GEOFFREY MASON*

S O M E T R A N S L A T I O N P L A N E S O F O R D E R 7 2

W H I C H A D M I T SL2(9 )

1. I N T R O D U C T I O N

In [3], Ostrom considered the possible groups G which can occur in the

translation complement of a translation plane H which is of dimension 2 over its kernel. The two 'largest' possibilities, in case [ G[ and the character- istic p are coprime, can be described as follows: assuming that G = G', we

may have G/Z(G) ~ A6, or G/Q ~- A 5 for a certain normal subgroup Q of G of order 32. The latter case has been studied in [1] and [2], several new

planes being constructed in the second paper. Both in [3] and in conversa- tions with the present author, Ostrom has asked about the existence of

planes of the first type of characteristic distinct from 3. Apparently there are

a paucity of examples, only one of order 25 being known up to the present.

In the present work we construct further examples of order 49. The inter- ested reader might like to compare the present construction with that of

I-2]. There is a strong similarity between the two, leading one to suspect

that they may well share a common geometric origin. Moreover, we see no reason why one cannot construct translation planes of higher order, as in

[2], admitting SL2(9 ). We refrain from such a discussion here.

To describe our results we use the

HYPOTHESIS . H is a translation plane of dimension 2 over its kernel. Moreover, the translation complement has a (necessarily normal) subgroup K such that K/Z(K) ~- A 6 .

T H E O R E M . There are exactly two isomorphism classes of planes FI with kernel GF(7) which satisfy the hypothesis. I f G is the translation complement

of H and D the kernel of I-I, then in one case we have G/D ~- A 6 , while in the second we have G/D ~ E 6 .

Notat ion is standard, and follows that of [2]. In Section 2 we study the group K and its representations, while Sections 3 and 4 are devoted to existence and uniqueness of the planes H in question.

*Supported in part by a grant from the National Science Foundation.

Geometriae Dedicata 17 (1985) 297-305. 0046--5755/85/0173~297501.35. © 1985 by D. Reidel Publishing Company.

298 G E O F F R E Y MASON

2. SOME P R O P E R T I E S OF SL2(9 )

We use the following notation throughout the paper: K = SL2(9 ) = 2-fold cover of A 6 .

G = K ( f ) , wherefinduces the Frobenius automorphism of GF(9). R = RIR 3 is a Sylow 3-subgroup of G, R~, R 3 being subgroups of R of

order 3 which represent the two G-classes of subgroups of order 3.

Ri = (ri) for i -- 1, 3.

LEMMA 2.1. (i) G has a

(ii)

(iii)

The followin9 hold: unique class of cyclic subgroups of order 12, and two classes of

elements of this order. Each such element is conjugate to its inverse. G has a unique class of cyclic subgroups of order 6 NOT contained in K,

and two classes of elements of order 6 NOT contained in K. We may choose notation in such a way that R 1 is contained in a cyclic

subgroup of order 6 not contained in K, and R 3 is contained in a cyclic subgroup of order 12.

LEMMA 2.2. Let H be a 9roup described (up to isomorphism) as follows:

2-Sylows of H are Q16, I H I = 48, and F(H) ~ Q8. Then G has exactly three classes of subgroups isomorphic to H, moreover we can choose representatives HI, H2, and H3 of these three classes in such a way that the following hold:

(i) H1 and H a ~ K, H 2 ~ K. (ii) R 1 ~ H1, R 3 <. H E and H 3. (Here we are using the notation of Lemma

2.1(iii).) Proof To begin with, it is readily checked that K has just two classes of

subgroups of order 48, with representatives Ha, H3, say. Furthermore, we can take Ri ~< Hi for i = 1, 3 and we have H i ~ H. It is thus sufficient to prove that G contains just one more class of subgroups isomorphic to H (necessarily not contained in K), and that Sylow 3-subgroups of this class

are conjugate to R 3 . Now let T be a Sylow 2-subgroup of G, with K c~ T = Q ~ Q 1 6 ; T

contains an involution x such that CQ(x)~ Qs, so T is described by the generators and relations

(x , y, zlX2 = y 8 = Z 4 = Ix, z] = 1, yx = yS, y== y -1)

where (y, z ) = Q. From this we calculate that T has just two subgroups isomorphic to QI6 , namely Q itself and (xy, yz) = QI, say. Note that Q~ n K = (y2, y z ) ~- Q8, so that if K~ = O2(NK(Q1 n K)) then K 1 ~- SL2(3) and KIQ 1 is a group isomorphic to H. It is therefore enough to show that a Sylow 3-subgroup of K1 is conjugate to R3.

S O M E T R A N S L A T I O N P L A N E S O F O R D E R 7 2 299

To see this, note that Cr(Q1 ~ K) = ( x y 3) _~ Z 4. Since this g roup is

necessarily centralized by any Sylow 3-subgroup of K1, the desired asser-

tion follows from the various parts of L e m m a 2.1.

We retain the nota t ion of L e m m a s 2.1 and 2.2 in the following, turning next

to the s tudy of certain representat ions of G. It is readily verified that the

following hold :

L E M M A 2.3. K has exactly two inequivalent, complex, irreducible charac- ters of degree 4, call them 7.1 and Z3. Moreover,

(i) each Zi is rational; (ii) each Xi is faithful;

(iii) /.l and Z3 differ only on 3-singular elements, and we may take

z,(r,) = 1, z,(r/) = - 2 for ! g= j.

L E M M A 2.4. G has exactly four inequivalent complex irreducible characters

of degree 4. We may denote them by Z1, •2, Z3, Z4, where Zi and Zi+a both extend the character Zi of Lemma 2.3 for i = 1, 3. Moreover, the following hold:

(i) Zi and )~i+1 are Galois conjugates for i = 1, 3. (ii) Z1 vanishes on G\K except on the elements of order 6, and we may take

~1(S1) = (.O - - (..0 2.

Here, co is a primitive cube of root of unity and s, e G\K satisfies s~ = rl. (Cf. L e m m a 2.1(iii).)

(iii) ;~a vanishes on G\K except on the elements of order 12, and we may take

Z3(t3) = "C + T - 1 .

Here, z is a primitive twelve root of unity and t 3 ~ G\K satisfies t34 = ra. (Cf. L e m m a 2. l(iii).)

L E M M A 2.5. Let p >>- 7 be a prime and regard the characters Zi as lying in an algebraic closure of GF(p), that is the values of zi lie in this field. Then the

following hold: (i) Zi I K lies in GF(p).

(ii) Z~ lies in GF(p)/f , and only if, p - l (mod 3). (iii) Z3 lies in GF(p)/ f , and only if, p =_ +_ l (mod 12). Proof (i) follows f rom L e m m a 2.3. As for (ii), the only irrational character

value of Zx is ~o - 00 2 (or its conjugate) by L e m m a 2.4(ii), and if e = ~o - co 2

300 G E O F F R E Y M A S O N

then ~2 = _ 3. So Xl lies in GP(p) if, and only if, - 3 is a quadratic residue

(rood p), so that (ii) holds. Finally, (iii) follows similarly using Lemma 2.4(iii).

We now specialize to the case p = 7. So let V be 4-dimensional GF(7)-

space. After Lemma 2.5 we may take G ~< GL(V), and we may take V to be the GF(7)G-module which affords the character •1- After Lemmas 2.3 and 2.4 we get

LEMMA 2.6. The following hold:

(i) Cv(r x) is 2-dimensional.

(ii) Cv(Sl) is 1-dimensional.

(iii) C~(r3) has 2-Sylow of type Q8 and Cv(r3) = O.

Finally we note the following property of K :

L E M M A 2.7. Let x e K and let a be either r I or r 3 . Then (a, a ~) ~- Z 3,

Z 3 x Z3, SL2(3) or SL2(5 ).

3. T H E SPREADS

We let V be as above, retaining other notation from Section 2 when appro- priate. Set

5Po = {Cv(Ro) IR o is a K-conjugate of R1}.

Note that 500 consists of 2-spaces by Lemma 2.6(i).

LEMMA 3.1. 5Co is a partial spread with 20 components.

Proof Since I K : Nr(ROI = 20, the second part of the assertion is cer- tainly true. Now choose S, T ~ 6go with A, B being K-conjugates of R1 which centralize S, T respectively. If S ~ T ~ 0, then (A, B) centralizes S c~ T, so (A, B) has odd order since the only involution of K inverts V.

By Lemma 2.7 we get (A, B) - Z 3 or Z 3 x Z 3. But a 3-Sylow of K cen- tralizes only 0, so (A, B) ~ Z 3 . Thus A = B and S = T, and the lemma

follows.

Next, let H 2 and H 3 be as in Lemma 2.2. From the proof of that lemma it is clear that we may take H2 c~ H 3 = J , say, with J ~ SL2(3). In this notation

we next prove

L E M M A 3.2. The following hold:

(i) As H Fmodule there is a decomposition

v = V l j ® v2j ( j = 2, 3)

S O M E T R A N S L A T I O N P L A N E S OF O R D E R 72 301

where each V~j is an irreducible, 2-dimensional GF(7)H imodule. More-

over, Vlj ;g Vzi. (ii) H j fixes just two 2-dimensional subspaces of V, namely Vl j and Vz j . Proof First note that a 3-Sylow subgroup of J may be taken to be R 3 by

Lemma 2.2. After Lemma 2.6(iii) we know that R a fixes no nonzero vectors of V. Thus either V is irreducible as Hi-module or V is a direct sum of two

2-dimensional Hi-modules. Now the character table of Hj is readily computed, and we find that Hj

has a unique 4-dimensional complex irreducible, say X. Moreover, x(r3) = 1, so a 3-element of Hj has nontrivial fixed-points on the corresponding module. After our earlier comments it follows that V is not absolutely irreducible as Hfmodule . But the two faithful, irreducible 2-dimensional representations of Hj are Galois conjugates of each other and can be written in GF(7); it follows that V has a decomposition as indicated in (i).

Next, the character X~ vanishes on elements of G of order 8. From this it is readily verified that as Himodule , V is the sum of the two irreducible, faithful Hfmodules. In particular V,j ;~ V2j.

Finally, (ii) is an immediate consequence of V,a ~ V2j, and all parts of the lemma hold.

LEMMA 3.3. J fixes just eight 2-spaces in V. Proof First note that J fixes each V~j, and by a standard argument either

the lemma holds or J fixes just two 2-spaces. In the latter case we would have V12 = V13, say, so (H2, Ha) fixes V12. However, we see that (H2, Ha) = N~(J) has Fitting subgroup isomorphic to Q8 * Z4 (cf. Lemma 2.4(iii)) and hence cannot act faithfully on Vt2. This contradiction proves the lemma.

LEMMA 3.4. Let J be the eight 2-spaces invariant under J. Then (H2, H3) = N~(J) acts on J as follows:

(i) an orbit {Vl2, V22}; (ii) an orbit {V13, V23};

(iii) an orbit of length 4. Proof We have already seen that (H2, H3) fixes no 2-spaces in V, so

each orbit on J has length 2 or 4. In fact, H2, H3 and a subgroup iso- morphic to SL2(3) , Z4 are the only subgroups of (H2, Ha) of index 2 and again the latter group cannot on a 2-space in V. Now the present lemma follows from Lemma 3.2.

LEMMA 3.5. The following hold: (i) The elements of V12 and V22 a r e covered by 5o0 .

302 G E O F F R E Y M A S O N

(ii) I f W is a J-invariant 2-space in V distinct from 1/12 and V22, then

W ~ S = O f o r a l I S e 5 0 0 . Proof. Again take R 3 ~< J, and pick a primitive cube root of unity co. Let

E be the ~o-eigenspace of r 3 acting on V. We have dim E -- 2 by L e m m a 2.6(iii), in part icular there are just eight R3-invariant 1-spaces on which r 3 acts with eigenvalue co. After L e m m a 3.3 it is clear that each J - invar ian t

2-space contains a unique such ~o-eigenspace for r 3 .

N o w R 3 fixes just two elements of 500, namely the fixed-spaces of R 1 and R~, where R~ is the (unique) K-conjugate of R 1 in R and distinct from R 1.

After the previous pa rag raph it follows that there are just two J- invar iant

2-spaces W satisfying W ~ S ~ 0 for some S e 500. Since J preserves 500 it follows that 50o covers W for such a W. Moreover , it follows from L e m m a

3.4 that the two J - invar ian t 2-spaces in question comprise an ( H z, H3)-o rb i t in f . It thus suffices to show that it is the orbit (i) of L e m m a 3.4

which meets 50o. If false, then we get 1/13 ~ S ~ 0 for some S ~ 50o . In fact the first para-

graph shows that we m a y take S to be R-invariant , with 1/13 c~ S the

co-eigenspace of r 3 in both 1/"13 and S. However , R fixes just two componen t s

of 50o, so as NG(R)/CG(R) ~- Z4 acts on these two componen t s then NK(R ) fixes each component . Since H 3 ~< K fixes 1/'13 then it follows that Nn3(R3) fixes bo th 1113 and S. As Nn3(R)/CR3(R) ~ Z 2 then Nn3(R ) is irreducible on

V13, so V13 = S. But then S admits ( H 3 , R ) = K, an absurdity. The l emma is proved.

We are now ready to define the spreads. We set

503 = {G-conjugates of 1713 }

504 = {K-conjugates of W, W some fixed element of the ( H 2 ,

H3)-o rb i t of length 4 in J (cf. L e m m a 3.4)}

P R O P O S I T I O N 3.6. 50o u 503 and 50o u 504 are spreads in V. Proof As a consequence of Lemmas 3.5 and 3.4 we certainly have

S c~ T = 0 for S e 50o, T e 50i (i = 3 or 4). Moreover , 15011 = 30, so it

suffices to show that 50i is a partial spread. So assume that there is T s 5 0 1 and 9 e G so that T c~ T o = D is a

1-space (if we are dealing with 504 then 9 ~ K). N o w we may take J to

normalize T and R 3 to normal ize D. N o w T contains both D and D °-x. We claim that without loss we may

take D °-~ to be R3-invariant. If i = 3 then the stabilizer of T in G is transitive on 1-spaces of T, so D ~ = D o- ~ for some t e stabs(T). So in this case replacing 9 by t9 does the trick. If i = 4 we do not have transit ivity on

SOME T R A N S L A T I O N P L A N E S OF O R D E R 7 z 303

1-spaces of T, but there is t ~ stabG(T) such that D o it-1 admits R3. So

again replace g by tg. Operat ing under the assumption that R 3 acts on D O 1, we see that

(R3 , R~) acts on D. By Lemma 2.7 we get R3 = R~, that is g E N(R3).

Suppose first that D = D o 1. Then (R 3, g ) acts on D, which forces

g e ( + R 3 ) , whence T = T °, contradiction. So D ~ D °-1. Note also that if

i = 3 we may always assume D = D 0-1 by a previous argument, so in fact

we are now in the case i = 4. N o w g e NK(R3). If g has order 4 then g 2 = - 1 , so g normalizes T =

(D, D9-1), an impossibility since Nj(R3) = ( + R 3 ) . So g has order 3 or 6,

so g centralizes R3. But then r 3 acts as a scalar on T = (D, Dg-1), again a

contradiction. The proposi t ion is proved.

R E M A R K . It is clear that G acts on Yo w 5"3 but not on 5% u Y4 . This

implies the existence part of the theorem of the Introduct ion.

4. SL2(9 ) AS A C O L L I N E A T I O N G R O U P

In this final section we assume the following: H is a translation plane of

dimension 2 over its kernel, and the translation complement has a section

isomorphic to A 6. Let the kernel be GF(q), (q, 30) = 1, with V the corre-

sponding 4-dimensional GF(q)-space and 5 p the corresponding spread.

L E M M A 4.1. The translation complement has a subgroup isomorphic to K

( - SL2(9)). Proof. By hypothesis there is a section isomorphic to A 6 , so there is a

subgroup isomorphic to Ko for some covering group K o of A 6 . AS we

remarked in Section 2, A 6 has no faithful 4-dimensional GF(q)-module, so

K o ;~ A 6 . If 3 [] Z(Ko)[ then 3-Sylow subgroups of K o are extra-special of

order 27. Again these groups have no 4-dimensional representation in

which Z(Ko) is a nonidenti ty scalar, so this case cannot occur. As [ Z(Ko) [

divides 6, the lemma follows.

In the following we again use K to denote the subgroup of Lemma 4.1. In

fact K is the commuta to r subgroup of the translation complement.

L E M M A 4.2. K contains exactly 20 subgroups of order 3 which induce homologies.

Proof. Let R be a 3-Sylow subgroup of K, so that R _-__ Z 3 x Z 3 . Since [5~1 - 2(mod 3) there is certainly an R-invariant component , say S. Of

course, S is a 2-dimensional GF(q)R-module, so if R is not faithful on S we

certainly can conclude that R contains a homology with axis S.

304 GEOFFREY MASON

We claim that in fact R is not faithful on S. If q = 2(mod 3) then Sylow

3-subgroups of GL2(q) are cyclic and the result is clear. If q = l(mod 3) we

can choose a subgroup R o of order 3 in the kernel of 17, so that R o R is elementary abelian of order 27. Again this ensures that R 0 R is not faithful on S, and as the group of homologies with axis S is cyclic, then the kernel H of the action of R o R on S has order 3.

Next, H fixes just two components of 50, so the same is true o fR o R. Now

a 2-Sylow subgroup of N~(R) is cyclic of order 8 and permutes the two R o R-invariant components, so each element of NK(R ) of order 4 fixes S.

Thus if y is such an element of order 4 then H A ( H , y). Since y inverts R

and centralizes R o then either H = Ro or H ~< R. As Ro is fixed-point-free on V we conclude that H ~< R and indeed R is not faithful on S.

Finally, H has just 20 conjugates in K and the lemma follows easily.

Proof of Theorem. Existence was established in the previous section. As for uniqueness, let 50 be the corresponding spread, so that 150[ = 50. Let 50o be the 20 components corresponding to the homologies of order 3

guaranteed by Lemma 4.1. Thus our group K acts on the remaining 30 components, call them 501.

First assume that K is not transitive on 50~. It is then immediate that K

has just two orbits on 501, each of length 15, each with isotropy group of order 48. Moreover, a Sylow 3-subgroup of such an isotropy group does not induce homologies. Now apply Lemma 2.2 to see that the isotropy

group in question is a K-conjugate of H3, so that in this case 501 is the partial spread 503 of the previous section.

Finally, assume K is transitive on 501, with an isotropy group isomorphic

to SL2(3). Again this must be K-conjugate to the group J of Section 3, so

501 corresponds to 504 in this case. The theorem is proved.

REMARK. It follows from Lemma 2.6(ii) that involutions in G\K induce

Baer involutions on the spread 500 w 503.

REFERENCES

1. Mason, G., 'Orthogonal Geometries over GF(2) and Actions of Extra-Special 2-groups on Translation Planes', Eur. d. Comb. 4 (1983), 347-357.

2. Mason, G. and Ostrom, T., 'Some Translation Planes of Order pZ and of Extra-Special Type', Geom. Dedicata 17 (1985), 307-322.

3. Ostrom, T., 'Collineation Groups whose Order is Prime to the Characteristic', Math. Z. 156 (1977), 56-71.

SOME T R A N S L A T I O N P L A N E S OF O R D E R 7 2 305

(Received February 3, 1984)

Author's address:

G. Mason, Dept. of Mathematics, University of California, Santa Cruz, CA 95064, U.S.A.