THE GROUPS DETERMINED BY THE RELATIONS

S= T’= (S-T-ST)v= 1

PART I

B H. S. M. COXETER

In workin out the commutator subgroups of the finite roups enerated byreflections, I came across a roup of order 288 havin the abstract definition

Sa= T= (S-T-1ST):- 1.

When I sent this result to Dr. Sinkov, he replied that he was making a specialstudy of such groups. So we agreed to write consecutive papers, his abstracttreatment to follow my geometrical treatment.

Groups of the form S= T= (ST) 1, considered for the sake of analogy

A triangle of angles -/1, r/m, /n can be drawn on a sphere, or in the euclideanplane, or in the hyperbolic plane, according as the number 1/l 1/m 1In isgreater than, equal to, or less than unity. By reflecting this triangle in its sidesrepeatedly, we fill the whole sphere or plane with such triangles, which may beshaded or left white, according to their orientation. Dyck showed that thewhite (or shaded) triangles correspond to the operators of the abstract group

S- T (ST)"= 1.

It follows that this group is finite when

1/l + 1/m + 1In > 1,

and infinite otherwise. More precisely, its order is

21/1 + 1/m - 1In 1

whenever this number is positive, and is infinite otherwise. Milleff proved thateach infinite group has an infinite number of finite factor groups.Very little is known about the infinite groups, save in the euclidean case

1/1 -q- 1/m -q- 1In 1.

This case is manageable on account of the presence of self-conjugate subgroupsgenerated by translations, whose quotient groups are obtained by identifying

Received April 11, 1985.W. Dyck, Gruppentheoretische Studien, Math. Ann., vol. 20 (1882), pp. 1-44.G. A. Miller, Groups defined by the orders of two generators and the order of their product,

Amer. Jour. of Math., vol. 24 (1902) pp. 96-100.61

62 I. S. M. COX:ETER

points of the plane that occupy corresponding positions in a network of periodparallelograms. These quotient groups are as follows:

Sa= Ta= (ST)a= (T-1)b(-lT)c-- 1,

S= T= (ST) (ST-=I)b(S-1T)= 1,

of order 3(b bc c2);

of order 4(b - c)

S T (ST) (TST-1S-1)b(ST-1S-IT) 1, of order 6(b - bc + c).

In particular (putting b c in the first two cases, and b 0 in the third),

S= Ta= (ST)a= (ST-IS-IT)’ 1 is of order 9p,S= T= (ST)= (T-I-IT)p 1 is of order 8p,Sa= T--- (ST)2= (ST-IS-T)p 1 is of order 6p.

This suggests the following theorem, which is not strictly relevant to our mainpurpose, but we shall state and prove it, on account of the close analogy withTheorem 4 below (where the geometry is in three dimensions instead of two).THEOttEM 1. In the group S T (ST) 1 with 1/l 1/m 1In <- 1,

the commutator of the generators is of infinite period.LEMMA 1.1. On a sphere, or in the euclidean or hyperbolic plane, the continued

product of the reflections in the sides of a triangle is an operation that leaves no pointinvariant.

Let R, R., Ra denote the reflections in the sides of a triangle AAAa. Ifpossible, let the point P be invariant under the operation RRRa, so that PP. RR2Ra, i.e., P. RaReR1 P. If P does not lie in the side A1A, let P.so that P’.RR P. Since RR leaves As invariant, AaP’ AaP. Hencelies on the perpendicular bisector of PP’, which is AIA (by definition of P’).On the other hand, if P lies in AA, we must have P.RR P, which makes Pcoincide with Aa (or its antipodes). In either case we are led to the absurd con-clusion that Aa lies in AA.. Therefore P cannot exist.LEMMA 1.2. For any finite set of (actual) points in the hyperbolic plane, we can

define a unique "centroid", which is invariant under all permutations of the points.Just as we may represent the points of the elliptic plane by concurrent lines in

ordinary space, so also we may represent the points of the hyperbolic plane bytime-like lines through a fixed point 0 of Minkowski three-space. There is,however, one important difference. In the latter case the representative linesare directed (in virtue of the "before-after" relation), and so can be replaced bypoints, equidistant from O, set off along them, either all "before" O or all"after" O. In other words, a "sphere" of time-like radius resembles a hyper-boloid of two sheets, and either of the sheets provides a (1, 1) mapping of the

W. Burnside, Theory of Groups of Finite Order, Cambridge, 1911, p. 419.This is a departure from the usual terminology. It seems desirable to speak of the

order of a group, but the period of an operator.Cf. Theorem 10 of Coxeter, Discrete groups generated by reflections, Annals of Math.

vol. 35 (1934), p. 602.

GROUPS 63

hyperbolic plane. Let G denote the centroicl of those points of Minkowski spacewhich represent the given set of points of the hyperbolic plane. Then therequired centroid of the given points is that point of the hyperbolic plane whichis represented by the line OG.

Proof of Theorem 1. Consider the larger group

(1.3) R R R (R2R3)"- (R3R1)"- (RIR.)- 1,

in which the given group is a subgroup of index 2, generated by S RR2,T R2R3. The generators R1, R., R are reflections in the sides of a triangle ofangles r/m, r/n, ’/1 (in the euclidean or hyperbolic plane, by virtue of theinequality).By Lemma 1.1, the operation RIR.Ra leaves no point invariant. If this opera-

tion were of finite period, its powers would transform any given point into afinite set, whose centroid would be invariant. Therefore RRR3 is of infiniteperiod. But (RRR3) ST-IS-T. ttence ST-IS-T is of infinite period.The same result could have been obtained trigonometrically, by showing that

the commutator is a rotation through , where

cos (/4) cos r/l - cos r/m + cos -/n - 2 cos -/1 cos rim cos /n.

When 1/1 - 1/m - 1In ,4 1, b is pure imaginary, or rather it is a hyperbolicargument instead of an angle.By virtue of Theorem 1, we should expect a great variety of factor groups of

S- T"= (ST)n= 1 (1/l + 1/m- 1In < 1)

to be obtainable by fixing the period of the commutator. Although some prog-ress has been made along these lines, the known results lack generality. Thereis, however, a general geometrical treatment for the case when we fix the periodof the commutator but leave the product (ST) unrestricted. In this respect,the product and commutator exchange rhles in a remarkable manner.

Schwarz used this operation (RRRa) to determine the triangle of minimum perimeterhaving one vertex on each side of a given triangle. Gesammelte math. Abhandlungen,vol. 2 (1890), p. 344.

We often find it convenient to write the commutator in this form, instead of the ortho-dox S-T-ST. In statements of period, this clearly makes no difference, since each ofthese operators is conjugate to the inverse of the other.

Putting 2r/1, 2/m and cos X (cos r/l cos -/m cos -/n)/sin -/l sin ./min formula (1) of G. de B. Robinson, The real representation of the commutator S-T-ST infour dimensions, Proc. Camb. Phil. Soc., vol. 26 (1930), p. 305.

H. R. Brahana, Certain perfect groups generated by two operators of orders two and three,Amer. Jour. of Math., vol. 50 (1928), pp. 345-356.

A. Sinkov, A set of defining relations for the simple group of order 1092, Bull. Amer. Math.Soc., vol. 41 (1935), p. 42.

Burnside (op. cir., p. 422) gives the symmetric group of degree 5 in the form S T(ST) (ST-ST) 1; S T (ST) (ST-ST) 1 is equally valid.

64 . s. M. COXETER

The case when T is involutory

When n 2, the group (1.3) becomes [1, m], the complete symmetry group ofthe regular polyhedronl {1, m}. The subgroup generated by R1R. and R2Rais the rotation group [1, m] r. When m is even, the operators RIR. and Ra generateanother subgroup (likewise of index 2), which we call [l’, m]. Writing S RR.,T R, we find (since R1, R are commutative) S-TST (RR3). Thus thegenerators of [l’, 2p] satisfy

(2.1) Sz= T2--- (S-1TST) 1.

They may possibly satisfy other relations, independent of these, but we canassert that [l’, 2p] is at least a factor group of (2.1).To show that [l’, 2p] is in fact the whole group (2.1), we observe11 that the

operators s S-, s’ TST of (2.1) satisfy

(2.2) st-- s’Z= (ss’) 1.

Since the group (2.2) is invariant under T, it is a subgroup of index 2 in (2.1).Similarly it is of index 2 in

S 2 (St)2p 1 (s’= tst).

This last group is [/, 2p]’.Thus (2.2) is of index 4 in [1, 2p], and of index 2 in (2.1). But [l’, 2p] is of

index 2 in [1, 2p], and is a factor group of (2.1). Therefore [l’, 2p] and (2.1) arethe same group.

Expressing this result in geometrical terms, we haveTHEOREM 2. The group [l’, 2p] has for fundamental region an isosceles triangle

of angles 2-/1, -/2p, 7/2p. It is generated by rotation through 2-/1 about the apex,and reflection in the base. Its abstract definition is

St= T2---- (S-ITST) 1.

In the figure on page 67, let CAA’ be this isosceles triangle, C’ the image of theapex C in the base AA’, and B the mid-point of AA’ (or of CC’). Then ABC is afundamental region for [1,, 2p], while CAA’ and ACC’ are alternative funda-mental regions for [1, 2p]’. T is the reflection in AA’, S or s-1 is the rotationthrough 2r/l about C, s’ is the opposite rotation about C’, ss’ or S-TST is therotation through 2/p about A, and is the rotation through about B.By evaluating the side BC of the triangle ABC, we obtain the followingCOROLLARY. The group [l’, 2p] is generated by rotation through 2r/1 about a

point, and reflection in a line distant from this point, where

sin -/1 cos/}, cos 7/2p;lo Bounded by/-gons, m at each vertex.1 For this remark I am indebted to Dr. Sinkov.In other words, this group is generated by rotations about the centers of the faces of the

polyhedron {1, 2p}, and reflections in its edges.

GROUPS 65

k 1, 0 or i according to the sign of 2/1 - 1./p 1, the plane being spherical,euclidean, or hyperbolic, in the three cases.Thus the group is finite only when

(2.3) 2/l + lip > 1,

its order then being 4/(2/1 - lip 1).[l’, 2] is the equatorial group C (occurring in crystallography when 2, 3,

4 or 6). It is the direct product of the cyclic group of order with the groupof order 2 generated by the equatorial reflection.

[2’, 2p] is the dihedral alternating group D. When p is odd, this is the directproduct of the dihedral group [2, p]’ with the group of order 2 generated by thecentral inversion. In particular, the rhombohedral group [2’, 6] can be obtainedby adjoining the central inversion to the trigonal dihedral group.

[3’, 4] is the pyritohedra113 group Th. This is the direct product of the tetra-hedral group [3, 3]’ with the group of order 2 generated by the central inversion.

If 2/1 lip 1, there are two groups illustrated on pages 66, 67. Theseare PSlya’s1 D and D, Niggli’s15 , and i. In both cases, (S-IT) and(ST) are translations, and we have the finite factor groups6

S T (S-TST) (S-T)b(ST)c 1, of order 8(b - c);

S T (S-1TST) (S-T)b(ST) 1, of order 6(b . bc - c).

The subgroups generated by S- and TST are two of Burnside’s groups men-tioned above.

The general case

Let [kl, ks, ka] denote the group

R R R R (RR.)k’ (R.R)k (RR)k(3.1)

(RR) (RR4) (RR) 1.

Since every generating relation here involves an even number of generators,there must be a self-conjugate subgroup of index 2, say [/, ks,/3]’, consisting of

a A. F. MSbius, Symmetrische Figuren, Gesammelte Werke, vol. 2 (1886), p. 672.G. PSlya, ber die Analogie der Kristallsymmetrie in der Ebene, Zeitschr. fiir Kristal-

log., vol. 60 (1924), p. 281.P. Niggli, Die Fldchensymmetrien homogener Diskontinuen, ibid., p. 291.

* These will be studied at greater length in Part II.

66 H. S. /. COXETER

T

T3T 8T

T3"T

GROUPS 67

A

(2’

T

68 H. S. M. COXETER

all those operators of [kl, k2, k3] which are products of even numbers of R’s.[kl, k2, k3]’ is clearly generated by the operators

TI RIR2, T2 R2R3, T3 RR4,

which satisfy7

T’= T,= Tk,= (T,T)2= (TT2T)= (TT3)= 1

If k. is even, all these relations involve T2 an even number of times. Theremust then be a self-conjugate subgroup of index 2, say [kl, k2, k3]", consistingof all those operators of [kl, k2, k]’ which involve T2 an even number of times.By repeated application of the relations

TT1 T T1,

TT- TTIT, T.T T TaT2,

all T’s that occur to an odd power (in the expression for any operator) can becollected in pairs; thus [k, k., ka]" is generated by T1, Ta, and T.

Since TITaTT- R.RR3R4RR.RRa R.RaR.R T, the operators Tand Ta generate [k, k2, ka]’ or [1, k, k] ’p according as k is odd or even.TM Inthe latter case they satisfy

Since we are chiefly concerned with the case when k. is even, it is convenientto write [/, 2p, m] instead of [kl, k,., k]. We shall also write S for T, and T forT-1, so that S RR, T RR3. Since [1, 2p, m]" and [m, 2p, 1]" are identical,we can assume that > m.

Geometrically,TM [1, 2p, m] is generated by reflections in the faces of a "double-rectangular" tetrahedron, whose dihedral angles are

(1 2) -/l, (2 3) -/2p, (3 ) r/m,

(1 3) r/2, (1 4) r/2, (2 4) r/2.

This tetrahedron will generally have to be in hyperbolic space, but it will be inspherical or euclidean space when l, m, p are sufficiently small.

Since the subgroup [1, 2p, m]" is of index 4, its fundamental region must bemade up of four such tetrahedra. From one such tetrahedron, the other threeare conveniently derived by reflecting in the faces 1 and . The whole funda--mental region is then a tetrahedron in which two opposite edges are perpendicu-

Cf. J. A. Todd, The groups of symmetries of the regular polytopes, Proc. Camb. Phil.Sot., vol. 27 (1931), p. 217.

Todd (ibid., p. 229) proved that it is impossible to generate [3, 4, 3]’ by two operators.Todd, ibid., pp. 214, 225.

GaouPs 69

lar, all the others being equal.are

Calling the faces 2, 3, 2’, 3’, the dihedral angles

z’) (s 3’) 2 /n,

3) (2’ 3) (2 S’) (2’ 3’) -/2p.

U

In the diagram I is BDD’, 2 is ADD’, 2’ is A’DD, 3 is AAD, 8 is AA’D’,is AA’C.The generators are rotations around the two perpendicular edges DD’, AA’"

S carries face 2’ into the position previously occupied by 2; T carries 3’ into.theposition previously occupied by 3. If we associate the original fundamentalregion with the operator 1, the surrounding regions, beyond faces ,, 3,correspond to the operators S, T, S-’, T-, respectively. (Face of tetrahedron1 is face 2’ of tetrahedron S, and so on.)We have already seen that the rotations S, T satisfy the relations

(3.2) S’= T"= (S-’T-’ST)" 1,

and that they suffice to generate the whole group [1, 2p, m]". We shall nextprove that every relation satisfied by these rotations is a consequence of (3.2).To find the operator that corresponds to a given region (i.e., that transforms

region I into the given region), we consider any path from a point within regionI to a point within the given region. Since we are taking products of operatorsfrom left to right, the successive steps along the path have to be written fromright to left. (E.g., we pass through face 3 of region 1 into region T, throughface 2’ of region T into region S-1T, through face 3 of S-IT into TS-T, andthrough face 3 of TS-1T into TS-T.)Any relation satisfied by the generators provides two different symbols for one

region, and so corresponds to closed path. The situation is easily visualized by

0 Cf. Annals of Math., vol. 35 (1934), p. 599, where, unhappily, I adopted the oppositeconvention.

70 H. S. M. COXETER

thinking of the path as an elastic string, threaded through a network of rigidwires forming the edges of all the tetrahedra. The path can be shrunk .to apoint by allowing it to slip through these edges, one at a time, the correspondingrelation being simplified, at each stage, by means of the generating relationS 1 or T 1 or (S-T-1ST) 1, according to the type of edge throughwhich the path slips. The sufficiency of these generating relations thus followsfrom the simple connectivity of the spherical, euclidean, or hyperbolic space, andwe haveTHEOREM 3. The group [1, 2p, m]" has for fundamental region a tetrahedron

with dihedral angles 2/1, 2/m at two opposite edges, the four remaining dihedralangles being r/2p. It is generated 5y rotations about the two special edges? Itsabstract definition is S T (S-T-IST)’ 1.The distance between the opposite edges DD’, AA’ of the tetrahedron AA

is just the length of the edge BC of the double-rectangular tetrahedron ABCD.Evaluating this by spherical trigonometry, we obtain the followingCOROLLARY. The group [1, 2p, m]" is generated by rotations throuffh 2-/1,

2/m about two perpendicular lines, distant apart, where

sin -/1 sin -/m cos k cos -/2p;

k 1, O, or i, according to the sign of sin r/1 sin -/m cos r/2p, the space beingspherical, euclidean, or hyperbolic, in the three cases.Thus the group is finite only when

(3.3) sin -/1 sin -/m > cos -/2p.

When m 2, this condition reduces to (2.3); in fact [1, 2p, 2]" [l’, 2p]. Whenp 1, we have the direct product of cyclic groups of orders l, m: [1, 2, m]" [/]’X [m] p. The remaining case, when m 3 and p 2, appears to be a newdiscovery: [3, 4, 3] p’ is of order 288, since [3, 4, 3] is of order 1152.23 (Actually, itis the commutator subgroup of [3, 4, 3].)

There are no possibilities in the critical case when sin -/1 sin -/m cos -/2p,save such as have m 2. When m 2 and 2/1 - lip 1, the fundamentalregion becomes a baseless prism whose cross-section is the isosceles triangle con-sidered in the two-dimensional representation. In other cases where m 2,the fundamental region has a pair of antipodal vertices (ideal, in the hyperbolicC3Se24).The trigonometry involved in proving the above corollary takes no account of

1 In other words, this group is generated by rotations about the edges of the two recip-rocal polytopes {m, 2p, 1}, {1, 2p, m}. For the theory of infinite regular polytopes, seeCoxeter, Proc. Camb. Phil. Soc., vol. 29 (1933), pp. 1-7.

This is, of course, the condition for the polytope {l, 2p, m} to be finite.23 E. Goursat, Sur les substitutions orthogonales et les divisions rgulires de l’espace,

Ann. Sci. de l’]cole Norm. Sup., (3), vol. 6 (1889), p. 87.The fundamental region has two ideal vertices whenever 2/l - 1/p < 1. All four

vertices are ideal if in addition 2/m T 1/p < 1.

GROUPS 71

the rationality of l, m, p. We can therefore state that the commutator of rota-tions 0 and about perpendicular lines distant h apart is a rotation , where

cos (/4) sin 0/2 sin /2 cos

Translating this result (in the spherical case) into terms of euclidean four-space,the commutator of pure rotations 0 and q about (not absolutely) perpendicularplanes inclined at angle k is a pure rotation , where5

cos (b/4) sin 0/2 sin q/2 cos

THEOREM 4.6 In the group S T (S-1T-1ST)v 1 with sin -/1 sin -/m__< cos -/2p, the product of the generators is of infinite period.LEMMA 4.1. In spherical, euclidean, or hyperbolic space, the continued product

of the reflections in the faces of a tetrahedron is an operation that leaves no pointinvariant.

This is analogous to Lemma 1.1, from which it easily follows.There is also a precise analogue of Lemma 1.2 in hyperbolic three-space

(proved by considering concurrent time-like lines in Minkowski four-space).Proof of Theorem 4. We have seen that the generators R1, R, R, R4 of

[1, 2p, m] are reflections in the faces of a tetrahedron. By Lemma 4.1, the opera-tion R1R.R4R (=ST) leaves no point invariant. Hence, the space beingeuclidean or hyperbolic (in virtue of the inequality), ST is of infinite period.

Clearly this holds also for ST-.The same result could have been obtained trigonometrically,:7 by showing

that ST (or ST-) is a double rotation of angles x, x’, where cosare the roots of the equation

(x cos //)(x cos -/m) x cos -/2p.

Elliptic space

THEOREM 5. Let G denote the dihedral alternating group of any odd degree, orthe pyritohedral group, or the new group [3, 4, 3]". Then G has a central of order2, generated by the central inversions (ST)1/2, where h is the period of ST.The only case that presents any difficulty is the last. In [3, 4, 3], the central

Cf. G. de B. Robinson, Proc. Camb. Phil. Soc., vol. 26 (1930), p. 309. On replacingby 2, we see that our formula is equivalent to his sin(h/4) sin(0/2) sin(/2)

(1 P[ P) with P 0 and P sinCf. Theorem 1.In the notation of F. N. Cole, On rotations in space offour dimensions, Amer. Jour. of

Math., vol. 12 (1890), pp. 205-208, the product of pure rotations 0 and about planes(0, 0, 1, 0, 0, 0) and (P, P, 0, P, P, P) is a double rotation of angles x, x’, where, sincea tan (0/2), b c f g h t t’ 0, D 1, B" BB’, and t" af’,sec (x/2) see (x’/2) see (/2) see (/2), tan (x/2) tan (x’/2) Pa tan (/2) tan (,/2).It follows that cos (x/2) cos (x’/2) cos (0/2) cos (/2) and cos (x/2) + cos (x’/2)

Pa) sin (t/2) sin (/2) cos (0/2) + cos" (/2) -[-cos (/2) + cos (/2) + (cos (/4). For this calculation I am indebted to Dr. Robinson.s Coxeter, Annals of Math., vol. 35 (1934), p. 606.

72 H. S. M. COXETER

inversion can be expressed in the form29 (R1R2R4R3)6; this is the operator (ST)of [3, 4, 3]".The central quotient groups, 1/2G, can be regarded as operating in elliptic space.

Abstractly, they are given by inserting the extra relation (ST)ih 1. Goursat3has enumerated all the crystallographic groups in elliptic three-space. Amongthese we easily pick out XX as the central quotient group of [3, 4, 3]". (It hasthe right order: 144.)1We thus have the following simple isomorphisms:

1/2[2, 2p, 21" 1/212’, 2p] [2, p]’

1/2[3, 4, 2]" 1/213’, 4] [3, 3]’

(p odd; h 2p),(h 6),

1/2[3, 4, 3]" [3, 3]’ X [3, 3]’ (h 12).

An infinite group in which both product and commutator have specifiedperiods

Let (1, m, n; p) denote the group S T (ST) (ST-S-T) 1. Thisis not altered by permuting l, m, n, since it can be put into the symmetrical formS T U" STU (SUT)’ 1. After comparing Theorems 1 and 4,it is natural to wonder whether (1, m, n; p) is necessarily finite. We shall showthat this is not so, since in fact (6, 6, 2; 2) is infinite.THEOREM 6. The group Se-- T6= (ST)2-" (ST-l)3q (ST-S-1T)2= 1

is of order 96 qa.LEMMX 6.1. The group S"= T- (ST)2= (ST-)’= (ST-IS-T) 1

is a subgroup of index 2 in s (st) (st-)2’ (st-Is-It) 1.This is easily proved by writing S s2t st-is- T =-t- so that

ST g, ST- st-s-lt and ST-S-T st-s-tt-ls2t- (st-)LEMM/k 6.2. The group s (st) (st-) 1 is infinite, and has a

representation in euclidean three-space in which the operation (st-s-t) is atranslation.We know that the group [4, 3, 4] (defined in (3.1)) is the complete symmetry

group of the cubic lattice in ordinary space.33 It has an involutory auto-morphism R2, such that

R4-- R2RR2.:9 Ibid., p. 608, (vi).30 Loc. cit., p. 66.Dr. Sinkov will clinch the matter by proving abstractly that 1/2 [3, 4, 3]" is the direct

product of two tetrahedral groups. He will also consider other factor groups of [3, 4, 3]",obtained by assigning a smaller period for ST.

2 This result is of special interest in view of the fact that (7, 6, 2; 2) is finite (oforder 2184). Another example is given, in effect, by H. R. Brahana, On the groups generatedby two operators of orders two and three whose product is of order eight, Amer. Jour. ofMath., vol. 53 (1931), p. 901. His results show that (3, 2, 8; 6) is infinite. Elsewhere, weshall prove that there are infinitely many infinite groups (l, m, n; p).

3 Coxeter, The densities of the regular polytopes, Proc. Camb. Phil. Soc., vol. 27 (1931), p.202 (3).

GROUPS 73

(Geometrically, this is the rotation through about the line joining the points(1/2, 1/4, 0) and (0, -, 1/2);it interchanges two reciprocal polytopes /4, 3, 4 }.) Byadjoining R2 to [4, 3, 4] we derive the group

’R3) (R1R3) (RIR’2R3R) 1

Writing s RR2, RR, we obtain a subgroup of index 2"

s4= t6= (st)2 (st-1)4= 1.

Clearly st-ls-lt (RR2Ra)2= RR2RR.

We nov make use of Theorem 13 of Discrete groups generated by reflections,3

which tells us that, of the cycles in which the operation RRR4R3 of [k,permutes the vertices of the polytope {kl, k, 1}, one is the cycle of vertices of aPetrie polygon. Now, the Petrie polygon of the net of cubes/4, 3, 4} is a helicalpolygon, whose sides take the three principal directions in turn, proceeding(say) from the origin to the points (1, 0, 0), (1, 1, 0), (1, 1, 1), (2, 1, 1),(2, 2, 1), Hence the operation RIRR4R or st-ls-t is a trigonal screw.Proof of Theorem 6. The translation (st-s-lt) and its conjugates generate a

three-dimensional lattice-group. The quotient group s (st) (st-)(st-Is-It) 1 is of order 192 (by direct calculation).35 By taking a longertranslation, we see that the group s "(st) (st-) (st-ls-lt) 1is of order 192 qa. The theorem now follows from Lemma 6.1.

In a somewhat similar manner, using the infinite groups

R R R] R (RR3) (R2R)

(RR)= (R2R)= (RR) (R4R)- 1,

s- t= (st) (st-) 1,

we my prove that the group S T (ST) (ST-) (ST-S-T)is of order 192 q.

TIINITY COLLEGE CAMBRIDGE.

Coxeter, Annals of Math., vol. 35 (1934), p. 605.By virtue of Lemma 6.1 we need only verify that the subgroup S T (ST)

(ST-) (ST-S-T) 1 is of order 96.