Z V O N I M I R J A N K O A N D T R A N V A N T R U N G

T H E F U L L C O L L I N E A T I O N G R O U P O F A N Y

P R O J E C T I V E P L A N E O F O R D E R 12 IS A { 2 , 3 } - G R O U P

We prove here the following result.

MAIN T H E O R E M . The full collineation 9roup of any projective plane of order 12 is a {2, 3}-group.

We shall prove this major result in a sequence of theorems.

T H E O R E M 1. A projective plane of order 12 has no homology of order 11. Proof. Let P be a projective plane of order 12 which possesses a homology

p of order 11. Let O be the center o f p and let o = {1, 2,...,13} be the axis of p, where 1, 2,...,13 are the points of o (incident with o). Then the 13 lines Pi = Oi, i = 1, 2,...,13 through the center O are p-invariant. Let 11 be any fixed line through l which is different from o and Pl" Set 11 c~pi = {io} for i = 2 , 3 , . . . , 1 3 , 12=2 ,3o , 12c~pl={lo} and i p.J = t j + 1 " for i = 1 , 2 , . . . , 1 3 and j = 0, 1, 2 . . . . . 10, where 10 + 1 = 0 (and so the indices j are to be considered as integers mod 11).

Since ]12 ~Pkl = 1 for k = 4, 5 . . . . . 13, I 2 # o and 0¢12, it follows that all orbit numbers 4, 5 . . . . . 13 appear in 12 ,i.e., we have"

12 = {2, lo, 3o, 4a, 5b, 6c,7a, 8e, 9i ' 10g, 11 h, 12 i, 13j},

where {a, b . . . . . j} = {1, 2 . . . . . 10}. This last fact follows from l l zc~l~ ' l = 1, for s = 0, 1 . . . . ,10. Since the orbit numbers 4, 5 . . . . . 13 are still 'free', we may permute them so that we may set:

l 2 = {2, 1o, 30,41,54, 6s, 79, 83,92, l0 s , 11 lo, 127,136}'

Similarly, for any line l~ = loi , 3 ~< i ~< 13, we have:

l~ = {i, lo, 20, 3 b . . . . , (i - 1)e, (i + l )f , . . . , 12j, 13k} ,

where {a, b, c, . . . , k} = {0, 1 . . . . . 10}. In particular, we have:

13 = {3, lo, 2~, 4b, 5c, 6a, 7e, 8y, 99, 10 h, 11i, 12i, 13k},

where {a, b, . . . , k} = {0, 1, . . . , 10} which gives:

(1) a + b + c + d + e + f + g + h + i + j + k = O ( m o d 11).

We have further 113 ~ lg~l = 1 for k = 0, 1 . . . . . 10 if and only if:

(2) b - 1, c - 4 , d - 5, e - 9 , f - 3, g - 2, h - 8 , i - 1 0 , j - 7, k - 6

are all ~ 0 (mod 11) and are all pairwise different (mod 11). Summing up in (2) we get:

(3) b + c + d + e + f + g + h + i + j + k = - O ( m o d 11)

Geometriae Dedicata 12 (1982) 101-110. 0046-5755/82/0121-0101 $01.50. Copyright ~) 1982 by D. Reidel Publishing Co., Dordrecht, Holland, and Boston, U.S.A.

102 Z V O N I M I R J A N K O A N D T R A N V A N T R U N G

and so we have:

(4) a - 0 (mod 11)

and

(5) {b,c ,d ,e , f ,g ,h , i , j ,k}={1,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 ,10} .

Summing up the squares of numbers (5) and (2) gives:

b2 +c2 +d2 +e2 + f2 +g2 +h2 +i2 +j2 +k2=

= 12 +22 + . . . + 10 2 ~ 0 ,

(b - 1) 2 + (c - 4 ) 2 + (d - 5) 2 + (e - 9 ) 2 + ( f - 3) 2 + (# - - 2) 2

+ (h - 8) 2 + (i - 10) 2 + (j - 7) 2 + (k - 6) 2 = 0 (mod 11).

The above two relations give at once:

- 2 b - 8 c - 1 0 d - 1 8 e - 6 f - 4 # - 1 6 h - 2 0 i - 14 j - 1 2 k - 0

and this last relation together with (3) gives:

(6) g - 6b + 7c + 5e + 3 f + h + 9i + 8j + 4k (mod 11),

(7) d-- 4b+ 3 c + 5 e + 7 f + 9 h + i + 2 j + 6k (rood 11).

These two relations (6) and (7) are used in determining all possibilities for 13 with the help of a computer. But we can still restrict 'the explosion for 13' with the following method.

We define the following two permutations:

c~ = (0)(1,2,4, 8,5, 10,9, 7, 3,6),

where ~ acts on the set of indices {0, 1, 2 . . . . . 10} and

~* = (0)(1)(2)(3)(4, 9, 5, 10, 6, 11, 7, 12, 8, 13),

where here ~* acts on the set of orbit numbers {0, 1,2, . . . , 13}. We define now a permutation n on all points of P as follows:

O" = 0 , i " = i ~e f o r i = l , 2 . . . . . 13 and (iy=(i~*)j~ f o r i = l , 2 . . . . . 13 and j = 0 , 1 . . . . ,10.

We see that I~ = 12 and also the permutation n fixes the set of all previously constructed lines:

{O'Pl'P2 .... P13 11 ,Iv, IP'° . . . . . . . , ,I 2 , l~, . . . , l~l°}.

It follows that in the set S of all possibilities for 13 we may choose only one representative from each <rt>-orbit, where S is obviously 7t-invariant. In this way we get only 363 possibilities S 1 , S 2 . . . . . $363 for 13. For example the case $1 is:

l 3 = {3, lo, 20,42,58,61o, 7 7 , 8 6 , 9 4 , 1 0 5 , 119,123, 131 }.

C O L L I N E A T I O N G R O U P OF O R D E R 12 P R O J E C T I V E PLANES 103

F r o m these 363 possibilities for 13 in only 45 cases we get (at most two) possibilities for a consistent 14. But then in any case some /~'s, 12/> i/> 4, (which are consistent with previously constructed lines) do not exist• This contradic t ion completes the p roo f of Theo rem I.

T H E O R E M 2. A projective plane of order 12 has no collineation of order 11. Proof. By Theo rem 1, we may assume that a projective plane P of order

12 has a coll ineation p of order 11 which is not a homology. Then p fixes exactly one triangle in P. Let 00~, 002, 00(33 be the three points in P fixed by

p. Then l - - o o 2 0 0 a , l o o 2 - - 0 0 1 0 0 3 and I = 00~002 are the three lines fixed by p.

Let 1 o be any point on l which is not fixed by p. Set (lo)P~= lk, k = 0 , 1 , . . . , 10, so tha t :

I~1 = { 0 0 2 , 0 0 3 , 10, 1 1 . . . . . 11o}•

Let 11 = : ~ 1 o and denote the other 11 points on 11 with 40,5 0 . . . . . 14 o. Also set i k = (lO) for all i = 4, 5, 14 and k = 0, 1, 10. Here is the list of all 11 lines th rough 00 1 which are not fixed by p:

11 = {001, l o , 4 o , 5 o . . . . . 14o}

l~ = {001,11,41,51 . . . . . 141}

lp,O={ool 110 41o,51o, 14~o)•

Set 12 ---- 00220, where 2 0 is a point on Io~ 2 which is not fixed by p. We set also 2 k ---- (2o) pk, k = 0, l, 2 . . . . . 10. The line l 2 intersects each of the 1 1 lines pk • • •

11 , k = 0, 1, 10 in exactly one point which has to be one of the points 4i ,5 i . . . . . 1 4 i , i = 0 , 1 . . . . . 10. Since 12c~1P2 k = {002} for all k - 1,2 . . . . . 10, it follows that all 11 different orbit numbers 4, 5, . . . , 14 and also all 11 different indices 0, 1, . . . , 10 must appear in the list of points of 12• Hence we get:

l 2 = {002,2o, 4a, 5b, 6c, 7a, 8 , 9S, 10g, l la , 12i, 13i, 14k},

where {a, b . . . . . k} = {0, 1 . . . . . 10}. Finally, set 13 = 00330, where 30 is a point on 1o 3 which is not fixed by p.

We set also 3 k = (3o)P", k = 0, 1 . . . . ,10. Since l a must intersect each of the lines pk , • , ,

11 , k = 0, 1, 10 exactly once, it follows as above:

l 3 = {003,30, 4t, 5 ,6 , ,7o ,8p ,9q , 10r, l l s , 12t, 13 ,14v},

where {l, m . . . . . v} = {0, 1, •. . , 10}. Let ~ be a permuta t ion of all points of P given as follows: ~ fixes each of

the points ool , 002, °°3 and (ij) ~ = iff for all 'orbi t numbers ' i = 1, 2, . . . , 14 a n d j ~ is given by the following action on the set of indices O, 1 . . . . . 10,

~z = (0)(1, 2, 4, 8, 5, 10, 9, 7, 3, 6).

104 Z V O N I M I R J A N K O A N D T R A N V A N T R U N G

We see that p~ = ~- lp~ = t92 and so ~ normalizes ( p ) . Any line which does not pass through any of the three points ooi, °°2, 0%

is mapped by ( p > onto one of the line lo2~, i = 0, 1, . . . , 10. Since there are exactly 11 (p>-orbi t s of lines in P which do not pass through any of the points ool, °°2, °°3, it follows that the 11 lines 14+i= l o 2 ~ , i = 0 , 1 , . . . , 10 lie in pairwise distinct (p ) -o rb i t s . It remains to determine these lines 14, 15,...,114.

We have 14+ ~c~ I o 3 = {3v} , where z t (t = 0, 1 . . . . . 10) is a permutat ion defined on the set {0, 1 . . . . . 10}. Since [14+inl~kl = 1 for all k = 0 , 1, ..., 10, it follows that for 0 ~< i ~< 10:

r ¢ t t

14+ i = { lo, 2i, 3v, 11,22,33, . . . , 1 0 1 0 } ,

where 1', 2', . . . . 10' are (not necessarily all pairwise different) orbit numbers from the set {4, 5 . . . . . 14}.

Since I/4+ic~(14+fl= 1, k = 1,2 , . . . ,10, we see that we have for the orbit numbers 1', 2', . . . . 10', one of the following two possibilities:

( typelO) 14+i={ lo ,2 i , 3 i , 1' 3' 3' 4' 4' ' , .,l'b,2'c,2'a, e ' f ' . ' h ' 5 k ' 5 ' l } o r

(type 7) 14+i {10,2i,3i,, 1' ' 1' ' 2' a' Y 4' 5' ' = , ' lb' c '2d . . . . f ' o' h, k,6z},

where this time 1', 2', . . . , 6' are pairwise different orbit numbers from the set {4, 5 . . . . ,14}, {a, b . . . . , l} = {1, 2 . . . . . 10} and in the case of type 10 we have:

{t(a, b), t(c, d), t(e,f) , t(9, h), t(k, l) } = { 1, 2, 3, 4, 5}

and in the case of type 7 we have:

{t(a, b), t(a, c), t(b, c), t(d, e), t ( f , 9)} = {1, 2, 3, 4, 5},

where t(x, y), x ¢ y, x, ye { 1, 2 . . . . . 10} is the uniquely determined number s t { l , 2, 3, 4, 5} such that either x - y - s or y - x -- s (mod 11).

The list of 25 possibilities for the indices {a, b}, {c, d}, { e , f } , {0, h}, {k, l} in the case of type 10 gives us the index list of type 10. Our group ( e ) acting on the set of indices {0, 1 . . . . . 10} permutes those 25 possibilities in five orbits and we choose the following representatives in those five orbits:

Case AI: {1, 2}, {4, 6}, {7, 10}, {5, 9}, {3, 8}; Case a 2 : {1, 2}, {7, 9}, {3, 6}, {4, 8}, {5, 10}; Case A 3 : {2, 3}, {5, 7}, {6, 9}, {8, 1}, {10, 4}; Case A 4 : {2, 3}, {6, 8}, {7, 10}, {1, 5}, {4, 9}; Case As: {5, 6}, 110, 1}, {4, 7}, {9, 2}, {3, 8}.

The list of 520 possibilities for the indices {a, b, c}, {d, e}, {f, 9}, h, k, l in the case of type 7 gives us the index list of type 7. For example, here is one possibility (line) of type 7:

{1, 2, 4}, {3, 7}, {5,10}, 6, 8, 9.

C O L L I N E A T I O N G R O U P OF O R D E R 12 P R O J E C T I V E P L A N E S 105

Let i:/:j, i, je{O, 1, 2, . . . , 10}. Then we investigate the 11 intersections of 14+ i with lP4~+i, k -~ 0, 1, . . . , 10. We have:

14+ i-- {lo,2i ,3i . . . . . }

14+ J = {10,2/,3J . . . . . }

P ' -J= {1 2 i . . . . } 14+j i j ' 3j'+O-j)'

pi ' - j ' 14+ j = {li, j, 2~+(i, j,),3 i . . . . . }

and so three of those 11 intersections are the points 1 o, 2 i, 3 r. We must have obviously that: i - j - ~ i ' - j ' (mod 11), where our indices are regarded

p k . . . , as integers mod ll . Since we need between 14+ i and 14+j, k--- l, 2, l0 another eight intersections, it follows that the scalar product between two rows corresponding to the lines 14+ ~ and 14+ j in the orbit matrix M is 8. Here the orbit matrix M is defined as follows. This is an 11 x l l matrix M = ( m i j ) , i , j = 4 , 5 , . . . , 14, where mij is the multiplicity (non-negative integer) with which the line I~ contains the orbit number j. Hence mij~{O , l, 2, 3} and each row in M is of type 7 or l0 corresponding to the lines I i of type 7 or 10. The sum of coefficients in any row (or column) is 10. In fact each row in M is either of the form:

{2 ,2 ,2 ,2 ,2 ,0 ,0 ,0 ,0 ,0 ,0} (type 10) o r

{3 ,2 ,2 ,1 ,1 ,1 ,0 ,0 ,0 ,0 ,0} (type7),

where of course the coefficients come is any order. The sum of the squares of coefficients in any row is 20. The scalar product of any two distinct rows is 8. We have M ' M ' = 121 + 8J, where I is the identity matrix and J is the 11 × 11 matrix of whose coefficients are 1. We see that det (MM') = 25.31° × × 411 and so M is non-singular (over rationals). We get M M ' M = 12M + + 8JM = 12M + 8MJ and so M ' M = 12I+ 8J. Hence, the sum of the squares of coefficients in every column is 20 and the scalar product of any two distinct columns in 8. This at once gives that also every column in M is either of type 10 or 7.

We turn now to the determination orbit matrices M using all above facts. We shall classify all possible orbit matrices up to the permutations of rows and columns. Permuting the rows in M corresponds to the renaming the lines 14, 15 . . . . . 114 (or permuting them). Permuting the columns in M corres- ponds to permutations of the orbit numbers 4, 5 . . . . . 14. According to our construction, this is allowed since our orbit numbers 4, 5 . . . . . 14 are still 'free'. The result is that there are exactly three such matrices M1, M2, M 3.

Instead of giving these matrices, we shall write the corresponding lines l~ indicating only the orbit numbers 4, 5 . . . . . 14 (without indices).

II II

II II

II II

II II

II II

II ~

II II

II II

IJ II

II II

II It

II ~

II II

II II

II II

II II

II II

II

°.

N

©

Z

Z

©

Z

Z

COLLINEATION GROUP OF ORDER 12 PROJECTIVE PLANES 107

We see that we have at least one line of type 10 in any of the three cases M~, M 2 , M 3. In what follows we shall construct 'the lines' l~ without points 21, 31, (in which l i intersects the lines l~o 2 and 13 ). Hence, for the line l 4 we may take each one of the five possibilities A,, A 2 . . . . . A 5 (which correspond to the previously listed five (~ >-representatives in the index list of type 10). For example, in Case A, we have:

14 = { 1 0 , 4 , , 42 ,54 , 56 ,67 , 6 ,0 , 75 , 79 , 83 , 88},

where the involved orbit numbers may be chosen to be 4, 5 . . . . . 8 (in that order).

The computer runs through the 520 lines of the index list of types 7 and supplies each line of that list with 1 o in front and with the orbit numbers 2', 3', 4', 5', 6', 7' according to the rule:

where (part 1)

or (part 2)

or (part 3)

li= 1 o 2 '2 '2 ' 3'3' 4'4' 5' 6' 7',

2' = O, {Y, 4'} _ S = {4, 5, 6, 7, 8},

5 ' = 6 ' = 7 ' = O :

2'= X~S= {4,5,6, 7,8},3 '=4'=0,

{5', 6', 7'} = {Y, O, O} (any ordering),

YeS\{X} ;

2' = O, {3', 4'} = {X, O} (any order),

XeS = {4, 5, 6, 7, 8}, {5', 6', 7' l = { Y, Z, O} (any order),

{Y,Z}_ S\{X};

so that the condition Ill c~ l~l ~ 1 is satisfied for all k = O, 1 . . . . . I0. Computer then runs through the 25 lines of the index list of type 10 and

supplies each line of that list with 1 o in front and with the orbit numbers 2', 3', 4', 5', 6' according to the rule:

1i=1 o 2'2' 3'3' 4'4' 5'5' 6'6',

where (part 4) any two of the symbols 2', 3', 4', 5', 6' are any two different elements from the set S = {4, 5, 6, 7, 8} and the other three symbols are replaced with 0 so that the condition [1 ic~ l~kl~< 1 is satisfied for all k = = 0 , 1 . . . . . 10.

Of course in all four parts we have to replace later all symbols O with the other orbit numbers 9, 10, 11, 12, 13, 14 in all possible ways. In this way we obtain the complete list of 'lines' I i (without points 21, 3r,) which could be compatible with the given 'start line' 14.

108 Z V O N I M I R J A N K O A N D T R A N V A N T R U N G

We get a complete list of such the orbit matrix is M~ and the lines:

14 -- 15 = 16 = 17 =

18 =

'line' 14, 15 . . . . . 114 only in one case, where start line 14 is in Case As.Here are these

19

/ l o =

111 =

112 =

/13 =

/14 =

{lo, 45, 46, 51o, 51, 64, 67, 79, 71, 8 3, 88}, {10, 95, 96,1010,101, 74, 77, 49, 42,113,118}, {10, 55, 56,1210,121, 44, 47, 99, 92,133,138}, {10, 65, 66, 410, 41,134, 137,149,142,103,108}, {10,115,116,1410,141,124, 127, 89, 82, 43, 48},

= {10,105,106, 810, 81, 94, 97, 5 9 , 52,143,148}, {lo,125,126,111o,111, 54, 57,109,102, 63, 68}, {lo,145,146 , 71o, 71,114, 117,139,132 , 53 , 58}, {lo,135,136 , 61o, 61, 84, 87,119,112, 93, 98}, {10, 62, 69, 75, 76, 91o, 91,144,147,123,128}, {1 o, 73, 78, 85, 86,104,107, 122,129,131,131o},

Applying our automorphism group ( p ) (i.e., adding successively 1 to the indices mod 11) we get from the above 11 lines exactly 121 lines of a partial plane! But unfortunately there is no way of extending those lines with points 2i, 3 v to obtain complete lines of a projective plane of order 12. This contradiction completes the proof of Theorem 2.

T H E O R E M 3. A projective plane of order 12 has no collineation o f order 5. Proof Let P be a projective plane of order 12 which possesses a collinea-

tion p of order 5. Then the fixed structure of p are seven points ~ 1 , 0(32 . . . . . 00 7 which form a subplane Po of order 2 of P. For the seven fixed

! ! t ,

{~1 , ~ 2 , °°3,10, 11,12, 13, 14, 10, 11,12, 13, t4} t , t t !

{ ~ 1, °°4, ~ s , 20, 21,2z, 23,24, 20' 21 ,2 l , 23' 24} , t

f3 = {~1 , ~ 6 , ~ 7 , 30, 31, 32, 33, 3,, 3 o,3 ' 1 , 32 , 33 , 34} f~ = {oo2, 004, oo 6 , 4o, 4, , 42 , 43 , 44, 4 o, 4' 1 , 4' 2 , 4' 3 , 44}

f5 = {°°2, °°5, °c7, 5o, 51, 52, 53, 54, 54, 51, 51, 54, 54} t ,

f6 = { ° ° 3 , ° ° 4 , °©7,6o, 61, 62, 63, 6,, 6o,6i, 62,63,64} t , t t

f7 = {°°3, °°5, °°6,7o, 71,72,7a, 74, 7o, 7,, 72' 73' 74}

where our automorphism p acts on the above points as follows:

i~ ~ = ik, (io) °~ = i' k, k = 0, 1, 2, 3, 4,

for all 'orbit' numbers ie{1,2 . . . . . 7} and i 'e{l ' ,2 ' , . . . . 7'}. Let {jk, k = = 0, 1, 2, 3, 4} be the other 16 (p ) -o rb i t s of points for the orbit numbers

j = 8, 9 . . . . . 22, 23. For the action of p we set again (jo) °~ = Jk, k = O, 1, 2, 3, 4. There are exactly 16 (p ) -o rb i t s of lines in P which do not pass through any

lines we may set:

f l = f2 =

C O L L I N E A T I O N G R O U P O F O R D E R 12 P R O J E C T I V E P L A N E S 109

of the seven points oo 1, 0 0 2 . . . . . 00 7. Let 11,12 . . . . . 116 be representative lines in those 16 (p ) -o rb i t s of lines. In what follows we shall write only points on lines I i with their orbit numbers (without indices). Since li must intersect any one of the lines f l , f2, ' " ,f7 in exactly one point, it follows that l~ contains exactly one of the orbit numbers i or i' for all i = 1, 2 . . . . . 7. There remain six points on l~ which come from the other orbits 8, 9 . . . . . 23.

The fact that [1~ c~ I ~ [ = 1 for all k = 1, 2, 3, 4 gives at once that the other six points on l~ are:

{r,r,s,s,t,u}, where r, s, t, u are pairwise different orbit numbers from the set {8, 9 . . . . ,23}. The fact that [li ~ l~[ = 1 for i @ j and k = 0, 1, 2, 3, 4 gives at once that the scalar product of any two distinct rows in the corresponding orbit matrix M is 5. Here the orbit matrix M is difined similarly as in the proof of Theorem 2. These facts lead to the determination of such 'lines' I r Of course we use the 'free' choice for the orbit numbers whenever we can (without loss of generality). For example in one case we get:

/ 1=1 2 3 4 5 6 7 8 8 9 9 10 11

12=1 2' 3' 4' 5' 6' 7' 8 8 12 12 13 14

13=1' 2 3 4 5' 6' 7' 15 15 16 16 8 17

14=1' 2 3' 4' 5 6 7' 18 18 19 19 8 17

l s = l ' 2' 3' 4' 5' 6' 7 9 9 17 17 20 21

l 6 = 1' 2' 3 4 5 6 7 12 12 20 20 15 18

17=1 2' 3 4 5 6 7' 17 17 22 22 13 14

18=1 2 3 4' 5' 6' 7' 10 10 20 20 19 22

19=1 2 3' 4' 5 6 7 13 13 16 16 20 21.

But there is no way to get all 16 lines I~, 12 . . . . . 116 in this case or in any other case. This contradiction completes the proof of Theorem 3.

Proof of the Main Theorem

Let P be a projective plane of order 12. Let p be an automorphism of P of prime order p. Suppose p > 13. I fp has no fixed point, then p[ 157 ( = 122 + + 12 + 1) and so p = 157 because 157 is a prime. But then p is a Singer cyclus (Abelian) which is not possible by p. 266 of [1]. Suppose that p has exactly one fixed point. Then p[ 156 = 4.3.13 which is a contradiction. Hence, p must have at least two fixed points A and B. Let l = AB be the line joining A and B. Then p fixes I pointwise and so p is a perspectivity. Ifp is a homolo~,y, then p = 11 and if p is an elation, then p[ 12. This is a contradiction in both cases. We have proved that p ~< 13. Suppose that p = 7. Then p has at least three fixed points A, B, O. Let s = AB and since p is not a perspectivity, so p fixes exactly six points on s:A, B, C, D, E, F. But then p fixes at least

110 Z V O N I M I R J A N K O A N D T R A N V A N T R U N G

four points G, H . . . . outside of s. Then A B G is a triangle. The fixed line GC has a fixed point J which is different from G and C. Then ABGJ is a quad- rangle in the fixed structure i f(p) of p. Hence if(p) is a proper subplane of P. Since i f (p) contains at least 10 points, so i f(p) must be a subplane of order 3. But 1 5 7 - 13 = 144 = 122 is not divisible by 7, a contradiction.

The case p = 13 is not possible according to[2]. The cases p = 5 and 11 are not possible by our Theorems 1,2 and 3. Hence, the full collineation group of P is a {2, 3}-group. The Main Theorem is proved.

B I B L I O G R A P H Y

1. Hughes, D. R. and Piper, F. C. : Projective Planes, Springer-Verlag, Heidelberg, 1973. 2. Janko, Z. and van Trung, Tran : 'On Projective Planes of Order 12 with an Automorphism

of Order 13, Part II : Orbit Matrices and Conclusion' , Geom. Dedicata 12, (1982) 87-99.

Authors' address."

Zvonimir Janko and Tran Van Trung, Mathematisches Institut, Universiffit Heidelberg, Im Neuenheimer Feld 288, 6900 Heidelberg 1, W. Germany

(Received May 5, 1980)