STEVE WILSON

A P P L I C A T I O N S A N D R E F I N E M E N T S O F V I N C E ' S

C O N S T R U C T I O N

ABSTRACT. This paper describes Vince's construction of regular maps using linear groups. We explain the details of the construction, and show how it can be modified to produce examples of regular maps having specified properties.

1. INTRODUCTION

Regular maps are a generalization of the regular polyhedra (the Platonic

solids). First studied in the 1880s by Dyck, later by Heffter, Brahana, and, most notably, Coxeter, regular maps have been examined as natural embodiments of reflection groups, as factors of groups of linear fractional transformations of the complex plane, as embodiments of Riemann surfaces, and as combinatorial objects in their own right. We will take our definitions

in this paper from [CM]. In 1974, Griinbaum conjectured that, for every hyperbolic type { p , q } ,

there is a finite regular map of that type [G]. Vince published in 1983 a proof of a generalization of Griinbaum's conjecture [V]. Implicit in Vince's existence proof is a construction, first examined explicitly in [W3].

Stripping away the details gives this as the bare bones of the construction: Any group generated by R and S in which

(*) I R S - ~I = 2

is the group of a regular map of type { p, q } where p = IRI, q = ISI. See [GW]. If a, b, c, A, B, C are elements of any field K of characteristic # 2 satisfying

(1) a z - - bc = 1

(2) A 2 - - B C = 1

(3) b C + c B = 2 a A ,

then the matrices

considered as elements of LF(2, K), satisfy (*), and so generate the rotation group of a regular map. The construction consists of determining

Geometriae Dedicata 48: 231-242, 1993. © 1993 Kluwer Academic Publishers. Printed in the Netherlands.

232 S T E V E W I L S O N

a, b, c, A, B, C in order to get a map of a predetermined type. To do that, we

need to review the relationship between powers of R (and of S) and the Chebyshev polynomials T, and U..

2. CHEBYSHEV POLYNOMIALS AND 2 x 2 MATRICES

In this paper, it will be convenient to use the following modifications of the

Chebyshev polynomials: Let A n ( X ) = 2T,(x/2), B n ( x ) = Un_x(X/2). These functions satisfy the recursions:

(4) An(x ) = x A . _ l(x) - A n_ 2(x)

(5) B. (x ) = x B n_ l(x) - B ._ 2(x)

and the first few polynomials are shown in Table I.

T A B L E I

n A.(x) B.(x)

0 2 0

1 x 1

2 x 2 - 2 x

3 x 3 - 3 x x 2 - 1

4 x 4 - 4 x 2 + 2 x 3 - 2 x

5 x 5 - 5 x 3 + 5 x x a - 3 x 2 + 1

6 x 6 - 6x '* + 9 x z - 2 x 5 - 4 x 3 + 3 x

Now, if M is any 2 x 2 matrix of determinant 1 and trace t, M satisfies its

own characteristic equation; i.e.

(6) M 2 - t M + I = O.

From this it is not hard to show by induction, using the recursion (5), that

(7) M" = Bn( t )M - Bn-x(t)I .

Let n be the smallest integer such that Bn(t ) = 0. Then M n = - B , _ l(t)I,

and since M has determinant 1, so does Mn; it follows that B,_ ~(t) = _+ 1, and so in LF(2, F), M n = I. Conversely. if M" = I, then Bn(t) = 0. Thus, for any M with entries from a finite field F, the order of M in LF(2, F) depends only on the trace t of M. In fact, it is the smallest n such that Bn(t) = 0. This is the number that Rankin [R] call the Chebyshev order of t.

If k divides n then Bk(x) divides Bn(x); this follows from the definition of the Chebyshev polynomials. If we remove from B,(x ) any common factors with

APPLICATIONS OF VINCE'S CONSTRUCTION 233

earlier B's, we produce a reduced polynomia l C,(x). Each element t of a finite

field is a roo t of some Bk; if B, is the first of these, then t is a roo t of C,. The

first few C, 's are shown in Table II.

T A B L E I I

n C.(x)

0 0

1 1

2 x

3 x 2 - 1

4 x 2 - 2

5 x 4 - 3 x z + 1

6 x 2 - 3

7 x 6 - 5 x 4 + 6 x 2 - 1

8 x 4 - 4 x z + 2

9 x 6 - 6 x 4 + 9 x 2 - I

1 0 x '~ - - 5 x 2 + 5

Observe that the C, 's are all monic and that, except for n = 2, they are all even functions; i.e. each one except C z is a po lynomia l in x z. In every case,

then, if e is a roo t of C., then - e is also a root.

3 . T H E C O N S T R U C T I O N

Given p and q, we can construct a regular m a p of type {p, q} by the following

process:

(1) Choose an odd pr ime P. We m a y not choose a p roper factor o fp or q to

be P, but otherwise, our choice is unrestricted.

(2) Find a, A in some extension F of Zp such that 2a is a root of Cp(x) and 2A is a roo t of Cq(x). Then Bp(2a) and Bq(2A) are bo th zero rood P.

(3) Choose any non-zero element of F to be b. (If a 2 = 1, then b may be

chosen to be 0.) (4) With b chosen, solve Equat ions (1), (2), (3) above for c, B, and C. This is

always possible, though the solutions m a y lie in E, a quadrat ic

extension of F. (5) F o r m the matr ices R and S as above. Then R p, by (7), is

Bp(2a)R--Bp_l(2a)I = - B p 1(2a)I. Because d e t ( R ) = 1, this last mat r ix must have de te rminant 1. Thus, By_ a(2a) = _ 1, and so ]R] = p.

Similarly, IS] = q, and so the m a p is of type {p, q}.

234 STEVE WILSON

As an example, let us find a map of type {5, 13}. The smallest prime for

which Cs and C13 have solutions is P = 79 I-D1]. Mod 79, solutions are a - - +15, +_25, and A = _+3, +9, _+17, _+20, +24, +_34.

For each P, the construction gives at least one map of type { p, q }, usually

more. In our example, choosing a = 15, A = 24, b = 1 gives c = - 13, B = 1, C = 22; the quadratic of step 4 can be solved in the base field. In the resulting

group, the order of RS is 13. Changing the choice of A from 24 to 3 gives B = 3 3 + 12i, C = 4 5 - 2i, where i is a root of x 2 + 1 in a quadratic

extension of 7179. In this map, RS has order 40; thus the two choices give non-

isomorphic maps. The group of the first map is LF(2, 79); for the second map, it is a conjugate of LF(2, 79) inside LF(2, 792). We will see later that although

the matrices come from LF(2, E), all the computat ions can be done within F.

Let X be the matrix

°1} Then X 2 = I, X R X = R - 1 and X S X = S - 1. Thus conjugation by X is the

automorphism required by Lemma 2 of [GW] and it follows that the map is reflexible. When X is in ( R , S } , the map is the smooth twofold orientable

cover of a non-orientable map [GW].

4. TRACES AND HOLES

So, not only does there always exist a map of type { p, q }, there are an infinite number of them. We wish to refine the construction to allow us to choose

from among this multitude a map which will fulfill some other requirement.

We have seen that the order of a matrix in the group generated by R and S depends only on its trace. This trace, surprisingly, depends on neither the

choice of b in step 3 nor the choice of sign in solving the quadratic in step 4. Given the product in R and S, the trace depends only on A and a. We will support this statement in a later section on computation, but for now let us

take the matrix RS as an example:

[: a°+b 1 RS = = cA + aC cB + aA

and its trace, by (3) above, is: aA + bC + cB + aA = aA + 2aA + aA = 4aA.

RS in the map is motion one step along what Coxeter calls a 'hole'. See [C1]. In general, if W is a word in R and S, the trace of the corresponding product is

a polynomial in a and A, Pw(a, A), determined by W. We use this property of the trace to refine the construction. As a simple

A P P L I C A T I O N S OF V I N C E ' S C O N S T R U C T I O N 235

example, suppose we want a m a p of type {4, 6} with a specified length of hole. C4(x ) = x 2 - 2 and C6(x ) = x 2 - 3, and we have seen above that the trace of

RS is 4aA. Since 2a is a roo t of C4 and 2A is a roo t of C6, we get that 8a2A 2 = 3. Using Equat ions (5) and (7), and the remarks that follow (7), with

t = the trace = 4aA, we get the polynomials shown in Table III .

T A B L E I I I

n B,,(4aA)

0 0

l 1

2 4aA

3 16a2A 2 - I = 6 - 1 = 5

4 20aA - 4aA = 16aA

5 64a2A 2 - 5 = 2 4 - 5 = 19

6 76aA -- 16aA = 60aA

7 240a2A 2 - 19 = 9 0 - 19 = 71

8 284aA - 60aA = 224aA

9 896a2A 2 - 71 = 336 - 71 = 265

10 1060aA - 224aA = 836aA

11 3344a2A 2 - 265 = 1254 - 265 = 989

etc.

We can see that to force holes to have length 5 we only need choose P = 19,

then Bs(4aA ) would be 0 and RS would have order 5. To get a hole of length 8, choose P - 7 (224 = 7.32). To have holes of length 11, choose P = 23 or

43 (since 989 = 23.43). Since this recursion produces the unreduced B,(4aA), we mus t be careful that our choice of pr imes does not divide some earlier B , ;

for example, at n = 10, B~o(4aA ) = 836aA = 4 -11 .19aA. Choosing P = 11 does give holes of length 10, but P = 19, as we have just seen, gives holes of length 5.

5 . T H E M A P { p , q } r A N D I T S G R O U P G p ' q ' r

In [C2], Coxeter discusses the group G p'q'r, which is the group of the m a p

{p, q},. This m a p is formed f rom the tessellation {p, q} of the hyperbol ic plane by identifying two points if they are at distance r apar t along a Petrie pa th (this is a zig-zag pa th through the edges of the tessellation, and mot ion one step a long such a pa th is the symmet ry T = RS-1X). In 1939, Coxeter found a n u m b e r of the {p, q}r's to be finite. In 1961 one more was found ({3, 7}16). The ones which are known to be finite are listed in Table 8 of [CM] .

The smallest interesting case is the m a p {3, 7},, which is known to collapse

236 STEVE W I L S O N

i f r < 8 o r r = 10, 11 and to be finite and unco l lapsed for r = 8, 9, 12, 13, 14,

15, 16. F o r general r > 16, the result is not known, but for even r, some

progress has been made. In June of 1988, La r ry G r o v e a n d Jane~ M c S h a n e

showed tha t G 3'7'20 is inf ini te , using the compu te r l anguage C A Y L E Y

successfully where ;o thers had failed. [They also showed G a'7~r to be infinite

for r = 18, 24, and 32.] In 1990, [ H P ] gives a p r o o f tha t G 3'7'2k is infinite for

all k t> 9 except the r emote poss ib i l i ty of k = 11. Their p r o o f relies on finding

a finite image of the group, which they do by p e r m u t a t i o n d iagrams. We will

use the cons t ruc t ion to find finite images in mat r ix groups.

F o r example , we will p roduce a finite m a p of type {3, 7}20. If we first ask for

a m a p of type {3,7}, we have C3(2a ) = 4a 2 - 1 = 0 and

C7(2A ) = 64A 6 - 8 0 A 4 + 24A 2 - 1 = 0. Now, m o t i o n two steps a long a Petr ie

pa th is T 2 = R S - 1 R - I S . The t race of this p roduc t is:

t = 4 a 2 - t - 4 A 2 - 2 = 4 A 2 - 1.

To simplify, let K = (2A) z. Then C7(2A ) = K 3 - 5 K 2 + 6 K - 1 and,

app ly ing E q u a t i o n (5) as before, we get

n B~(t)

0 0

1 1

2 K - - 1

3 K 2 - 2 K

4 K 3 -- 3K 2 + K + 1

Because C7(2A ) = K s - 5K 2 + 6K - 1 = 0, this last reduces to

2K 2 - 5K + 2. S imi lar ly reducing at each step, we get:

4 2K 2 - 5K + 2

5 2K 2 -- 3K

6 3K 2 -- 4 K

7 6K 2 -- l l K + 3

8 10K 2 -- 18K + 3

9 16K 2 -- 28K + 4

10 26K 2 -- 46K + 9

Since R S - 1 R - 1 S is m o t i o n two steps a long a Petr ie path , which we want to

have length 20, the o rde r of R S - 1 R - 1 S mus t be 10, and so

Blo(t) = 26K 2 - 46K + 9 mus t be 0. To find a s imul taneous so lu t ion to

A P P L I C A T I O N S OF VINCE'S CONSTRUCTION 237

Blo(t) = 0 and C7(2A ) = 0, find the resultant of these two as polynomials in K. See [D1]. This is the determinant of

1 - 5 6 - 1 0

0 1 - 5 6 - 1

26 - 46 9 0 0

0 26 - 46 9 0

0 0 26 - 4 6 9

This matrix

-1 0 0

0 1 0

0 0 1

0 0 0

0 0 0

integer row-reduces to

0 - 1 6 -

0 8

0 - 4

1 2

0 41

This shows that the determinant is 41, so any common solution must be mod 41. Moreover, reading the fourth row as an equation, we see that K + 2 is 0, i.e. K = -2 (mod41) . This gives A = + 15 (mod41). Meanwhile, the roots of C3(2a ) rood41 are + 20. Solving Equations (1), (2), (3), we find that the matrices

R = 13 20 ' S = 18 15

generate the group of a map M of type {3, 7}2 o.

Since M is a map of type {3, 7}1o, the map {3, 7}2o does not collapse. In fact, the full symmetry group of M, which is G(M) = PGL(2, 41), must be a factor group of the full symmetry group of {3, 7}2o, which is G a'7'/°.

In Table IV we summarize the results of applying the construction to the problem of finding a map of type { p, q }r, where r is an even number from 8 to 30 and {p, q} is one of the types {3, 7}, {4, 5}, {4, 6}, {4, 7}. I fa solution exists, the prime or primes are shown. 'NS' indicates that there is no solution; this happens if the resultant is a power of 2 or if the solutions actually yield a map of smaller type.

An asterisk in Table IV indicates a type for which the map {p, q}r is listed in Table 8 of [CM].

In the examples above, we simplified the algebraic difficulties by choosing to tackle problems in which p = 3 or 4. Since C3(2a ) and C4(2a ) are quadratic, tr(T 2) reduces to a quadratic in A, and so we only needed to consider the

238 STEVE W I L S O N

TABLE IV

r (3, 7} {4, 5} (4, 6} {4, 7}

6 NS *5 NS "13 8 *7 *3 7 7

10 NS NS 5, 11 29 12 "13 11 NS 13 14 "13 29 13, 29 13 16 *NS 31 47 239 18 NS 19 17, 19 71 20 41 19 41 379 22 43 11, 131 89, 199 43, 769 24 NS NS 23 167 26 13 79, 131 233, 521 181, 549 28 29 419 281 NS 30 29 29 31, 61 29

resultant of two polynomials in A alone. When Cp(2a) is of higher degree, the

problem is solvable, but the procedure is a little more complicated: if r = 2n, we first treat Cp(tr(T2)) as a polynomial in whose coefficients are polynomials

in A. The resultant of this and Cp(2a) is then itself a polynomial in A, D(A). Then the resultant of D(A) and C~(2A) is a number, E, and the factors of E

give us the primes we want.

For example, to construct a map of type {5, 7}6, abbreviate 4a 2 by K, 4A 2

by L; then C~(2a) = K 2 - 3K + 1, and C7(2A ) = L 3 - 5L 2 + 6L -- 1. The trace of T 2 is K + L - 2, and C3(K + L-- 2) = K 2 + (2L - 4)K + (L 2 -- 4L + 3).

The resultant of K 2 - 3 K + 1 and K 2 + ( 2 L - 4 ) K + ( L 2 - 4 L + 3 ) as

quadratics in K is L 4 - 2L 3 - 3L 2 + 4L - 1. The resultant of this with

L 3 - - 5 L 2 -}- 6L -- 1 is 29. M o d 29, these equat ions have a unique solution

L = 5, K = --4, and from these, values for a and A are easily found.

5.1, Self-Dual Maps

Another si tuation in which we can restrict our at tention to polynomials of a

single variable is that of finding self-dual maps, which have type {P,P}2r. Whenever A = a, the resulting map is self-dual; the converse of that statement is an open question. As above, we let K = (2a) 2 -- (2A)2; the trace of T 2 is

4a 2 + 4A 2 -- 2 = 2K -- 2. The construct ion will yield a self-dual regular map

of type {P,P}zr for every simultaneous solution to Cp(2a) (which is a polynomial in K) and Cr(2K - 2). These polynomials are listed in Table V.

Of these {5, 5}s and {7, 7}8 yield no solution, though self-dual maps of both types are known. M a n y yield a single solution ({10, 10}16 has only one

A P P L I C A T I O N S OF VINCE 'S C O N S T R U C T I O N 239

s o l u t i o n , K = 366 m o d 1861), wh i l e s o m e h a v e m u l t i p l e s o l u t i o n s ({7,7}14

h a s t h e s o l u t i o n K = 16 m o d 4 1 a n d t he s o l u t i o n s to K 2 - 6 K - 1 m o d 13,

n a m e l y K = 9, 10 m o d 13).

TABLE V

p Cp(2a) r C,(2K - 2)

5 K 2 -- 3K + 1 3 6 K - 3 4 7 K 3 -- 5K 2 + 6K - 1 5 8 K 2 - 4 K + 2 6 9 K 3 - 6K 2 + 9K - 1 7

10 K 2 - 5K + 5 8 9

10

4K 2 - 8K + 3 4 K 2 - 8 K + 2 16K 4 - 6 4 K 3 + 84K 2 - 4 0 K + 5 4K 2 - 8K + 1 64K 6 - 384K 5 + 880K'* - 960K 3 + 504K 2 -- 112K + 7 16K 4 - 64K 3 + 80K 2 - 32K + 2 64K 6 - 384K 5 + 864K 4 - 896K a + 420K ~ -- 72K + 7 16K 4 - 64K 3 + 76K 2 -- 24K + 1

5.2. C a n t a n k e r o u s M a p s

T h e p a p e r [ W 2 ] i n t r o d u c e s a k i n d o f m a p ca l l ed ' c a n t a n k e r o u s ' ; t h i s is a n o n -

o r i e n t a b l e r e g u l a r m a p w h i c h h a s a n o r i e n t a t i o n - r e v e r s i n g cycle of l e n g t h 2.

I n t h a t p a p e r , (1) we s h o w t h a t a c a n t a n k e r o u s m a p m u s t b e se l f -Pe t r ie , a n d

so i ts o p p o s i t e is se l f -dua l (see [ W 1 ] for t h e s e te rms) , a n d (2) we i n d i c a t e t h a t

s u c h m a p s a re r e l a t i ve ly scarce . W e wi sh to use V i n c e ' s c o n s t r u c t i o n to

p r o d u c e c a n t a n k e r o u s m a p s , a n d in p a r t i c u l a r , t o i n v e s t i g a t e a c o n j e c t u r e

t h a t c a n t a n k e r o u s m a p s of all t ypes exist .

So, s u p p o s e M is t h e o p p o s i t e of a c a n t a n k e r o u s m a p of t y p e {p, 2r}p. T h e n

M is se l f -dua l of t y p e {p, P}2r a n d sat isf ies

(**) I = R T r - I S T ~ 1

As a b o v e , we c a n let A = a, a n d t h e n Cp(2a) m u s t be zero . I f we let

J = 2 (K - 1) = 8a 2 - 2 = 8A 2 - 8, t h e n a g a i n Cv(2a ) is a p o l y n o m i a l in J ,

a n d a m o d e r a t e l y c o n v o l u t e d i n d u c t i o n s h o w s t h a t if (**) ho lds , t h e n

Er(J ) = 0, w h e r e

~'A, if r = 2n E , l B , + B . _ I i f r = 2 n + l .

B e c a u s e t h e s e Er ' s a re n o t r e d u c e d , we h a v e to c h e c k t h a t a c o m m o n s o l u t i o n

o f Cp a n d E , d o e s n o t a l so sa t i s fy Cp,, E,, for s o m e s m a l l e r p ' , r'.

T h e s e p o l y n o m i a l s sa t i s fy t h e r e c u r s i o n E r ( J ) = J E r_ 2(J) - E~_4(J ). I f p is

fixed, t h e n t he r e s u l t a n t H , of Cp w i t h E~ for r = n o r for r = 2n a l so sat isf ies

240 STEVE W I L S O N

some recursion, and the quest ion of the existence of a can tankerous m a p of type {p, 2r}~ might be settled by showing that H , has some pr ime factor which does not divide some previous H, .

As an example, we examine the case p = 8, r = 2n + 1. Then the recursion is:

H 2 = H _ t = H o = H 1 = 1, H , + 4 = - 4 H n + 3 - 2 2 H , + 2 - 4 H , + 1 - H . .

The first few terms of the sequence Hn are:

n: 1 2 3 4 5 6 7

Hn: 1 - 3 1 97 289 - 3 1 6 7 5953 44609

The following appea r to be true abou t this sequence:

(1) If n > l t h e n H . # l .

(2) If p is a pr ime dividing H , for some n, then p __+ 1 (mod 16).

(3) F o r each n > 1, there is a pr ime p such that p divides H . but p does not divide H,. if 1 < m < n.

Conjecture (3) holds for all n < 200 or more, so there are finite cantanker -

ous maps of type { 8, 6} s, {8, 10} 8, { 8, 14} 8 . . . . . {8,402} 8,- -- • We need to prove conjecture (3) in order to establish the general case p = 8, r odd. Unfor tuna te - ly, even conjecture (1) resists a t tempts at proof. If the reader can produce any results, part ial results, ideas or conjectures on this sequence, please c o m m u -

nicate them to the au thor at the address given.

6. COMPUTATION

In applying the ideas of this paper, especially on a computer , we wish to keep the compu ta t ions as simple as possible. In part icular , when F is the field 2p,

we want to deal only with numbers f rom 7/e, and not f rom any quadrat ic extension, if that ' s possible. We can accomplish this restriction in two

different ways. One is by recursively determining the trace of a p roduc t by regarding it as a

word in R, and S. Then the following hold:

tr(UV) = tr(VU)

tr(gRR) = 2a t r (gR) -- tr(U)

tr(USS) = 2A tr(US) -- tr(U)

tr(URSR) = 4aA t r(UR) + tr(US) - 2A tr(U)

APPLICATIONS OF VINCE'S CONSTRUCTION 241

Using these rules, the trace of a word can be expressed in terms of traces of

smaller words, finally in terms of a and A. The benefit is avoiding computat ion in E, the quadratic extension of F; the price is a tree-like

structure for the decomposition of a product like R S R R S into words of length 1.

Another idea of computational simplification is this: any matrix U in the

group generated by the matrices R and S may be written in the form

U = A I I + AzR + A3S + A4SR -1,

where each At is a polynomial in a and A. If U is so written then

UR = - - A 2 I + (2aA z + AOR + (2aA3 + A 4 ) S -- A 3 S R - 1,

and

US = - -A3I + (2AA z - A 4 ) R + (2AA 3 + A1)S + A2SR x.

Computat ion may be done by regarding the matrices as 4-tuples in a and A, multiplying 4-tuples by using these rules and their consequences. Again, the

benefit is avoiding computat ion in E, the quadratic extension ofF; the price is a slightly more messy form of matrix multiplication.

Both of these are useful in the symbolic stage as well as the computational.

Suppose we require a map in which R 2 S 3 = I. The product R R S S S evaluates to

A1 + (aA 3 - 2aA)R + (1 - A2)S + (aA 2 -- a)SR-1.

If this is to be the identity then aA 3 - 2aA, 1 - A 2, and aA 2 - a must all be

zero, from which it follows that A = _ 1, a = 0 is the only solution.

7. OTHER APPLICATIONS

Even from this brief list of applications, I hope the reader will see that Vince's

construction is of tremendous use in deciding questions of existence of regular maps satisfying various conditions. It is moderately straightforward, and easy to implement on a computer. In turn, the construction poses problems which are challenging and enlightening.

REFERENCES

[C1] Coxeter, H. S. M., 'Regular skew polyhedra in three and four dimensions, and their topological analogues', Proc. Lond. Math. Soc. (2), 43 (1937), 33 67.

[C2] Coxeter, H. S. M., 'The abstract groups G '~'"'p', Trans. Arner. Math. Soc. 45 (January 1939).

242

[CM]

[D1] [D2] [G]

[GW]

[HP]

[R]

[v] [Wl] I-w23

l-W3]

STEVE WILSON

Coxeter, H. S. M. and Moser, W. O. J., Generators and Relations for Discrete Groups, Springer-Verlag, 1972. Dickson, Leonard, First Course in the Theory of Equations, Wiley, 1922. Dickson, Leonard, Linear Groups, Dover, 1958. Griinbaum, Branko, Geometry and Combinatorics of Complexes, Univ. of Washington Notes, 1974. Gray, A. and Wilson, S. E., 'A more elementary proof of Grunbaum's Conjecture', Congr. Numer. 72 (1990), 25-32. Holt, D. F. and Plesken, W., 'A cohomological criterion for a finitely presented group to be infinite', Durham Conference on Groups and Combinatorics (to appear in Congr. Numer.). Rankin, R. A., 'Chebyshev polynomials and the modulary group of level p', Math. Scand. 2 (1954), 315-326. Vince, Andrew, 'Regular combinatorial maps', J. Combin. Theory, 35, No. 3 (1983). Wilson, S. E., Operators over Regular Maps, Pacific J. Math. 81 No. 2 (1979), 559-568. Wilson, S. E., 'Cantankerous maps and rotary embeddings of K,', J. Combin. Theory, Set. B 47, No. 3 (1989). Wilson, S. E., 'A construction technique from an existence proof', Congr. Numer. 50 (1985), 25-30.

Author's address:

Steve Wilson, Dept. of Mathematics, Northern Arizona University, Flagstaff, AZ 86011, U.S.A.

(Received, June 16, 1992)