L \ ( L n i r 0 ) and concentrate only on the case where G does act regularly. We shall prove (eventually) that K is central in this case also. 4. The case of G regular For the remainder of this paper we shall assume that G acts regularly on L\(L,n7r 0 ). Consider the following elements of G: gi = (Ai,B 1 , Cx) where Bx: 1—*bly t—»af and g2 = (A2, B2, Cz) where B 2 : l-»6 2 , t - • /3 + ort.
106
STEPHEN D. COHEN AND MICHAEL J. GANLEY
Now G also contains the p-element g3 = (B3, B 3 , B3), where B 3 : 1—>1 and t -> a"1(3 +1. The action of gig3 on L is 6 ^ 3 and BlB3. 1->•{>! and
t^>P + at.
So BXB3 and B2 have the same effect on t. But G acts regularly on L\(Lri7To) and so B1B3 = B 2 . In particular we must have b^ = b2. Now, clearly, if g = (A, B,C)sG is such that IB = b and tB = ^ + at, then b is determined uniquely by a and 0. However the working above shows that, in fact, b depends only on a. Consider now the elements g4 = (A4, B 4 , C4), g5 = (A5, B 5 , C5) where B,: 1-•*>,, t^-Oit (i=4,5). Then B4B5: l^>b4bs and B 4 B 5 : t-> (a 4 a s )t. If we write b, = b(aj) for i = 4, 5, then we have shown that b(a4)b(a5) = b(a4a5), so that b is a multiplicative mapping from K"\{0} to itself. Let OJ be a generator of K, and suppose 6())r = (w')r = (o>r)*- So, we have proved: LEMMA 4.1. Suppose G acts regularly on l^\(L,r\TT0). Let g = (A,B,C)eG such that B:l->b and t^-p + at. Then b = a' for some (fixed) s in the range 0 « S s < p - l .
A similar result holds for the c's (where 1C = c). With the notation as at the beginning of the section we have that BiB3 = B2, so that the autotopism gig 3 g2' = ( . . . , / , . . . ) and hence, by the assumption that G acts faithfully on L, gigsg^1 is the identity in G, so gig3 = g2. Thus 1C,C 3 = IC2, so that c, = c2. As before this shows that if g = (A, B, C) e G is such that l C = c and B: t—*(i + at, then c depends only on a. The rest of the proof then follows through and we have LEMMA 4.2. // G acts regularly on L\ (L n ir0) and ifg = (A, B,C)eG is such that C: 1 —» c, B: t —> |3 + a£ then c = a' for some (fixed) r in the range 0 ^ r < p — 1.
We now consider the multiplication in Q. Begin by writing t2 = v + ut for some u,veK, where vj=0. Let g = (A, B, C)eG, where lB = b, 1C = c, tB = p+at. Then t2C = (v + ut)C = vc + u(tC) = vc + u(b~jcp + b~lcat) i.e.
(tA)(B + at) = (vc + ub~xcP) + (ub~lca)t.
From the proof of Lemma 3.2 we know that tA=/(b- 1 )c + fe-2cO + al), so cp + (b-2ca)r)Q3 + at) = (vc + ub~
SOME CLASSES OF TRANSLATION PLANES
107
i.e.
((fib'1) + b~2p) + (b~2a)t)(p + at) = (v + ub~xp) + {ub'la)t,
i.e.
(fib'1) + b"2/3)(/3 + at) + (b~2a)(tip + at)) = {v + ub~xp) + (ub~la)t,
i.e.
(fib'1) + b-2p)p + {(Jib'1) + b~2p)a)t + (b~2a)(t(p + at))
i.e.
(b~2a)(t(p + at)) = (v + ub'1^ -
i.e.
(p
)
+ (ub-b2f(b-1)-p)t. Now consider the product (8 + yt)(& + at) = 8p + (8a)t + y(t(& + at)) = (Sj3 + ya-\vb2+
upb -
+ (8a-py + yub-yb2f(b-1))t.
(4.1)
Before writing down the final form for the multiplication in Q we find an expression for fib'1). By Lemma 3.2 we have f(k) = kbf(b-1) + b~2af(kb)
(VfceK)
(4.2)
where we now know that b = a* for some s (see Lemma 4.1). With k = 1 in the above equation we have
So (4.2) becomes f(k) =
a~2'+1f(ka1)-a-2'+1kf(a'),
i.e. a2"1f(k)=f(ka-)-kf(a').
(4.3)
Note that (4.3) is true for all a,keK, with a ^ O . p_t Now f: K^*K and so may be written f(k)= £ /,k' for /„, A , . . . , /p-i €K. Then (4.3) becomes '"°
a23'1 "f ftkl = ' I Uka')1 - kf(a'). j-0
(-0
Equating coefficients of k we have
a2"lfl=f1a--f(a'). Replace a by a" 1 and recall that b = a', to obtain f(b-1) = (a--a~2-+1)U. Combining equations (4.1) and (4.4) gives our next result.
(4.4)
108
STEPHEN D. COHEN AND MICHAEL J. GANLEY
THEOREM 4.3. Provided G acts regularly on L\(Lriir 0 ), multiplication in Q has the following form
+ya-1(va2* + upa' - (a' - a)fS yua'-(a'-a)fiy)t,
provided a^O.
when a =0. (Here / j , u, veK, O « s < p - 1 and the mapping / is as in Lemma 3.1). 5. The centralhy of K (revisited)
Of course simply writing down the multiplication in Theorem 4.3 does not guarantee that Q is a quasifield. We must still check that the various axioms for a quasifield are satisfied. This is a simple matter for all the axioms, except for the one which states that the equation cx = d must have a unique solution x e Q, for all c, deQ with c ^ 0. Consequently we must consider the following equation. t),
(5.1)
where a, /3, 7, 5 eK, 8 + ytj=O. We need that (5.1) has a unique solution (x, y) 6 K2. There are 3 cases to consider. Case (i), 7 = 0. Then (5.1) reduces to 8y + (Sx)t = (i + at, which yields the unique solution (x, y) = (a8~1, P8~l). Case (ii), 7 ^ 0 and y2f(ay~l) ^ py — aS. Firstly observe that in this case there is no solution to (5.1) having x = 0. For if there were, we would have Sy + 7f(y) = 0 and 7y = a, but that would lead to y = ay'1 and Say~l + yfiay'1) = 0, which contradicts our assumption. So, we assume x ^ 0 . Then (5.1) yields 8y+ 7X~1(x2lu+ x ' y u - ( x * - x ) / i y - y 2 ) = ft and a.
(5.2)
Multiply the first of these equations by x and the second by y and then subtract to obtain px
(5.3)
Multiply (5.2) by a and substitute for ay from (5.3) to obtain vx2* + («7"x)(u -h)x' + (ay-'f, + 7 - 2 (8a - yp))x = (a7"1)2Write ay~l = i; and y~2{a8 — fiy) = ri and we have Ux
2
' + (u-/ 1 )4x'+(4f l + -n)x = ^ .
(5.4)
Replace x by x~l and s by p — l — t ( 0 < t « p - l ) then (5.4) may be
SOME CLASSES OF TRANSLATION PLANES
109
written Ux
2t+1
+ ( u - / 1 ) | x ' + 1 - | 2 x = - ( ^ 1 + T,).
(5.5)
We require (5.5) to have a unique solution x^O for every ^, TjeK, provided T J ^ / ( | ) . Write F(x, Z) = vx2'+l + (u-fl)&'+1-£2x, so (5.5) becomes )=-(€f 1 + T?).
(5.6)
We return to (5.6) shortly, but first we consider Case (Hi), Y ^ 0 and y2f(ay~1) = /3-y - a8. This time it is simple to check that (5.1) has a solution (x, y) = (0, a7 - 1 ). We must therefore check that (5.1) has no solution with x^O. Following the working for case (ii) we end up with F(x, £ ) = - ( & + /(£))
(5.7)
where we want that (5.7) has no solution x^O for each £ e K Now let £ be a fixed value, say £ = £0 ar»d let i7o = /(£o)- (5.6) requires that F(x, | 0 ) = — (£o/i + TJ) has a unique solution x ^ 0 for each TJ ^ TJ0 and (5.7) requires that F(x, £0) = ~(£o/i + Vo) has no solution x^O. The upshot of this is that the equation has a unique solution x =£ 0 if and only if T\ ^ TJ0 and so we must have that F(0, fo) = -«ofi + T)O) = -(&/!+/(&))• But obviously F(0, $ 0 ) = 0 and so we have /(£0) = —fi€oIn particular, if we put £0 = 1. w e n a v e / i = 0, so that /(^0) = 0 V £0 £ K. As remarked earlier this means that iC is central in Q. THEOREM
5.4. K is central in Q (or equivalently f(k) = 0 V k e K).
We also have the following THEOREM 5.5. The multiplication for Q exhibited in Theorem 4.3 is a quasifield multiplication if and only if F(x, £) is a permutation polynomial for each £eK, where F(x, £) = ux 2l+1 + u£c t+1 -£ 2 x for some u,ve and some t in the range ( X f S p —1.
6. Examples In view of Theorem 5.5 we need only look for permutation polynomials of the form F(x, £). Permutation polynomials of this form are few and far between.
110
STEPHEN D. COHEN AND MICHAEL J. GANLEY
EXAMPLE
1. Let u = 0, t = —— and let v be a non-square in K. We
have F(x, £) = uxp — £2x which is a permutation polynomial for all £ e K The corresponding value of s is also —— . The multiplication for Q now becomes (8 + -yt)(/3 + at) = (5/3 + ya~\v - /S2)) + (8a - &y)t, provided a f 0. However this is simply the multiplication for a Hall quasifield, so in this case we do not obtain any new planes. The following result shows that there are no other examples having u=0. THEOREM
6.1. When u = 0, F(x, £) is a permutation polynomial for all
£eK=GF(p)
if and only if t=^—- (where 0 < t < p - l ) .
Proof. Suppose the result is false. We can assume K t
2 and u^O, any further example must satisfy \Jp
