GENERATORS AND RELATIONS FOR SPACE GROUPS
|
|
- Ada Shelton
- 6 years ago
- Views:
Transcription
1 GENERATORS AND RELATIONS FOR SPACE GROUPS Eric Lord Bangalore 2010
2 Contents Introduction Notation PART I. NON-CUBIC GROUPS Explanation of the figures Triclinic Monoclinic Orthorhombic Tetragonal.... Trigonal Hexagonal PART II. CUBIC GROUPS Explanation of the figures Cubic Appendix : matrix methods
3 GENERATORS AND RELATIONS FOR SPACE GROUPS Introduction In their book Generators and Relations for Discrete Groups Coxeter and Moser listed a set of algebraic relations satisfied by a minimal set of generators for seventeen abstract groups isomorphic to the seventeen wallpaper groups. The tabulation provided here does the same thing for the 230 crystallographic space groups. The list of generators given for each space group in the International Tables for Crystallography is extensive and, of course, highly redundant. What is given here is a minimal set of generators for each group and the algebraic relations they satisfy. Choosing a minimal set of generators for each of the space groups involves considerable arbitrariness. This arbitrariness is reduced by making use of the space-filling polyhedra known as asymmetric units. These are space-filling polyhedra that are related to each other in the tiling of 3D space that they produce, by the various Euclidean transformations of the group (rotations, reflections, inversions, rotary-inversions, glidereflections and screw transformations). For some of the space groups the choice of asymmetric unit is not unique. An asymmetric unit is essentially a Voronoi region surrounding a point of general position, and "some general positions are more general than others". For example, the number of facets of the unit is usually reduced if the point for which it is a Voronoi region is chosen to lie on the axis of a screw transformation or rotary-inversion, or on a glide plane. These and other considerations have occasionally led me to a different choice of asymmetric unit from the one proposed in the International Tables. The minimal sets of generators have been chosen from those that map an asymmetric unit to a contiguous asymmetric unit sharing a face. However the unit is chosen, the number of generators of the minimal set is unaffected that is an invariant, characteristic of the space group. For each group our tabulation gives: (i) the number assigned to the group in the International Tables for Crystallography; its Hermann-Mauguin symbol; {a list of the chosen generators a minimal set followed by additional (redundant) generators that extend the minimal set to a set that relates an asymmetric unit to all contiguous unit}; generators indicated in the H-M symbol, expressed in terms of the chosen set; (ii) a set of generating relations that define the abstract group; (iii) translations expressed in terms of the chosen generators; (iv) a particular realization of the generators in terms of Euclidean transformations; specified in terms of the image of a general point [x, y, z]. For trigonal and hexagonal groups we employ a hexagonal coordinate system (x and y axes at 2π/3, z axis perpendicular to them); (v) a simple diagram of the asymmetric unit. 1
4 Notation The type of a generator will frequently be indicated by the letter that denotes it: X, Y, Z: translations [x + 1, y, z], [x, y + 1, z], [x, y, z + 1]. W : a centering translation. W = [x + ½, y + ½, z + ½] for body-centred. The notations W 1 = [x, y + ½, z + ½], W 2 = [x + ½, y, z + ½], and W = W 3 = [x + ½, y + ½, z] will be used for the face-centred cases. The base-centering transformation W means [x + ½, y + ½, z] unless otherwise specified. T : the translation [x + 2 / 3, y + 1 / 3, z + 1 / 3 ] along an edge of the rhombohedral cell in rhombohedral groups. S : a screw transformation. A, B, C, D, N : glide reflections. The nature of a generator of a cyclic subgroup will be indicated by the letter used to denote it, according to the following scheme. To avoid constant repetition the generating relations for these cyclic subgroups will be omitted from the tabulation but they are always implied: I : an inversion I 2 = E. Unless otherwise stated, I = [ x, y, z ] R : a diad rotation R 2 = E. M : a reflection M 2 = E. F : fourfold rotation. F 4 = E. Unless otherwise stated, F = [ y, x, z]. F : 4 transformation. F 4 = E. Unless otherwise stated, F = [y, x, z ]. Q : [ y, x y, z] in trigonal or hexagonal groups; [z, x, y] in cubic groups. Q 3 = E. In the cubic groups X = QZQ 1, Y = QXQ 1, Z = QYQ 1. Q : A 3 transformation. [y, y x, z] in trigonal or hexagonal groups; [ z, x, y ] in cubic groups. Q 6 = E. Note that Q 2 = Q and Q 3 = I. H : sixfold rotation [x y, x, z]. H 6 = E. H : A 6 transformation [y x, x, z]. H 6 = E 2
5 Explanation of the Figures PART I. NON-CUBIC GROUPS The figure given with each of the non-cubic space groups represents an asymmetric unit viewed perpendicularly to the xy plane. The units are various kinds of prisms. The generators that relate a unit to a contiguous unit are indicated as follows: a diad axis (R) parallel to the xy plane. When this symbol occurs as an edge of the figure the axis is half-way between the upper and lower faces of the unit (typically at z = 0 and z = ½, so that the axis would then be at z = ¼). When it bisects the figure, it represents two diad axes, on the upper and lower faces. a mirror (M) perpendicular to the xy plane a glide plane perpendicular to the xy plane a c-glide plane a d-glide plane a 2 1 screw axis parallel to the xy plane (located half way between the upper and lower faces of the unit). The symbols and are multi-purpose symbols; their meaning is given by context: at the mid-point of an edge of the figure represents a diad axis perpendicular to the xy plane. In the centre of the figure it represents a screw axis perpendicular to the xy plane, of a type indicated by the overall symmetry of the figure. (Our figures do not distinguish between enantiomorphic groups P4 1 and P4 3 for example have the same figure. This causes no confusion because the name of the group makes the distinction!) At a vertex of the figure represents a rotation axis perpendicular to the xy plane: 3-fold, 4-fold or 6- fold according to whether the angle at the vertex is 120º, 90º or 60º. at the mid-point of an edge of the figure represents an inversion centre halfway between the upper and lower faces of the unit. In the centre of the figure it represents two inversion centres, on the upper and lower faces. At a vertex of the figure it represents the centre of a rotary-inversion half way between the upper and lower faces of theunit; the axis is perpendicular to the xy plane. The transformation is a 3, a 4 or a 6 according to whether the angle at the vertex is 60º, 90º or 120º. A 4 2 axis is produced by two generators a 2-fold rotation and a 4 2 screw transformation, with a common axis. A special symbol is used to denote this. It was needed for only three of the groups (77, 94 and 102). (6 2, 6 3 and 6 4 axes are also produced by two generators but, luckily, no situations arose requiring symbols for these.) When a figure is shaded in grey, this denotes the existence of two mirrors parallel to the xy plane, on the upper and lower faces of the unit. 3
6 Finally, the existence of a translation as one of the generators relating the unit to a contiguous unit is indicated by a parallel pair of unmarked (very pale grey) edges of the figure, or, for the Z translation, by the empty interior of the figure. All this may seem complicated at first. However, once one has become familiar with these conventions, the extreme simplicity of the figures will be appreciated. All of the information about a space group is encapsulated, in a coordinate-free manner, in the figure representing its asymmetric unit. In fact, the whole of the unit cell diagram for a space group, given in the International Tables (many of which look quite daunting), can be rapidly reconstructed from the figure we have assigned to it, once our notation has become familiar. All this, of course, applies only to the non-cubic groups. The asymmetric units of the cubic groups are more complicated and do not lend themselves to representation in a 2D figure. 4
7 Triclinic (1) P1 {X, Y, Z} YZY 1 Z 1 = ZXZ 1 X 1 = XYX 1 Y 1 = E (2) P 1 {X, Y, I 1, I 2 } XYX 1 Y 1 = (I 1 X) 2 = (I 1 Y) 2 = (I 2 X) 2 = (I 2 Y) 2 = E Z = I 2 I 1 I 1 = [ x, y, z ] I 2 = [ x, y, z + 1] Monoclinic (3) P2 { X, Z, R 1, R 2 } ZXZ 1 X 1 = XR 1 X 1 R 1 = (ZR 1 ) 2 = XR 2 X 1 R 2 = (ZR 2 ) 2 = E Y = R 2 R 1 R 1 = [x, y, z ] R 2 = [x, y + 1, z ] (4) P2 1 {X, Z, S} ZXZ 1 X 1 = XSXS 1 = ZSZS 1 = E Y = S 2 S = [ x + ½, y + ½, z + ½] (5) C2 {Z, R, S, R 2 = SRS 1 } RS 2 RS 2 = (ZR) 2 = ZSZS 1 = E X = R 2 R Y = S 2 W = SR R = [ x, y, z + ½] S = [ x + ½, y + ½, z + ½] R = [1 + x, y, z + ½] 5
8 (6) Pm {X, Z, M 1, M 2 } ZXZ 1 X 1 = M 1 XM 1 X 1 = M 2 XM 2 X 1 = M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = E Y = M 2 M 1 M 1 = [x, y, z] M 2 = [x, y + 1, z] (7) Pc {X, Y, C} XYX 1 Y 1 = CXC 1 X 1 = CYC 1 Y = E Z = C 2 C = [x, y, z + ½] (8) Cm {Z, M, A, M 2 = AMA 1 } MA 2 MA 2 = AZA 1 Z 1 = MZMZ 1 = E X = A 2 Y = M 2 M W 3 = AM M = [x, y, z] A = [x + ½, y + ½, z] M 2 = [x, y + 1, z] (9) Cc {W, C, W a = CWC 1 } WW a W 1 W 1 a = C 2 WC 2 W 1 = E 1 X = W W a Y = W W a Z = C 2 C = [x, y, z + ½] W a = [x + ½, y ½, z] (10) P2/m {Z, M 1, M 2, R 1, R 2 } M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = (R 1 Z) 2 = (R 2 Z) 2 = (M 1 R 1 ) 2 = (M 1 R 2 ) 2 = (M 2 R 1 ) 2 = (M 2 R 2 ) 2 = E X = R 2 R 1 Y = M 2 M 1 M 1 = [x, y, z] M 2 = [x, y + 1, z] R 1 = [ x, y, z ] R 2 = [ x + 1, y, z ] 6
9 (11) P2 1 /m {Z, M, I 1, I 2, M 2 = I 1 MI 1 } S = MI 1 M 2 I 2 MI 2 = (ZI 1 ) = (ZI 2 ) = ZMZ 1 M = ZM 2 Z 1 M 2 = E X = I 2 I 1 Y = M 2 M M = [x, y, z] I 1 = [ x, y, z ] I 2 =[ x + 1, y, z ] M 2 = [x, y + 1, z] (12) C2/m {Z, R, M, I, R 2 = IRI } (MR) 2 = (IR) 2 = MZMZ = (RZ) 2 = (IZ) 2 = E X = (IR) 2 Y = M 2 M W = IRM M = [x, y, z] R = [ x, y, z ] I = [ x + ½, y + ½, z ] R 2 = [ x + 1, y, x ] (13) P2/c {Y, R 1, R 2, C} CYC 1 Y = (R 1 C) 2 = (R 2 C) 2 = R 1 YR 1 Y 1 = R 2 YR 2 Y 1 = E X = R 2 R 1 Z = C 2 R 1 = [ x, y, z + ½] R 2 = [ x + 1, y, z + ½] C = [x, y + 1, z + ½] (14) P2 1 /c {X, S, C} (C 1 S) 2 = XSXS 1 = XCX 1 C 1 = E Y = S 2 Z = C 2 S = [ x + 1, y + ½, z + ½] C = [x, y + ½, z + ½] 7
10 (15) C2/c {R, C, S, R 2 = SRS 1 } (RC) 2 = (SC) 2 = (SC 1 ) 2 = RS 2 RS 2 = E X = R 2 R Y = S 2 Z = C 2 W = SR R = [ x, y, z ] S = [ x + ½, y + ½, z ] C = [x, y, z + ½] R 2 = [ x, y, z ] Orthorhombic (16) P222 {Z, R 1, R 2, R 3, R 4 } (R 1 R 4 ) 2 = (R 4 R 2 ) 2 = (R 2 R 3 ) 2 = (R 3 R 1 ) 2 = (R 1 Z) 2 = (R 2 Z) 2 = (R 3 Z) 2 = (R 4 Z) 2 X = R 2 R 1 Y = R 4 R 3 R 1 = [ x, y, z ] R 2 = [ x + 1, y, z ] R 3 = [x, y, z ] R 4 = [x, y + 1, z ] (17) P222 1 {X, Y, R 1, R 2 } S = R 2 R 1 (R 2 X) 2 = R 1 XR 1 X 1 = (R 1 Y) 2 = R 2 YR 2 Y 1 = E Z = S 2 R 1 = [x, y, z ] R 2 = [ x, y, z + ½] (18) P {Z, S, R, R 2 = SRS } S 2 = SR R 2 2 = RZRZ 1 = SZS 1 Z = E X = S 2 2 = R 2 R Y = S 2 R = [ x, y, z] S = [ x + ½, y + ½, z ] R 2 = [ x + 1, y, z] 8
11 (19) P {S 1, S 3, S 4 = S 3 S 1 1 S 3 } S 2 = S 3 S 1 S 2 1 S 2 4 = S 2 2 S 3 S 2 2 S 1 3 = S 2 3 S 1 S 2 3 S 1 1 = S 2 1 S 2 S 2 1 S 1 2 = E X = S 1 Y = S 2 Z = S 3 S 1 = [x + ½, y + ½, z ] S 3 = [ x, y, z + ½] S 4 = [x + ½, y ½, z ] (20) C222 1 {R 1, R 2, W, W 2 = R 1 WR 1 } S = R 2 R 1 (S 2 R 1 ) 2 = (S 2 R 2 ) 2 = WSWS 1 = W 2 SW 2 S 1 = W 2 2 WW 2 2 W 1 = W 2 W 2 W 2 W 1 2 = E 1 X = WW 2 Y = WW 2 Z = S 2 R 1 = [x, y, z ] R 2 = [ x, y, z + ½] (21) C222 {Z, R 1, R 2, R 3 } (R 1 R 2 ) 2 = (ZR 1 ) 2 = (ZR 2 ) 2 = ZR 3 Z 1 R 3 = (R 3 R 1 R 3 R 2 ) 2 = E X = (R 3 R 2 ) 2 Y = (R 3 R 1 ) 2 W 3 = R 3 R 2 R 1 R 1 = [x, y, z ] R 2 = [ x, y, z ] R 3 = [ x + ½, y + ½, z] (22) F222 {R 1, R 2, R 3, R 4, R 5 = R 3 R 1 R 3, R 6 = R 2 R 4 R 2 } (R 1 R 2 ) 2 = (R 2 R 5 ) 2 = (R 3 R 4 ) 2 = (R 3 R 6 ) 2 = (R 5 R 1 R 4 ) 2 = (R 4 R 6 R 1 ) 2 = R 1 R 4 R 5 R 6 = E X = (R 3 R 2 ) 2 Y = R 5 R 1 Z = R 6 R 4 W 1 = R 4 R 1 W 2 = R 3 R 4 R 2 W 3 = R 3 R 2 R 1 R 1 = [x, y, z + ½] R 2 = [ x, y, z + ½] R 3 = [ x + ½, y + ½, z] R 4 = [x, y + ½, z ] R 5 = [x, y + 1, z + ½] R 6 = [x, y + ½, z + 1] 9
12 (23) I222 {R 1, R 2, S, R 3 = SR 1 S, R 4 = SR 2 S } R 5 = R 1 R 2 R 3 2 = R 4 2 = R 5 2 = (R 1 R 2 ) 2 = (R 1 R 4 ) 2 = (R 2 R 3 ) 2 = (R 3 R 4 ) 2 = E X = R 3 R 1 Y= R 4 R 2 Z = S 2 W = R 3 R 4 R 5 R 1 = [ x, y, z ] R 2 = [x, y, z ] S = [ x + ½, y + ½, z + ½] R 3 = [ x + 1, y, z ] R 4 = [x, y + 1, z ] (24) I {R 1, R 2, R 3, R 4 = R 1 R 3 R 1, R 5 = R 2 R 4 R 2, R 6 = R 2 R 3 R 2 } S 1 = R 3 R 2 S 2 = R 3 R 1 S 3 = R 1 R 2 S 2 1 R 1 S 2 1 R 1 = S 2 2 R 2 S 2 2 R 2 = S 2 3 R 3 S 2 3 R 3 = E X = S 1 Y = S 2 Z = S 3 W = S 3 R 4 R 1 = [x, y, z + ½] R 2 = [ x, y, z ] R 3 = [ x + ½, y + ½, z] R 4 = [ x + ½, y ½, z] R 5 = [ x ½, y ½, z] [ x ½, y + ½, z] (25) Pmm2 {Z, M 1, M 2, M 3, M 4 } R = M 1 M 2 (M 1 M 2 ) 2 = (M 2 M 3 ) 2 = (M 3 M 4 ) 2 = M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = M 3 ZM 3 Z 1 = M 4 ZM 4 Z 1 = E X = M 4 M 2 Y = M 3 M 1 M 1 = [x, y, z] M 2 = [ x, y, z] M 3 = [x, y + 1, z] M 4 = [ x + 1, y, z] (26) Pmc2 1 {Y, M 1, M 2, C} S = M 1 C M 1 CM 1 C 1 = M 2 CM 2 C 1 = M 1 YM 1 Y 1 = M 2 YM 2 Y 1 = CYC 1 Y = E X =M 2 M 1 Z = C 2 M 1 = [ x, y, z] M 2 = [ x + 1, y, z] C = [x, y, z + ½] 10
13 (27) Pcc2 {X, C, R 1, R 2 } C 2 = R 1 C R 1 CR 1 C 1 = R 2 CR 2 C 1 = (R 1 X) 2 = (R 2 X) 2 = CXC 1 X = E Y = R 2 R 1 Z = C 2 C = [ x, y, z + ½] R 1 = [ x, y, z] R 2 = [ x, y + 1, z] (28) Pma2 {Z, M, R 1, R 2, M 2 = R 1 MR 1 } A = R 1 M MZMZ 1 = M 2 ZM 2 Z 1 = R 1 ZR 1 Z 1 = R 2 ZR 2 Z 1 = R 2 R 1 MR 1 R 2 M = E X = M 2 M Y = R 2 R 1 M = [ x, y, z] R 1 = [ x + ½, y, z] R 2 = [ x + ½, y + 1, z] M 2 = [ x + 1, y, z] (29) Pca2 1 {Y, C, A} S = A 1 C ACAC 1 = CYC 1 Y 1 = AYA 1 Y = E X = A 2 Z = C 2 A = [x + ½, y, z] C = [ x + ½, y, z + ½] (30) Pnc2 {X, C, R, R 2 = CRC 1 } N = CR (RX) 2 = CXC 1 X 1 = C 2 RC 2 R = E Y = R 2 R Z = C 2 R = [ x, y, z] C = [x, y + ½, z + ½] R 2 = [ x, y + 1, z] 11
14 (31) Pmn2 1 {Y, M, S, M 2 = SMS 1 } N = SM SYS 1 Y = S 2 MS 2 M = MYMY 1 = E X = M 2 M Z = S 2 M = [ x, y, z] S = [ x + ½, y + ½, z + ½] M 2 = [ x + 1, y, z] (32) Pba2 {Z, B, A} R = A 1 B AZA 1 Z 1 = BZB 1 Z 1 = (AB) 2 = (AB 1 ) 2 = E X = A 2 Y = B 2 B = [ x + ½, y + ½, z] A = [x + ½, y + ½, z] (33) Pna2 1 {A, S, A 2 = SA 1 S 1 } N = SA A 2 SA 2 S 1 = S 2 AS 2 A 1 = E X = A 2 1 Y = AA 2 Z = S 2 A = [x + ½, y + ½, z] S = [ x + ½, y + ½, z + ½] A = [x + ½, y ½, z] (34) Pnn2 {Z, N, R} N 2 = NR NZN 1 Z 1 = RZRZ 1 = E X = Z 1 2 N 2 Y = Z 1 N 2 R = [ x, y, z] N = [x + ½, y + ½, z + ½] 12
15 (35) Cmm2 {Z, M 1, M 2, R} (M 1 M 2 ) 2 = RZRZ 1 = M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = (M 2 RM 1 R) 2 = E X = (RM 1 ) 2 Y = (RM 2 ) 2 W = RM 1 M 2 M 1 = [ x, y, z] M 2 = [x, y, z] R = [ x + ½, y + ½, z] (36) Cmc2 1 {M, C, B, M 2 = BMB 1 } MCMC 1 = BCBC 1 = B 2 MB 2 M = E X = BMB 1 M Y = B 2 Z = C 2 W = BM M = [ x, y, z] C = [x, y, z + ½] B = [ x + ½, y + ½, z] M 2 = [ x + 1, y, z] (37) Ccc2 {R 1, R 2, C, R 3 = CR 2 C 1 } C 2 = CR 1 CR 1 C 1 R 1 = C 2 R 2 C 2 R 2 = (R 1 R 2 R 3 ) 2 = E X = R 2 R 3 Y = R 3 R 1 R 2 R 1 Z = C 2 W = R 2 R 1 R 1 = [ x, y, z] R 2 = [ x + ½, y + ½, z] C = [x, y, z + ½] R 3 = [ x ½, y + ½, z] (38) Amm2 {M 1, M 2, M 3, C, M 4 = CM 2 C 1 } R = M 1 M 2 (M 1 M 2 ) 2 = (M 2 M 3 ) 2 = (M 3 M 4 ) 2 = (M 4 M 1 ) 2 = CM 1 C 1 M 1 = CM 3 C 1 M 3 = E X = M 3 M 1 Y = M 4 M 2 Z = C 2 W = CM 1 R M 1 = [ x, y, z] M 2 = [x, y, z] M 3 = [ x + 1, y, z] C = [x, y + ½, z + ½] M 4 = [x, y + 1, z] 13
16 (39) Abm2 {M, R 1, R 2, C, M 2 = R 1 MR 1 } B = R 1 M M 2 CMC 1 = M 2 C 1 MC = M 2 R 2 MR 2 = R 1 CR 1 C 1 = R 2 CR 2 C 1 = C 2 MC 2 M = E X = R 2 R 1 Y = B 2 Z = C 2 W 1 = MC M = [x, y, z] R 1 = [ x, y + ½, z] R 2 = [ x + 1, y + ½, z] C = [x, y + ½, z + ½] M 2 = [x, y + 1, z] (40) Ama2 {M, R, S, M 2 = RMR, R 2 = SRS 1 } A = MR M 2 SMS 1 = (MSMRS 1 ) 2 = S 2 RS 2 R = S 2 MS 2 M = (MR) 2 S(MR) 2 S 1 = E X = M 2 M Y = R 2 R Z = S 2 W 1 = SR R = [ x + ½, y, z] M = [ x, y, z] S = [ x + ½, y + ½, z + ½] M 2 = [ x + 1, y, z] R 2 = [ x + ½, y + 1, z] (41) Aba2 {B, A, C} R = A 1 C ACA 1 C 1 = BCBC 1 = (AB 1 ) 2 = (BA 1 ) 2 = E X = A 2 Y = B 2 Z = C 2 W 1 = BC A = [x + ½, y + ½, z] B = [ x + ½, y + ½, z] C = [ x + ½, y, z + ½] (42) Fmm2 {M 1, M 2, R, C, M 3 = RM 1 R} M 3 CMC 1 = (M 1 M 2 ) 2 = M 2 CM 2 C 1 = CRC 1 R = M 1 C 2 M 1 C 2 = RM 1 RCM 1 C 1 = E X = (RM 1 ) 2 Y = (RM 2 ) 2 Z = C 2 W 1 = RM 2 C W 2 = CM 1 W 3 = RM 1 M 2 M 1 = [ x, y, z] M 2 = [x, y, z] R = [ x + ½, y + ½, z] C = [ x + ½, y, z + ½] M 3 = [ x + 1, y, z] 14
17 (43) Fdd2 {D 1, D 2, R 1 = D 1 2 D 1, S = D 2 D 1, R 2 = SR 1 S 1, D = SD 1 S 1 } (D 1 D 1 2 ) 2 = (D 2 D 1 ) 2 (D 1 D 2 ) 2 = D 2 1 D 2 2 D 2 1 D 2 2 = D 2 1 (D 2 D 1 D 2 )D 2 1 (D 2 D 1 D 2 ) 1 = D 2 2 (D 1 D 2 D 1 )D 2 2 (D 1 D 2 D 1 ) 1 = E X = D 3 2 (D 1 D 2 D 1 ) 1 Y = D 3 1 (D 2 D 1 D 2 ) 1 Z = S W 1 = D 1 W 2 = D 2 W 3 = (D 1 D 2 D 1 ) 1 2 D 2 D 1 D 1 = [ x + ¼, y + ¼, z + ¼] D 2 = [x + ¼, y + ¼, z + ¼] R 1 = [ x, y, z] S = [ x + ½, y, z + ½] R 2 = [ x + 1, y, z] D = [ x + ¾, y ¼, z] (44) Imm2 {M 1, M 2, S, M 3 = SM 1 S 1, M 4 = SM 2 S 1 } R = M 1 M 2 (M 1 M 2 ) 2 = S 2 M 1 S 2 M 1 = S 2 M 2 S 2 M 2 = (M 4 M 1 ) 2 = (M 3 M 2 ) 2 = E X = M 3 M 1 Y = M 4 M 2 Z = S 2 W = SM 1 M 2 M 1 = [ x, y, z] M 2 = [x, y, z] S = [ x + ½, y + ½, z + ½] M 3 = [ x + 1, y, z] M 4 = [x, y + 1, z] (45) Iba2 {A, B, S} R = A 1 B (AB) 2 = (AB 1 ) 2 = SAS 1 A = SBS 1 B = (S 1 BSA) 2 = (S 1 ASB) 2 = E X = A 2 Y = B 2 Z = S 2 W = BSA A = [x + ½, y + ½, z] B = [ x + ½, y + ½, z] S = [ x + ½, y + ½, z + ½] (46) Ima2 {M, R, C, M 2 = RMR, R 2 = CRC 1 } A = RM CMC 1 M = C 2 RC 2 R = CM 2 C 1 M 2 = E X = M 2 M Y = R 2 R Z = C 2 W = CRM M = [ x, y, z] R = [ x + ½, y, z] C = [x, y + ½, z + ½] M 2 = [ x + 1, y, z] R 2 = [ x + ½, y + 1, z] 15
18 (47) Pmmm {M 1, M 2, M 3, M 4, M 5, M 6 } (M 1 M 3 ) 2 = (M 2 M 3 ) 2 = (M 3 M 4 ) 2 = (M 4 M 1 ) 2 = (M 1 M 5 ) 2 = (M 2 M 5 ) 2 = (M 3 M 5 ) 2 = (M 3 M 5 ) 2 = (M 4 M 5 ) 2 = (M 1 M 6 ) 2 = (M 2 M 6 ) 2 = (M 3 M 6 ) 2 = (M 4 M 6 ) 2 = (M 5 M 6 ) 2 = E X = M 2 M 1 Y = M 3 M 4 Z = M 6 M 5 M 1 = [ x, y, z] M 2 = [ x + 1, y, z] M 3 = [x, y + 1, z] M 4 = [x, y, z] M 5 = [x, y, z ] M 6 = [x, y, z + 1] (48) Pnnn { R 1, R 2, I 1, I 2, R 3 = I 1 R 1 I 2, R 4 = I 1 R 2 I 2 } N 1 = I 2 R 1 N 2 = I 2 R 2 N 3 = I 2 R 1 R 2 R 2 3 = R 2 4 = (R 1 R 2 ) 2 = I 1 I 2 R 1 R 2 I 2 I 1 R 1 R 2 = (N 1 N 2 N 3 ) 2 = (N 3 N 2 N 1 ) 2 = E X = R 3 R 1 Y = R 4 R 2 Z = I 2 I 1 (N 2 1 = YZ N 2 2 = ZX N 2 3 = XY) R 1 = [x, y, z ] R 2 = [ x, y, z ] I 1 = [ x + ½, y + ½, z ½] I 2 = [ x + ½, y + ½, z + ½] N 1 = [ x + ½, y + ½, z + ½] N 2 = [x + ½, y + ½, z + ½] N 3 = [x + ½, y + ½, z + ½] R 3 = [x + 1, y, z ] R 4 = [ x, y + 1, z ] (49) Pccm { R 1, R 2, R 3, R 4, M, M 2 = R 1 MR 1 } C 1 = R 1 M C 2 = R 2 M M 2 R 3 MR 3 = (R 1 R 2 ) 2 = (R 2 R 3 ) 2 = (R 3 R 4 ) 2 = (R 4 R 1 ) 2 = (R 1 R 2 M) 2 = (R 2 R 3 M) 2 = (R 3 R 4 M) 2 = (R 4 R 1 M) 2 = E X = R 3 R 1 Y = R 4 R 2 Z = C 2 1 = C 2 2 = M 2 M R 1 = [ x, y, z + ½] R 2 = [x, y, z + ½] R 3 = [ x + 1, y, z + ½] R 4 = [x, y + 1, z + ½] M = [x, y, z ] M 2 = [x, y, z + 1] 16
19 (50) Pban {Z, R 1, R 2, I} B = IR 1 A = IR 2 N = IR 1 R 2 (R 1 R 2 ) 2 = (AB 2 = (ABN) 2 = (IZ) 2 = (R 1 Z) 2 = (R 2 Z) 2 = E X = B 2 Y = A 2 R 1 = [ x, y, z ] R 2 = [x, y, z ] I = [ x + ½, y + ½, z ] (51) Pmma {M 1, M 2, M 3, R 1, R 2, M 4 = R 1 M 2 R 1 } A = R 1 M 2 (M 1 M 2 ) 2 = (M 2 M 3 ) 2 = (M 3 M 4 ) 2 = (M 4 M 1 ) 2 = (R 1 M 1 ) 2 = (R 1 M 3 ) 2 = (R 2 M 1 ) 2 = (R 2 M 3 ) 2 = M 4 R 2 M 2 R 2 = E X = M 4 M 2 Y = M 3 M 1 Z = R 2 R 1 M 1 = [x, y, z] M 2 = [ x, y, z] M 3 = [x, y + 1, z] R 1 = [ x + ½, y, z ] R 2 = [ x + ½, y, z + 1] M 4 = [ x + 1, y, z] (52) Pnna {R 1, R 2, I 1, I 2, R 3 = R 2 R 1 R 2, R 4 = I 1 R 2 I 1 } N 1 = R 3 I 1 N 2 = I 1 R 2 R 1 A = I 1 R 2 R 4 I 2 R 2 I 2 = R 3 R 4 R 1 R 4 = R 4 R 2 R 1 R 2 R 4 R 1 = (R 1 I 1 I 2 ) 2 = E X = R 4 R 2 Y = R 3 R 1 Z = I 2 I 1 R 1 = [x, y, z + ½] R 2 = [ x, y + ½, z] I 1 = [ x + ½, y + ½, z ] I 2 = [ x + ½, y + ½, z + 1] R 3 = [x, y + 1, z + ½] R 4 = [ x + 1, y + ½, z] (53) Pmna {Y, M, R 1, R 2, M 2 = R 1 MR 1 } N = R 1 R 2 YM A = R 1 M (MR 2 ) 2 = YMY 1 M = YR 1 Y 1 R 1 = (YR 2 ) 2 = (M 2 R 2 ) 2 = E X = M 2 M Z = (R 1 R 2 ) 2 M = [ x, y, z] R 1 = [ x + ½, y, z + ½] R 2 = [x, y + 1, z ] M 2 = [ x + 1, y, z] 17
20 (54) Pcca {R 1, R 2, R 3, C, R 4 = R 2 R 1 R 2 ] C 2 = R 3 C A = C 1 R 1 R 4 CR 1 C 1 = (R 1 R 3 R 4 ) 2 = C 2 R 2 C 2 R 2 = C 2 R 3 C 2 R 3 = E X = R 4 R 1 Y = R 2 R 3 Z = C 2 R 1 = [ x, y, z + ½] R 2 = [ x + ½, y, z] R 3 = [ x + ½, y + 1, z] C = [ x + ½, y, z + ½] R 4 = [ x, y + 1, z + ½] (55) Pbam {B, A, M 1, M 2 } M 1 AM 1 A 1 = M 2 AM 2 A 1 = M 1 BM 1 B 1 = M 2 BM 2 B 1 = (AB) 2 = (AB 1 ) 2 = E X = A 2 Y = B 2 Z = M 2 M 1 B = [ x + ½, y + ½, z] A = [x + ½, y + ½, z] M 1 = [x, y, z ] M 2 = [x, y, z + 1] (56) Pccn {C, R 1, R 2, I, I 2 = CIC] C 2 = CR N = C 1 R 2 CI I 2 2 = CR 1 C 1 R 1 = CR 2 C 1 R 2 = (IR 1 R 2 ) 2 = (IC 2 ) 2 = E X = I 2 I Y = R 2 R 1 Z = C 2 C = [ x + ½, y, z + ½] R 1 = [ x + ½, y ½, z] R 2 = [ x + ½, y + ½, z] I 2 = [ x + 1, y, z ] 18
21 (57) Pbcm {M, R, I 1, I 2, M 2 = I 1 MI 1, R 2 = I 1 RI 1 ] B = I 1 R C = I 1 M M 2 RMR = M 2 I 2 MI 2 = M 2 R 2 MR 2 = R 2 I 2 RI 2 = RR 2 M 2 M = E X = I 2 I 1 Y = B 2 Z = C 2 M = [x, y, z ] R = [x, y, z + ½] I 1 = [ x, y + ½, z + ½] I 2 = [ x + 1, y + ½, z + ½] M 2 = [x, y, z + 1] R 2 = [x, y + 1, z + ½] (58) Pnnm {M, S 1, S 2, M 2 = S 1 MS 1 1 } N 1 = S 2 M N 2 = S 1 M M 2 S 2 MS 1 2 = (S 1 1 S 2 ) 2 = (S 1 S 2 ) 2 = (S 1 S 2 M) 2 = (S 1 1 S 2 M) 2 = E 2 2 X = S 1 Y = S 2 Z = M 2 M M = [x, y, z ] S 1 = [x + ½, y + ½, z + ½] S 2 = [ x + ½, y + ½, z + ½] M 2 = [x, y, z + 1] (59) Pmmn {Z, M 1, M 2, I} N = IM 1 M 2 (M 1 M 2 ) 2 = (IM 1 IM 2 ) 2 = (IM 1 M 2 IM 1 IM 2 ) 2 = M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = (IZ) 2 = E X = (IM 1 ) 2 Y = (IM 2 ) 2 M 1 = [ x, y, z] M 2 = [x, y, z] I = [ x + ½, y + ½, z ] (60) Pbcn {B, S 1, S 2 } C = S 2 B N = S 2 B 1 S 1 (S 2 S 1 ) 2 = (S 2 S 1 1 ) 2 = BS 2 BS 1 2 = (BS 1 ) 2 = (BS 1 1 ) 2 = E 2 X = S 1 Y = B 2 2 Z = S 2 S 1 = [x + ½, y + ½, z ] S 2 = [ x + ½, y + ½, z + ½] B = [ x + ½, y + ½, z] 19
22 (61) Pbca {B, C, A} BCBC 1 = CACA 1 = ABAB 1 = (ABC) 2 = (CBA) 2 = E X = A 2 Y = B 2 Z = C 2 A = [x + ½, y, z + ½] B = [ x + ½, y + ½, z] C = [x, y + ½, z + ½] (62) Pnma {M, I, S, M 2 = IMI, I 2 = SIS } N = MS A = SI I 2 2 = M 2 SMS 1 = (I 2 M 2 M) 2 = MI 2 I 1 MI 1 I 2 ) = (IS 2 ) 2 = S 2 MS 2 M = E X = I 2 I Y = M 2 M Z = S 2 M = [x, y, z] I = [ x, y + ½, z + ½] S = [ x + ½, y + ½, z + ½] M 2 = [x, y +1, z] I 2 = [ x + 1, y + ½, z + ½] (63) Cmcm {M 1, M 2, R, I, M 3 = IM 2 I } C = M 2 R M 3 RM 2 R = (M 1 M 2 ) 2 = (M 1 M 3 ) 2 = (M 1 R) 2 = (IM 1 IR) 2 = E X = (IM 1 ) 2 Y = (RI) 2 Z = M 2 M 3 W = IM 1 R M 1 = [ x, y, z] M 2 = [x, y, z ] R = [x, y, z + ½] I = [ x + ½, y + ½, z + ½] M 3 = [x, y, z + 1] (64) Cmca {M, C, R, I, R 2 = IRI} A = CIM IRICRC = (RM) 2 = (IRIM) 2 = CMC 1 M = CIC 1 I = (C 2 RIR) 2 = E X = (IM) 2 Y = (IR) 2 Z = C 2 W = IMR M = [ x, y, z] C = [x, y + ½, z + ½] R = [x, y, z + ½] I = [ x + ½, y + ½, z + ½] R 2 = [x, y + 1, z + ½] 20
23 (65) Cmmm {M 1, M 2, M 3, M 4, R} (M 1 M 2 ) 2 = (RM 1 RM 2 ) 2 = (RM 3 ) 2 = (RM 4 ) 2 = (M 1 M 3 ) 2 = (M 2 M 3 ) 2 = (M 1 M 4 ) 2 = (M 2 M 4 ) 2 = E X = (RM 1 ) 2 Y = (RM 2 ) 2 Z = M 4 M 3 W = RM 1 M 2 M 1 = [ x, y, z] M 2 = [x, y, z] M 3 = [x, y, z ] M 4 = [ x, y, z + 1] R = [ x + ½, y + ½, z] (66) Cccm {M, R 1, R 2, R 3, M 4 = R 1 MR 1 } C 1 = R 2 MR 1 R 2 C 2 = R 1 MR 1 R 2 M 4 R 2 MR 2 = (R 1 R 2 ) 2 = (MR 1 R 2 ) 2 = (MR 3 ) 2 = (MR 4 ) 2 = (R 3 R 1 R 3 R 2 ) 2 = E X = (R 3 R 1 ) 2 Y = (R 3 R 2 ) 2 Z = M 4 M W = R 3 R 1 R 2 M = [x, y, z ] R 1 = [ x, y, z + ½] R 2 = [x, y, z + ½] R 3 = [ x, y, z] M 4 = [x, y, z + 1] (67) Cmma {Z, M 1, M 2, R 1, R 2 } A = R 1 M 1 (M 1 M 2 ) 2 = (R 1 R 2 ) 2 = (M 1 R 2 ) 2 = (M 2 R 1 ) 2 = (ZR 1 ) 2 = (ZR 2 ) 2 = ZM 1 Z 1 M 1 = ZM 2 Z 1 M 2 = E X = (R 1 M 1 ) 2 Y = (M 2 R 2 ) 2 W = R 1 R 2 M 1 M 2 M 1 = [ x, y, z] M 2 = [x, y, z] R 1 = [ x + ½, y, z ] R 2 = [x, y + ½, z ] (68) Ccca {R 1, R 2, R 3, C, R 4 = R 3 R 2 R 3 } C 2 = CR 3 A = CR 3 R 1 R 4 CR 2 C = (R 1 R 2 ) 2 = (R 1 R 4 ) 2 = CR 3 C 1 R 3 = (C 2 R 1 ) 2 = (C 2 R 2 ) 2 = E X = (R 3 R 1 ) 2 Y = (R 3 R 2 ) 2 Z = C 2 W = R 3 R 2 R 1 R 1 = [ x, y, z ] R 2 = [x, y, z ] R 3 = [ x + ½, y + ½, z] C = [x, y + ½, z + ½] R 4 = [x, y + 1, z ] 21
24 (69) Fmmm {M 1, M 2, M 3, R 1, R 2, M 4 = R 1 M 3 R 1 } M 4 R 2 M 3 R 2 = (M 2 M 3 ) 2 = (M 3 M 1 ) 2 = (M 1 M 2 ) 2 = (R 1 R 2 ) 2 = (R 1 M 2 ) 2 = (R 2 M 1 ) 2 = (R 1 R 2 M 3 ) 2 = E X = (R 1 M 1 ) 2 Y = (R 2 M 2 ) 2 Z = M 4 M 3 W = R 1 R 2 M 2 M 1 M 1 = [ x, y, z] M 2 = [x, y, z] M 3 = [x, y, z ] R 1 = [ x + ½, y, z + ½] R 2 = [x, y + ½, z + ½] M 4 = [x, y, z + 1] (70) Fddd {D 1, D 2, D 3, S = D 2 D 1, R 1 = D 1 2 D 1, R 2 = SR 1 S 1, R 3 = D 1 3 D 2, R 4 = D 3 D 1 1, D = SD 1 S 1 } SR 3 R 4 = (D 2 D 1 3 ) 2 = (D 3 D 1 1 ) 2 = (D 1 D 1 2 ) 2 = (D 1 D 2 D 1 3 ) 2 = (D 1 D 2 D 3 ) 2 = (D 2 D 1 D 3 ) 2 = D 2 2 D 2 3 D 2 2 D 2 3 = D 2 3 D 2 1 D 2 3 D 2 1 = D 2 1 D 2 2 D 2 1 D 2 2 = E X = (D 2 D 3 ) 2 Y = (D 3 D 1 ) 2 Z = (D 1 D 2 ) 2 2 W 1 = D 1 2 W 2 = D 2 2 W 3 = D 3 D 1 = [ x + ¼, y + ¼, z + ¼] D 2 = [x + ¼, y + ¼, z + ¼] D 3 = [x + ¼, y + ¼, z + ¼] R 1 = [ x, y, z] S = [ x + ½, y, z + ½] R 2 = [ x + 1, y, z] R 3 = [x, y, z ] R 4 = [ x + ½, y, z + ½] D = [ x + ¾, y ¼, z] (71) Immm {M 1, M 2, M 3, I, M 4 = IM 3 I } (M 1 M 2 ) 2 = (M 2 M 3 ) 2 = (M 3 M 1 ) 2 = (IM 1 IM 2 ) 2 = (IM 2 IM 3 ) 2 = (IM 3 IM 1 ) 2 = E X = (IM 1 ) 2 Y = (IM 2 ) 2 Z = M 4 M 3 W = IM 3 M 2 M 1 M 1 = [ x, y, z] M 2 = [x, y, z] M 3 = [x, y, z ] I = [ x + ½, y + ½, z + ½] M 4 = [x, y, z + 1] 22
25 (72) Ibam {M, R 1, R 2, I, M 2 = IMI } B = IR 2 A = IR 1 M 2 R 1 MR 1 = M 2 R 2 MR 2 = (R 1 R 2 ) 2 = (MR 1 R 2 ) 2 = (R 1 R 2 IR 1 IR 2 I) 2 = (R 1 IR 1 (IM) 2 ) 2 = (R 2 IR 2 (IM) 2 ) 2 = (IMR 2 IR 1 ) 2 =E X = A 2 Y = B 2 Z = M 2 M W = IMR 2 R 1 M = [x, y, z ] R 1 = [ x, y, z + ½] R 2 = [x, y, z + ½] I = [ x + ½, y + ½, z + ½] M 2 = [x, y, z + 1] (73) Ibca {R 1, R 2, C, I, R 3 = IR 1 I } B = IR 1 A = IR 2 R 3 R 2 R 1 R 2 = R 3 CR 1 C = CR 2 C 1 R 2 = BCBC 1 = CACA 1 = ABAB 1 = (ABC) 2 = (CBA) 2 = (CI) 2 = E X = A 2 Y = B 2 Z = C 2 W = IC 1 BA R 1 = [x, y, z + ½] R 2 = [ x, y + ½, z] C = [x, y + ½, z + ½] I = [ x + ½, y + ½, z + ½] R 3 = [x + 1, y, z + ½] (74) Imma {M 1, M 2, R 1, R 2, M 3 = R 1 M 1 R 1, M 4 = R 2 M 2 R 2 } A = R 1 M 1 (M 1 M 2 ) 2 = (M 2 M 3 ) 2 = (M 3 M 4 ) 2 = (M 4 M 1 ) 2 = (R 1 M 2 ) 2 = (R 1 M 4 ) 2 = (R 2 M 1 ) 2 = (R 2 M 3 ) 2 = E X = M 3 M 1 Y = M 4 M 2 Z = (R 1 R 2 ) 2 W = R 1 R 2 M 1 M 2 M 1 = [ x, y, z] M 2 = [x, y, z] R 1 = [ x + ½, y, z + ½] R 2 = [x, y + ½, z ] M 3 = [ x + 1, y, z] M 4 = [x, y + 1, z] 23
26 Tetragonal (75) P4 {Z, R, F} ZFZ 1 F 1 = ZRZ 1 R = (RF 1 ) 4 = E X = RF 2 Y = FRF R = [ x + 1, y, z] (76) P4 1 {X, S, Y = SXS 1 ) XS 2 XS 2 = E Y = SXS 1 Z = S 4 S = [ y, x, z + ¼] (77) P4 2 {R 1, R 2, S, Y = SR 2 R 1 S 1 } R 1 SR 1 S 1 = R 2 R 1 S 2 R 2 R 1 S 2 = (R 2 R 1 SR 2 S 1 ) 2 = E X = R 2 R 1 Y = SR 2 R 1 S 1 Z = R 1 S 2 R 1 = [ x, y, z] R 2 = [ x + 1, y, z] S = [ y, x, z + ½] (78) P4 3 {X, S, Y = SX 1 S 1 } XS 2 XS 2 = E Y = SX 1 S 1 Z = S 4 S = [y, x, z + ¼] 24
27 (79) I4 {F, S, F 2 = SFS 1 } (FF 2 ) 2 = S 2 FS 2 F 1 = E X = F 2 F 1 Y = F 2 1 F Z = S 2 W = SF 2 S = [ x + ½, y + ½, z + ½] F 2 = [ y + 1, x, z] (80) I4 1 {R, S, R 2 = SRS 1, R 3 = S 1 RS, R 4 = S 2 RS 2 } R 1 R 2 R 3 R 4 = (RR 3 R 2 ) 2 = RS 4 RS 4 = E X = R 2 R Y = R 4 R Z = S 4 W = S 2 R R = [ x, y, z] S = [ y + ½, x, z + ¼] R 2 = [ x + 1, y, z] R 3 = [ x + 1, y + 1, z] R 4 = [ x, y + 1, z] (81) P 4 { F, Z, R} (R F ) 4 = (R F ) 2 (R F 1 ) 2 = RZRZ 1 = F Z F 1 Z = E X = R F 2 Y = R 2 F 2 R = [ x + 1, y, z] (82) I 4 { F, W, R, F 2 = W F W 1 } (R F ) 4 = (W F ) 4 = R(W F 2 ) 2 R(W F 2 ) 2 = F (W F 2 ) 2 F 1 (W F 2 ) 2 = (W F 2 ) 2 ( F 2 W) 2 = E X = R F 2 Y = F R F Z = (W F 2 ) 2 R = [ x + 1, y, z] F 2 = [y, x + 1, z ] 25
28 (83) P4/m {F, M 1, M 2, R} (RF) 4 = (M 1 F) 4 = (M 2 F) 4 = (M 1 R) 2 = (M 2 R) 2 = FM 1 F 1 M 1 = FM 2 F 1 M 2 = E X = RF 2 Y = R 2 F 2 Z = M 2 M 1 M 1 = [x, y, z ] M 2 = [x, y, z + 1] R = [ x + 1, y, z] (84) P4 2 /m { F, R, M 1, M 2 } (RM 1 ) 2 = (RM 2 ) 2 = ( F 2 M 1 ) 2 = ( F 2 M 2 ) 2 = M 1 M 2 F M 1 M 2 F 1 = (R F ) 4 = E X = R F 2 Y = F 1 R F 1 Z = M 2 M 1 F = [y, x, z + ½] R = [ x + 1, y, z] M 1 = [x, y, z ] M 2 = [x, y, z + 1] (85) P4/n {Z, F, I} N = IF 2 FZF 1 Z 1 = (IZ) 2 = (FI) 4 = E X = F 1 IFI Y = FIF 1 I I = [ x + ½, y + ½, z ] (86) P4 2 /n { F, I 1, I 2, F 2 = I 2 F I 1 } N = I 1 F 2 S = F I 2 R = I 1 I 2 S 2 (I 1 I 2 S 2 ) 2 = F I 1 I 2 F 1 I 1 I 2 = ( F 2 I 1 F I 2 F I 1) 2 = ( F 2 I 2 F I 1 F I 2 ) 2 = E X = I 2 F 1 I 1 F Y = I 1 F I 2 F 1 Z = I 2 I 1 F = [y, x, z + ½] I 1 = [ x + ½, y + ½, z ] I 2 = [ x + ½, y + ½, z + 1] F 2 = [y, x + 1, z + ½] 26
29 (87) I4/m {F, M, I, M 2 = IMI} (IF) 4 = MFMF 1 = M 2 FM 2 F 1 = (FIF 1 M 2 M) 2 = E X = IFIF 1 Y = IF 1 IF Z = M 2 M W = IMF 2 M = [x, y, z ] I = [ x + ½, y + ½, z + ½] M 2 = [x, y, z + 1] (88) I4 1 /a {S, F, F 2 = S F S } A = F S X = A 2 Y = F 2 F Z = S 4 W = F S 2 F S = [ y, x + ½, z + ¼] F = [y, x, z ] F 2 = [y, x + 1, z ] (89) P422 {Z, R 1, R 2, R 3 } F = R 3 R 1 (R 1 R 2 ) 2 = (R 2 R 3 ) 4 = (R 3 R 1 ) 4 = (R 2 R 3 R 1 ) 2 = (R 1 Z) 2 = (R 2 Z) 2 = (R 3 Z) 2 = E X = R 2 R 3 R 1 R 3 Y = R 3 R 2 R 3 R 1 R 1 = [x, y, z ] R 2 = [ x, y, z ] R 3 = [y, x, z ] (90) P {Z, F, R} S = RF (RZ) 2 = (RF) 4 = FZF 1 Z = (F 1 RFR) 2 = E X = (RF 1 ) 2 Y = S 2 R = [ y + ½, x + ½, z ] 27
30 (91) P {X, R 1, R 2, Y = R 2 X 1 R 2 } S = R 2 R 1 (R 1 X) 2 = (XR 2 ) 2 (X 1 R 2 ) 2 = (X(R 2 R 1 ) 2 R 2 ) 2 = E Y = R 2 X 1 R 2 Z = S 2 R 1 = [ x, y, z ] R 2 = [ y, x, z + ¼] (92) P {S 1, S 2, S 3 = S 2 1 S 1 2 S } R = S 2 S 1 S 2 2 S 2 3 = (S 1 S 1 2 ) 2 = (S 1 S 2 ) 2 = (S 1 S 1 3 ) 2 = (S 1 S 3 ) 2 = S 4 1 S 2 2 S 4 1 S 2 2 = E X = S 1 S S 1 Y = S 2 Z = S 1 S 1 = [ y, x, z + ¼] S 2 = [ x + ½, y + ½, z + ½] S 3 = [ x + ½, y ½, z + ½] (93) P {R 1, R 2, R 3, R 4, R 5, R 6 = R 2 R 5 R 1 } S = R 5 R 2 (R 2 R 5 R 1 ) 2 = (R 1 R 2 ) 2 = (R 2 R 3 ) 2 = (R 3 R 4 ) 2 = (R 4 R 1 ) 2 = (R 3 R 4 R 5 ) 2 = (R 3 R 4 R 6 ) 2 = (R 3 R 6 R 5 ) 2 = R 4 R 2 R 5 R 2 R 5 R 2 R 4 R 5 R 2 R 5 = E X = R 3 R 1 Y = R 4 R 2 Z = R 5 R 6 R 1 = [ x, y, z ] R 2 = [x, y, z ] R 3 = [ x + 1, y, z ] R 4 = [x, y + 1, z ] R 5 = [y, x, z + ½] R 6 = [y, x, z ½] (94) P {R 1, R 2, R 3, S} (R 1 R 2 ) 2 = (R 3 R 2 R 3 R 1 ) 2 = SR 3 S 1 R 3 = R 1 R 3 R 2 S 2 R 1 R 3 R 2 S 2 = R 1 R 2 R 3 S 2 R 1 R 2 R 3 S 2 = E X = R 3 R 2 R 1 Y = R 1 R 3 R 2 Z = S 2 R 3 R 1 = [y, x, z ] R 2 = [ y, x, z ] R 3 = [ x + 1, y, z] S = [ y + ½, x ½, z + ½] 28
31 (95) P (X, R 1, R 2, Y = R 2 XR 2 } S = R 2 R 1 (R 1 X) 2 = (X 1 R 2 ) 2 (XR 2 ) 2 = (X 1 (R 2 R 1 ) 2 R 2 ) 2 = E Y = R 2 XR 2 Z = (R 2 R 1 ) 2 R 1 = [ x, y, z ] R 2 = [y, x, z + ¼] (96) P {S 1, S 2, S 3 = S 2 1 S 1 2 S } R = S 2 S 1 S 2 2 S 2 3 = (S 1 S 1 2 ) 2 = (S 1 S 2 ) 2 = (S 1 S 1 3 ) 2 = (S 1 S 3 ) 2 = S 4 1 S 2 2 S 4 1 S 2 2 = E X = S 1 S S 1 Y = S 2 Z = S 1 S 1 = [y, x, z + ¼] S 2 = [ x + ½, y + ½, z + ½] S 3 = [ x + ½, y ½, z + ½] (97) I422 {F, R 1, R 2, F 2 = R 1 F 1 R 1 } (FF 2 ) 2 = (FR 2 ) 2 = (FR 1 R 2 R 1 ) 2 = E X = (R 1 F 1 ) 2 Y = (R 1 F) 2 Z = (R 2 R 1 ) 2 W = R 2 R 1 F 2 R 1 = [ y + ½, x + ½, z + ½] R 2 = [y, x, z ] F 2 = [y, x + 1, z ] (98) I {R 1, R 2, R 3, R 4 = SR 3 S 1, R 5 = S 2 R 3 S 2, R 6 = S 1 R 3 S} S = R 2 R 1 R 3 R 4 R 5 R 6 = (R 1 R 4 ) 2 = (R 1 R 6 ) 2 = (R 4 R 3 R 6 ) 2 = E X = R 3 R 4 Y = R 3 R 6 Z = S 4 W = R 3 S 2 R 1 = [ y, x, z ] R 2 = [x, y, z + ¼] R 3 = [ x + ½, y + ½, z] R 4 = [ x ½, y + ½, z] R 5 = [ x ½, y ½, z] R 6 = [ x + ½, y ½, z] 29
32 (99) P4mm {Z, M 1, M 2, M 3 } F = M 1 M 3 (M 1 M 2 ) 2 = (M 2 M 3 ) 4 = (M 3 M 1 ) 4 = M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = M 3 ZM 3 Z 1 = E X = M 3 M 2 M 3 M 1 Y = M 3 M 1 M 3 M 2 M 1 = [ x, y, z] M 2 = [x, y, z] M 3 = [ x + ½, y + ½, z] (100) P4bm {Z, M, F} B = MF (MFMF 1 ) 2 = ZMZ 1 M = ZFZ 1 F 1 = E X = (MF 1 ) 2 Y = B 2 M = [ y + ½, x + ½, z] F = [ y, x, z] (101) P4 2 cm {M, R, C} S = CM CRC 1 R = (MCMC 1 ) 2 = (RCMC 1 RM) 2 = E X = RC 1 RMCM Y = MRC 1 RMC Z = C 2 M = [y, x, z] R = [ x + 1, y, z] C = [x, y, z + ½] (102) P4 2 nm {M 1, M 2, R, S} (M 1 M 2 ) 2 = SRS 1 R = (RM 2 RM 1 ) 2 = RM 1 M 2 S 2 RM 1 M 2 S 2 = M 2 RM 1 S 2 M 2 RM 1 S 2 = E X = RM 1 M 2 Y = M 1 RM 2 Z = RS 2 M 1 = [y, x, z] M 2 = [ x, y, z] R = [ x + 1, y, z] S = [ y + ½, x + ½, z + ½] 30
33 (103) P4cc {F 1, F 2, C} CF 1 C 1 F 1 = CF 2 C 1 F 2 = F 1 F 2 F 2 1 F 2 F 1 F 2 2 = E 1 X = F 2 F 1 Y = F 1 2 F 1 Z = C 2 F 1 = [ y, x, z] F 2 = [ y + 1, x, z] C = [y, x, z + ½] 2 (104) P4nc {F 1, F 2, C} N = CF 1 CF 1 C 1 F 2 = CF 2 C 1 F 1 = F 1 F 2 F 2 1 F 2 F 1 F 2 2 = E 1 X = F 2 F 1 Y = F 1 2 F 1 Z = C 2 F 1 = [ y, x, z] F 2 = [ y + 1, x, z] C = [ y + ½, x + ½, z + ½] (105) P4 2 mc {M 1, M 2, C, M 3 = CM 1 C 1, M 4 = CM 2 C 1 } S = CM 2 (M 1 M 2 ) 2 = (M 1 CM 1 C 1 ) 2 = (M 2 CM 2 C 1 ) 2 = M 1 C 2 M 1 C 2 = M 2 C 2 M 2 C 2 = E X = M 4 M 1 Y = M 3 M 2 Z = C 2 M 1 = [ x + 1, y, z ] M 2 = [x, y, z ] C = [y, z, z + ½] M 3 = [x, y + 1, z ] M 4 = [ x, y, z ] (106) P4 2 bc {A, C, B = CA 1 C 1 } C 2 AC 2 A 1 = A 2 CA 2 C 2 = E X = A 2 Y = B 2 Z = C 2 A = [ x + ½, y + ½, z] C = [ y + ½, x + ½, z + ½] B = [x + ½, y + ½, z + ½] 31
34 (107) I4mm {M 1, M 2, C, M 3 = CM 1 C 1 } F = M 2 M 1 (M 2 M 1 ) 4 = (CM 1 C 1 M 1 ) 2 = CM 2 C 1 M 2 = C 2 M 1 C 2 M 1 = E X = CM 1 C 1 M 2 M 1 M 2 Y = M 2 CM 1 C 1 M 2 M 1 Z = C 2 W = CM 1 M 2 M 1 M 1 = [x, y, z] M 2 = [y, x, z] C = [ y + ½, x + ½, z + ½] M 3 = [ x, y, z] (108) I4cm {F, C, M} CMC 1 M = FCFC 1 = (FMF 1 M) 2 = E X = (MF 1 ) 2 Y = (MF) 2 Z = C 2 W = F 2 CM C = [y, x, z + ½] M = [ y + ½, x + ½, z] (109) I4 1 md {M, S, M 2 = SMS 1, M 3 = S 2 MS 2, M 4 = S 1 MS } D = MS (MSMS 1 ) 2 = S 4 MS 4 M = E X = S 2 MS 2 M Y = S 1 MS 2 MS 1 Z = S 4 W = D 2 M = [ x, y, z] S = [ y + ½, x, z + ¼] M 2 = [x, y, z] M 3 = [ x + 1, y, z] M 4 = [x, y + 1, z] (110) I4 1 cd {A, S, B = S 1 AS} (AB) 2 = (AB 1 ) 2 = S 4 AS 4 A 1 = E X = A 2 Y = B 2 Z = S 4 W = AB 1 S 2 = ASA 1 S A = [x + ½, y + ½, z] S = [y + ½, x, z + ¼] B = [ x + ½, y + ½, z] 32
35 (111) P 4 2m {Z, R 1, R 2, M} F = MR 1 (R 1 R 2 ) 2 = (R 1 M) 4 = (R 2 M) 4 = (R 1 Z) 2 = (R 2 Z) 2 = MZMZ 1 = E X = R 1 MR 2 M Y = R 2 MR 1 M R 1 = [ x + 1, y, z ] R 2 = [x, y, z ] M = [y, x, z] (112) P 4 2c {R 1, R 2, C, R 3 = CR 2 C 1, R 4 = CR 1 C 1 } F = R 2 C (C 2 R 1 ) 2 = (C 2 R 2 ) 2 = (R 1 C) 4 = (R 2 C) 4 = E X = R 1 R 3 Y = R 4 R 2 Z = C 2 R 1 = [ x + 1, y, z + ½] R 2 = [x, y, z + ½] C = [y, x, z + ½] R 3 = [ x, y, z + ½] R 4 = [ x + 1, y, z + ½] (113) P m { F, Z, M} S = M F 1 (M F M F 1 ) 2 = MZMZ 1 = Z F 1 Z F = E X = (M F ) 2 Y = (M F 1 ) 2 M = [ y + ½, x + ½, z] (114) P c { F, C, F 2 = C F C} S = C F (C F C F 1 ) 2 = F C 2 F 1 C 2 = E X = (C F ) 2 Y = (C F 1 ) 2 Z = C 2 C = [ y + ½, x + ½, z + ½] F 2 = [ y, x + 1, z] 33
36 (115) P 4 m2 {Z, M 1, M 2, R} F = M 1 R (M 1 M 2 ) 2 = (M 1 R) 4 = (M 2 R) 4 = M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = (RZ) 2 = E X = M 2 RM 1 R Y = RM 2 RM 1 M 1 = [x, y, z] M 2 = [ x + 1, y, z] R = [y, x, z ] (116) P 4 c2 { F, R, C} RCRC 1 = F 2 C F 2 C 1 = ( F R) 4 = ( F C) 2 = E X = F 2 R Y = F R F Z = C 2 R = [ x + 1, y, z] C = [ x, y, z + ½] (117) F 4 b2 {Z, F, R, } B = R F 1 ( F R F 1 R) 2 = F Z F 1 Z = (RZ) 2 = E X = (R F ) 2 Y = B 2 R = [ y + ½, x + ½, z ] (118) P 4 n2 { F, R 1, R 2, F 2 = R 2 F 1 R 1 } N = F R 2 ( F R 1 R 2 ) 4 = ( F R 1 F 1 R 2 ) 2 = F R 2 R 1 F 1 R 2 R 1 = F R 1 R 2 F 1 R 1 R 2 = E X = R 2 F R 1 F Y = F 1 R 2 F R 1 F 2 Z = R 2 R 1 R 1 = [ y + ½, x + ½, z ½] R 2 = [ y + ½, x + ½, z + ½] F 2 = [ y, x + 1, z] 34
37 (119) I 4 m2 {M, R 1, R 2, R 3 = R 2 R 1 R 2, M 2 = R 1 MR 1 } F = R 1 M M 2 R 3 MR 3 = (MM 2 ) 2 = (R 1 M) 4 = (R 3 M) 4 = (R 3 M) 4 = (R 2 M) 4 = (R 2 M 2 ) 4 = M(R 2 R 1 ) 2 M(R 1 R 2 ) 2 = E X = MR 2 M 2 R 2 Y = M 2 R 2 MR 2 Z = R 3 R 1 W = R 2 MR 1 M M = [ x, y, z] R 1 = [y, x, z ] R 2 = [ y + ½, x + ½, z + ½] R 3 = [y, x, z + 1] M 2 = [x, y, z] (120) I 4 c2 { F, R 1, R 2, R 3 = R 2 R 1 R 2 } C = R 1 F 1 R 3 F R 1 F = ( F R 1 ) 2 = ( F R 3 ) 2 = ( F R 2 F 1 R 2 ) 2 = E X = (R 2 F ) 2 Y = (R 2 F 1 ) 2 Z = R 1 R 3 W = R 1 R 2 F 2 F = [ y, x, z + ½] R 1 = [y, x, z ] R 2 = [ y + ½, x + ½, z + ½] R 3 = [y, x, z + 1] (121) I 4 2m {M, R, C, R 2 = CRC} F = MR (C 2 R) 2 = (MR) 4 = (RC) 4 = (CMRM) 4 = MC 2 MC 2 = MRMC 2 MRMC 2 = E X = R 2 MRM Y = R 2 MR 2 R Z = C 2 W = CRMR M = [y, x, z] R = [x, y, z ] C = [ y + ½, x + ½, z + ½] R 2 = [ x + 1, y, z ] (122) I 4 2d { F, R, R 2 = F R F 1, F 2 = RR 2 F R 2 R} D = R F F 2 R F R F 1 R F 2 R F 1 R F R = E X = ( F R F ) 2 Y = (R F 2 ) 2 Z = (R F R F ) 2 W = D 2 F = [y, x, z + ¼] R = [x, y + ½, z ] R 2 = [x, y + ½, z + ½] F 2 = [y + 1, x, z + ¼] 35
38 (123) P4/mmm {M 1, M 2, M 3, M 4, M 5 } F = M 1 M 2 (M 2 M 3 ) 4 = (M 3 M 1 ) 2 = (M 1 M 2 ) 4 = (M 1 M 4 ) 2 = (M 2 M 4 ) 2 = (M 3 M 4 ) 2 = (M 1 M 5 ) 2 = (M 2 M 5 ) 2 = (M 3 M 5 ) 2 = E X = M 2 M 3 M 2 M 1 Y = M 3 M 2 M 1 M 2 Z = M 5 M 4 M 1 = [ x, y, z] M 2 = [y, x, z] M 3 = [x, y + 1, z] M 4 = [x, y, z ½] M 5 = [x, y, z + 1] (124) P4/mcc {M, R 1, R 2, R 3, M 2 = R 1 MR 1 } F = R 1 R 3 C 1 = R 1 M C 2 = R 3 M M 2 R 2 MR 2 = M 2 R 3 MR 3 = (R 1 R 2 ) 2 = (R 1 MR 1 R 3 ) 4 = (R 3 R 1 ) 4 = (R 3 R 2 ) 4 = (R 1 R 2 M) 2 = (R 1 MR 1 MR 1 ) 2 = (R 2 MR 1 MR 1 ) 2 = (R 3 M R 1 MR 1 ) 2 = E X = R 3 R 2 R 3 R 1 Y = R 2 R 3 R 1 R 3 Z = M 2 M M = [x, y, z ] R 1 = [ x, y, z + ½] R 2 = [x, y + 1, z + ½] R 3 = [y, x, z + ½] M 2 = [x, y, z + 1] (125) P4/nbm {Z, M, R 1, R 2 } F = R 1 R 2 N = MFR 1 B = MF (R 1 R 2 ) 4 = (R 1 M) 4 = (R 2 M) 2 = MZMZ 1 = (R 1 Z) 2 = (R 2 Z) 2 = E X = (MR 2 R 1 ) 2 Y = B 2 M = [ y + ½, x + ½, z] R 1 = [ x, y, z ] R 2 = [y, x, z ] (126) P4/nnc {R 1, R 2, C, R 3 = CR 1 C} F = R 1 R 2 N 1 = CRF 1 N 2 = CF (R 1 R 2 ) 4 = (R 1 C) 4 = (R 2 C) 2 = (R 1 C 2 ) 2 = E X = (R 2 CR 1 ) 2 Y = R 1 CFC R 2 Z = C 2 R 1 = [ x, y, z ] R 2 = [y, x, z ] C = [ y + ½, x + ½, z + ½] R 3 = [x, y + 1, z ] 36
39 (127) P4/mbm {F, M 1, M 2, M 3 } B = M 2 F (M 2 M 3 ) 2 = (M 1 M 2 ) 2 = M 1 FM 1 F 1 = (M 2 FM 2 F 1 ) 2 = (FM 3 ) 4 = FM 3 F 1 M 3 = E X = (M 2 F 1 ) 2 Y = (M 2 F) 2 Z = M 3 M 1 M 1 = [x, y, z ] M 2 = [ y + ½, x + ½, z] M 3 = [x, y, z + 1] (128) P4/mnc {F, M, R, M 2 = RMR } N = RMF C = RM (FRF 1 R) 2 = FMF 1 M = FRMRF 1 RMR = E X = (RF 1 ) 2 Y = (RF) 2 Z = M 2 M M = [x, y, z ] R = [ y + ½, x + ½, z + ½] M 2 = [x, y, z + 1] (129) P4/nmm {Z, M 1, M 2, R} F = M 2 M 1 N = R 1 FM 2 RM 2 RR 2 MR 2 = (RM 1 ) 2 = (RM 2 ) 4 = (R 2 M 1 ) 2 = (M 1 RM 2 R) 2 = (M 1 M 2 ) 4 = (R RM 2 R) 4 = (R 2 RM 2 R) 4 = (M 2 RM 2 R) 4 = R 2 RM 2 RR 2 M 2 = E X = M 1 RM 2 RM 1 M 2 Y = M 1 M 2 M 1 RM 2 R M 1 = [y, x, z] M 2 = [ x, y, z] R = [ y + ½, x + ½, z ] R 2 = [ y + ½, x + ½, z + 1] (130) P4/ncc {F, R, C} N = RF 2 C (RC 2 ) 2 = FC 2 F 1 C 2 = F 2 CF 2 C 1 = F 2 (RFRF 1 R)F 2 (RF 1 RFR) = E X = (RF) 2 Y = (FR) 2 Z = C 2 = C 2 2 R = [y + ½, x + ½, z + ½] C = [y, x, z + ½] 37
40 (131) P4 2 /mmc {M 1, M 2, M 3, R, M 4 = RM 3 R} C = RM 3 (M 1 R) 4 = (M 2 R) 4 = (M 1 M 2 ) 2 = (M 1 M 3 ) 2 = (M 2 M 3 ) 2 = (M 1 M 4 ) 2 = (M 2 M 4 ) 2 = (RM 4 ) 2 = (M 1 M 2 M 3 ) 2 = (M 1 M 2 M 4 ) 2 = E X = RM 2 RM 1 Y = RM 1 RM 2 Z = C 2 M 1 = [ x, y, z] M 2 = [x, y, z] M 3 = [x, y, z ] R = [y + ½, x + ½, z + ½] M 4 = [x, y, z + 1] (132) P4 2 /mcm {M 1, M 2, R 1, R 2, M 3 = R 1 M 2 R 1 } C = R 1 M 2 S = M 1 C M 3 R 2 M 2 R 2 = (R 1 R 2 ) 2 = (M 1 M 2 ) 2 = (M 1 M 3 ) 2 = (R 1 M 1 ) 4 = (R 2 M 1 ) 4 = (R 1 R 2 M 2 ) 2 = E X = R 2 M 1 R 1 M 1 Y = R 1 M 1 R 2 M 1 Z = C 2 M 1 = [y, x, z] M 2 = [x, y, z ] R 1 = [x, y, z + ½] R 2 = [ x + 1, y, z + ½] M 3 = [x, y, z + 1] (133) P4 2 /nbc {R 1, R 2, C, R 3 = CR 2 C} N = C 1 R 2 R 1 R 2 B = C 1 R 2 R 1 (CR 2 C) 2 = (R 1 C) 2 = (R 2 C 2 ) 2 = E X = C 2 R 1 R 2 R 1 Y = C 2 R 1 R 3 R 1 Z = C 2 R 1 = [y, x, z ] R 2 = [x, y, z + ½] C = [ y + ½, x + ½, z + ½] R 3 = [x, y + 1, z + ½] (134) P4 2 /nnm {M, R 1, R 2, R 3, R 4 = R 3 R 1 R 2 } N 1 = R 3 R 1 MR 1 N 2 = R 1 R 3 M (R 1 R 4 ) 2 = (R 1 M) 4 = (R 4 M) 4 = (R 2 M) 2 = (R 3 M) 2 = (R 1 R 2 R 3 ) 2 = (MR 1 MR 2 R 3 ) 2 = E X = R 4 MR 1 M Y = MR 4 MR 1 Z = R 3 R 2 M = [y, x, z] R 1 = [x, y, z + ½] R 2 = [ y + ½, x + ½, z ] R 3 = [ y + ½, x + ½, z + 1] R 4 = [ x + 1, y, z + ½] 38
41 (135) P4 2 /mbc { F, M, R, M 2 = RMR} S = F 1 M B = R F 1 C = RM M 2 F M F 1 = ( F 2 M) 2 = F C 2 F 1 C 2 = F 2 (R F R F 1 R) F 2 (R F 1 R F R) = (RC 2 ) 2 = E X = (R F ) 2 Y = B 2 Z = C 2 M = [x, y, z ] R = [ y + ½, x + ½, z + ½] M 2 = [x, y, z + 1] (136) P4 2 /mnm { F, M 1, M 2, M 3 = F M 2 F 1 S = F 1 M 2 N = M 1 F 1 M 2 (M 1 M 2 ) 2 = (M 1 M 3 ) 2 = ( F M 1 F ) 2 = ( F 2 M 2 ) 2 = ( F 2 M 3 ) 2 = F 2 (M 1 F M 1 F 1 M 1 ) F 2 (M 1 F 1 M 1 F M 1 ) = (M 1 M 2 M 3 ) 2 = (M 1 F M 2 F M 2 ) 2 = E X = (M 1 F ) 2 Y = (M 1 F 1 ) 2 Z = M 3 M 2 F = [y, x, z + ½] M 1 = [ y + ½, x + ½, z] M 2 = [x, y, z ] M 3 = [x, y, z + 1] (137) P4 2 /nmc {R, M, C, M 2 = CMC 1 } S = MC 1 N = RMCM (RM) 4 = (RC) 2 = (MM 2 ) 2 = (C 2 MR) 4 = MC 2 MC 2 = E X = (MCR) 2 Y = (RMC) 2 Z = C 2 R = [ y + ½, x + ½, z + ½] M = [ x, y, z] C = [y, x, z + ½] M 2 = [x, y, z] (138) P4 2 /ncm {R, C, M, R 2 = CRC, M 2 = CMC 1 } S = CM N = MR 1 R 2 (R 1 R 2 ) 2 = (RM) 2 = (R 2 M 2 ) 2 = (MM 2 ) 2 = (RC 2 ) 2 = MC 2 MC 2 = (MRC 2 ) 2 = E X = (MC 1 R) 2 Y = (RMC) 2 Z = C 2 R = [y, x, z + ½] C = [x, y, z + ½] M = [ y + ½, x + ½, z] R 2 = [ y, x, z + ½] M 2 = [ y + ½, x ½, z] 39
42 (139) I4/mmm {M 1, M 2, M 3, R} F = M 2 M 1 (RM 1 ) 4 = (RM 2 ) 2 = (M 1 M 2 ) 4 = (M 1 M 3 ) 2 = (M 2 M 3 ) 2 = (M 1 RM 2 M 1 M 3 RM 2 ) 2 = E X = (RM 2 M 1 ) 2 Y = (M 2 M 1 R) 2 Z = (RM 3 ) 2 W = RM 1 M 2 M 1 M 3 M 1 = [x, y, z] M 2 = [y, x, z] M 3 = [x, y, z ] R = [ y + ½, x + ½, z + ½] (140) I4/mcm {M 1, M 2, R 1, R 2, M 3 = R 1 M 1 R 1 } F = R 1 R 2 C = R 1 R 2 M M 3 R 2 M 1 R 2 = (R 2 M 2 ) 2 = (R 1 R 2 ) 4 = (M 1 M 2 ) 2 = E X = M 2 R 2 R 1 R 2 M 2 R 1 Y = M 2 R 1 M 2 R 2 R 1 R 2 Z = M 3 M 1 W = R 2 R 1 R 2 R 1 M 1 M 1 = [x, y, z ] M 2 = [ y + ½, x + ½, z] R 1 = [ x, y, z + ½] R 2 = [y, x, z + ½] M 3 = [x, y, z + 1] (141) I4 1 /amd { R 1, R 2, M, M 2 = SMS 1, M 3 = S 2 MS 2, M 4 = S 1 MS} S = R 2 R 1 A = R 2 SM D = AR 2 (R 1 M) 2 = (R 1 M 3 ) 2 = (R 2 M) 4 = (R 2 M 2 ) 4 = (R 2 M 3 ) 4 = (R 2 M 4 ) 4 = (MM 2 ) 2 = E X = A 2 Y = R 2 A 2 R 2 Z = S 4 W = D 2 R 1 = [x, y + ½, z ] R 2 = [y, x, z + ¼] M = [ x, y, z] M 2 = [x, y, z] M 3 = [ x + 1, y, z] M 4 = [x, y + 1, z] (142) I4 1 /acd {R 1, R 2, B, A 1 = R 2 BR 2 } S = R 2 R 1 A = R 2 BS C = B 1 S 2 D = BS S 4 C 2 = (BR 1 ) 2 = (BR 2 ) 4 = B(R 2 R 1 R 2 )B 1 (R 2 R 1 R 2 ) = AC 2 A 1 C 2 = CA 2 C 1 A 2 = BA 2 B 1 A 2 = (R 2 C) 4 = E X = A 2 Y = B 2 Z = C 2 W = D 2 R 1 = [x, y + ½, z ] R 2 = [y, x, z + ¼] B = [ x + ½, y + ½, z] A 1 = [x + ½, y + ½, z] 40
43 Trigonal (143) P3 {Z, Q 1, Q 2 } (Q 1 Q 2 ) 3 = ZQ 1 Z 1 Q 1 1 = ZQ 2 Z 1 Q 1 2 = E X = Q Q 2 Y = Q 2 Q 1 Q 1 = [ y + 1, x y, z] Q 2 = [ y + 1, x y + 1, z] (144) P3 1 {X, S, Y = SXS 1, XY } XYX 1 Y 1 = S 3 XS 3 X 1 = E Z = S 3 S = [ y, x y, z + 1 / 3 ] (145) P3 2 {X, S, Y = S 1 XS XY } XYX 1 Y 1 = S 3 XS 3 X 1 = E Z = S 3 S = [y x, x, z + 1 / 3 ] (146) R3 {S, Q, Q 2 = SQS 1, Q 3 = S 1 QS} QS 3 Q 1 S 3 = Q 2 QQ 2 1 Q 1 = (QQ 2 ) 3 = (QS) 2 QS 2 = E X = Q 2 Q 1 Y = QQ 2 Q Z = S 3 T = SQ 1 S = [ 2 / 3 y, 1 / 3 + x y, 1 / 3 + z] Q 2 = [ y + 1, x y, z] Q 3 = [ y + 1, x y + 1, z] 41
44 (147) P 3 {Z, Q, I} ( Q I) 3 = (ZI) 2 = Q Z Q 1 Z = E X = Q I Q 2 Y = Q 1 IQ 2 Q = [y, y x, 1 / 3 z] I = [1 x, 1 y, 1 / 3 z] (148) R 3 { Q, S, I = S Q S} (S Q S) 2 = (IQ ) 3 = (S Q 1 ) 2 = (S 2 Q ) 2 = E X = I Q 1 IQ Y = I Q I Q 1 Z = S 3 T = IQ 1 S S = [ 2 / 3 y, 1 / 3 + x y, 1 / 3 + z] I = [1 x, 1 y, z] (149) P312 {Z, R 1, R 2, R 3 } Q = R 2 R 1 (R 2 R 3 ) 3 = (R 3 R 1 ) 3 = (R 1 R 2 ) 3 = (R 1 Z) 2 = (R 2 Z) 2 = (R 3 Z) 2 = E X = R 3 R 1 R 3 R 2 Y = R 3 R 2 R 3 R 1 R 1 = [x, x y, z] R 2 = [y x, y, z] R 3 = [1 y, 1 x, z] (150) P321 {Q, R 1, R 2, Q 2 = R 1 Q 1 R 1 } Q 2 R 2 QR 2 = (QQ 2 ) 3 = E X = Q 1 Q 2 Y = Q 2 Q 1 Z = R 2 R 1 Q = [1 y, x y, z] R 1 = [y, x, z] R 2 = [y, x, 1 z] Q 2 = [1 y, 1 + x y, z] 42
45 (151) P {X, R 1, R 2, Y = R 2 R 1 XR 1 R 2, XY} (R 2 X) 2 = (X 1 R 1 ) 2 (XR 1 ) 2 = (XR 1 R 2 R 1 R 2 R 1 ) 2 = E Y = R 2 R 1 XR 1 R 2 Z = (R 2 R 1 ) 3 R 1 = [x, x y, z] R 2 = [ x + y, y, 1 / 3 z] (152) P {X, R 1, R 2, Y = R 1 XR 1, XY} (R 1 X) 2 (R 1 X 1 ) 2 = X(R 1 R 2 ) 2 X 1 (R 1 R 2 ) 2 = E Y = R 1 XR 1 Z = (R 2 R 1 ) 3 R 1 = [y, x, z] R 2 = [ x, x + y, 1 / 3 z] (153) P {X, R 1, R 2, Y = R 1 R 2 XR 2 R 1, XY } (R 1 X) 2 = (R 2 X) 2 (R 2 X 1 ) 2 = (XR 2 R 1 R 2 R 1 R 2 ) 2 = E Y = R 1 R 2 XR 2 R 1 Z = (R 2 R 1 ) 3 R 1 = [ x + y, y, z] R 2 = [x, x y, 1 / 3 z] (154) P {X, R 1, R 2, Y = R 2 XR 2, XY } (R 2 X) 2 (R 2 X 1 ) 2 = X(R 2 R 1 ) 2 X 1 (R 2 R 1 ) 2 = E Y = R 2 XR 2 Z = (R 2 R 1 ) 3 R 1 = [ x, x + y, z] R 2 = [y, x, 1 / 3 z] 43
46 (155) R32 {Q R 1, R 2, Q 2 = R 2 Q 1 R 2, Q 3 = R 1 Q 2 1 R 1 } QR 1 Q 1 R 1 = (QQ 2 ) 3 = (QQ 2 Q)R 1 R 2 R 1 R 2 R 1 (QQ 2 Q) 1 R 1 R 2 R 1 R 2 R 1 = E X = Q 2 Q 1 Y = QQ 2 Q Z = (R 2 R 1 ) 3 R 1 = [y, x, z] R 2 = [ 2 / 3 x, 1 / 3 x + y, 1 / 3 z] Q 2 = [1 y, x y, z] Q 3 = [1 y, 1 + x y, z] (156) P3m1 {Z, M 1, M 2, M 3 } Q = M 2 M 1 (M 2 M 3 ) 3 = (M 3 M 1 ) 2 = (M 1 M 2 ) 2 = M i ZM i Z 1 = E X = M 1 M 3 M 1 M 2 Y = M 2 M 3 M 2 M 1 M 1 = [x, x y, z] M 2 = [ x + y, y, z] M 3 = [1 y, 1 x, z] (157) P31m {Z, M, Q} (QMQ 1 M) 3 = QZQ 1 Z 1 = MZMZ 1 = E X = QMQ 1 MQ Y = (Q 1 M) 2 M = [x y, y, z] Q = [1 y, x y, z] (158) P3c1 {C, Q 1, Q 2 } B = (Q 1 Q 2 ) 1 (Q 1 Q 2 ) 3 = Q 1 CQ 1 C 1 = Q 2 CQ 2 C 1 = E 1 1 X = Q 1 Q 2 Y = Q 2 Q 1 Z = C 2 C = [1 y, 1 x, ½ + z] Q 1 = [1 y, x y, z] Q 2 = [1 y, 1 + x y, z] 44
47 (159) P31c {C, Q, Q 2 = CQ 1 C 1 } (QQ 2 ) 3 = C 2 QC 2 Q 1 = E X = Q 1 Q 2 Y = Q 2 Q 1 Z = C 2 C = [y, x, ½ + z] Q = [1 y, x y, z] Q 2 = [1 y, 1 + x y, z] (160) R3m {M, S, M 2 = SMS 1, M 3 = S 1 MS} Q = S 2 MS 2 M MS 3 MS 3 = (M 2 M) 3 = E X = MSMS 1 MS 1 MS Y = SXS 1 Z = S 3 T = SMS 1 MS M = [x, x y, z] S = [ 2 / 3 y, 1 / 3 + x y, 1 / 3 + z] M 2 = [ y, x, z] M 3 = [ x + y, y, z] (161) R3c {S, C, S 2 = CSC 1, Q = S 1 S 2, Q 2 = S 2 S 1 } (Q 1 Q 2 ) 3 = S 3 C 2 = E X = Q 1 Q 2 Y = Q 2 Q 1 Z = C 2 T = S 2 2 S 1 S = [ 2 / 3 y, 1 / 3 + x y, ½ + z] C = [x, x y, ½ + z] S 2 = [ 2 / 3 x + y, 1 / 3 x, ½ + z] Q = [1 y, x y, z] Q 2 = [1 y, 1 + x y, z] (162) P 3 1m {Z, M, R 1, R 2 } Q = MR 2 (R 1 R 2 ) 3 = (MR 1 ) 2 = (MR 2 ) 6 = (ZR 1 ) 2 = (ZR 2 ) 2 = MZMZ 1 = E X = R 1 R 2 (MR 2 ) 2 Y = R 2 MR 2 R 1 R 2 M M = [x y, y, z] R 1 = [1 x + y, y, z] R 2 = [x, x y, z] 45
48 (163) P 3 1c {C, R 1, R 2, R 3 = CR 1 C} Q = C 1 R 1 R 3 C 1 R 2 C = (R 2 R 3 ) 3 = (R 3 R 1 ) 3 = (R 1 R 2 ) 3 = (CR 2 ) 2 = (R 1 C) 6 = E X = R 2 R 3 R 2 R 1 Y = R 2 R 1 R 2 R 3 Z = C 2 C = [y, x, ½ + z] R 1 = [ x + y, y, ½ z] R 2 = [ y, x, ½ z] R 3 = [x, x y, ½ z] (164) P 3 m1 { M 1, M 2, R 1, R 2, M 3 = R 1 M 1 R 1 } Q = M 1 R M 3 R 2 M 1 R 2 = (M 2 M 3 ) 3 = (M 3 M 1 ) 3 = (M 1 M 2 ) 3 = (M 1 R 1 ) 6 = (M 2 R 1 ) 2 = (M 2 R 2 ) 2 = E X = M 1 M 2 M 1 M 3 Y = M 2 M 3 M 2 M 1 Z = R 2 R 1 M 1 = [x, x y, z] M 2 = [1 y, 1 x, z] R = [y, x, z] R 2 = [y, x, 1 z] M 3 = [ x + y, y, z] (165) P 3 c1 {R, C, Q} Q = RQ 1 CQ (QRQ 1 R) 3 = (RQ 1 CQ) 2 Q = RCRC 1 = CQC 1 RQR = C 2 QRQC 2 (QRQ) 1 = E X = (Q 1 R) 2 Y = (RQ 1 ) 2 Z = C 2 R = [y, x, ½ z] C = [1 y, 1 x, ½ + z] Q = [1 y, x y, z] (166) R 3 m {M, R 1, R 2, M 2 = R 1 MR 1, M 3 = R 2 MR 2 } Q = MR 1 (MR 2 R 1 R 2 ) 2 = (MR 1 R 2 R 1 ) 2 = (MR 2 ) 6 = (MR 1 ) 6 = E X = MM 3 MM 2 Y = M 3 M 2 M 3 M Z = (R 2 R 1 ) 3 T = R 2 MR 1 M M = [x, x y, z] R 1 = [y, x, z] R 2 = [ 1 / 3 + x y, 2 / 3 y, 1 / 3 z] M 2 = [ x + y, y, z] M 3 = [1 y, 1 x, z] 46
49 (167) R 3 c { Q, I, R 1, R 2, } ( Q I) 6 = ( Q 2 IQ I Q I) 2 = Q R 2 Q 1 R 1 = (I(R 2 R 1 ) 3 ) 2 = Q 2 IQIR 2 R 1 Q 2 IQ 1 I R 1 R 2 = E X = I Q 1 IQ Y = I Q I Q 1 Z = (R 2 R 1 ) 3 T = R 2 R 1 Q 2 I = [1 x, 1 y, z] R 1 = [ 2 / 3 x, 1 / 3 + y x, 1 / 6 z] R 2 = [ 1 / 3 + x y, 2 / 3 y, 1 / 6 z] 47
50 Hexagonal (168) P6 {Z, H, Q} QZQ 1 Z 1 = HZH 1 Z 1 = (Q 1 H) 2 = E X = QH 2 Y = (QH 2 Q) 1 Q = [1 y, x y, z] (169) P6 1 {X, S, Y = S 2 XS 2 XY = SXS 1 } S 3 XS 3 X = X(SXS 1 )X 1 (SXS 1 ) 1 = E Y = S 2 XS 2 Z = S 6 S = [x y, x, z + 1 / 6 ] (170) P6 5 {X, S, Y = S 2 XS 2 XY = SXS 1 } S 3 XS 3 X = X(SXS 1 )X 1 (SXS 1 ) 1 = E Y = S 2 XS 2 Z = S 6 S = [x y, x, z 1 / 6 ] (171) P6 2 {R, S, R 2 = SRS 1, R 3 = S 1 RS} (RR 3 S) 2 = (RR 2 R 3 ) 2 = S 3 RSRS 3 RS 1 R = E X = R 2 R Y = RR 3 Z = S 3 R = [1 x, 1 y, z] S = [ x + y + 1, x + 1, z + 1 / 3 ] R 2 = [2 x, 1 y, z] R 3 = [1 x, y, z] 48
51 (172) P6 4 {R, S, R 2 = SRS 1, R 3 = S 1 RS } (RR 3 S) 2 = (RR 2 R 3 ) 2 = S 3 RSRS 3 RS 1 R = E X = R 3 R Y = RR 2 Z = S 3 R = [1 x, 1 y, z] S = [ y + 1, x y, z + 1 / 3 ] R 2 = [1 x, y, z] R 3 = [2 x, 1 y, z] (173) P6 3 {Q, S, Q 2 = SQS 1 } (QQ 2 ) 3 = S 2 QS 2 Q 1 = E X = Q 1 Q 2 Y = Q 2 Q 1 Z = S 2 Q = [1 y, x y, z] S = [1 x, 1 y, ½ + z] Q 2 = [1 y, 1 + x y, z] (174) P 6 {Z, H 1, H 2 } ( H 1 H 2 ) 3 = ( H 4 1 H 2 ) 6 = ( H 4 2 H 1 ) 6 = H 3 1 H 3 2 = H 1 Z H 1 1 Z = H 2 Z H 1 2 Z = H 2 1 ( H 1 H 2 ) 1 H 2 ( H 2 H 1 ) 1 = E 1 X = H 1 H 2 Y = H 1 2 H 1 H 1 = [1 + y x, 1 x, z] H 2 = [y x, 1 x, z] (175) P6/m {Z, H, R} (R H ) 6 = (R H 2 ) 6 = (R H 3 ) 2 = ( H 4 R H R) 6 = H Z H 1 Z = H 3 R H 3 R = = ZRZ 1 R = ( H 2 R( H 1 R) 2 ) 2 = E X = H R H 1 R Y = R H 1 R H H = [1 + y x, 1 x, z] R = [1 x, 1 y, z] 49
52 (176) P6 3 /m { H 1, H 2, I, I 2 = H 3 1 I H 3 1 } H 3 1 H 3 2 = ( H 1 H 2 ) 3 = ( H 1 H 2 2 ) 6 = H 2 1 I 2 I H 1 1 I 2 I = H 2 2 I 2 I H 1 2 I 2 I = E 1 X = H 1 H 2 Y = H 1 2 H 1 Z = I 2 I H 1 = [1 + y x, 1 x, ½ z] H 2 = [y x, 1 x, ½ z] I = [1 x, 1 y, z] I 2 = [1 x, 1 y, 1 z] (177) P622 {Z, R 1, R 2, R 3 } H = R 1 R 2 (R 2 R 3 ) 3 = (R 3 R 1 ) 2 = (R 1 R 2 ) 6 = (R 1 Z) 2 = (R 2 Z) 2 = (R 3 Z) 2 = E X = R 2 R 3 H 2 Y = R 1 XR 1 R 1 = [y, x, z] R 2 = [x, x y, z] R 3 = [1 x, 1 y, z] (178) P {X, R 1, R 2, Y = S 1 XS, XY} S = R 2 R 1 (R 1 X) 2 = (XR 2 ) 2 (X 1 R 2 ) 2 = SR 2 XSR 2 SX 1 = E Y = S 1 XS Z = S 6 R 1 = [x y, y, z] R 2 = [x, x y, 1 / 6 z] (179) P {X, R 1, R 2, Y = S 1 XS, XY} S = R 1 R 2 (R 1 X) 2 = (XR 2 ) 2 (X 1 R 2 ) 2 = SR 2 XSR 2 SX 1 = E Y = S 1 XS Z = S 6 R 1 = [x y, y, 1 / 6 z] R 2 = [x, x y, z] 50
53 (180) P {R 1, R 2, R 3, R 4 = R 1 R 3 R 1, R 5 = R 2 R 4 R 2 } (R 2 R 3 ) 2 = (R 1 R 5 ) 2 = ((R 2 R 1 ) 3 R 3 ) 2 = (R 3 R 4 R 5 ) 2 = E X = R 5 R 3 Y = R 3 R 4 Z = (R 2 R 1 ) 3 R 1 = [x, x y, z] R 2 = [1 y, 1 x, 1 / 3 z] R 3 = [1 x, 1 y, z] R 4 = [1 x, y, z] R 5 = [2 x, 1 y, z] (181) P {R 1, R 2, R 3, R 4 = R 1 R 3 R 1, R 5 = R 2 R 4 R 2 } (R 2 R 3 ) 2 = (R 1 R 5 ) 2 = ((R 2 R 1 ) 3 R 3 ) 2 = (R 3 R 4 R 5 ) 2 = E X = R 5 R 3 Y = R 3 R 4 Z = (R 1 R 2 ) 3 R 1 = [x, x y, 1 / 3 z] R 2 = [1 y, 1 x, z] R 3 = [1 x, 1 y, z] R 4 = [1 x, y, z] R 5 = [2 x, 1 y, z] (182) P { R 1, R 2, R 3, R 4, R 5 = R 1 R 4 R 1 } (R 2 R 3 ) 3 = (R 3 R 1 ) 3 = (R 1 R 2 ) 3 = R 5 R 4 R 2 R 1 R 3 R 1 R 4 R 5 R 1 R 3 R 1 R 2 = E X = R 3 R 1 R 3 R 2 Y = R 1 R 2 R 3 R 2 Z = R 5 R 4 R 1 = [1 y, 1 x, ½ z] R 2 = [y x, y, ½ z] R 3 = [x, x y, ½ z] R 4 = [y, x, z] R 5 = [y, x, ½ z] (183) P6mm {Z, M 1, M 2, M 3 } H = M 1 M 2 (M 2 M 3 ) 3 = (M 3 M 1 ) 2 = (M 1 M 2 ) 6 = M 1 ZM 1 Z 1 = M 2 ZM 2 Z 1 = M 3 ZM 3 Z 1 = E X = M 2 M 3 (M 2 M 1 ) 2 Y = (M 3 M 1 M 2 ) 2 M 1 = [y, x, z] M 2 = [x, x y, z] M 3 = [1 y, 1 x, z] 51
54 (184) P6cc {H, C, R} C 2 = HC RCRC 1 = (RH) 3 = (H 2 RHRHR) 2 = R(C 2 H)R(C 2 H) 1 = E X = RC 2 1 RC 2 Y = RC 1 HRH 1 C Z = C 2 C = [y, x, ½ + z] R = [1 x, 1 y, z] (185) P6 3 cm {Q, C, M} MCMC 1 = (QMQ 1 M) 3 = C 2 (QMQ)C 2 (QMQ) 1 = E X = (Q 1 M) 2 Y = (MQ 1 ) 2 Z = C 2 Q = [1 y, x y, z] C = [1 y, 1 x, ½ + z] M = [y, x, z] (186) P6 3 mc {M 1, M 2, C, M 3 = CM 2 C 1 } (M 1 M 2 ) 3 = M 1 CM 1 C 1 = (M 2 CM 2 C 1 ) 3 = M 2 C 2 M 2 C 1 = E X = M 1 M 2 M 3 Y = M 3 M 2 M 3 M 1 Z = C 2 M 1 = [1 y, 1 x, z] M 2 = [x, x y, z] C = [y, x, ½ + z] M 3 = [1 x, 1 y, z] (187) P 6 m2 {M 1, M 2, M 3, M 4, M 5 } Q = M 3 M 2 M 4 (M 1 M 2 ) 3 = (M 2 M 3 ) 3 = (M 3 M 1 ) 3 = (M 1 M 4 ) 2 = (M 2 M 4 ) 2 = (M 3 M 4 ) 2 = (M 1 M 5 ) 2 = (M 2 M 5 ) 2 = (M 3 M 5 ) 2 = E X = M 1 M 2 M 1 M 3 Y = M 3 M 2 M 3 M 1 Z = M 5 M 4 M 1 = [1 y, 1 x, z] M 2 = [x, x y, z] M 3 = [ x + y, y, z] M 4 = [x, y, z] M 5 = [x, y, 1 z] 52
55 (188) P 6 c2 {M, R 1, R 2, R 3, M 2 = R 1 MR 1 } H = MR 1 R 3 C = MR 1 M 2 R 2 MR 2 = M 2 R 3 MR 3 = (R 1 R 2 ) 3 = (R 2 R 3 ) 3 = (R 3 R 1 ) 3 = R 1 MR 1 = R 2 MR 2 = R 3 MR 3 = E R 1 MR 1 = R 2 MR 2 = R 3 MR 3 X = R 1 R 2 R 1 R 3 Y = R 3 R 2 R 3 R 1 Z = M 2 M 1 M = [x, y, z] M 2 = [x, y, 1 z] R 1 = [x, x y, ½ z] R 2 = [1 y, 1 x, ½ z] R 3 = [ x + y, y, ½ z] (189) P 6 2m {M, M 1, M 2, Q} H = QMQ 1 M 1 (QMQ 1 M 1 ) 6 = (QMQ 1 M 2 ) 6 = (MM 1 ) 2 = (MM 2 ) 2 = QM 1 Q 1 M 1 = QM 2 Q 1 M 2 = (MQMQ 1 ) 3 = E X = (Q 1 M) 2 Y = (MQ 1 ) 2 Z = M 2 M 1 Q = [1 y, x y, z] M = [y, x, z] M 1 = [x, y, z] M 2 = [x, y, 1 z] (190) P 6 2c {M, R, Q} H = QRQ 1 RM C = RM QMQ 1 M = (QRQ 1 R) 3 = Q(RMR)Q 1 (RMR) = E X = (Q 1 R) 2 Y = (RQ 1 ) 2 Z = C 2 M = [x, y, z] R = [y, x, ½ z] Q = [1 y, x y, z] (191) P6/mmm {M 1, M 2, M 3, M 4, M 5 } H = M 1 M 2 (M 2 M 3 ) 3 = (M 3 M 1 ) 2 = (M 1 M 2 ) 6 = (M 4 M 1 ) 2 = (M 4 M 2 ) 2 = (M 4 M 3 ) 2 = (M 5 M 1 ) 2 = (M 5 M 2 ) 2 = (M 5 M 3 ) 2 = E X = M 2 M 3 H 2 Y = M 1 XM 1 Z = M 5 M 4 M 1 = [y, x, z] M 2 = [x, x y, z] M 3 = [1 y, 1 x, z] M 4 = [x, y, z] M 5 = [x, y, 1 z] 53
56 (192) P6/mcc {M, R 1, R 2, R 3, M 2 = R 1 MR 1 } H = R 1 R 2 C 1 = M 1 R 2 R 1 R 2 C 2 = M 1 R 1 (R 2 R 3 ) 3 = (R 3 R 1 ) 2 = (R 1 R 2 ) 6 = MR 2 R 3 MR 3 R 2 = MR 3 R 1 MR 1 R 3 = MR 1 R 2 MR 2 R 1 = E X = R 2 R 3 H 2 Y = R 1 XR 1 Z = M 2 M M = [x, y, z] M 2 = [x, y, 1 z] R 1 = [y, x, ½ z] R 2 = [x, x y, ½ z] R 3 = [1 x, 1 y, ½ z] (193) P6 3 /mcm {M 1, M 2, R 1, R 2, M 3 = R 1 M 2 R 1 } C = R 1 M 2 M 3 R 2 M 1 R 2 = (R 1 R 2 ) 3 = (M 1 R 1 ) 6 = (M 1 R 2 ) 2 = (M 1 M 2 ) 2 = (M 1 M 3 ) 2 = M 1 C 2 M 1 C 2 = (R 2 C 2 ) 2 = E X = (R 2 R 1 M 1 ) 2 Y = (M 1 R 2 R 1 ) 2 Z = C 2 M 1 = [y, x, z] M 2 = [x, y, z] M 2 = [x, y, 1 z] R 1 = [x, x y, ½ z] R 2 = [1 y, 1 x, ½ z] (194) P6 3 /mmc {M 1, M 2, M 3, R, M 4 = RM 3 R} C = RM 3 (M 1 M 2 ) 3 = (M 1 R) 6 = (M 2 R) 2 = (M 3 M 1 ) 2 = (M 3 M 2 ) 2 = E X = (M 2 M 1 R) 2 Y = (RM 2 M 1 ) 2 Z = C 2 M 1 = [x, x y, z] M 2 = [1 y, 1 x, z] M 3 = [x, y, z] R = [y, x, ½ z] M 4 = [x, y, 1 z] 54
57 PART II. CUBIC GROUPS As in part I, our tabulation gives: (i) the number assigned to the group in the International Tables for Crystallography; its Hermann-Mauguin symbol; {a list of the chosen generators a minimal set followed by additional (redundant) generators that extend the minimal set to a set that relates an asymmetric unit to all contiguous unit}; generators indicated in the H-M symbol, expressed in terms of the chosen set; (ii) a set of generating relations that are sufficient to define the abstract group; (iii) translations expressed in terms of the chosen generators; (iv) a particular realization of the generators in terms of Euclidean transformations; specified in terms of the image of a general point [x, y, z]. (v) a diagram of the asymmetric unit. Explanation of the Figures Each figure illustrates an asymmetric unit. The cube outlined in grey is an eighth of a unit cell, with the axes like this: Twofold axes are indicated in red, threefold axes in green and fourfold axes in blue. Centres of 3 and 4 transformations are indicated by and, respectively. Mirror faces of the units are unmarked. 55
58 (195) P23 {Q, R 1, R 2, R 3 = R 1 R 2 } K 1 = QR 1 L 1 = Q 1 R 1 K 2 = QR 2 L 2 = Q 1 R 2 (R 1 R 2 ) 2 = (QR 3 ) 3 = (L 1 K 2 ) 2 = (K 1 K 2 ) 3 = (L 1 L 2 ) 3 = (K 1 L 2 1 ) 3 = K 1 3 L 2 3 = QR 3 K 2 2 L 1 K 1 = K 1 2 L 1 2 K 1 L 1 2 K 1 2 L 1 = K 2 2 L 2 2 K 2 L 2 2 K 2 2 L 2 = E Z = QR 3 L 1 R 1 = [x, 1 y, z] R 2 = [1 x, y, z] R 3 = [1 x, 1 y, z] (196) F23 {Q, Q 2, R} (Q 1 Q 2 ) 2 = (QQ 2 ) 3 = (QR) 3 = (Q 2 R) 3 = E Z = (Q 2 RQ 1 ) 2 W = RQQ 2 Q Q 2 = [ z, x, y] R = [½ x, ½ y, z] (197) I23 {S 1, S 2, R = S 1 S 2 } Q = S 1 2 S 1 K = QR = S 1 2 S 2 1 S 2 L = Q 1 R = S 1 1 S 2 2 S 1 (S 1 S 2 ) 2 = (S 1 2 S 1 ) 3 = (S 2 1 S 2 2 ) 2 = (S 2 1 S 2 2 ) 3 = S 3 1 S 3 2 S 3 1 S 3 2 = K 2 L 2 KL 2 K 2 L = E Z = (S 1 1 S 2 2 ) 2 W = S 3 1 Z S 1 = [½ + y, ½ z, ½ x] S 2 = [½ + z, ½ x, ½ y] R = [1 x, y, z] Axes of S 1 = 3 2 (λ,λ 1 / 3, 2 / 3 λ)[ 1 / 6, 1 / 6, 1 / 6 ] and S 2 = 3 2 (λ, 2 / 3 λ, λ 1 / 3 )[ 1 / 6, 1 / 6, 1 / 6 ] marked in turquoise. 56
59 (198) P2 1 3 {Q, S 1, S 2, Q 2 = S 1 Q, Q 3 = S 2 Q, Q 4 = QS 1 } S 1 S 2 2 S 1 1 S 2 2 = S 2 S 2 1 S 1 2 S 2 1 = QS 2 2 Q 1 S 2 1 = E 2 Z = S 2 S 1 = [½ + x, ½ y, z] S 2 = [½ x, 1 y, ½ + z] Q 2 = [½ + z, ½ x, y] Q 3 = [½ z, 1 x, ½ + y] Q 4 = [ z, ½ + x, ½ y] S 1 and S 2 are the the 2-fold screw transformations indicated in red. The two blue shaded facets are related through the threefold rotation Q 3 = 3(λ, 1 λ, ½ λ). The two yellow shaded facets are similarly related through Q 4 = 3(λ, λ + ½, λ). (Axes of Q 3 and Q 4 have been omitted in the figure.) (199) I2 1 3 {Q, R 1, R 2, S, R 3 = QR 2 Q 1 } S 2 = R 3 R 1 K 1 = QR 1 L 1 = Q 1 R 1 K 2 = QR 2 L 2 = Q 1 R 2 L 1 K 2 1 K 2 L 2 2 = K 3 2 S 3 = K 1 L 1 R 2 R 1 = (K 2 L 1 ) 2 S 2 = K 2 2 L 2 L 1 = (L 1 K 2 ) 2 (K 2 L 2 ) 2 = L 1 K 1 L 2 1 K 2 2 = (K 1 K 2 2 ) 2 = S 2 2 R 2 S 2 2 R 2 = K 2 1 L 2 1 K 1 L 2 1 K 2 1 L 1 = K 2 2 L 2 2 K 2 L 2 2 K 2 2 L 2 = E 2 Z = S 2 W = K S 2 R 1 = [½ x, y, z] R 2 = [1 x, ½ y, z] S = [z, 1 x, ½ y] R 3 = [x, 1 y, ½ z] The two shaded triangular facets are related through S = 3 1 (λ, 5 / 6 λ, λ 1 / 6 )[ 1 / 6, 1 / 6, 1 / 6 ] (200) Pm 3 {Q, M 1, M 2, M 3 = Q 1 M 1 Q} Q = (M 2 Q) 3 Q (M 1 M 2 ) 2 = (M 2 M 3 ) 2 = (M 1 M 2 M 3 ) 2 = (M 1 QM 1 Q 1 ) 2 = (M 2 QM 2 Q 1 ) 2 = (M 2 Q) 6 = E Z = QM 1 Q 1 M 2 M 1 = [x, 1 y, z] M 2 = [x, y, z] M 3 = [1 x, y, z] 57
60 (201) Pn 3 { Q 1, Q 2, R = Q 2 1 Q 1 } N = R Q 1 3 Q = Q 1 2 Q 2 = Q 2 2 K = QR L = Q 1 R K 2 = Q 2 R L 2 = Q 2 1 R ( Q 1 RQ 1 1 R) 2 = RQ 1 1 Q 2 RQ 2 1 Q 1 = ( Q 1 3 RQ 1 R) 3 = K 2 L 2 KL 2 K 2 L = K 2 2 L 2 2 K 2 L 2 2 K 2 2 L 2 = E Z = ( Q 1 2 RQ 1 1 R) 2 Q 1 = [½ z, ½ x, ½ y] Q 2 = [½ + z, ½ + x, ½ y] R = [1 x, y, z] (202) Fm 3 {Q, M, R} Q = (MQ) 3 Q (RQ) 3 = (RM) 2 = (MQ) 6 = (MQMQ 1 ) 2 = E Z = (Q 1 RQM) 2 W = R(QM) 2 Q R = [½ x, ½ y, z] M = [x, y, z] (203) Fd 3 { Q 1, Q 2 } D = Q 1 Q 2 2 Q 1 2 ( Q 1 Q 1 2 ) 2 = (Q 2 1 Q 2 2 ) 2 = ( Q 2 1 Q 2 2 ) 3 = E Z = Q Q 2 W = D 2 Q 1 = [¼ z, ¼ x, ¼ y] Q 2 = [¼ + z, ¼ x, ¼ + y] 58
61 (204) Im 3 { Q, M, M 2 = Q M Q 1 } I = Q 3 Q = Q 4 (QM) 6 = (MQMQ 1 ) 2 = (IMIQMQ 1 ) 2 = E Z = (M Q 3 ) 2 W = QZQ 1 (MQ) 3 Q = [½ z, ½ x, ½ y] M = [x, y, z] M 2 = [1 + x, y, z] (205) Pa 3 { Q 1, Q 2, Q = ( Q 1 Q 2 ) 1 } A = ( Q 2 2 Q 1 ) 1 Q = Q 3 2 ( Q 1 Q 1 2 ) 2 Q Q 1 6 = Q 2 6 = (Q 1 Q 2 ) 3 = (Q 1 2 Q 2 ) 2 = ( Q 1 2 Q 2 2 ) 2 = ( Q 1 Q 2 1 ) 2 Q 1 3 Q 2 3 = Q 1 2 (Q 2 1 Q 1 Q 2 1 )(Q 1 1 Q 2 Q 1 1 )Q 2 2 (Q 1 1 Q 2 Q 1 1 )(Q 2 1 Q 1 Q 2 1 ) = E Z = ( Q 2 1 Q 1 ) 2 Q 1 = [½ + z, x, ½ y] Q 2 = [z, ½ x, ½ + y] The asymmetric unit has six facets, related in pairs through Q, Q 1 = 3 (½ λ, ½ + λ, λ; ½, ½, 0) and Q 2 = 3 ( λ, ½ λ, λ; 0, ½, 0) (206) Ia 3 { Q, R, I 2 = ( Q 2 R) 2 Q 1 R } A = Q 2 RQ Q = Q 2 I = Q 3 K = Q R L = Q 1 R K 6 = L 6 = (KL) 2 K 3 L 3 = QI 2 Q 1 I 2 = Q RILKL = Q(RI) 2 Q 1 (RI 2 ) 2 = K 2 L 2 KL 2 K 2 L = E Z = (KL) 2 W = I 2 I R = [x, ½ y, z] I 2 = [½ x, ½ y, ½ z] 59
62 (207) P432 {F, R 1, R 2 } Q = F 1 R 2 K = QR 1 L = Q 1 R 1 (R 1 R 2 ) 4 = (R 2 F) 3 = (QR 2 ) 4 = (F 2 R 1 ) 2 = (F 2 R 2 ) 4 = (QFR 1 ) 2 = (FQR 1 ) 4 = (FR 1 F 1 R 1 ) 2 = (FR 1 FR 2 R 1 R 2 ) 2 = K 2 L 2 KL 2 K 2 L = (KL 2 KL 3 ) 2 = E Y = F 2 R 2 R 1 R 2 F = [1 + y, x, z] R 1 = [1 x, y, z] R 2 = [1 x, z, y] (208) P {R 1, R 2, R 3, R 4 } Q = R 2 R 1 Q 2 = R 4 R 3 R 5 = R 1 R 3 R 6 = R 2 R 4 (R 1 R 2 ) 3 = (R 3 R 4 ) 3 = (R 1 R 3 ) 2 = (R 2 R 4 ) 2 = R 1 R 3 R 2 R 4 = E X = R 2 (R 1 R 4 ) 3 R 1 R 1 = [½ z, ½ y, ½ x] R 2 = [½ x, ½ z, ½ y] R 3 = [½ + z, ½ y, ½ + x] R 4 = [½ x, ½ + z, ½ + y] (209) F432 {F, R 1, R 2 } Q = R 1 F 1 (R 1 R 2 ) 2 = (FR 1 ) 3 = (FR 2 ) 3 = (R 1 F 2 ) 4 = (R 2 F 2 ) 4 = (R 1 R 2 F) 4 = E Y = (R 1 R 2 F 2 ) 2 W = R 2 F 2 R 1 F 2 F = [x, z, y] R 1 = [y, x, z] R 2 = [½ y, ½ x, z] 60
63 (210) F {Q, R 1, R 2, R 3 = Q 1 R 2 } S = R 3 R 1 (QR 1 ) 3 = (QR 2 ) 2 = (QR 2 R 1 Q(R 2 R 1 ) 3 ) 2 = ((R 2 R 1 ) 3 Q) 2 = E Z = S 4 W = R 2 R 1 R 2 QR 1 Q 1 R 1 = [x, y, z] R 2 = [¼ z, ¼ y, ¼ x] R 3 = [¼ y, ¼ x, ¼ z] (211) I432 {R 1, R 2, R 3 } Q = R 2 R 1 F = QR 3 Q (R 2 R 1 ) 3 = (QR 3 ) 4 = (FR 1 F 1 R 1 ) 2 = (Q(FR 1 ) 2 ) 2 = (R 1 FQF) 4 = E Z = (R 1 R 2 R 1 R 3 ) 2 W = Q 1 R 1 R 3 F 2 R 1 = [½ z, ½ y, ½ x] R 2 = [½ x, ½ z, ½ y] R 3 = [y, x, z] (212) P {R 1, R 2, R 3, S, Q 2 = R 3 R 1 } Q = R 2 R 1 G = QR 3 J = Q 1 R 3 (R 1 R 2 ) 3 = (R 1 R 3 ) 3 = S 4 Q 2 S 2 Q 1 = S 4 (Q 1 2 Q) 2 = S 2 G 2 = S 4 Q 1 (R 2 R 3 ) 2 Q = (QS 4 Q 1 )S 2 (QS 4 Q 1 )S 2 = (Q 1 S 4 Q)S 2 (Q 1 S 4 Q)S 2 = (G 2 J 3 ) 2 = E Z = S 4 R 1 = [¼ z, ¼ y, ¼ x] R 2 = [¼ x, ¼ z, ¼ y] R 3 = [¾ x, ¼ + z, ¼ + y] S = [ ¼ + y, ¾ x, ¼ + z] Q 2 = [½ + z, ½ x, y] The two shaded triangular facets are related through the fourfold screw transformation S = 4 1 (¼, ½, z)[0, 0, ¼], indicated in blue 61
64 (213) P {R 1, R 2, R 3, S, Q 2 = R 1 R 3 } Q = R 1 R 2 G = QR 3 J = Q 1 R 3 (R 1 R 2 ) 3 = (R 1 R 3 ) 3 = S 4 Q 1 2 S 2 Q = S 4 (QQ 1 2 ) 2 = S 2 J 2 = S 4 Q(R 2 R 3 ) 2 Q 1 = (QS 4 Q 1 )S 2 (QS 4 Q 1 )S 2 = (Q 1 S 4 Q)S 2 (Q 1 S 4 Q)S 2 = (G 2 J 3 ) 2 = E Z = S 4 R 1 = [¼ x, ¼ z, ¼ y] R 2 = [¼ z, ¼ y, ¼ x] R 3 = [¼ + z, ¾ y, ¼ + x] S = [¾ y, ¼ + x, ¼ + z] Q 2 = [ z, ½ x, ½ + y] The two shaded triangular facets are related through the fourfold screw transformation S = 4 1 (½, ¼, z)[0, 0, ¼], indicated in blue (214) I {R 1, R 2, R 3, R 4, R 5 = R 2 R 3 R 2, Q = R 1 R 5, S = R 2 R 3 } (R 1 R 2 ) 2 = (R 1 R 3 ) 3 = (R 1 R 4 ) 3 (R 4 R 3 ) 3 = S 4 Q 1 (R 4 R 2 ) 4 Q = (S 3 R 1 S 2 R 1 ) 2 = E Z = (R 2 R 4 ) 4 W = S 4 Q(R 1 R 4 ) 3 Q 1 R 1 = [¼ x, ¼ z, ¼ y] R 2 = [x, y, ½ z] R 3 = [ ¼ + z, ¼ y, ¼ + x] R 4 = [¼ + y, ¼ + x, ¼ z] R 5 = [¼ z, ¼ y, ¼ x] S = [ ¼ + z, ¼ + y, ¼ x] Shaded facets related through S = 4 1 (0, y, ½)[0, ¼, 0] 62
65 (215) P 4 3m {M 1, M 2, R, M 3 = RM 1 R} F = M 1 R Q = M 1 M 2 K = QR L = Q 1 R (M 1 M 2 ) 3 = (M 3 M 2 ) 3 = (M 1 R) 4 = (M 2 R) 4 = (M 3 R) 4 = K 2 L 2 KL 2 K 2 L = E Y = F 2 M 2 RM 2 M 1 = [y, x, z] M 2 = [x, z, y] R = [1 x, y, z] M 3 = [1 y, 1 x, z] F = [y, 1 x, z] (216) F 4 3m {M 1, M 2, M 3, M 4 } F = M 1 M 2 M 3 Q = M 3 M 1 (M 1 M 2 ) 2 =(M 2 M 3 ) 3 = (M 3 M 1 ) 3 = (M 1 M 4 ) 3 = (M 2 M 4 ) 3 = (M 3 M 4 ) 2 = (M 1 M 2 M 3 ) 4 = (M 2 M 3 M 4 ) 4 = (M 3 M 1 M 4 ) 4 = (M 1 M 2 M 4 ) 4 = E X = M 4 M 1 M 2 M 4 M 3 M 1 M 2 M 3 W = M 4 M 1 M 2 M 3 M 2 M 1 M 1 = [x, z, y] M 2 = [x, z, y] M 3 = [y, x, z] M 4 = [½ y, ½ x, z] (217) I 4 3m { F, M, M 2 = F 1 M F } Q = MM 2 G = F M J = F 1 M K = Q F 2 L = Q 1 F 2 (MM 2 ) 3 = ( F 2 M) 4 = QG 3 Q 1 J 3 = J 3 G 3 J 3 G 3 = M( F Q) 2 M( F Q) 2 = K 2 L 2 KL 2 K 2 L = E Z = ( F Q) 2 W = ZJ 3 F = [½ z, ½ y, ½ + x] M = [y, x, z] M 2 = [x, z, y] 63
66 (218) P 4 3n {Q, F 1, F 2 } N = F F 2 2 K 1 = Q F 1 L 1 = Q F 1 K 2 = Q F 2 L 2 = Q 1 2 F 1 ( F 2 1 F 2 2 ) 2 = (Q F 2 1 F 2 2 ) 3 = (Q F 1 ) 4 = (Q 1 F 2 ) 4 = ( F 1 Q 1 ) 2 ( F 2 Q) 2 = (L 1 K 2 ) 2 = (K 1 K 2 ) 3 = (L 1 L 2 ) 3 = (K 1 L 1 2 ) 3 = K 3 1 L 3 2 = K 2 1 L 2 1 K 1 L 2 1 K 2 1 L 1 = K 2 2 L 2 2 K 2 L 2 2 K 2 2 L 2 = E Z = (Q 1 F 1 ) 2 F 1 = [½ + z, ½ y, ½ x] F 2 = [½ + y, ½ x, ½ z] 1 (219) F 4 3c { F 1, F 2 } Q = F 2 F 1 C = F 2 F 1 1 F 2 2 F 1 ( F 2 F 1 1 ) 3 = ( F 1 F 2 ) 3 = ( F 2 1 F 1 2 ) 4 = ( F 2 2 F 1 1 ) 4 = E Z = C 2 W = (F 1 2 F 2 1 ) 2 F 1 = [½ x, z, y] F 2 = [y, ½ x, z] (220) I 4 3d {Q, F 1, F 2, S, R = Q F 2 2 Q 1 } D = R F 2 2 K 1 = Q F 1 L 1 = Q F 1 K 2 = Q F 2 L 2 = Q 1 2 F 2 ( F 1 F 1 2 ) 3 = (Q F 1 ) 4 = (Q 1 F 2 ) 4 = ( F 1 S) 4 = ( F 2 S) 4 = K 3 1 L 3 2 = K 3 2 Q 1 L 3 1 Q = K 2 1 L 2 1 K 1 L 2 1 K 2 1 L 1 = K 2 2 L 2 2 K 2 L 2 2 K 2 2 L 2 = E Z = (L 1 K 1 ) 2 W = D 2 F 1 = [¼ + z, ¾ y, ¼ x] F 2 = [¼ + y, ¾ x, ¼ z] S = [z, 1 x, ½ y] R = [x, 1 y, ½ z] Shaded triangular facets related through Shaded triangular facets related through S = 3 1 (λ, 5 / 6 λ, λ 1 / 6 )[ 1 / 6, 1 / 6, 1 / 6 ] S = 3 1 (λ, 5 / 6 λ, λ 1 / 6 )[ 1 / 6, 1 / 6, 1 / 6 ]??? 64
DIFFRACTION METHODS IN MATERIAL SCIENCE. PD Dr. Nikolay Zotov Lecture 4_2
DIFFRACTION METHODS IN MATERIAL SCIENCE PD Dr. Nikolay Zotov Email: zotov@imw.uni-stuttgart.de Lecture 4_2 OUTLINE OF THE COURSE 0. Introduction 1. Classification of Materials 2. Defects in Solids 3. Basics
More informationChapter 1. Introduction
Chapter 1. Introduction 1. Generation of X-ray 1) X-ray: an electromagnetic wave with a wavelength of 0.1 100 Å. 2) Types of X-ray tubes a) A stationary-anode tube b) A rotating-anode tube 3) X-ray tube:
More informationAppendixes POINT GROUPS AND THEIR CHARACTER TABLES
Appendixes APPENDIX I. POINT GROUPS AND THEIR CHARACTER TABLES The following are the character tables of the point groups that appear frequently in this book. The species (or the irreducible representations)
More information3.1. Space-group determination and diffraction symbols
International Tables for Crystallography (2006). Vol. A, Chapter 3.1, pp. 44 54. 3.1. Space-group determination and diffraction symbols BY A. LOOIJENGA-VOS AND M. J. BUERGER 3.1.1. Introduction In this
More informationInfluence of Chemical and Spatial Constraints on the Structures of Inorganic Compounds
381 Acta Cryst. (1997). B53, 381-393 Influence of Chemical and Spatial Constraints on the Structures of Inorganic Compounds I. D. BROWN Brockhouse Institute of Materials Research, McMaster University,
More informationMultipole Superconductivity and Unusual Gap Closing
Novel Quantum States in Condensed Matter 2017 Nov. 17, 2017 Multipole Superconductivity and Unusual Gap Closing Application to Sr 2 IrO 4 and UPt 3 Department of Physics, Kyoto University Shuntaro Sumita
More informationTranslational symmetry, point and space groups in solids
Translational symmetry, point and space groups in solids Michele Catti Dipartimento di Scienza dei Materiali, Universita di Milano Bicocca, Milano, Italy ASCS26 Spokane Michele Catti a = b = 4.594 Å; Å;
More informationCrystallographic Symmetry. Jeremy Karl Cockcroft
Crystallographic Symmetry Jeremy Karl Cockcroft Why bother? To describe crystal structures Simplifies the description, e.g. NaCl structure Requires coordinates for just 2 atoms + space group symmetry!
More informationOverview - Macromolecular Crystallography
Overview - Macromolecular Crystallography 1. Overexpression and crystallization 2. Crystal characterization and data collection 3. The diffraction experiment 4. Phase problem 1. MIR (Multiple Isomorphous
More informationresearch papers Homogeneous sphere packings with triclinic symmetry 1. Introduction 2. Sphere packings corresponding to lattice complex P11 1a
Acta Crystallographica Section A Foundations of Crystallography ISSN 008-7673 Received 3 May 2002 Accepted 7 July 2002 Homogeneous sphere packings with triclinic symmetry W. Fischer and E. Koch* Institut
More informationPX-CBMSO Course (2) of Symmetry
PX-CBMSO Course (2) The mathematical description of Symmetry y PX-CBMSO-June 2011 Cele Abad-Zapatero University of Illinois at Chicago Center for Pharmaceutical Biotechnology. Lecture no. 2 This material
More informationSPACE GROUPS. International Tables for Crystallography, Volume A: Space-group Symmetry. Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain
SPACE GROUPS International Tables for Crystallography, Volume A: Space-group Symmetry Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain SPACE GROUPS Crystal pattern: Space group G: A model of the
More informationFundamentals. Crystal patterns and crystal structures. Lattices, their symmetry and related basic concepts
Fundamentals. Crystal patterns and crystal structures. Lattices, their symmetry and related basic concepts Didactic material for the MaThCryst schools, France massimo.nespolo@univ-lorraine.fr Ideal vs.
More informationINTERNATIONAL SCHOOL ON FUNDAMENTAL CRYSTALLOGRAPHY
INTERNATIONAL SCHOOL ON FUNDAMENTAL CRYSTALLOGRAPHY SPACE-GROUP SYMMETRY (short overview) Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain SPACE GROUPS Crystal pattern: infinite, idealized crystal
More informationTables of crystallographic properties of double antisymmetry space groups
Tables of crystallographic properties of double antisymmetry space groups Mantao Huang a, Brian K. VanLeeuwen a, Daniel B. Litvin b and Venkatraman Gopalan a * a Department of Materials Science and Engineering,
More informationThe structure of liquids and glasses. The lattice and unit cell in 1D. The structure of crystalline materials. Describing condensed phase structures
Describing condensed phase structures Describing the structure of an isolated small molecule is easy to do Just specify the bond distances and angles How do we describe the structure of a condensed phase?
More informationLecture 2 Symmetry in the solid state -
Lecture 2 Symmetry in the solid state - Part II: Crystallographic coordinates and Space Groups. 1 Coordinate systems in crystallography and the mathematical form of the symmetry operators 1.1 Introduction
More informationCrystallography Reading: Warren, Chapters 2.1, 2.2, 2.6, 8 Surface symmetry: Can be a clue to underlying structure. Examples:
Crystallography Reading: Warren, Chapters 2.1, 2.2, 2.6, 8 Surface symmetry: Can be a clue to underlying structure. Examples: Snow (SnowCrystals.com) Bismuth (Bao, Kavanagh, APL 98 66103 (2005) Hexagonal,
More informationCrystallography basics
Crystallography basics 1 ? 2 Family of planes (hkl) - Family of plane: parallel planes and equally spaced. The indices correspond to the plane closer to the origin which intersects the cell at a/h, b/k
More informationSymmetry in 2D. 4/24/2013 L. Viciu AC II Symmetry in 2D
Symmetry in 2D 1 Outlook Symmetry: definitions, unit cell choice Symmetry operations in 2D Symmetry combinations Plane Point groups Plane (space) groups Finding the plane group: examples 2 Symmetry Symmetry
More informationBasics of crystallography
Basics of crystallography 1 Family of planes (hkl) - Family of plane: parallel planes and equally spaced. The indices correspond to the plane closer to the origin which intersects the cell at a/h, b/k
More informationM\1any arguments have been concerned with what these symbols mean, and how they
SOME DESIRABLE MODIFICATIONS OF THE INTERNATIONAL SYMMETRY SYMBOLS* BY MARTIN J. BUERGER MASSACHUSETTS INSTITUTE OF TECHNOLOGY Communicated August 21, 1967 With the publication of Hilton's Mathematical
More informationCrystallographic Point Groups and Space Groups
Crystallographic Point Groups and Space Groups Physics 251 Spring 2011 Matt Wittmann University of California Santa Cruz June 8, 2011 Mathematical description of a crystal Definition A Bravais lattice
More informationWALLPAPER GROUPS. Julija Zavadlav
WALLPAPER GROUPS Julija Zavadlav Abstract In this paper we present the wallpaper groups or plane crystallographic groups. The name wallpaper groups refers to the symmetry group of periodic pattern in two
More informationIntroduction to crystallography The unitcell The resiprocal space and unitcell Braggs law Structure factor F hkl and atomic scattering factor f zθ
Introduction to crystallography The unitcell The resiprocal space and unitcell Braggs law Structure factor F hkl and atomic scattering factor f zθ Introduction to crystallography We divide materials into
More informationWednesday, April 12. Today:
Wednesday, April 2 Last Time: - The solid state - atomic arrangement in solids - why do solids form: energetics - Lattices, translations, rotation, & other symmetry operations Today: Continue with lattices,
More informationCrystal Symmetry Algorithms in a. Research
Crystal Symmetry Algorithms in a High-Throughput Framework for Materials Research by Richard Taylor Department of Mechanical Engineering and Materials Science Duke University Date: Approved: Stefano Curtarolo,
More informationSymmetry mode analysis in the Bilbao Crystallographic Server: The program AMPLIMODES
Symmetry mode analysis in the Bilbao Crystallographic Server: The program AMPLIMODES http://www.cryst.ehu.es AMPLIMODES Symmetry Modes Analysis Modes in the statics of low-symmetry distorted phases: Distorted
More informationTHE FIVE TYPES OF PLANAR 2-D LATTICES. (d) (e)
THE FIVE TYPES OF PLANAR 2-D LATTICES (a) (d) (b) (d) and (e) are the same (e) (c) (f) (a) OBLIQUE LATTICE - NO RESTRICTIONS ON ANGLES BETWEEN THE UNIT CELL EDGES (b) RECTANGULAR LATTICE - ANGLE BETWEEN
More informationResolution of Ambiguities and the Discovery of
ISST Journal of Applied hysics, Vol. 6 No. 1, (January - June), p.p. 1-10 ISSN No. 0976-90X Intellectuals Society for Socio-Techno Welfare Resolution of Ambiguities and the Discovery of Two New Space Lattices
More informationINTERNATIONAL SCHOOL ON FUNDAMENTAL CRYSTALLOGRAPHY AND WORKSHOP ON STRUCTURAL PHASE TRANSITIONS. 30 August - 4 September 2017
INTERNATIONAL SCHOOL ON FUNDAMENTAL CRYSTALLOGRAPHY AND WORKSHOP ON STRUCTURAL PHASE TRANSITIONS 30 August - 4 September 2017 ROURKELA INTERNATIONAL CRYSTALLOGRAPHY SCHOOL BILBAO CRYSTALLOGRAPHIC SERVER
More informationSymmetry Crystallography
Crystallography Motif: the fundamental part of a symmetric design that, when repeated, creates the whole pattern In 3-D, translation defines operations which move the motif into infinitely repeating patterns
More informationSPACE GROUPS AND SYMMETRY
SPACE GROUPS AND SYMMETRY Michael Landsberg Electron Crystallography Workshop C-CINA, Basel, 1-7 Aug 2010 m.landsberg@uq.edu.au Averaging Why single molecule EM techniques are far superior in resolution
More informationNove fizickohemijske metode. Ivana Radosavljevic Evans Durham University, UK
Nove fizickohemijske metode Ivana Radosavljevic Evans Durham University, UK Nove fizickohemijske metode: Metode zasnovane na sinhrotronskom zracenju Plan predavanja: Difrakcione metode strukturne karakterizacije
More information5 Symmetries and point group in a nut shell
30 Phys520.nb 5 Symmetries and point group in a nut shell 5.1. Basic ideas: 5.1.1. Symmetry operations Symmetry: A system remains invariant under certain operation. These operations are called symmetry
More informationTILES, TILES, TILES, TILES, TILES, TILES
3.012 Fund of Mat Sci: Structure Lecture 15 TILES, TILES, TILES, TILES, TILES, TILES Photo courtesy of Chris Applegate. Homework for Fri Nov 4 Study: Allen and Thomas from 3.1.1 to 3.1.4 and 3.2.1, 3.2.4
More informationTim Hughbanks CHEMISTRY 634. Two Covers. Required Books, etc.
CHEMISTRY 634 This course is for 3 credits. Lecture: 2 75 min/week; TTh 11:10-12:25, Room 2122 Grades will be based on the homework (roughly 25%), term paper (15%), midterm and final exams Web site: http://www.chem.tamu.edu/rgroup/
More informationSymmetry mode analysis in the Bilbao Crystallographic Server: The program AMPLIMODES
Symmetry mode analysis in the Bilbao Crystallographic Server: The program AMPLIMODES Acknowledgements: Bilbao: M. Aroyo, D. Orobengoa, J.M. Igartua Grenoble: J. Rodriguez-Carvajal Provo, Utah: H.T. Stokes,
More informationAlgebra I. Book 2. Powered by...
Algebra I Book 2 Powered by... ALGEBRA I Units 4-7 by The Algebra I Development Team ALGEBRA I UNIT 4 POWERS AND POLYNOMIALS......... 1 4.0 Review................ 2 4.1 Properties of Exponents..........
More informationBasic Crystallography Part 1. Theory and Practice of X-ray Crystal Structure Determination
Basic Crystallography Part 1 Theory and Practice of X-ray Crystal Structure Determination We have a crystal How do we get there? we want a structure! The Unit Cell Concept Ralph Krätzner Unit Cell Description
More information2016 Annual Implementation Plan: for Improving Student Outcomes. John Monash Science School Based on Strategic Plan
885 J M Scc Sc B Src -9 G fc ffr wr, fr rr b f fr r Vcr r c y. fr rr r: Excc c r rf r c fr r Cy r. Sx c-b c fy ffc, r c-b r w w ccy r r c. r c w fr -w rr, fw wy ( rfr Frwrk fr r S Oc: G fr c): Er rry Er
More informationChapter 2 Introduction to Phenomenological Crystal Structure
Chapter 2 Introduction to Phenomenological Crystal Structure 2.1 Crystal Structure An ideal crystal represents a periodic pattern generated by infinite, regular repetition of identical microphysical structural
More informationtrawhmmry ffimmf,f;wnt
r nsr rwry fff,f;wn My 26, $51 Swe, k "Te Srwberry Cp f e Vr,, c) [ re ers 6 (, r " * f rn ff e # s S,r,* )er*,3n*,.\ ) x 8 2 n v c e 6 r D r, } e ;s 1 :n..< Z r : 66 3 X f; 1r_ X r { j r Z r 1r 3r B s
More informationLecture 1 Symmetry in the solid state -
Lecture 1 Symmetry in the solid state - Part I: Simple patterns and groups 1 Symmetry operators: key concepts Operators: transform (move) the whole pattern (i.e., the attributes, or content, of all points
More information2016 Annual Implementation Plan: For Improving Student Outcomes Guide to developing the Annual Implementation Plan: for Improving Student Outcomes
: Fr rv S Oc G v : fr rv S Oc fc ffr wr, fr rr v b f fr r Vcr vr c y. fr rr r: 478 Ovrr rry Sc Excc c r rf r v c fr r Cy r. Sx vc-b v c fy ffcv, rv vc-b r w w ccy rv rv c. v r c w fr -w rr, fw wy ( rfr
More informationChemical Crystallography
Chemical Crystallography Prof Andrew Goodwin Michaelmas 2014 Recap: Lecture 1 Why does diffraction give a Fourier transform? k i = k s = 2π/λ k i k s k i k s r l 1 = (λ/2π) k i r l 2 = (λ/2π) k s r Total
More informationVERTEX-TRANSITIVE MAPS ON A TORUS. 1. Introduction
VERTEX-TRANSITIVE MAPS ON A TORUS ONDREJ ŠUCH Abstract. We examine FVT (free, vertex transitive) actions of wallpaper groups on semiregular tilings. By taking quotients by lattices we then obtain various
More informationCrystal Structure. Dr Bindu Krishnan
Solid State Physics-1 Crystal Structure Dr Bindu Krishnan CRYSTAL LATTICE What is crystal (space) lattice? In crystallography, only the geometrical properties of the crystal are of interest, therefore
More informationThe structure of planar defects in tilted perovskites
The structure of planar defects in tilted perovskites RichardBeanland Department of Physics, University of Warwick, Coventry, CV4 7AL, UK. E-mail: r.beanland@warwick.ac.uk Synopsis Continuity of octahedral
More informationRoger Johnson Structure and Dynamics: The 230 space groups Lecture 3
Roger Johnson Structure and Dnamics: The 23 space groups Lecture 3 3.1. Summar In the first two lectures we considered the structure and dnamics of single molecules. In this lecture we turn our attention
More informationISOTROPY. Tutorial. July 2013
ISOTROPY Tutorial July 2013 Harold T. Stokes, Dorian M. Hatch, and Branton J. Campbell Department of Physics and Astronomy Brigham Young University with a contribution by Christopher J. Howard University
More informationUNIT I SOLID STATE PHYSICS
UNIT I SOLID STATE PHYSICS CHAPTER 1 CRYSTAL STRUCTURE 1.1 INTRODUCTION When two atoms are brought together, two kinds of forces: attraction and repulsion come into play. The force of attraction increases
More information35H MPa Hydraulic Cylinder 3.5 MPa Hydraulic Cylinder 35H-3
- - - - ff ff - - - - - - B B BB f f f f f f f 6 96 f f f f f f f 6 f LF LZ f 6 MM f 9 P D RR DD M6 M6 M6 M. M. M. M. M. SL. E 6 6 9 ZB Z EE RC/ RC/ RC/ RC/ RC/ ZM 6 F FP 6 K KK M. M. M. M. M M M M f f
More informationON DIVISION ALGEBRAS*
ON DIVISION ALGEBRAS* BY J. H. M. WEDDERBURN 1. The object of this paper is to develop some of the simpler properties of division algebras, that is to say, linear associative algebras in which division
More informationMATRIX CALCULUS APPLIED TO CRYSTALLOGRAPHY. (short revision) Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain
MATRIX CALCULUS APPLIED TO CRYSTALLOGRAPHY (short revision) Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain INTRODUCTION TO MATRIX CALCULUS Some of the slides are taken from the presentation Introduction
More informationCRYSTALLOGRAPHIC SYMMETRY OPERATIONS. Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain
CRYSTALLOGRAPHIC SYMMETRY OPERATIONS Mois I. Aroyo Universidad del Pais Vasco, Bilbao, Spain SYMMETRY OPERATIONS AND THEIR MATRIX-COLUMN PRESENTATION Mappings and symmetry operations Definition: A mapping
More informationGeometry JWR. Monday September 29, 2003
Geometry JWR Monday September 29, 2003 1 Foundations In this section we see how to view geometry as algebra. The ideas here should be familiar to the reader who has learned some analytic geometry (including
More informationArithmetic Operations. The real numbers have the following properties: In particular, putting a 1 in the Distributive Law, we get
MCA AP Calculus AB Summer Assignment The following packet is a review of many of the skills needed as we begin the study of Calculus. There two major sections to this review. Pages 2-9 are review examples
More informationCrystal Chem Crystallography
Crystal Chem Crystallography Chemistry behind minerals and how they are assembled Bonding properties and ideas governing how atoms go together Mineral assembly precipitation/ crystallization and defects
More informationIMO Training Camp Mock Olympiad #2 Solutions
IMO Training Camp Mock Olympiad #2 Solutions July 3, 2008 1. Given an isosceles triangle ABC with AB = AC. The midpoint of side BC is denoted by M. Let X be a variable point on the shorter arc MA of the
More informationQUESTION BANK ON STRAIGHT LINE AND CIRCLE
QUESTION BANK ON STRAIGHT LINE AND CIRCLE Select the correct alternative : (Only one is correct) Q. If the lines x + y + = 0 ; 4x + y + 4 = 0 and x + αy + β = 0, where α + β =, are concurrent then α =,
More informationRaman and IR spectroscopy in materials science. Symmetry analysis of normal phonon modes Boriana Mihailova
University of Hamburg, Institute of Mineralogy and Petrology Raman and IR spectroscopy in materials science. Symmetry analysis of normal phonon modes Boriana Mihailova Outline. The dynamics of atoms in
More informationExcellence in teaching and learning
: fr rv S Oc 8 M Ez SC B Src 5-8 Er G v : fr rv S Oc fc ffr wr, fr rr v b f fr r Vcr vr c y. fr rr r: Excc c r rf r v c fr r Cy r. Sx vc-b v c fy ffcv, rv vc-b r w w ccy rv rv c. v r c w fr -w rr, fw wy
More informationLecture course on crystallography, 2015 Lecture 9: Space groups and International Tables for Crystallography
Dr Semën Gorfman Department of Physics, University of SIegen Lecture course on crystallography, 2015 Lecture 9: Space groups and International Tables for Crystallography UNIT CELL and ATOMIC POSITIONS
More informationTHE MIDWAY & GAMES GRADE 4 SOCIAL STUDIES DEEP IN THE HEART OF TEXAS THE TEXAS STAR ILLUMINATED
THE MIDWY & GMES GRDE 4 SOCI STUDIES DEEP IN THE HERT OF TEXS THE TEXS STR IUMINTED T Mw TECHER G F SOCIIES STUD Dp H f Tx T Tx S Im I w: z pc mb p f. Smmz f S. U v pm c c cq fm b Tx. C w f Tx S., c v
More informationAn eightfold path to E 8
An eightfold path to E 8 Robert A. Wilson First draft 17th November 2008; this version 29th April 2012 Introduction Finite-dimensional real reflection groups were classified by Coxeter [2]. In two dimensions,
More informationMineralogy Problem Set Crystal Systems, Crystal Classes
Mineralogy Problem Set Crystal Systems, Crystal Classes (1) For each of the three accompanying plane patterns: (a) Use a ruler to draw solid lines to show where there are mirror planes on the pattern.
More informationNozha Directorate of Education Form : 2 nd Prep
Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep Nozha Language Schools Geometry Revision Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. In the parallelogram, each
More informationThe Licking County Health Department 675 Price Rd., Newark, OH (740)
T Liki y Drm 675 Pri R. Nrk O 43055 (740) 349-6535.Liki.r @iki.r A R r # W Ar Pbi Amim i Liki y : U P Sri m LD ff i ri fr fbk iify ri f Br i y fr r mmi P. Imrvm R. J b R.S. M.S. M.B.A. Liki y r mmii m
More informationCondensed Matter Physics Prof. G. Rangarajan Department of Physics Indian Institute of Technology, Madras
Condensed Matter Physics Prof. G. Rangarajan Department of Physics Indian Institute of Technology, Madras Lecture - 03 Symmetry in Perfect Solids Worked Examples Stated without prove to be in the lecture.
More informationUnderstanding Single-Crystal X-Ray Crystallography Exercises and Solutions
Understanding Single-Crystal X-Ray Crystallography Exercises and Solutions Dennis W. Bennett Department of Chemistry and Biochemistry University of Wisconsin-Milwaukee Chapter Crystal Lattices. The copper
More informationCRYSTAL CELLS IN GEOMETRIC ALGEBRA
Proceedings of the International Symposium on Advanced Mechanical Engineering Between University of Fukui Pukyong National University, November 27, 2004, University of Fukui, Fukui, Japan, Organized by
More informationWeek 4-5: Binary Relations
1 Binary Relations Week 4-5: Binary Relations The concept of relation is common in daily life and seems intuitively clear. For instance, let X be the set of all living human females and Y the set of all
More informationTopic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths
Topic 2 [312 marks] 1 The rectangle ABCD is inscribed in a circle Sides [AD] and [AB] have lengths [12 marks] 3 cm and (\9\) cm respectively E is a point on side [AB] such that AE is 3 cm Side [DE] is
More informationIntroduction to Crystallography and Mineral Crystal Systems by Mike and Darcy Howard Part 6: The Hexagonal System
Introduction to Crystallography and Mineral Crystal Systems by Mike and Darcy Howard Part 6: The Hexagonal System Now we will consider the only crystal system that has 4 crystallographic axes! You will
More informationNeutron Powder Diffraction Theory and Instrumentation
NTC, Taiwen Aug. 31, 212 Neutron Powder Diffraction Theory and Instrumentation Qingzhen Huang (qing.huang@nist.gov) NIST Center for Neutron Research (www.ncnr.nist.gov) Definitions E: energy; k: wave vector;
More informationKey: Town of Eastham - Fiscal Year :52 am
//7 SQ #:, RR WR PR SS SS SRP R Key: own of astham - Fiscal Year 8 : am S WY c/o SM RWR S WY SM, M 6 8-8- S WY M-S M of 7 RSFR SRY S WY SR W V & JY /SF/ bhd F F J S SF F P V R M J V S 97,, -.8,7,6 S 6/8/
More informationApplications of X-ray and Neutron Scattering in Biological Sciences: Symmetry in direct and reciprocal space 2012
Department of Drug Design and Pharmacology Applications of X-ray and Neutron Scattering in Biological Sciences: Symmetry in direct and reciprocal space 2012 Michael Gajhede Biostructural Research Copenhagen
More informationLattices and Symmetry Scattering and Diffraction (Physics)
Lattices and Symmetry Scattering and Diffraction (Physics) James A. Kaduk INEOS Technologies Analytical Science Research Services Naperville IL 60566 James.Kaduk@innovene.com 1 Harry Potter and the Sorcerer
More information" W I T H M C A L I C E T O " W ^ V H, 3 D N O N E ^ N D O H A - R I T Y F O R A. L L. NOBODY'S 6LAIM, deaths off Wm. itafftkon and Mrs. Kennodyb..
~ M M U M M «««M URZZ F)R U V Q Y \ $ R YJ UMUM!!!!!! MD M C C V 3 D D R Y F R V CUY MC MRC 2 89 39 C F C D J M R Y F! < «F C V C F :: $6 FC MC VR D UCQ C x M M R q R Y Y J C [ CM F FURUR D UDRR JD J YDR
More informationPART 1 Introduction to Theory of Solids
Elsevier UK Job code: MIOC Ch01-I044647 9-3-2007 3:03p.m. Page:1 Trim:165 240MM TS: Integra, India PART 1 Introduction to Theory of Solids Elsevier UK Job code: MIOC Ch01-I044647 9-3-2007 3:03p.m. Page:2
More information2007 Hypatia Contest
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 007 Hypatia Contest Wednesday, April 18, 007 Solutions c
More informationSymmetry. 2-D Symmetry. 2-D Symmetry. Symmetry. EESC 2100: Mineralogy 1. Symmetry Elements 1. Rotation. Symmetry Elements 1. Rotation.
Symmetry a. Two-fold rotation = 30 o /2 rotation a. Two-fold rotation = 30 o /2 rotation Operation Motif = the symbol for a two-fold rotation EESC 2100: Mineralogy 1 a. Two-fold rotation = 30 o /2 rotation
More informationCBSE Sample Papers for Class 10 SA2 Maths Solved 2016 Set 3
CBSE Sample Papers for Class 10 SA Maths Solved 016 Set Answers: Section A 1.Determine the value of k for which the indicated value of x is a solution: x + kx- 4 = o; X = -4. x = - 4 is a solution x +
More informationAustralian Intermediate Mathematics Olympiad 2014
5. Let 1 a + 1 b = 1 20, where a and b are positive integers. Find the largest value of a + b. [4 marks] A u s t r a l i a n M at h e m at i c a l O ly m p i a d C o m m i t t e e a d e p a r t m e n t
More informationn-dimensional, infinite, periodic array of points, each of which has identical surroundings.
crystallography ll Lattice n-dimensional, infinite, periodic array of points, each of which has identical surroundings. use this as test for lattice points A2 ("bcc") structure lattice points Lattice n-dimensional,
More informationTHE BILBAO CRYSTALLOGRAPHIC SERVER EXERCISES
INTERNATIONAL TABLES FOR CRYSTALLOGRAPHY Volume A: Space-group Symmetry Volume A1: Symmetry Relations between Space Groups THE BILBAO CRYSTALLOGRAPHIC SERVER EXERCISES Mois I. Aroyo Departamento Física
More informationWeek 4-5: Binary Relations
1 Binary Relations Week 4-5: Binary Relations The concept of relation is common in daily life and seems intuitively clear. For instance, let X be the set of all living human females and Y the set of all
More informationBY MolmS NEWMAN (1011) (11
THE STRUCTURE OF SOME SUBGROUPS OF THE MODULAR GROUP BY MolmS NEWMAN Introduction Let I be the 2 X 2 modular group. In a recent article [7] the notion of the type of a subgroup A of r was introduced. If
More informationMolecular Spectroscopy. January 24, 2008 Introduction to Group Theory and Molecular Groups
Molecular Spectroscopy January 24, 2008 Introduction to Group Theory and Molecular Groups Properties that define a group A group is a collection of elements that are interrelated based on certain rules
More informationChapter 4. Crystallography. 4.1 The crystalline state
Crystallography Atoms form bonds which attract them to one another. When you put many atoms together and they form bonds amongst themselves, are there any rules as to how they order themselves? Can we
More informationThe 17 Plane Symmetry Groups. Samantha Burns Courtney Fletcher Aubray Zell Boise State University
The 17 Plane Symmetry Groups Samantha Burns Courtney Fletcher Aubray Zell Boise State University INTRODUCTION Our paper is going to be about the 17 plane Symmetry Groups, also called the wallpaper groups,
More informationOfficial Solutions 2014 Sun Life Financial CMO Qualifying Rêpechage 1
Official s 2014 Sun Life Financial CMO Qualifying Rêpechage 1 1. Let f : Z Z + be a function, and define h : Z Z Z + by h(x, y) = gcd(f(x), f(y)). If h(x, y) is a two-variable polynomial in x and y, prove
More informationAlgorithm for conversion between geometric algebra versor notation and conventional crystallographic symmetry-operation symbols
Algorithm for conversion between geometric algebra versor notation and conventional crystallographic symmetry-operation symbols Eckhard Hitzer and Christian Perwass June, 2009 Introduction This paper establishes
More informationSTRAIGHT LINES EXERCISE - 3
STRAIGHT LINES EXERCISE - 3 Q. D C (3,4) E A(, ) Mid point of A, C is B 3 E, Point D rotation of point C(3, 4) by angle 90 o about E. 3 o 3 3 i4 cis90 i 5i 3 i i 5 i 5 D, point E mid point of B & D. So
More informationVectors - Applications to Problem Solving
BERKELEY MATH CIRCLE 00-003 Vectors - Applications to Problem Solving Zvezdelina Stankova Mills College& UC Berkeley 1. Well-known Facts (1) Let A 1 and B 1 be the midpoints of the sides BC and AC of ABC.
More informationPOINT SYMMETRY AND TYPES OF CRYSTAL LATTICE
POINT SYMMETRY AND TYPES OF CRYSTAL LATTICE Abdul Rashid Mirza Associate Professor of Physics. Govt. College of Science, wahdatroad, Lahore. 1 WHAT ARE CRYSTALS? The word crystal means icy or frozen water.
More information4.3 Equations in 3-space
4.3 Equations in 3-space istance can be used to define functions from a 3-space R 3 to the line R. Let P be a fixed point in the 3-space R 3 (say, with coordinates P (2, 5, 7)). Consider a function f :
More informationJunior Soldiers Unit 13 : Lesson 1
J S U 1 : L 1 R Bb PURPOSE: T v c pp xp pc f Bb f fm B Pc. Ev bf m, G v c C b f. Ep 1: NLT) C Pp R: Ep 1:-10 NLT) 1 C 8:6 CEV) Ep :-5 CEV) T Bb c b cf m, pc fc m p f. T v c B Pc m f Bb. W vc f G W, p v
More informationInorganic materials chemistry and functional materials
Chemical bonding Inorganic materials chemistry and functional materials Helmer Fjellvåg and Anja Olafsen Sjåstad Lectures at CUTN spring 2016 CRYSTALLOGRAPHY - SYMMETRY Symmetry NATURE IS BEAUTIFUL The
More information