American Mineralogist, Volwne 61, pages 145-165, 1976
A derivationof the 32 crystallographicpoint groupsusing elementary group theory
MoNrn B. Borsnu.Jn., nNn G. V. Gtsss Virginia PolytechnicInstitute and State Uniuersity Blacksburg,Virginia 2406I
Abstract
A rigorous derivation of the 32 crystallographic point groups is presented.The derivation is designed for scientistswho wish to gain an appreciation of the group theoretical derivation but who do not wish to master the large number of specializedtopics used in other rigorous group theoreticalderivations.
Introduction have appealedto a number of interestingalgorithms basedon various geometricand heuristic arguments This paper has two objectives.The first is the very to obtain the 32 crystallographicpoint groups.In this specificgoal of giving a complete and rigorous deri- paper we present a rigorous derivation of the 32 vation of the 32 crystallographicpoint groups with crystallographicpoint groups that usesonly the most emphasisplaced on making this paper as self-con- elementarynotions of group theory while still taking tained as possible.The second is the more general advantageof the power of the theory of groups. objective of presenting several concepts of modern The derivation given here was inspired by the dis- mathematics and their applicability to the study of cussionsgiven in Klein (1884),Weber (1896),Zas- crystallography.Throughout the paper we have at- senhaus(1949), and Weyl (1952).The mathematical tempted to present as much of the mathematical the- approach usedin their discussionsis a blend of group ory as possible,without resortingto long digressions, theory and the theory of the equivalencerelation' so that the reader with only a modest amount of Each of these mathematical concepts is described mathematical knowledge will be able to appreciate herein and is used in the ways suggestedby these the derivation. Those facts that are used but not authors. However we depart from their approach in proven in this paper will be clearly identified and the determinationof the interaxialangles. While they referenceswill be given to sources where the appro- appeal to a study of Platonic solids,we continuewith priate discussionscan be found. the theory of groups and equivalencerelations in our Crystal symmetry was studied during the nine- determinationof theseangles. teenth century largely from a geometricalviewpoint. In the first part of the paper we discussthe defini- A comprehensiveoutline of the more important con- tion and propertiesof point isometriesand show that tributions to this subject by such men as Hessel, everypoint isometryis eithera proper or an improper Bravais, Miibius, Gadolin, Curie, Federov, Min- rotation. If a rotation leavesa lattice invariant, then nigerode, Schoenflies, and Miers is presented by it is shown that its turn anglemust be one of the eight Swartz (1909). It appears that Minnigerode (1884), followingpossibilities:0o, +60o, +90o, +120o, 180o. Schoenflies(1891), and Weber (1896)were the first to The method for composing two point isometriesis recast the problem of classifying crystal symmetry in discussed.The propertiesdiscovered about this conr- terms of group theory. Recent treatmentsmake ex- position motivate the definition of an abstractgroup, tensive use of specializedtopics in group theory in the which is then stated. We then show that the set of all derivation of the crystallographicpoint groups (see point isometriesthat leavesa given lattice invariant for example Seitz, 1934; Zachariasen, 1945; Burck- forms a finite group. These are the crystallographic hardt, 1947;Zassenhaus, 1949; Lomont, 1959;Weyl, point groups. In the secondpart, the proper crystal- 1952;McWeeny, 1963;Altmann, 1963;Yale, 1968; lographic groups are discussed.The cyclic monaxial Benson and Grove, l97l; Janssen, 1973; Coxeter, crystallographic groups are treated first. Then the 1973;andSenechal, 1976). In addition, severalwork- notion of an equivalencerelation and equivalence ers (for example,Donnay, 1942,1967; Buerger, 1963) classesare presentedin conjunction with a group t45 t46 M. B. BOISEN, JR,, AND G. V. GIBBS
z Point isomefiies ln our study of point groups we are interestedin a special kind of transformation known as a point isometry. The mapping O from R3onto RBis a point isometry if and only if it satisfiesthe following four properties:
(l) Given any element4e R3,its image O($ is a ' uniquely R3. (This - ->\y determined element of means that O is a mapping from Rato R3.)* Ftc l. Right-handed cartesian coordinate system with unit (2) Given any element4€ R3,there exists an ele- vectors ryr{ directed along the coordinate axesX, Y, and Z, respec- ment ,r E R3such that O(q) : r,. (This means tively, with the vector r expressedas a linear combination j of i, that O is a mapping from R3 R3.) and k onto (3) The magnitude of r is equal to the magni- tude of the image of r under O for all4€ R3, i.r., : (This meansthat o pre- theoretical development of the basic ideas leading lltll llotdll. servesmagnitudes and henceangles, sizes, and to the derivation of all the proper polyaxial shapes. crystallographicgroups togetherwith their interaxial ) (4) Givenany two elements4s g R3,O(r *-r) : angles.In the final part of the paper, the improper O(9,)+ O(s) and given any real numberx, crystallographic groups are constructed from the o(x4) : In particular,if4= yi-+ proper crystallographicgroups. xO(4). x!.+ z\,then o(d : x
__/.,,1"""",sin2p P(g)=(cos2p,sin2p,O) ,/>" iI .,
g=P(d= (cosp,sinp,O) '-=,,,o,o
o. i b. X Ftc. 2. (a) Diagram showing how the turn angle p of the rotation isometry P is measured with respectto the rotation axis{; (b) a drawing of the basesvectors {/J} of f' viewed down the rotation axis of P where f is the set of all vectors in I perpendicular to the aiii df P. It is assumedwithout loss ofgen- erality that the unit length along the X axis is llrll. The vectorr is the imageofr under P. Accord- ingly, s lies in the XY plane, making an angle of p with respect to X. BecausePG) = P(P(r)) : P'(1) and because P' defines a rotation of 2p about Z, P(s) must lie in the XY plane, maling an aigle of 2i with respecttothepositiveXaxis.sincePleavesf-invariant,P(0€fandsoP(s)ff'.Sincerandsare ', bases vectors for I' and since P(s) 6 f Pf-d : u1 + us.for some integers u and u. I48 M B. BOISEN."/R.. AND G. V. GIBBS
z z axis is the X axis, leavesVtO(g), which is equal to i | | unmoved. HenceV,Vt(O(r)) :-i andVzV,(O(i )) :7 j'l:, i:fnri:":l,x;:ufidinx*i fF]ril; conveniencewe will label V. From above, it there- fore follows that uroQ) : i and vo(4) :j. Two casesare possible for the image of tD(ft) under V, : : -4 namelyVo(4) 4 o. wo(q) (re" Fig.3c). z z Wefirst consider the case where VO(ft) : k. Because ; .roq) : l: /(Q,vo(4 :i: I(D'^naVo(t): (!) .r.9,f - i,,-.,:,,,*rzt$(l]* 4: il4l. ie s..ihut i.J:'u"d(havithe sameimages i' '.':-..ffi.i. inO.rVo as theyOi ilnOeill Hencewe conchide :J-Y that VO : 1. Accordingly, O is the inverseof a ' ' " ""\'V rotation;rolallon' this,rnls, as previouslyprevlouslynoted,noleo' implieslmpllesthattnat Oq' isls a f=V,6flf't''+ ?,StI'r- 'L f V,-ft1 F(k) .i '' * rotation.Next we considerthe casewhereVO(k) : ^.- -L (a (3b) UV periormrng a hall-turn 2 rotatron wrth a turn angleof 180' ) about the Z axis,we have2VA(i ) z z : -i,2vo( il: -i and2va(k): -k- Hence,the | | -r isomltrytDvrrrvLrJ .r2v-E Y map11.rrrqlJr t torv for-all-tt3r,. r in' R'. we call k 3t". vzvrE(I)=! I this mappingthe inuersiordenoted 6y i (i.e.,i(l : [,,,*,,,=,+ + 1;r::t"iif:''o):"fr:"Jffiil':"i i,.,[';;'6l: y 1,,,,.r,ir_i 'i ,pzY+*t!=t JEW{(l)=t y (2q/) is ile compositionof the rotation(2v)-' I andalru thefrlv inversion,rrrvvrJrvllr l'.r. A ^ point isometrylJvllrwrrJ formed byuJ therllw {,,-, .,.".r,{. | - -g Pvlrr! ...,.Y2Yt9\))=I YzYr9tJJ=tt%V,f t9= compositionof a rotationand the inversion is calleda X- Xt rotoinuersion-^1^:,-..^-^;^.- Weu/^ haveL^.,^:,,^+ just shown^L^,,,- that+L^+ every^.,^-., point..^:-l ( 3c ) isometry is either a rotation or a rotoinversion.Since RTGHT-HANDED LEFT-HANDED a rotoinversionmaps a right-handedcoordinate sys- (!, (!, (1t (.1_),(!, tem into a left-handedone (and vice versa),it is often { f f f } { f f d( p- I referred to as an improper rotation. On the other FIc 3'Grvenapointisometryo,either!o,Q'o,Q'.o.(4))rorms hand, because-' rotations do not changethe hand of a righrhandedcoordinate system (depicteO in thtj left column) .; . the system_._theyare sometimes referred to as proper or a left-handedone (depictedin the right column).t"l ,rr"-.',nJ originalposrtions of oO, o(t), and o6l.tUl showsthe result rotations. We have also shown that, if O is an im- of applyingV", to o(;ioQ, ana 6g,j fne rotationrlr, was proper rotation, then there exists a proper rotation PsuchthatO:Pi.SinceP(t):P( zll(stippled)plane;sucharotarioncanbeconstiu""lgr,,ll',1!1.:'."1.,",iip3(4toi,whereuponoQ.ando(k)mapintorhe .{):P((-lE): -p(r)forall4e-'-;, - R3andsinciipl4)jj(pg)):-ltrl as the rotationaxis the line throughthe origin perpendicular " _ to a planedetermined by i andor4. rn" r"i. for.alll..: R3, we see that Pd: tP and so i com- ""g,JJq",'l, simplythe anglebetween r'iioo14. @; rho,rorthe resulr of apply- mutes with every proper rotation. rng q/, to rf,Of4,U.,O(i), and Ur, (l) Closure:If O,O 6 K,then OO e K.(This can be seenby observingthat o@(I) : O(O(I)) : o(r) : I.) (2) Associatiuelaw: If O,O,V 6 K, then O(OV) : (OO)V. (This was observedearlier.) points) in spaceconsisting of the end-pointsof vec- (3) Existenceof an identity element:There exists'an tors in I radiating from the origin, O. As stated elementl € K suchthatl€) : €D1: O forall before, we will not distinguishbetween a vector and O € K. (The identity mapping discussedearlier its correspondinglattice point. has this property and is in K because/(I) : I (in fact, I(t) : I for all t € I) ) I Lattice inuariant isometries (4) Existence iJ anlnuersefdr each element' Given lf tD is a point isometry, then we say that I is an element@ g K, thereexists an elementO-' t 10 invariant under rD in casef : O(I) where we define € K suchthat < 180o.It is provenin a numberof placesthat sinceP Then, as demonstratedin Figure 2b,{: (cosp,sinp,0) leavesI invariant, the only possibleturn anglerepre- and P(s) : (cos2p,sin2p,0).Since P(s)6 I', it must be sentativesof p must be one of the angles0o, +60o, an intelral combination of l{,E|,i.ei, P1s; : u7+ ug '- + 90o, + I 20o, I 80o (seefor a matrix argument,Seitz, where u and u are integers.Therefore. 1935;Burckhardt, 1947;' Lomont, 1959;and Janssen, (cos2p,sin 2p,0) : u(1,0,0) + u(cosp, sin p,0) 1913; and see, for a geometric argument, Hilton, :(u*ucosp,usinp,0) 1903; Robertson, 1953; Buerger, 1963;and Bloss, l97l). We will presenta nbw and different proof o[ from which it follows that sin2p : usinp. Replacing this fundamental property of a rotation that leavesa sin2p by 2sinpcosp, we see that 2sinpcosp : usinp. lattice invariant. First we observethat in the caseof Sincep I 0 and p + l80,sirip I 0and so cosp: u/2. any such non-identity rotation, P, there must exist a Becausecosp is a function bounded by t I and Dis an two-dimensionalsublattice I' of I in the plane per- integer, the only possible solutions are cosp : 0, pendicular to the rotation axis of P (cf Seitz, 1935; +l/2, +1. Hence p is one of the following eight Zachariasen,1945; and Boisen and Gibbs, in prepa- possibleturn angles:0, +60o, +90o, +120o, 180". ration). Boisenand Gibbs show that if r is a shortest Recall that a positive p angle is measured in a nonzero vector in I' and if .s.is a shoitest nonzero counterclockwisedirection. The symbol and name vector in I' noncollinearwith r. then the set {/,J} correspondingto the rotations associatedwith each ', -l servesas basisgenerating f that is 7' : Ptr us I of the eight turn anglesare given in Table l. u,n€ Zl (the fact that suchvectors r and s exiit is also The symbols listed in the table describeonly the shown by Boisenand Gibbs). Let ibe a Jhortestnon- turn angles of correspondingrotations but lack zero vector in l'. Without loss of generality,we may information about the orientation of the rotation place the X axis of our cartesiancoordinate system axis. Hence, when specifyinga non-identity rotation along r and the Z axis along the rotation axis of P using this symbolism,a descriptionof the orientation sinceit is perpendicularto f'. This situation is shown of the rotation axis is required. When discussinga in Figure 2b. If we definethe unit length along the X point group that leaves a particular lattice I in- axisto be llrll, then4: (1,0,0).Let s be the image variant, it is customaryto choosea conventionalunit of {,under P]that is P(r) : s. Hence,Fe I', and llrll cefl (Table 2.2.2,International Tables, vol. I) outlined : llrll If the turn-angli of F ir 0' o.l80o, then p-is by the cell edgevectors {a,b,cl (note that {a,b,c.}is not one-of the possibleturn angle representativesenu- alwayschosen to be a primitiue triplet: tha*tlsla,!, 91 merated above and we would be done. On the other is sometimesnot a basis for f ). The conveniional hand, if p is not equalto 0o or 180o,thenrand-s are choice of q, b, and c (Table 2.3.1,International Ta- not collinear and s, being the same length as r. is bles, vol. tl ti.nr ouito be suchthat if / is the axis of clearly a shortestnin-rero vector in f' non-colliiear a rotation that leavesI invariant, then there existsa with r. Hence,r and s constitute a basis for I' and vector r = ua * ub * wc along I such thatu, u, andw eachv'ector / 6 l' must be an integralcombination of are inGgers'1see2acholriasen, 1945; Buerger, 1963). 4and s, t.".11: u1+ u4wherer.r1nd u are integers. Since the components u,u,w, of r completely define the orientation of the rotation axis with respectto la,b,cl, luuwl will be used as a left superscripton the Taslr 1. Symbolsand Names for Rotations rdiition symbolto specifythe orientationof the rota- Associatedwith the Eight PossibleTurn Angles tion axis. Hence,[u"] n where n : 2, 3, 4, or 6 + + wg. Symbol for rotation symbofizes a l/n turn about the line ua u! associatedwith Thus.t/m12 symbolizesa half-turn paralielto a,l0t0l4 Turn angle p turn angle Name of rotation s;nibolizes a quarter-turn parallel to b whereas llttlj-t symbolizes a negative third-turn farallel to 180' 2 half-turn -a-b*c. The one exceotionto our rule is when 1200 3 third-turn itiiion axis parallels-c in that caseno ori- 90" 4 quarter-turn tni to the rotation sym- 600 6 sixth-turn entation symbol is attached bol, hence4-t denotesa negativequarter-turn about I 0 identity c. We will always place the Z axis of our cartesian -60" 6-r negativesixth-turn Zoordinate systemalong c. For groups requiring ori- - 900 4-r negativequarter-turn entation svmbols.a and b will be chosenso that they -120" 3-r negativethird-turn ur" p"rp"ndiculaiio c,ind X will be chosen to li! GROUP THEORY DERIVATION OF THE 32 POINT GROUPS I5I along c. (The fact that a. and b can be placed Tnsr,n 2. Names and Symbols for the Rotoinversions perpend-icularto c is not ob-vious/nO ir explored by Associatedwith the Eight Turn Angles Seitz, 1935,and B'oisenand Gibbs, in preparation.) Symbol for roto- As observedearlier, if O denotesa rotoinversion, lnverslon assocl Name of then @ : Pd flor some proper rotation P. We will Turn angle p with turn angle rotoinversion assignas the turn angleof O the turn angle of P and 180" m: i2 : 2i:2 reflection as the rotation axis of O the rotation axisof P. Hence 120" T: B: 3i third-turn inversion the orientation symbol for O is the orientation sym- 900 4: i4:4i quarter-turninversion bol for P. The names and symbols for the 600 6:i6:6i sixth-turn inversion rotoinversions corresponding to the eight possible 0 i: it:Ii:1 inversion turn anglesare given in Table 2. Note that 2i : 2 can - 600 6-t:i6-':6-'i negativesixth-turn also be visualizedas a reflectiondenoted by rn, where inversion m refersto the mirror plane about which the reflec- - 900 4-t: i4-':4-'i negativequarter- tion occurs. The orientation symbol given to rz is turn inversion inheritedfrom the half-turn perpendicularto the mir- ]-':r3-':J-'i negativethird-turn inversion ror plane. For example,U0ol^ denotes a reflection whose mirror plane is perpendicular to q while ra denotes a reflection whose mirror plane is 63,6n,6u,6"1: 16,3,2,3-r,6-',Il and denote the group perpendicularto c. by 6. Note that (f-1) : 16-',6-",6-t,6-4,6-5,6-6l. : Since the order of listing the Proper crystallographic point groups l6-t,3-t,2,3,6,11. elementsis immaterial, we seethat (6) : (6-r) and Ifl G is a finite group consistingof rotations of the hence 6-1 is another generator of the group 6. A type listed in Table 1, then G may be shown to leave similar situation existsfor the other monaxial groups I (Boisen in prepa- some lattice invariant and Gibbs, as evincedby Table 3. In summary, we have shown proper ration). Consequently,the searchfor crystal- that all of the proper monaxial crystallographicpoint point groups is to searchfor lographic equivalent the groups are cyclic. Becauseof theoretical considera- groups rotations type those finite consistingof of the tions used later in this paper, we require a more listed in Table l. When all of the rotations of G have formal definition of a cyclic group. A group G is a the same rotation axis, we call G a proper monaxial cyclic group if and only if there exists an elementg e group. When G has two or more rotations with dis- G such that G : (g) : lg"ln e Z). Hereg is referredto tinct axes, then we call G a proper polyaxial group. as the generator of the group. In the casewhere G is a Proper monaxial groups finite group it is sufficientto consideronly the posi- tive powers of the generatoras borne out by our ex- It is evident that each of the following collectionsof amples above. In fact, if (g) is finite and t : #((g)), rotations crystallographic lll, 12,Il,13,3-' JL 14,2,4-', then I is the smallest positive integer such that gr forms proper monaxial 1), and {i6,3,2,3-',6-',1) a : 1. For example, in the caseof (4-1), the smallest group where all of rotations in each collection the positive integer I such that (4-L)t : / is t = 4, hence take place about the sameaxis. (Recall we needonly #((4-')): 4 asevinced by Table3. As seenin Table3, demonstrateclosure to concludethat eachis a group. there is only one distinct monaxial group of each For example, for we observethat each of {3,3-',1} order about a single axis. If G is a proper monaxial = I, compositionsSS:3-1,33-t: l,3I 3,3-'3: group of order n, then we will use the symbol n to 3-'3-' : 3,3-'I : 3-" lj: 3,/,3,' :3-'and II : 1 is in {3, 3-', l} and so we have demonstratedclosure for {3, 3-', II.) The list of monaxial groups given Tanr,r 3. The Proper Monaxial Groups, above exhauststhose that can be made from the ro- Their Generatorsand Elements tations presented in Table 1. This can be quickly Group symbol Group generatorsand elements checked by considering each of the remaining col- lections and showing that no other such group is I (r): {r} 2 : possible(e.g., \1,2,31is not closedbecause 23 : 6-' <2>u,2l -',6 -', 3 (3) : (3-') : J, J-1} e U,2,31).Since 16,3,2,3 I | : l$,62,63,6u,6u,6"1, {r, 4 we call |16,3,2,3-t,6-r,1)a cyclic group generatedby <4>:(4-'>:U,4,2,4-'l 6 <6): <6-'): lr,6, J,z, J-L 6. In group theoretic notation we write (6t : 16,6', t52 M. B. BOISEN. JR,. AND G. V. GIBBS designatethe point group. The axis common to all of Tasr.s 4. Group Multiplication Table for Group 322 : the rotations of a given monaxial group n (wheren lrool2 toLot 3-L 2 2, 3, 4, or 6) will be called an n-fold axrs. (When n is trool2 tlrot torot different from l, 2,3,4, and 6, a perfectlygood point I I 3 3-r 2 2 rrrot Iorol trool group results in the manner describedabove. How- 3 3 J-r 7 2 2 2 torol lrool lrrol ever, for such a point group no spacelattice is left in- 3-r ,-r I 3 2 2 2 tlool Irool Iorot variant and hence it is not a crystallographicpoint 2 2 2 I J-l J trlol Ilrol Iroot group. 2 2 2 3 I 3-l ) loLol lorot trrol 2 2 2 31 Proper polyaxial groups In our investigation of the polyaxial groups, we will be examining combinations of monaxial groups. assignedto a rotation symbol when the rotation oc- The task of finding the possible proper polyaxial curs about an axis other than Z), and the pole point groups will be considerablymore difficult than points belonging to the rotations of 322 are de- that of finding the monaxial point groups. It will picted in Figure 4b. The multiplication table for 322 require some fairly sophisticatedarguments and a given in Table 4 servesto define the structure of the considerableamount of notation which will be ex- group by enumerating the compositions of all plained as we go along. The main aim of this section possible pairs of group rotations. Since the only is to establishthe inequality which in looseterms will rotations that appear in- the table are the six rota- state that three monaxial groups with orders zr, zr, tions already listed for 322, we may conclude that and zr, respectively,may be used to form a proper 322 is closed under composition and hence is a polyaxial point group only if group. Letp denotea pole point of G. Then the collection l/v,1- l/v" -l l/y") l. of all rotations of G that have p as a pole point is designatedby Go, i.e., Go: {ge Olg(p) : pl.* lf To facilitate our proof that will establish this in- gy gz e Gp, then g,gr(p) = g,@,(p)) : S,(p) : p. equalitywe will considerthe surfaceE of a unit sphere Since g, and gz were arbitrarily chosen rotations centered aL O. Any point isometry O acting on E of Go and since grgz € Gp, we conclude that Go is maps E onto itself. In fact, tD is completely closed under the composition and hence a sub- determined by its action on 3 becausethe effect of O group of G. Since the rotations of G, leavep fixed, tD(i), on E determinesthe end-points of O(7) and they all have the line Qp containing Q and p as their O@) and thereforethe componentsof dQ, o() and rotation axis. Consequently,G, is a proper monaxial O(4). A non-identity proper rotation ft about the group and thereforeone of the cyclic groups listed in rotation axis I leavesexactly two antipodal points on Table 3. E unmoved. These points are preciselythe points at : In the caseof 322,(322)p^ (322)p"z: {'oo12,ll, which I and E intersect and are called the pole points (322)es,: (322)e"s: {"ol2,ll, (322)e22: (322)ou,= belongingto ft. If points thesetwo are labelledp and \t0t0l2,11and (322)o,, : (322)or": lJ,3-',/,}. We ob- q, then they are the only points'x on E that satisfy the servein this examplethat two pole points are associ- : equalityh(x) x. Let G be a proper polyaxial group. ated with eachmonaxial group (Fig. a). lt is also evi- The trivial group t has already been treated as a dent that the pole points pzr, pzz,pza, pn, prr, and pn monaxial group and since it has no pole points we are each associatedwith the group 2 where the rota- deliberately exclude it from consideration in the tion axis of each has a different orientation. The re- remainder of this discussion.Hence we will assume maining pole points p ' andpr" are similarly associated > that l(G) 2. To help illustrate a number of the with group 3 (seeTable 3). (The significanceof the conceptsgiven below, we will on occasion use the double subscriptsattached to eachpole point belong- proper polyaxial group 322 as an example.However, ing to 322will be discussedlater. For the time being, none of the theoreticalconcepts that follow depend the subscriptsmay serveto distinguishdistinct pole on the assumption lhat 322 exists.The rotations of points.) 322 are ll,3i-'!nl2lttol2lorol27. Since 322 con- sistsof 6 distinct rotations, by definition #(322) : 6. The orientation of each of these rotations is shown * The subset Go of G, consisting of all g E G for which g(p) = in Figure 4a (recall that the left superscriptis only p, is a subgroup of G sometimes called the stabilizer of p. GROUP THEORY DERIVATION OF THE 32 POINT GROUPS 153 ,(3221p"={ t,troola, ,322)p22= lt'totolzl (322)p33= lt,[oro]il (322)psr=lt,[rro]21 (322\^ P2l I L{szz)',.=1r,3,3-rl z \ l/ ./ FIc. 4. The orientations of the rotation axes and the associatedpole points for point group 322 are shown in (a) and (b), respectively.The bases vectorsa,b,c lie along X, Y, and Z, respectively,and so are not shown. Note that in this figure X, Y, and Z are not mutually perpendicular,since the angle between X and Y is 120" (the ansles between X and Z and )/ and Z are both 90'). 154 M. B. BOISEN, JR., AND G. V. GIBBS be ints belo to dicted from Theorem I that there will be no group the elementin 322that mapspzt to pr becausetheir cor- rotation eG respondinggroups (322)e.,and (322)o' are of differ- groups e same ent orders.This is easilyverified by consideringthe rotationsin 322one at a time and observingthat in- deednone of them mappzr to ptr. Proof. From the discussionabove we know that Go We will now investigatean important relationship and Gn are monaxial and hencecyclic groups.Let i that existsbetween the pole points of any givenrota- be a generatorof Goand let n : and,m: #(c) #(G). tion group G. We define0(G) to be the setof all pole We want to show that m : r. As observedin the pointsbelonging to the rotationsof G, i.e.,0(G) -- sectionon propermonaxial groups, n is the smallest e Sl thereexists at leastone g€ G suchthat g * I positiveinteger such that h" : 1. Considerthe rota- {p andg(p): p|. For example,in thecase of 322,0(322) tion ghg-t where g(p) : q (such a g existsby the We now define a hypothesisof thetheorem). Srnce g-'(q): p andh(p) lPrrrPrrrPrr,Pr2,p2s,psbps2rprrl. relation on 0(G) by definingfor eachp"p" 0(G) : p, we observeIhat ghg-t(q) : : € S(h(g-'(q))) that is equivalentto p, (symbolized - pr) if and : g(p): Henceghg-' q demonstrat- h h dh(p)) 4. @): only if thereexists a rotation € G suchthat g(p') = ing that ghg-' Go. We next show that gftg-r is a I e pr. Hence,if p' - pr, then from TheoremI we know. generatorof Go.Letf be an arbitraryrotation of Go. that Go, and Go, are cyclic groups of identical Theng-'/ge Go becauseg-'fg(p):g-'W@))): orders,i.e., #(Gpr): #(Go). S-'(^q)) : S-'@) : p. Therefore,since Go : (h), We observethe following important propertiesof g-'fg : it for someinteger I < t < n. Hencegh'g-' 0(G) with respectto the relation definedin the last : : Therefore S@-yg)S-t .f. paragraph: f: sh's-' (l) Reflexiue.'For all p e 0(G), p - p. (This is : Shh ... hg-, (wherel occurst times) becauseG has an identity elementI and 1(p) = p. = Sh(S-'S)h(S-'g)h.-. @-'glhg-'(sinceg-f : 1; Hencep - p.) : @hg-')@hy-r).. ' (ghg-') (by associativelaw) (2) Symmetric:For allpr, pr€9(G), if p, - p2,then : (ghg-')' pz - pr.(This is becauseif ge G suchthat g(p r) : p", g-'€ g-'(p") : pr. Consequently is a power of ghg-'. Since/ was an theng has an inverse G so that f - arbitrarily chosen rotation of Go, we have shown that Hencepz pr.) (3) pr, pr, 0(G), if p, - p, Go : @hg-t). Since m : #(Go), as observed earlier Transitiue.'For all Pse - - - : z is the smallestpositive integer such that (ghg-'1^ andp2 p", thenp, p3.(This is becauseif g(pt) p, : 1. However, since n : (ghg-') : gl2ng-r - and h@r) p, whereg, h e G, then hg e G and #(G), = : : - glg-' : 1, but m is the smallestsuch integer hence hg(p') h(g(p,)) h(p") p'. Henceh pg.) m 1 n. But (ghg-r)- : gh*g-' : 1 and accordingly Any relation definedon a set satisfyingthese three propertiesis called an equiualencerelation. Let p h- : g-'g: .1.However n: l(Go) is the smallest E positive integer such that h" : I. Hence n 1m. G(G), then we define the equiualenceclass determined : - Therefore, since n < m and m 1 n, we conclude that by p to be [p] {qe O(G)lp qI.* That is, [p] is the setof all polepoints of G equivalentto p. Sincep- p, m : n and so #(Gr) : #(G"). In our example 322, if we consider the cyclic we havep e lpl andso everyelement of 0(G) appears hence0(G) is the union** group (322)^t : ll,II101,2)we see Ihat ttt0lz(p2r): pr". in someequivalence class; Hence, according to Theorem I we can conclude that of all of its equivalenceclasses. Furthermore, if [p] and one or more elementsin common(sym- #((322)e) : f;\(322)o"). We observe that this is in- [4] have bolized bV g),*** then deedtrue because(322) ez2: glotol2Tand henceboth [p]O lql t [p] tql. Vml21prr7: sets contain two rotations. Similarl, * The subsetof S consistingof all g(p) asI'ranges over all G is pr2 and therefore we can conclude that (322)o' and sometimescalled the orbit of p. (322)pn have the same order. Again an examination ** The union of a collectionof given setsis the set consisting given of Figure 4 shows this to be the casesince (322)prr : of all thoseelements that appearin at leastone of the sets. The union of two setsA and B is symbolizedA l-l B and the ,3,3-'l and (322)rr" : . It may also be pre- ll 1l,3,3-'l symbolism1) A, -"un, Ar U A, U Ar. i-l *** The expression"A O B" denotesthe intersectionof sets i* Those familiar with group theory will recognize that this A and B, i.e., the set of all elementsthat are in both A and B. theorem shows that Go is isomorphic to Go since they are both The symbol0 denotesthe emptyset; i.e., the setcontaining no cychc. elements. GROUP THEORY DERIVATION OF THE 32 POINT GROUPS t55 Accordingly, distinct classesare disjoint, i.e., they H;y,then H11: lhi1,h1t,"',h'il.Moreover the axis haveno elementsin common.A thoroughdiscussion common to the rotations of Hlyis a vrfold axis (this of equivalenceclasses including proofs of our obser- axis of course passes through pit and the origin, vationsabove can be foundin mostmodern abstract Q). In the context of our example 322, we see alegbratexts (e.g., Larsen, 1969; Shapiro, 1975). that Hr, : ll ,3,3-'l,Hr, : U,3,3-t), Hr, : ll ,Uoo)21. In our exampleof 322,the elementsof the equiva- Hr" $,loto\)|, an u,ulol2L H''-: l[,utol2\ pt0l2\. lenceclass of p2,in 0(322) arelprrf : llapg(322)lp^ H", : ll,U00l2l, and Haa : 11 In addition : : : -Ptl: we observedthat #(H'') : #(Hr) t,r 3, #(Hrr) : : : |n e 0(322)l thereexists an elementge322 such that #(H",) : #(Hr') vz: 2, and l(H") #(H'r) #(H") g(pz,): Pul :vs:2, : lg(o,)lSe 3zzl We are about to present a lengthy derivation of some important relationshipsbetween many of the : (I(p"r),3(pr,), ,rr]-,,rour, conceptsdiscussed above. These relationships will be to'2(prr), totot used to develop facts about G which will ultimately "'ot 2(p"r), 2(prr)l enable us to determine all of the proper : l4"t, Pzz,Pza, Pzr, Pzz, Pzsl crystallographicpoint groups which we then list in Table 6. In an attempt to presentthis discussionin a : l\rr, Pzz,Pzal' more understandablefashion, each of the more im- Following the same procedure one may calculate portant relationshipswill be set off in a box at the each of the equivalenceclasses of 0(322). The results beginningof each argumentwhere the relationshipis = are: lp"rl : IP"rl: [prr] : lPzbPzz,Pzs],[pr'] : lP'") established. : and : : [prr] lprr,prr,p"rl [p1l lpr"] lprr,prrl.We t notice that it is indeed true that these equivalence 2(N-l):En;Qt-r) (l) classesare disjoint and that 0(322) is their union. d-l The pole point notation for 322 was deliberately : : nt : chosen at the outset so that points with the same Recall that N #(G), vi #(Hit), #(C{G)), classesof pole first subscript would be in the sameequivalence class. and t is the number of equivalence Equation Let t denote the number of distinct equivalence points. The basic strategyfor establishing counting the classesof 0(G)where C'(G), Cr(G),"',C'(G) denotes (l) will be to find two distinct ways of rotations of G. The result will be two the equivalenceclasses of 0(G), i.e., C{G) : [p] for nonidentity of non- somep€G(G),hence expressionseach equaling twice the number rotations of G. This will establishEquation t identity : g j. ( I ) sincethese two equal expressionswill be precisely -U c,(G) @(c) and c,(G)a cr(G): if i + those appearingin the equation. We begin by taking eachpole pointpi.rin 0(G) one at a time and counting It should be observed that 0(G) is a finite set because the number of non-identity rotations of G leavingpiy G is a finite set and each non-identity element of G fixed. The sum of these numbers taken over all the has only two pole points. Hence,each C1(G.) is a finite pole points in 0(G) will equal twice the number of setand we let nl denotethe number of points in C1(G); nonidentity rotations in G becauseeach ofthese rota- that is, n1 : fr(C{G)). Therefore, we may list the tions leavesexactly two pole points fixed and henceis elementsof C,(G) asp1, ptz,. . . n the caseof our ,ptn.l of non-identity rotations : : counted twice. The number example, C{322) {pr,ppl, C"(322) \pzr,pzz,pzel - : - leaving piifixed is f(H,7) | vt 1. Thus, for the and Cr(322) : lpu,psz,par|and as observed above, pole points in C'(G) we have eQ22) : c{322) : c1322) c,(322) E U U fi(non-identityrotations leaving - cs(322).In addition, notice that n, : #(CLQ22)): 2, p1 fixed): vr | : nz: #(Cz(322)) : 3 and n" : #(cs(322)) 3. f(non-identityrotations leaving Let Hiy denote G' the group of all rotations in G p', fixed)='vr - | that leavepu fixed. Sincep;y andp1p are both in Cr(G), nr equatlons #(H,t) : f(H1e) becausepu - p,n (seeTheorem l). Hence Hrr,Hrr,... , Hrntall have the same order which we signify as 2,. It follows that H1,is a cyclic l(non-identity rotations leaving group of order v1 and if we let hqbe the generator for pt", fixed) : vr - | 156 M B BOISEN. "/R.. AND G. V. GIBBS Tasls 5. Display of the Derivation of Equation (1) for Group 322 Notation for group G Notation for group 322 e(G) 0(322): l\rr, Pn' Pzt,Pzz, Pzz, Psr, Pzz, P,nl ,(G) c{322) c2Q22) caQ22) Pti Pn Pzt Pzz I'23 Ptt Pzz Pzs Gpii -- : {c € Gls@')(322),,, : (322)p,, : (7))\^ (322)e"" -- (322),"": (322)e", : (322),"" (322)""": : 'too'2] Ptil {r, J, J-'} {r, J,J-'} [ 1, lI,'otot2\ l1, "to'2]' | 1, "'o' 2l {.1, "oot2l {/,'o'o'z} \v; l) (3 - 1) (3- 1) (2- r) (2- r) (2- r) (2- r) (2- r) (2 r) n - l) 2(3- r): 4 1/?-l\-? 3(2-t1:3 "(u; - :10 f nn@, r) 4+ J + 3 2(#(G)- r) 2(#(322)- 1):2(6- l):10 Summing up thesenumbers we flnd that the contribu- Then for this particular i we want to show that N : tion from C,(G) is nr(u, - l). Following the same n;21.Consider the pole points belongingto C;(G).For argument, we seethat the contribution from C;(G) is eachptj, there existsat leastone rotation Xf€ G such - ni(vi 1) for each I < i < t. Adding the contribution thatS(pir) : Pii.We singleout one suchrotation for from each of the I equivalenceclasses, we find that eachTand label it g;. Then for each | < i < nt, g/Pir) : pii. By the way the gr's were chosen, the col- the sum is I n1@1- I ). SinceN- I is the numberof t=1 lection of pole points lg'(p,'), gr(Pi'), , gni non-identity rotations in G, it follows that twice the '',Pt,,l : (p^)l : lPi', Pn,' c,(G)' we now form number of non-identityrotations in G is 2(N-l) and the following list whereas before,hiris ageneratorof so we haveestablished Equation (l), namely Hi,Q.e., Hi, : lhir,hi1,''',hk\): 2(N- 1): t nili-r). (1) grhir,grhiJ,''',grh',1 grt"'g'hi"' " g"h'r\ Returning to our example 322, we observedearlier " : : : that N : 6, h : 2, n, : 3, fra 3, u, 3, uz 2, and ::: vt : 2. The notation usedin the derivationof Equa- g^h,,,g-h,"''''' gn'h'i!* tion (l ) is displayedfor 322in Table5. It is clearfrom the tablethat, as predictedby Equation(l),2(#(322) We now wish to show that every rotation in G - l): l0:nlvr - l)+ nr(vz- l)tne(us- l)' appearsin this listing exactlyonce so that the number : of entriesin the list equals#(G) N. Let g€ G, then, N : nivi for I < i < t (2) sinceg mapspil to an equivalentpole point,g(p;1)€ C;(G)and henceg(p1'): pii : g/p,,,) for some | < i < Since the followrng argument is quite tedious,the n;. Therefore, g1'g(pn) : pit. Accordingly, gi-'g e readermay chooseto skip to Equation (3) on first Hi, : {hiuh,?,"',hhland so gr'g : ftf for some reading.This can be done without sacrificinga basic understandingof the development.An argument es- * Those familiar with group theory will recognizethat eachrow tablis.hingEquation (2) using the following para- in our list is actually an enumeration of a coset of Hit Further- graph and the footnote is presentedfor thosewho are more, the rotations in G that roLalepit lo pih are all listed in the Hence each row correspondsto a pole point in C,(G) alreadyfamiliar with group theory.The roleof Equa- k'b row. and conversely Consequently there is a one-to-one corre- tion (2) is to facilitatethe analysisof the implications spondencebetween the number of pole points in C1(G)and cosets of Equation(l ). of H,,. Hence there are ni cosetsof Hit and since l(H,r) = v1wa Let i be any positive integer such that | < i < t. have, by Lagrange'stheorem, N : np1. GROUP THEORY DERIVATION OF THE 32 POINT GROUPS 157 I < k < /,. Hence,g : gth,l which appearsin our 2(N- 1): NQ - r/v) list. Sinceg is an arbitrarily chosenrotation of G, we E may concludethat every rotation in G appearsin the list at least once. Our next task is to show that no 2 - 2/N: - r/r). rotation of G appearsmore than once in the list. Sup- | 1r pose that g € G appears in the list as g = g,hfi and as BecauseH;7 is a ,uUgrolpof G, we canconclude g: grhnwith I 1k,m l vtand I ( r,s We 3 ni. that l(H;;) : vi ! #G) : N. Also, sinceHs7 is the set : wish to show under these circumstancesthat r s of all rotations in G that leavethe pole pointply fixed, = and k n. Sincep1 is a pole point associatedwith it must, by the definition of a pole point, contain at : : : hn, wa have g,hfrQtir) g"(p,r) pi, and g"hTr@,r) leastone rotation in addition to the identity. Hence : : g"(pt') pi,. Since g,hfr, g"hTr,by the definition of 2 1 vi 1N. Using the fact that zi must be greaterthan = equal mappitrss,pir: pis and so r s. With this re- or equal to 2, the right member of Equation (3) : sult we can now write g,hft g,hfi which implies by satisfiesthe inequaltiy the cancellation law that hrt : hts and so k : m since Itt | < k, m 1 vi. Therefore, r : s and k : m as re- Irr - r/v)t Irr - U2): E Itool 2 J-t, 2. 2-2/N: Itt -r/,n) "ott"'l i -l It is apparent that each rotation of 322 is listed - - exactly once and that v".ns : 2.3 : 6 : #(322). : (1 l/r,) * (l r/vz) , By a little algebraicmanipulation we find that -r/,0) 2-2/N: l1r (3) 2: N/u'l N/vz. i-1 From Equation (2) we have that N/vi = n1,and so the SinceN : ftrur,it followsthatn,: N/u,. Replacing above expressionsimplifies to n; in Equation(l) by N/vLwe obtain 2:nrln"' 2(N - l): I (N/v)(v,- t) Becausen, and r, are positive integers,we conclude 158 M. B. BOISEN, JR., AND G. V. GIBBS Tasrp 6. Possible Proper Crystallographic Point Groups SymbolforG:vp2v3 #(G): N #(C'(G): N/v' #(c,(G) -- N/r, #(c,(G)): N/vz 222 A 2 2 2 322 6 2 J J ^,,' 8 2 A A Dihedral groups 622 l2 2 6 6 ?1t+ 72 4 4 6 Tetrahedralgroup 432 24 6 8 L2 Octahedralgroup t The group 332is usuallydesignated by 23. that n, : flz: I is the only possiblesolution. Hence, which yields ni : N/vi. These ate also recorded in for a rotation group with t : 2, we have two Table 6. equivalenceclasses consisting of one pole point each. Determination of the interaxial angles Altogether G has a total of two pole points, which accordingly define one and only one rotation axis. We have shown above that there are only I I pos- Therefore,those groups with two equivalenceclasses sibleproper crystallographicpoint groups,namely l, of pole points must be the monaxial groups given in 2, 3, 4, 6, 222, 322, 422, 622, 332, and 432. The Table 3. The number of elementsin each of these monaxial groups I, 2, 3, 4, and 6 were established possiblemonaxial groups is equal to the order of the earlier. However, we have not yet establishedthe fact rotation axis, l(G) : ut : ttz. that each of the polyaxial combinations actually Casewhere t : 3. In this caseEquation (3) expands yields a group. Also, the possibility renlains that to some of thesecombinations may leadto two or more 2 - 2/N = (l - l/vr) + (l - l/vz)+ (l - l/us)' groups which may arise by assumingdifferent inter- sectionangles between the rotation axes(these angles Rewriting this resultwe seethat are called interaxial angles.).We will seelater in this | + 2/N : l/vr I l/vzl l/us ' (4) section,however, that each of the possibilitiesleads Since N > 2, it follows that I * 2/N > I and so to preciselyone point group. In the processwe will actually determine the interaxial anglesused in the l/v,*l/y"tl/ur)l' (5) formation of these point groups. Inherent in the For convenience,we can assumethat v, ) vz) vs. strategy we will follow in determining these inter- In constructing Table 6 we have considered all axial angles is the observation that before a poly- possiblecombinations of 21,az, and zr; selectedfrom axial combination qualifies as a group it must the permissiblevalues 2, 3, 4, and 6; and recorded satisfy the requirement that each of its rotations those that satisfy Inequality (5). Note that the other permutes the pole points within each of the three possibilitiesfor urvrv,are 632,442, 642, 662, 333, 433, equivalenceclasses. Hence a rotation must map a 633, 443, 643,663, 444,644,664, and 666. A quick given pole point to another pole point in the same check showsthat none of thesesets satisfy Inequality equivalenceclass. We recall that becauseequiva- (5). (Note that the derivationof Inequality(5) did lenceclasses are disjoint no two distinct equivalence not require uy u2,a(rd z, to be 2, 3, 4, or 6. Hence a classesshare a common pole point. This observa- large number non-crystallographicpoint groups may tion will alsobe of considerableuse in the determina- be formed that satisfy Inequality (5). For example, tion of the interaxial angles. In addition it will be r22 satisfies (5) for all n ) l.) * The ordersof eachof helpful to observethat if a half-turn maps the point the potential polyaxial crystallographicpoint groups p € S to 4 then its axis must bisect both of the arcs in Table 6 are found by substitutingtteu2, and z, into determined by p and q. Hence, if r is a pole point : Equation (4) and solving to obtain N : #(G) : of such a half-turn, then fr Q. 2vrv2vs/(vrv,I v2vst vsv, - vp2vs).The order of C,(G) : n; is found by appealing to Equation (2), Interaxial angles for the dihedral groups ur22 * See Klein, Weyl, or Zassenhausfor a derivation of the non- We begin by consideringthe interaxial anglesthat crystallographic rotation groups based on these ideas. must exist for a polyaxial combination of the form CROUP THEORY DERIVATION OF THE 32 POINT GROUPS 159 vr22 (when ur : 2, 3, 4, 6) to be a group. Table 6 tor so that the arc length between any two adjacent showsthat there are exactlytwo pole points which we points is 360"/vt. denoteprt and p* in the equivalenceclass Cr(r'22). We now observethat there is only one rotation lt Let lr be a non-identity rotation in G suchthath@r) associatedwith the pole points in C'(vr22)such that : pr, i.e., ftC Hrr. Then hftrz) must equal pn ot pn h(p) : p* wherep1 andp,2 are adjacent pole points sinceevery rotation in vQ2 must permute the equiva- in C"(ur22). If we let trool2 denote the half-turn uNl2(prr) lenceclass C1(vr22): lpnepl. Becauseit was assumed about the X axis then pzt, and above that h(prr) : p' and becausel cannot map the rotation ltltoo)2 has the property thathltnl2(prr) : two distinct pole points to the same pole point, we : pzz. However, lrltu)2 I fr becaus" i1 lruoa)2 conclude that h(prz) must not equal pll and hence ft, then \runl2 : 11 / which according to the h(p'r) : p'r. Thereforepll and Pu both lie on the axis cancellationlaw would yield the contradiction that of the rotation ft. Consequently,p, andp* are antip- It00l2 : 1. But h was the only rotation associated odal points and so there is exactly one z,-fold axis with Cr(vr22)which maps p2r to prr. Hence htt0012 associatedwith Cr(vr22).Also each of the equivalence must be a rotation associatedwith a pole point in classesassociated with the cyclic groups of order 2 either C2@r22)or Cs@'22).Therefore iltml2 must be contain z, pole points. Accordingly, we write Cr(vr22) a half-turn which maps pz, to pz" and whoserotation ... -- .'. : lpa.ezz, euI and C"(vr22) lp$esr, Aul . axis is perpendicularto the Z axis.Thus the axis of Moreover, since there is a total of ff(CdvQ2)) + iltm\2 must bisectthe arcs along the equator deter- pole point fi(Cs(vr22)) : 2v, pole points associated with the mined by p^ and pzz. Let q denote the 2-fold axes,there must be exactly v12-fold axesassoc- belonging to il|ul2 which bisectsthe shorter arc be- iated with Cz@r22)and C"(u'22).Now that we know tween p21 and pr". Sincepzr and p2 are adjacent pole the number of each kind of rotation axis present in points in C2@r22),q cannot be in Cr(v'22) and hence vr22, the task is to discover the appropriate inter- must be in Cs(vQ2).The imagesof 4 under the rota- axial anglesbetween these axes so asto form a group. tions about the z,-fold axis are the remaining pole If we place the Z axis along the v'-fold axis, then points in Ct(vr22).In summary, we have shown that, p' and ptz ?te the points where the Z axis intersects if v22 (whereut : 2, 3, 4, 6) is to be a group, thereis with the surfaceE of the unit sphere.The question only one possiblearrangement of rotation axes.The now is "Where can a 2-fold axis associatedwith say interaxial anglesfor this arrangementare such that pz, be placed so that the associatedhalf-turn g will the angle betweenthe z,-fold rotation axis and each : : of the 2-fold rotation axes associatedwith the pole permute lprr,prrl?" If g(ptt) /rr, then g(prr) pr". Sinceg has only two pole points, p21must be either points in Cz(v22) and Cs(u,22)is 90o and the angle pt ot prr, which is a contradiction because the betweenany two adjacent2-fold axesis 180'/z'. Re- equivalenceclasses of pole points are disjoint. Hence fgrring to Figure 4, we seethat our example group g(pi : prr, andp' bisects the arc ft,r. Therefore, 322 does indeed satisfly this interaxial criteria. The ;;;fid;": f,i,,' andsinceft,, +"'fr," : l8oo,it fact that vr22 actually forms a group when its inter- follows that ftrr: 90o. Accordingly, it follows that axial anglesare as describedabove can be shown by the 2-fold axis associated with p' must be preparing a multiplication table (note that this can perpendicularto the z,-fold axis.A similar argument now be easily done by those familiar with matrix showsthat the 2-fold axesassociated with the remain- theory since the matrix representationfor the rota- ing pole points in Cr(vr22) and Cs(vr22)are also per- tions in vr22 is completely determined by the inter- pendicularto the z'-fold axis.Thus, the z' 2-fold axes axial angles). in u'22 are all perpendicularto the v'-fold axis. Interaxial angles the tetrahedral group 332 ( i.e., 23 The remaining question is "What must the inter- for ) axial angles be betweenthese 2-fold axes?" Without Our next task is to determinethe interaxial angles loss of generality,because p' is on the equator (the for 332. It is clear from Table 6 that there are two great circle perpendicularto Z), we can place the X equivalenceclasses associated with the 3-fold axes axis of our coordinate system so that it passes containing4 points each,and one equivalenceclass of through pu. We know that the imagesofp, under the pole points associatedwith the 2-fold axescontaining rotations along the zt-fold axis are all in C2(ur22). 6 pole points. Accordingly, there are four 3-fold axes Moreoyer, becauseof the nature of the rotations and three 2-fold axes in 332. about the zr-fold axis, we obtain z, poirtts (henceall We begin by consideringthe placementof the three of the points in C2@r22))spaced equally on the equa- 2-fold axes. Let ps, denote a point in Cs(332).Since 160 M B. BOISEN, JR., AND G V. GIBBS CaQ32)contains all of the pole points associatedwith p"rplay the role of q in our observationin the previous the half-turns in G and sincethe antipodal point of paragraph,then we know that the 3-fold axis through prt is associatedwith the same half-turn associated p' is perpendicularto someplane containingp.' and with pr,, it follows that the antipodal point of pr,, two other pole points in C'(332). By inspectingall which we denote6y prr, is in Cr(332).Without lossof such planescontaining p' and rememberingthat the generality,we place Ihe Z axisalong the line segment 3-fold axis cannot be coincident with any of the 2- pt'1ttr.Let prr be any one of the remaining pole points fold axes,it is clear that the plane determinedbypr', in Cr(332).We will show that flp"" :90' by assum- pss,?ndpru (seeFig. 5) is the only possibility.There- ing that frp* + 90o and arriving at a contradiction. fore the third-turn associatedwith p'r mapsp31 to pas, tf"frp""?-90;,theneitherfr,, i 90"orfr,n < 90' p$ to p"u, and pss to p' requiring that Pr be where p3n is antipodal to pga. Without loss of equidistant from the points pu, pss,and pss(see Fig. = : generality,we may assumethat f ,j"" < 90'. If we let 6a). Therefore 6i,., Gp,, 6h". Solving in the 2 denote the half-turn along the Z axis (with pole standard way the sphericaltriangles that arise from pointsp31,p32), then 2(ps) : pq for somepsj€ Cs(332). theseref ationships, we conclude that fijr, : 54.74". Hence ffpr1 : frp* The imagesof p,i under the half-turnsin 332yield the observation on the effect of a half-turn, p' bisects four pole points belongingto C'(332) : lpt, pn, pn, hence 190' + 90' : 180'. Thereforepr; prnlas pictured in Figure 6a. The antipodal points to fi)rr, i|,pr" ' I pgr sinceGpr, = 180o.Moreover, it is clear that py those appearingin C1(332)are not in C'(332),but € \p"r,p"",p""|. Hence the rotation axis associatedwith they are the pole points that form Cr(332).The result- pr; is different from those associatedwith lltrr,prrl and ing set of all pole points belongingto 332 are shown Lptt,ptnl. Therefore, psj must be one of the remaining in Figure 6b. We have therefore shown that this pole points in Cr(332),say pat : ps,. We also observe, arrangementyields the only possibilityfor 332to be a since p' bisects 6prr, that Tpru Qp* and @"u group. If a multiplication table is prepared,it can be are coplanar where O is the origin of R3. Hence shown that this arrangement does indeed yield a the three axes associatedwith half-turns are all group. A list of selectedinteraxial angles between the coplanar. We now make an obvious but important rotation axesin 332 is given in Figure 6c, from which observationabout third-turns. lfg is a third-turnand the remaining anglesmay be deduced. 4 is a point not bn its rotation axis, then the points lqS@),g'(q)l determinea plane perpendicularto the axis of g. Hencegiven any point q in Cr(332)and any 3-fold axis belonging to 332, there exist two other pole points in Cr(332)such that the plane determined by q and these other two points is perpendicularto the 3-fold axis. But, under the hypothesisthat 6br" * 9O', we have shown that all the pole points of Ca(332)are coplanar and henceall of the 3-fold axes must be perpendicularto the plane determined by thesepole points. This meansthat 332can have only one 3-fold axis, which is a contradiction becausewe know that there are four such axes. Therefore, the anglefijr": 90o. Sincep' was an arbitrarily chosen point in Ca(332)and sincepse was selectedto be any pole point exceptp3land pr", we seeby similar reason- ing that 6b"" :90o and6]i"u :90o. Consequently, the three 2-fold axes are mutually perpendicular. Note that this says that 222 is contained in 332. Now we will turn to the placementof the four 3- fold axes.Let pt be a pole point in C1(332).Then p,, appearson E in one of the octants delineatedby the I Frc 5. The orientations of the 2-fold rotation axes for the three 2-fold axes.Without loss generality, of we can tetrahedral group 332 = 23 with their associatedpole points assumethat p' is located somewherein the first oc- shown as solid circles The surface area on the unit sphere where tant (denotedby the shadedregion in Fig. 5). If we let pole point prr must lie is shaded GROUP THEORY DERIVATION OF THE 32 POINT GROUPS Pe o til'. t3,'.' Pse o!31:Pa. 'B'; 'fi, Ftc. 6. (a.) Stereographic projection of the 2-fold rotation axes of the tetrahedral group and the pole points belonging to C'(332) where the solid circles denote points on the upper hemisphere and the open circles denote points on the lower hemisphere.(b.) Stereographic projection of all the pole points for the tetrahedral group. (c.) Stereographic projection ofthe rotation axes ofthe tetrahedral group and pole points appearing in the upper hemisphere. Selected interaxial angles are given by the following -- arclengths: pT,p,, : frh"" = fipru : flp,, fipr" = f?rp"o= 54.74";f1fo,, = 6A": fip,.: flp,, = -- 70.32';fiit,,: 6fu"0: 109.47";i?,p"" 90";6i",: 125.26'. Interaxial angles for the octahedral group 432 axesin 432 must also be arrangedin the same man- ner. That is to say that332 is containedin432. Hence The final casein our considerationof proper rota- we have placedall of the 4-fold and 3-fold axespres- tion groups is 432, which has three 4-fold axes,four ent in 432 (see Fig. 7a). Since there are half-turns 3-fold axes,and six 2-fold axes(see Table 6). Implicit implicit in the horizontally placed 4-folds and since in the three 4-fold axes are three 2-fold axes. Since these axesare perpendicularto the vertically placed there are more than one 3-fold axespresent, the argu- four-fold axis, we can concludethat 422 is contained ment given in our discussionof 332 allows us to in 432. Hence by our discussionof 422 we know that concludethat the three 4-fold axesare mutually per- there are 2-fold axes bisectingthe horizontal 4-fold pendicular.We use the sameorientation for these4- axes.Hence the point ps1placed as shown in Figure '7a fold axes as for the 2-fold axes in 332 (seeFig. 5). is a pole point belonging to Csg32). The Sincethe 6 pole points belongingto the 4-fold axesin remainingeleven pole points in C"(432)can be found 432 arearranged in exactlythe sameway as the 6 pole by locating the images of p., under the rotations of points belongingto the 2-fold axesin 332, the 3-fold 432 which have already been placed.The placement M. B. BOISEN, JR., AND G. V. GIBBS FIc. 7. (a.) Stereographic projection of the 3-fold and 4-fold rotation axes and pole point pr for the octahedral group 432. (b.) Stereographic projection of the rotation axes for the octahedral group. of alf the rotations of group 432 areshown in Figure Furthermore, all of the sets constructed from a 7b. The interaxial anglescan be deducedfrom those proper crystallographicgroup G as in (l) or (2) are given for 332. Also we note that by forming a multi- groups and hence improper crystallographicpoint plication table, one can verify that this arrangement groups. ofthe axesinvolved in 432 doesin fact yield a group. Before presentingthe proof of Theorem 2 we will In summary we have found interaxialangles for the use the theorem to derive the collection of all im- elevenproper crystallographic point groups 1,2,3,4, proper crystallographicpoint groups, give some ex- 6,222,322,422,622,332,and 432. It can be shown amples, and record in Table 7 all of the crystallo- that for eachofthese groups there existsa lattice that graphic point groups. is left invariant under each of the rotations of the If H is a subgroup of G such that : 2, group and so they are all bonafide crystallographic #(G)/#(H) then we say that H is a haluing group. This theorem point groups (Boisen and Gibbs, in preparation). shows that if we take each proper crystallographic Furthermore, we have shown that there.areno other point group at a time and first form G Gi proper crystallographicpoint groups. G one U and then H U (G\H)t for each of its halving groups Improper crystallographic point groups H (note that in certain casesG does not have any halving groups), then the resulting collection of In the previous section we showed that there are groups is preciselythe collectionof all improper crys- only I I proper rotation groups, and we derivedeach tallographic point groups. of them. In this sectionwe will show how eachof the point improper crystallographicpoint groups can be con- Note that the improper crystallographic structed from these proper crystallographic point groups of the form G U Gi have order 2f(G) and groups. This will be easily accomplished once we are those that contain the inversion (sinceI € G). have establishedthe following important theorem. Hence we have a total of eleven improper rotation groups that contain the inversion. When a group G Theorem 2. lf I is an improper crystallographic has a halving group H, then we can form the im- po proper crystallographic group H U (G\H)i which graphic point group G such that either has order #(G) and has the property that it does not contain the inversion (since/ € (G\H)). Table 7 (l) | : G where f is the inversion.(Here U 9i shows that there are ten such groups, bringing the Gt={gtlgeG}) total number of crystallographicpoint groups to 32. or If A is a rotoinversion such that O g G U Gi or (2)t:H U (G\H)t where H is a su ofG O e H U (G\H)i, then there exists an element g e G such that O : gi By the definition of the rotation axis of an improper rotation given earlier, GROUP THEORY DERIVATION OF THE 32 POINT GROUPS Taslr 7. The 32 Crystallographic Point Groups and Their Orders as Derived From the Proper CrystallographicPoint Groups The 11Proper Crystallographic Halving The 2l Improper CrystallographicPoint Groups Point Groups Groups* Containing i (centrosymmetrical) Not containing r' G # (G) H GUG' #(c u Gr) H \J (G\H)t #(H U (G\ H)r) 1 I none I 2 none 2 2 I 2/m 4 m 2 3 J none J 6 none 4 4 2 4/m 8 4 6 6 3 6/m 12 6 6 222 2 mmm 8 mm2 4 322 6 J 5z1m 12 3mm 6 4)) 8 A 4f mmm 16 4mk 8 222 42m 8 .A 622 t2 6 6f mmm L+ 6mm l2 322 62m l2 332: 23 t2 none 2imT 24 none 432 ai 23 4/m32/m 48 4zm 24 * The halving groups H of a given proper crystallographicpoint group G can be easily found by examining those groupsincolumnlthathaveonehalfthenumberofelementsasG. If Lisincolumnl suchthat# (L): # (G)/2, then L is a halving group ofG if (l) all of the rotations of L also appearin G and (2) the correspondinginteraxial angles betweenthese rotations are the samein L asin G. In certain caseslike 422 where222 occvs in more than one orientation, there is technically more than one group of the form H U (G\H)t for a given H. However, the construction of these groups results in the same point group but in different orientations, and consequentlywe do not make a distinction betweenthem. the rotation axis of @ is the sameas that of g. Hence, To illustratethe useof Theorem2 in the construc- the rotation axes appearing in G U Gt and H U tion of the impropercrystallographic groups, we will (G\H)i are the same as those in G Therefore it firstconsider the constructions G U Gi and H U (G\Hy follows that the interaxial anglesobtained for G ap- whereG is theproper monaxial group 6. For thiscase ply to both G U Gt and H l-J (G\H), as well. By the G U Gt becomes6 U 6, : U,6,3,2,3'',6-') U way Table 7 is constructedand becauseof the rela- u, 6, 3, 2, 3-',, 6-'li : ll, 6, 3, 2, 3-" 6-'l U tionship betweenthe proper and improper groups,all lti, 6i, 3i, 2i, 3-'.i, 6-'il U, 6, 3 3-" 6-" the crystallographicpoint groupswith the sameinter- i, 6,3, m, i-',6--'l *hich is denotedby 6/m. To axial anglesappear in the same row. We note again constructthe group H U (G\H)i, G must contain that in this paper we have not demonstratedfor each a subgroupH suchthat l(H) : #(G)/2.As notedin of these crystallographicgroups the existenceof a Table 7, if suchan H exists,it must appearin col- lattice I which is left invariant. However, this fact umn 1 of the table.In the caseunder consideration : can be shown and will be exploredin a future paper. #(6) = 6, hence#(H) 3. The only groupin column We will now do several examples of the I with order 3 that qualifiesas a possiblehalving construction described in Theorem 2 followed by groupin 6 is the propermonaxial group 3. Sinceall Table 7 showing the results we obtain by applying the rotationsin 3 are containedin 6, we conclude this constructionto all elevenproper crystallographic that 3 is the only halvinggroup in 6. Accordingly,it point groups. We recall that the orientation symbo is clear that H U (G\H), becomes3 U (6\3), : attached to a given rotation or rotoinversion is t1,3,3-1 U 16,2,6-Lli : \1,3,3-" 6i,2i, 'i1 dependent on the choice of the basis. In each of 6 : 11,3, 3-',6, m,d-'l denotedby 6-. our examples the basis will be chosen to be con- In our nextillustration, we will constructG U Gd sistent with the International Tables for X-ray and H U (G\H)t whereG is the proper polyaxial Crystallography (see Table 2.3.1 , Henry and Lons- group 322. In this caseG U Gi becomes(322) U -', ffil dale, 1952). We then conclude the paper with the (322)i : ll, 3, 3 U001 2, 11lol 2, l0t|l ), i, 3, 1-', l1 m, proof of Theorem 2. Ittllm,l0tllmldenoted Uy 32/m. The monaxialgroup r64 M. B. BOISEN, JR,, AND G. V. GIBBS 3 qualifiesas the only halving group in 322 because rotations lk$i,kzsJ,"',kosJl. By the cancellation #(3) : 3 and becauseall the rotationsin 3 arecon- law we know that theseare distinct rotations. Hence tained in 322. Thus, H U (G\H), becomes3 U we have found a list of n improper rotations in l. (322)\3)t = Il, 3, J-1, U00lp,tttl]m, t010lmlde- Therefore n < m. Now consider the set of rotations noted by 3mm. {(s'd)(s'i),(s'i)(srd),' " , (sriXs-i)}. Sincethe inversion Finallywe considerthe impropercrystallographic commutes with any proper rotation we have pointgroups that areconstructed from G = 422.The (s'd)(s;d): (r"Xls7) : s1fs1: srsr which is a proper G U Gi is constructedas before and resultsin the rotation. Hence we have that {s?,Jrf,z, . . . , Jp.} is group 4/ryrmm.Two proper groups4 and 222 in a set of z distinct (by the cancellationlaw) proper Table7 qualifyas halving groups in 422.Letting tt : rotations in I and so m 1 r. Hence m : n. Next we 4, we construct4 U (422)\4)i: U,4,2,4-', consider the set of proper rotations G : {ftr, ... , t}lm, t|lml l100lm,U l|t|lm, lf denotedby 4mm. Letting ko, sr, . ' . , J,l (remember that the q's are not H : 222, we construct(222) U ((422)\(222))i: in l). We will show that G is a group by showing II , 2, lt00l2,l0t0l2, f,7-r, lilol., ttllml denoted elosure. Let a, b e G. Then we have four cases by 42rn. to consider. First, if a and 6 are both /r's then Note that group 332lacks halving groups despite since they are proper rotations belonging to l, the the fact that groups322 and6 both haveorders one- composition ab is also a proper rotation in I and so half that of 332.This is becauiea six-foldaxis is not aD is one of the /r's listed in G. Second,if a : kt for containedin 332and the interaxialangles between some 1