American Mineralogist, Volwne 61, pages 145-165, 1976
A derivationof the 32 crystallographicpoint groupsusingelementarygroup 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
This paper has two objectives. The first is the very
specificgoal of giving a complete and rigorous derivation of the 32 crystallographicpoint groups with
emphasis placed on making this paper as self-contained as possible.The second is the more general
objective of presenting several concepts of modern
mathematics and their applicability to the study of
crystallography.Throughout the paper we have attempted to present as much of the mathematical theory as possible,without resortingto long digressions,
so that the reader with only a modest amount of
mathematical knowledge will be able to appreciate
the derivation. Those facts that are used but not
proven in this paper will be clearly identified and
referenceswill be given to sources where the appropriate discussionscan be found.
Crystal symmetry was studied during the nineteenth century largely from a geometricalviewpoint.
A comprehensiveoutline of the more important contributions to this subject by such men as Hessel,
Bravais, Miibius, Gadolin, Curie, Federov, Minnigerode, Schoenflies, and Miers is presented by
Swartz (1909). It appears that Minnigerode (1884),
Schoenflies(1891),and Weber (1896)were the first to
recast the problem of classifying crystal symmetry in
terms of group theory. Recent treatments make extensive use of specializedtopics in group theory in the
derivation of the crystallographicpoint groups (see
for example Seitz, 1934; Zachariasen, 1945; Burckhardt, 1947;Zassenhaus,1949;Lomont, 1959;Weyl,
1 9 5 2 ;M c W e e n y , 1 9 6 3 ;A l t m a n n , 1 9 6 3 ;Y a l e , 1 9 6 8 ;
Benson and Grove, l97l; Janssen, 1973; Coxeter,
1973;and Senechal,1976).In addition, severalworkers (for example,Donnay, 1942,1967;Buerger,1963)
have appealedto a number of interestingalgorithms
based on various geometric and heuristic arguments
to obtain the 32 crystallographicpoint groups. In this
paper we present a rigorous derivation of the 32
crystallographicpoint groups that usesonly the most
elementarynotions of group theory while still taking
advantageof the power of the theory of groups.
The derivation given here was inspired by the disc u s s i o n sg i v e n i n K l e i n ( 1 8 8 4 ) ,W e b e r ( 1 8 9 6 ) ,Z a s senhaus(1949), and Weyl (1952). The mathematical
approach usedin their discussionsis a blend of group
theory and the theory of the equivalencerelation'
Each of these mathematical concepts is described
herein and is used in the ways suggested by these
authors. However we depart from their approach in
the determinationof the interaxial angles.While they
appeal to a study of Platonic solids,we continue with
the theory of groups and equivalencerelationsin our
determination of these angles.
In the first part of the paper we discussthe definition and propertiesof point isometriesand show that
every point isometry is either a proper or an improper
rotation. If a rotation leavesa lattice invariant, then
it is shown that its turn angle must be one of the eight
f o l l o w i n g p o s s i b i l i t i e s : 0 o+, 6 0 o , + 9 0 o , + 1 2 0 o , 1 8 0 o .
The method for composing two point isometriesis
discussed.The propertiesdiscoveredabout this conrposition motivate the definition of an abstractgroup,
which is then stated. We then show that the set of all
point isometriesthat leavesa given lattice invariant
forms a finite group. These are the crystallographic
point groups. In the secondpart, the proper crystallographic groups are discussed.The cyclic monaxial
crystallographic groups are treated first. Then the
notion of an equivalencerelation and equivalence
classesare presented in 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:
-
'
->\y
Ftc l. Right-handed cartesian coordinate system with unit
vectors ryr{ directed along the coordinate axesX, Y, and Z, respectively, with the vector r expressedas a linear combination of i, j
and k
theoretical development of the basic ideas leading
to the derivation of all the proper polyaxial
crystallographicgroups together with their interaxial
angles. In the final part of the paper, the improper
crystallographic groups are constructed from the
p r o p e r c r y s t a l l o g r a p h i cg r o u p s .
Discussion of basic concepts
Three dimensional space
Our discussion of the 32 crystallographic point
groups will be conducted within the framework of
real three-dimensionalspace,R3,where R designates
the set of all real numbers. The elementsof RBcan be
describedgeometricallyby placing three unit vectors,
i,1 and k along three mutually perpendicularaxesX,
Y, and Z which form a right-handed cartesian coordinate system as in Figure 1. Accordingly, the vectors
i j,k form a basisfor R' such that Rsmay be viewed as
iiin-pty the collection of all linear combinationsof the
form r : Jri + yj + zk where x, y, z are elementsol
the sef R lsfrrbolized i,y,, e R). Note that x, y. and z
are the X,Y and Z components of r. Stated more
concisely,n' = {4t yj.+ t&lx,y,z e R} which is read
"R3 is the set of all vect6rs x1+ yj f zk such that x,y,z
are elements of the set n." ti is-impoiiunt to observe
that each vector in R3may be expressedin one and
only one way as a linear combination of {i,7',&}.
We
call the vector O : 0i + 0i + 0k the orisin"df'R'and
also useo to dJnotittre p'oint?t the oJgin. In general, unlessotherwise stated, we will not distinguish
between a vector and its end-point. lf r: xi -l yj.-f
z_(is some elementof R3,then its magnitudeor leng-th,
llfll , it defined to be
(l) Given any element4e R3,its image O($ is a
uniquely determined element of R3. (This
means that O is a mapping from Rato R3.)*
(2) Given any element4€ R3,there exists an element ,r E R3such that O(q) : r,. (This means
that O is a mapping from R3onto R3.)
(3) The magnitude of r is equal to the magnitude of the image of r under O for all4€ R3,
i.r., lltll : llotdll. (This meansthat o preservesmagnitudes
and henceangles,sizes,and
shapes.
)
(4) Given any two elements4s g R3,O(r *-r) :
O(9,)+ O(s) and given any real numberx,
o(x4) : xO(4). In particular,if4= x!.+ yi-+
z\,then o(d : x<D(i)+.yo(4) + zo(k). (This
meansthat O is a lineartranformationand that
O(4) is completelydeterminedby O(r), O(7')
and o(&).)
is that ib is a oneA consequence
of theseproperties
to-onemapping.That is, if r andj arein Rssuchthat
if o(4) : o({)
414, thena(L) f o(s). Equivalently,
for somer,seR3,then4 mustequals. A consequence
of
property(4) is that o(q,) : o(04 + E' + 04) : 0og)
+ 0O ("1)+ 0O (Q : 9.. Hencethe origin O of R3is
fixedunderQ, i.e.,O(g) : 9.(fne word "point" in
point isometryand point group alludesto the property that somepoint, in this casethe origin, is left
fixed-that is, unchanged
in position-by o.)
If €Dand O are two point isometries,
then we say
that they are equal-that is O : O-if O(r) : O(r)
for all4 € R'. Hence in view of (4), tio poin't
isometries€) and O are equalif and only if O(r) :
o(i), o0r) : o(/) and o(ft) : o(&).A trivialaut
imp'ortan?exampGof a poinl isometiyis the identity
mapping1 on R3definedbV I(D: r for all r 6 Rs.In
the case of the identity mapping,magnitudesare
preserved
sinceunder/ eachvectorin Rsis mapped
onto itself.Anotherexampleof a point isometryis a
rotation. For our purposesa rotation will mean a
turningofspaceabouta line,calledthe rotationaxis,
that goes through Q and has a positivedirection
* f n g e n e r a l ,a m a p p i n g a o f a s e t A i n t o a s e t B i s a r u l e w h i c h
assignsto each element in A a unique element in B. In this paper, R3
plays the role of both A and B, that is, the mapping we will be using
w i l l m a p R 3i n t o R 3 .
147
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
defined on it. Let P denote such a rotation and let I
denote its rotation axis. Then a turn angle p of P is
an angle measuring the action of P on a plane
perpendicularto /. The turn angle is consideredto be
positive if the action of the rotation is measuredin a
counterclockwisedirection on the plane when viewed
from the positive direction defined on / (seeFig.2a).
It may be noted that the turn angle of P is not
uniquely determined.In fact if p is a turn angle for P,
then so is p + k(360o) where k is any integer.Hence,
any specific value for p is merely a representative
of the turn angle of P. For example, the rotations
about the line I with turn angles 240o and -120o,
respectively, are really the same rotation P with two
different turn-angle representations.The identity can
be viewed as a rotation whoseturn angle is zero (or a
multiple of 360') and whose rotation axis may be
considerednon-existentor chosenarbitrarily, whichever is more convenient to the situation at hand. It
should be stressedagain that all of the rotations
consideredin this paper have axesthat passthrough
-o.
If *. are given any two point isometriesO and O,
we may define a new mapping called the composition
(or product) of @ and O denoted by @tD and defined
b y O o Q : O ( O ( { ) ) f o r a l l 4 e R 3 .I t c a n b e s h o w n
that @O is a poirit isometry. Likewise, it can be
shown that the composition of point isometries is
associatiue,i.e., O(IDV) : (OO)V for all point
isometriesO. @ and rlr. It can also be showR that
correspondingto each poini isometry O there existsa
unique point isometry O such that OO : O€D : /.
In this caseO is called the inuerseo/O and is denoted
b y O : @ - ' . B y u n i q u ew e m e a n t h a t i f @ O : O € D
: 1 and ov = vo : /, then o : rlr. The existence
of inverses implies the cancellation law which states
that if A, B and Q are any three point isometriessuch
that AB : Afl, then B : Q. If P' and P, are rotations
about the samerotation axis I with turn anglesp' and
pr, respectively,then the composition PzP' is simply
the rotation about I with a turn angle of p1 * p2.ln
particular, if pr: -pr, Ihen PrPr has a turn angle of
z e r o a n d s o P z P r: I ( i . e . ,P r : P r - t ) . H e n c e , b y t h e
uniquenessof an inverse,the inverseof a rotation is a
rotation. It can be shown that the composition of any
two rotations (possiblywith distinct rotation axes)is
again a rotation (seeSeitz, 1935).Sincethis last result
is intuitively obvious and becausethe proof does not
contribute to an understandingof the subject under
consideration,it will not be given here.
Proper and improper rotations.
In this sectionwe will show that every point isometry is either a rotation or a rotoinuersion, this latter
being the composition of a rotation with a special
isometry calledthe inversion,as definedbelow. Let O
denote a point isometry. We recall from property (4)
of a point isometry that the image of every vector
under O is completely determinedby tD(i), O(7-)and
o(k), the images of the basis vectors u-irder6. In
Figire 3(a) we have depicted the images of i, j-and k
under O. Since O is a point isometry and so
sin2p P(g)=(cos2p,sin2p,O)
__/.,,1"""",
,/>" iI
.,
g =P ( d = ( c o s p , s i n p , O )
o.
'-=,,,o,o
i
X
b.
Ftc. 2. (a) Diagram showing how the turn angle p of the rotation isometry P is measured with
respect to 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 ofgenerality that the unit length along the X axis is llrll. The vectorr is the imageofr under P. Accordingly, 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
fF]ril
dinx*i
;:ufi
conveniencewe will label V. From above, it therefore follows that uroQ) : i and vo(4) :j. Two
casesare possible for the image of tD(ft) under V,
z
namelyVo(4) : 4 o. wo(q) : -4 (re" Fig.3c).
Wefirstconsider
thecasewhereVO(ft) : k. Because
z
.roq) : l: /(Q,vo(4 :i:
;
(!)
.r.9,f i'
i,,-.,:,,,*rzt$(l]
*
'.':-..ffi.i.
:J-Y
''
't''+
. if=V,6flf
" ""\'V F(k)
f V,-f
*
' ' t1
?,StI
'r- 'L
^.(3b)
z
|
z
|
vzvrE(I)=!
Ik
,pzY+*t!=t
X-
t h a t V O : 1 . A c c o r d i n g l y , O i s t h e i n v e r s eo f a
previously noleo'
rotation; rnls,
rolallon'
this, as prevlously
noted, lmplles
implies tnat
that q'
O ls
is a
r o t a t i o n . N e x t w e c o n s i d e rt h e c a s ew h e r e V O ( k ) :
-L UV periormrng a hall-turn 2 (a rotatron wrth a
turn angle of 180' ) about the Z axis,we have2VA(i )
- k - H e n c e ,t h e
-i and2va(k):
: -i,2vo( il:
-r
for-all
- t t 3 rr, . in' R'. we call
tisomltry
D v r r r v L r J .2v-E
r
Y
rmap11.
r r q l J r t to
rv
3t".
this mappingthe inuersiordenoted6y i (i.e.,i(l :
+
[,,,*,,,=,
y
JEW{(l)=t
.,.".r,{.
|
I
-
-g
YzYr9tJJ=t
t%V,f t9=
...,.Y2Yt9\))=I
4: il4l. ie s..ihut i.J:'u"d(havithe sameimages
inO.rVo as theyOi ilnOeill Hencewe conchide
1;r::t"iif:''o):"fr:"J
ffiil':"ii,.,[';;'6l:
1,,,,.r,ir_i
y (2q/) 'i is ile composition
of the rotation(2v)-'
+
{,,-,
:
I(D'^naVo(t)
p
o i n t il Jsv ol l rm
r vv
v re
J rrvs
l l ir o n ,
^
w r re
J t r yf o r m e d b
u Jy t
r lh
l we
a
a l rn
ud t
f rh
l ve ir rn
rl .' . A
Pvlrr!
compositionof a rotationandthe inversionis calleda
-^1^:,-..^-^;^.u/^
L ^ . , ^ : , ,just
^+
^L^,,,+L^+
^.,^-., ..^:-l
point
rotoinuersion
We have
shown
that every
( 3c )
isometry is either a rotation or a rotoinversion.Since
a r o t o i n v e r s i o nm a p s a r i g h t - h a n d e dc o o r d i n a t es y s RTGHT-HANDED
LEFT-HANDED
(!,
tem into a left-handedone (and vice versa),it is often
(!,
(1t
(.1_),
(!,
p
{f
}
f
f
{f
f
d( - 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 change the hand of
Xt
a righrhandedcoordinatesystem(depicteO
in thtj left column)
or a left-handedone (depictedin the right column).t"l ,rr"-.',nJ
originalposrtionsof oO, o(t), and o6l.tUl showsthe result
of applyingV", to o(;ioQ,
ana 6g,j fne rotationrlr, was
.;
.
the system_._theyare sometimes referred to as proper
rotations. We have also shown that, if O is an improper rotation, then there exists a proper rotation
suchthatO:Pi.SinceP(t):P(
ne
d coo( kn )sm
a "p"ilngtro, ,rP
.{):P((-lE):
z1 l.l:('s. "t i1p.p, "l e, idi p
) p3l (a4nteo;is, w
u chhear reout p
ao
r ino on Q
c a. a
nb
tiu
l-hlp'e,(1r )! f o r a l l 4 e
-'-;, R3andsinciipl4)jj(pg)):-ltrl
as the rotation axis the line through the origin perpendicular
_
f"o r . a l l l . . :
R3,we see that Pd: tP and so i comt o a p l a n ed e t e r m i n ebdy i a n d o r 4 . r n " r " i .
" " g , J of
J qapply",'l,
mutes with every proper rotation.
simplythe anglebetween
r'iioo14. @; rho,ror
the resulr
rng q/, to rf,Of4,U.,O(i),
and Ur,<D(k). The rotation Vr, was
chosen to mapUnr<DQtoi. Note thatqr, is constructed in the same
w a y a s w a s U r , i n ( 3 b ) . S i n c e i ' i s p e r p e n d i c u l a rt o a p l a n e c o n tainingllf,rD[) andi the rotation axis of Vz can always be chosen
to be the X axis. Htince Vr%<D(i) : ,.
Space lattices
A crystallographicpoint group consistsof a set of
point isometriesthat leave a given spacelattice invariant. A subset I of R3is a spacelattice if and only if
preservesmagnitudes and angles, {O(i),O(7'),rD(k)} there exist three non-coplanar vectors1r, 1, and t,
also forms a set of three vectors of unit-lengih wtrGtr such that
are mutually perpendicular.We will analyzeO (Fig.
f = {a]:r+ u!: + wt,"l u, u, w e 7l
3a) by consideringtwo consecutiverotations V, followed by \r, such that V, maps @(i) to iw h e r eZ d e n o t e st h e i n t e g e r s( i . e . ,Z : t " ' , - 2 , - 1 ,
whereupon qr,(O(/)) and V,(rD(k)j must-lie in ihe 0 , 1 , 2 , " . ) ) . N o t i c e t h a t i n t h e d e f i n i t i o no f I o n l y
YZ plane (Fig. 3b)l-and W, is thelotation about the integers are allowed as coefficients of /r, t2, zr:,dl.s,
X axis that maps rlrl(O(/_))to jr.and, sinceits rotation whereasin the definition of Raanv realnrimber cdn
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
t49
(l) Closure:
If O,O 6 K,thenOO e K. (Thiscan
that o@(I) : O(O(I))
be seenby observing
: o(r) : I.)
points) in spaceconsisting of the end-points of vectors in I radiating from the origin, O. As stated
before, we will not distinguish betweena vector and
i t s c o r r e s p o n d i n gl a t t i c ep o i n t .
Lattice inuariant isometries
lf tD is a point isometry, then we say that I is
invariant under rD in casef : O(I) where we define
O(I) : {o(t) | { e I}; that is, o(l) is the set of all
rD(l) such that r is an elementof f . This is not to say
thaTt : @(l) for all r e I but merely that I and rD(l)
are eQualsetsofveclors. In other words thevectors in
conclusion is true for any linear transformation).
We will now show that there is only a finite number
of distinct point isometriesthat leavea given latticeI
invariant. Consider a sphereS centeredat O with a
large enough radius so that S contains all three of the
tD be an element
basisvectorstl,
*, and tt of I and let
of the set K of all point isometriesthat leavethe lattice
I invariant. Sinci O does not stretchvectors(magnitudes are preserved),O(tl), O(tr) and O(tr) must be
vectors in I that are cdntaineb within ihe sphere.
Moreover, since the radius of the sphereis finite, we
may assume that there is only a finite number of
lattice points of I in the sphere(a rigorous proof of
this intuitively obvious fact is given in a paper on the
derivation of the 14 Bravais lattices being prepared
by Boisenand Gibbs). Consequently,there are only a
finite number of choicesfor O(tr), O(tr), and O(t3),
i.e., there are only a finite numb-erof pissibilitiesTor
the imagesof the vectorst1,t2,and t, under @. Since,
as previously observed, 6 is comp-letelydetermined
by {tD(1,), O(tr), O(ts)}, we conclude that there is
only a frnite ndmber df distinct point isometriesin K
(i.e., thal leave I invariant).
Now we considerthe algebraicpropertiesof the set
of all point isometries K that leave a lattice I
invariant. The followins facts are observedabout K:
(2) Associatiuelaw: If O,O,V 6 K, then O(OV)
: (OO)V. (This was observedearlier.)
(3) Existenceof an identity element:There exists'an
e l e m e n t l € K s u c h t h a t l € ) : € D 1: O f o r a l l
O € K. (The identity mapping discussedearlier
has this property and is in K because/(I) : I
(in fact, I(t) : I for all t € I) ) I
(4) Existence iJ anlnuersefdr each element' Given
an element@ g K, there existsan elementO-'
10 : 1. (We have
t
€ K s u c ht h a t < D O : O
observed earlier that the point isometry @-1
e x i s t s .F u r t h e r m o r e ,O - ' € K s i n c e@ - ' ( f ) :
o - 1 ( o ( r ) )= o - ' o ( r ) : 1 ( r ): f . )
Any set of elements together with a composition
obeying the four axioms listed above is an algebraic
systemcalled a group. Hence K is a group, and sinceit
has only a finite number of elements,it is said to be a
finite group. The number of elementsN in K is called
the order of K and is designatedby N : l(K). The
observation that K is a finite group is important becausein generalif H is a finite non-empty subsetof a
group G, then H is itself a group (under the
composition of G) if and only if it satisfiesthe closure
axiom (Herstein, 1964). When a subset H of G is a
group under the operation of G , we call H a subgroup
of G. Hence, a nonempty subsetH of the finite group
G is a subgroup of G if and only if h'h, e H for all
hr,h, € H. Consequently,if T is any nonempty set of
point isometries that leave the lattice I invariant,
then, sinceT is a subsetof the finite group K, we need
only show closure to demonstratethat T is a group.
To distinguish these groups from groups of point
isometries in general, we call such groups
crystallographic point groups if a space lattice is left
invariant under every element of the group. (Note
that the crystallographic point groups are special
because,in general,given a point group G there may
not exist such a lattice). Moreover, if all of the elements of the group are proper rotations we call it a
proper crystallographic point group. An improper
crystallographic point group is one that contains at
l e a s to n e i m p r o p e r r o t a t i o n .
Possible turn anglesfor lattice
inuariant isometries
Let P denote a rotation that leaves a lattice I
invariant. For convenience,we will chooseas its turn
angle representativean anglep such that - 180' ( p
I5O
M. B BOISEN.,/R.. AND G. V. GIBBS
< 1 8 0 o .I t i s p r o v e n i n a n u m b e r o f p l a c e st h a t s i n c eP
leavesI invariant, the only possibleturn angle repres e n t a t i v e so f p m u s t b e o n e o f t h e a n g l e s0 o , + 6 0 o ,
+ 90o, + I 20o, I 80o (seefor a matrix argument,Seitz,
1 9 3 5 ;B u r c k h a r d t , 1 9 4 7 ;L' o m o n t , 1 9 5 9 ;a n d J a n s s e n ,
1913; and see, for a geometric argument, Hilton,
1 9 0 3 ; R o b e r t s o n , 1 9 5 3 ; B u e r g e r , 1 9 6 3 ;a n d B l o s s ,
l97l). We will present a nbw and different proof o[
this fundamental property of a rotation that leavesa
lattice invariant. First we observethat in the caseof
any such non-identity rotation, P, there must exist a
two-dimensionalsublattice I' of I in the plane perpendicular to the rotation axis of P (cf Seitz, 1935;
Zachariasen,1945;and Boisen and Gibbs, in prepar a t i o n ) . B o i s e na n d G i b b s s h o w t h a t i f r i s a s h o r t e s t
nonzero vector in I' and if .s.is a shoitest nonzero
v e c t o r i n I ' n o n c o l l i n e a rw i t h r . t h e n t h e s e t { / , J }
servesas basis generatingf ', that is 7' : Ptr -l us I
u,n € Zl (the fact that such vectorsr and s exiit is also
shown by Boisenand Gibbs). Let ibe a Jhortestnonzero vector in l'. Without loss of generality,we may
place the X axis of our cartesiancoordinate system
along r and the Z axis along the rotation axis of P
sinceit is perpendicularto f'. This situation is shown
in Figure 2b. If we definethe unit length along the X
. et s be the image
a x i st o b e l l r l l , t h e n 4 : ( 1 , 0 , 0 ) L
o f { , u n d e r P ] t h a t i s P ( r ) : s . H e n c e ,F e I ' , a n d l l r l l
: llrll If the turn-angli of F ir 0' o.l80o, then p-is
one-of the possible turn angle representativesenumerated above and we would be done. On the other
h a n d , i f p i s n o t e q u a l t o 0 o o r 1 8 0 o ,t h e n r a n d - s a r e
not collinear and s, being the same length as r. is
clearly a shortestnin-rero vector in f' non-colliiear
with r. Hence,r and s constitute a basis for I' and
each v'ector/ 6 l' must be an integral combination of
4 a n d s , t . " . 1 1 : u 1 + u 4 w h e r er . r 1 n du a r e i n t e g e r s .
Then, as demonstratedin Figure 2b,{: (cosp,sinp,0)
SinceP(s)6 I', it must be
and P(s) : (cos2p,sin2p,0).
an intelral combination of l{,E|, i.ei, P1s; : u7+ ug
'where u and u are integers.Therefore.
( c o s 2 p ,s i n 2 p , 0 ) :
u ( 1 , 0 , 0 ) + u ( c o sp , s i n p , 0 )
:(u*ucosp,usinp,0)
from which it follows that sin2p : usinp. Replacing
sin2p by 2sinpcosp, we see that 2sinpcosp : usinp.
Sincep I 0 and p + l80,sirip I 0and so cosp : u/2.
Becausecosp is a function bounded by t I and D is an
integer, the only possible solutions are cosp : 0,
+l/2, +1. Hence p is one of the following eight
p o s s i b l et u r n a n g l e s :0 , + 6 0 o , + 9 0 o , + 1 2 0 o , 1 8 0 " .
Recall that a positive p angle is measured in a
counterclockwisedirection. The symbol and name
correspondingto the rotations associatedwith each
o f t h e e i g h t t u r n a n g l e sa r e g i v e n i n T a b l e l .
The symbols listed in the table describe only the
t u r n a n g l e s o f c o r r e s p o n d i n gr o t a t i o n s b u t l a c k
information about the orientation of the rotation
axis. Hence, when specifyinga non-identity rotation
using this symbolism,a descriptionof the orientation
of the rotation axis is required. When discussinga
point group that leaves a particular lattice I invariant, it is customaryto choosea conventionalunit
cefl (Table 2.2.2,InternationalTables,vol. I) outlined
by the cell edgevectors {a,b,cl(note that {a,b,c.}is not
alwayschosento be a primitiue triplet: tha*tlsla,!, 91
is sometimes not a basis for f ). The conveniional
choice of q, b, and c (Table 2.3.1, International Tables, vol. tl ti.nr ouito be such that if / is the axis of
a rotation that leavesI invariant, then there existsa
vector r = ua * ub * wc along I such thatu, u, andw
are inGgers'1see2acholriasen,1945; Buerger, 1963).
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 orientation of the rotaAssociatedwith the Eight PossibleTurn Angles
tion axis. Hence, [u"] n where n : 2, 3, 4, or 6
symbofizes a l/n turn about the line ua + u! + wg.
Symbol for rotation
T h u s . t / m21 s y m b o l i z e sa h a l f - t u r n p a r a l i e lt o a , l 0 t 0 l 4
associatedwith
s;nibolizes a quarter-turn parallel to b whereas
Turn angle p
turn angle
Name of rotation
llttlj-t symbolizes a negative third-turn farallel to
180'
2
half-turn
-a-b*c.
T h e o n e e x c e o t i o nt o o u r r u l e i s w h e n
1200
3
third-turn
a x i s p a r a l l e l s- c i n t h a t c a s e n o o r i tni itiiion
quarter-turn
90"
4
entation symbol is attached to the rotation sym600
6
sixth-turn
bol, hence4-t denotesa negativequarter-turn about
0
I
identity
We will always place the Z axis of our cartesian
c.
-60"
6-r
negativesixth-turn
system along c. For groups requiring oriZoordinate
- 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!
I5I
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
along c. (The fact that a. and b can be placed
perpend-icularto c is not ob-vious/nO ir explored by
Seitz, 1935,and B'oisenand Gibbs, in preparation.)
As observedearlier, if O denotesa rotoinversion,
then @ : Pd flor some proper rotation P. We will
assignas the turn angle of O the turn angle of P and
as the rotation axis of O the rotation axis of P. Hence
the orientation symbol for O is the orientation symbol for P. The names and symbols for the
rotoinversions corresponding to the eight possible
turn angles are given in Table 2. Note that 2i : 2 can
also be visualizedas a reflectiondenotedby rn, where
m refers to the mirror plane about which the reflection occurs. The orientation symbol given to rz is
inherited from the half-turn perpendicularto the mirror plane. For example,U0ol^ denotes a reflection
whose mirror plane is perpendicular to q while ra
d en o t e s a r e f l e c t i o n w h o s e m i r r o r p l a n e i s
perpendicularto c.
Tnsr,n 2. Names and Symbols for the Rotoinversions
Associatedwith the Eight Turn Angles
Symbol for rotolnverslon assocl
Turn angle p
with turn angle
- 600
m: i2: 2i:2
T: B: 3i
4: i4:4i
6:i6:6i
i: it:Ii:1
6-t:i6-':6-'i
- 900
4-t:
180"
120"
900
600
0
i4-':4-'i
]-':r3-':J-'i
Name of
rotoinversion
reflection
third-turn inversion
quarter-turninversion
sixth-turn inversion
inversion
negativesixth-turn
inversion
negativequarterturn inversion
negativethird-turn
inversion
: 16,3,2,3-r,6-',Il and denote the group
63,6n,6u,6"1
by 6. Note that (f-1) : 16-',6-",6-t,6-4,6-5,6-6l.
: l6-t,3-t,2,3,6,11.Since the order of listing the
Proper crystallographic point groups
elementsis immaterial, we see that (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
some lattice I invariant (Boisenand Gibbs, in prepa- as evinced by Table 3. In summary, we have shown
ration). Consequently,the searchfor proper crystal- that all of the proper monaxial crystallographicpoint
lographic point groups is equivalentto the searchfor groups are cyclic. Becauseof theoretical considerathose finite groups consistingof rotations of the type 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 element g e
group. When G has two or more rotations with disG 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
group it is sufficientto consideronly the posifinite
Proper monaxial groups
tive powers of the generatoras borne out by our exIt is evident that each of the following collectionsof amples above. In fact, if (g) is finite and t :
#((g)),
c r y s t a l l o g r a p h i rco t a t i o n s l l l , 1 2 , I l , 1 3 , 3 -J' L 1 4 , 2 , 4 - ' , then I is the smallest positive integer such that gr
1 ) , a n d { i 6 , 3 , 2 , 3 - ' , 6 - ' , 1 )f o r m s a p r o p e r m o n a x i a l : 1. For example, in the case of (4-1), the smallest
group where all of the rotations in each collection
positive integer I such that (4-L)t : / is t = 4, hence
take place about the same axis. (Recall we need only
# ( ( 4 - ' ) ) : 4 a s e v i n c e db y T a b l e 3 . A s s e e ni n T a b l e 3 ,
demonstrateclosureto concludethat each is a group.
there is only one distinct monaxial group of each
For example, for {3,3-',1} we observe that each of
order about a single axis. If G is a proper monaxial
c o m p o s i t i o n s S S :3 - 1 , 3 3 - t : l , 3 I = 3 , 3 - ' 3 : I , 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
Tanr,r 3. The Proper Monaxial Groups,
for {3, 3-', II.) The list of monaxial groups given
Their Generatorsand Elements
above exhauststhose that can be made from the rotations presented in Table 1. This can be quickly
Group generatorsand elements
Group symbol
checked by considering each of the remaining colI
lections and showing that no other such group is
2
:
possible(e.g., \1,2,31is not closed because23
6-'
-',6 -',
(3) : (3-') : {r, J, J-1}
3
I | : l$,62,63,6u,6u,6"1,
e U,2,31).Since 16,3,2,3
4
< 4 > : ( 4 - ' >U: , 4 , 2 , 4 - ' l
we call |16,3,2,3-t,6-r,1)
a cyclic group generatedby
: <6-'): lr,6, J, z, J-L
6
<6)
:
(6t
6. In group theoretic notation we write
16,6',
(r): {r}
: u,2l
<2>
M. B. BOISEN. JR,. AND G. V. GIBBS
t52
designatethe point group. The axis common to all of
the rotations of a given monaxial group n (wheren :
2, 3, 4, or 6) will be called an n-fold axrs. (When n is
different from l, 2,3,4, and 6, a perfectlygood point
group results in the manner describedabove. However, for such a point group no spacelattice is left invariant and hence it is not a crystallographicpoint
group. )
Tasr.s 4. Group Multiplication Table for Group 322
3-L
I
3
,-r
I
3
3-r
tlool
trlol
2
2
loLol
2
Irool
2
Ilrol
lorot
2
2
3
J-r
I
Iorot
2
Iroot
2
trrol
2
lrool2
trool2
3-r
rrrot
7
2
torol
3
2
I
3
toLot
2
tlrot
2
Iorol
2
lrool
2
torot
2
trool
2
lrrol
2
J-l
J
I
3-l
31
Proper polyaxial groups
In our investigation of the polyaxial groups, we
will be examining combinations of monaxial groups.
The task of finding the possible proper polyaxial
point groups will be considerablymore difficult than
that of finding the monaxial point groups. It will
require some fairly sophisticatedarguments and a
considerableamount of notation which will be explained as we go along. The main aim of this section
is to establishthe inequality which in loose terms will
state that three monaxial groups with orders zr, zr,
and zr, respectively,may be used to form a proper
polyaxial point group only if
l/v,1-
l/v" -l l/y")
l.
To facilitate our proof that will establish this inequality we will considerthe surfaceE of a unit sphere
centered aL O. Any point isometry O acting on E
maps E onto itself. In fact, tD is completely
determined by its action on 3 becausethe effect of O
on E determines the end-points of tD(i), O(7) and
O@) and thereforethe componentsof dQ, o() and
O(4). A non-identity proper rotation ft about the
rotation axis I leavesexactly two antipodal points on
E unmoved. These points are preciselythe points at
which I and E intersect and are called the pole points
belongingto ft. If thesetwo points are labelledp and
q, then they are the only points'x on E that satisfy the
equality h(x) : x. Let G be a proper polyaxial group.
The trivial group t has already been treated as a
monaxial group and since it has no pole points we
deliberately exclude it from consideration in the
remainder of this discussion.Hence we will assume
that l(G) > 2. To help illustrate a number of the
concepts given below, we will on occasion use the
proper polyaxial group 322 as an example.However,
none of the theoretical conceptsthat follow depend
on the assumption lhat 322 exists.The rotations of
322 are ll,3i-'!nl2lttol2lorol27.
Since 322 consistsof 6 distinct rotations, by definition #(322) : 6.
The orientation of each of these rotations is shown
in Figure 4a (recall that the left superscript is only
assignedto a rotation symbol when the rotation occurs about an axis other than Z), and the pole
points belonging to the rotations of 322 are depicted in Figure 4b. The multiplication table for 322
given in Table 4 servesto define the structure of the
group by enumerating the compositions of all
possible pairs of group rotations. Since the only
rotations that appear in- the table are the six rotations already listed for 322, we may conclude that
322 is closed under composition and hence is a
group.
Letp denote a pole point of G. Then the collection
of all rotations of G that have p as a pole point is
d e s i g n a t e db y G o , i . e . , G o : { g e O l g ( p ) : p l . * l f
gy gz e Gp, then g,gr(p) = g,@,(p)) : S,(p) : p.
Since g, and gz were arbitrarily chosen rotations
of Go and since grgz € Gp, we conclude that Go is
closed under the composition and hence a subgroup of G. Since the rotations of G, leavep fixed,
they all have the line Qp containing Q and p as their
rotation axis. Consequently,G, is a proper monaxial
group and thereforeone of the cyclic groups listed in
Table 3.
In the case of 322,(322)p^ : (322)p"z: {'oo12,ll,
(322)es,: (322)e"s: {"ol2,ll, (322)e22: (322)ou,=
\t0t0l2,11and (322)o,, : (322)or": lJ,3-',/,}. We observein this example that two pole points are associated with each monaxial group (Fig. a). lt is also evident that the pole points pzr, pzz,pza, pn, prr, and pn
are each associatedwith the group 2 where the rotation axis of each has a different orientation. The remaining pole points p ' andpr" are similarly associated
with group 3 (seeTable 3). (The significanceof the
double subscriptsattachedto each pole point belonging to 322 will be discussedlater. For the time being,
the subscriptsmay serveto distinguish distinct pole
points.)
* The subset Go of G, consisting of all g E G for which g(p) =
p, is a subgroup of G sometimes called the stabilizer of p.
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
,(3221p"={ t,troola,
,322)p22=
lt'totolzl
(322)p33
= lt,[oro]il
( 3 2 2 ) p s=rl t , [ r r o ] 2 1
(322\^
P2l
I
L { s z z ) ' , . =1 r , 3 , 3 - r l
z
\
l/
./
FIc. 4. The orientations of the rotation axes and the associated pole 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
a r e n o t s h o w n . N o t e t h a t i n t h i s f i g u r e X , Y , a n d Z a r e n o t m u t u a l l y p e r p e n d i c u l a r ,s i n c e t h e a n g l e
between X and Y is 120" (the ansles between X and Z and )/ and Z are both 90').
153
154
M. B. BOISEN, JR., AND G. V. GIBBS
dicted from Theorem I that there will be no group
elementin 322 that mapspzt to pr becausetheir correspondinggroups(322)e.,and(322)o' are of differeG
groups
e same
ent orders.This is easilyverifiedby consideringthe
rotationsin 322 one at a time and observingthat indeed none of them map pzr to ptr.
Proof. From the discussionabovewe know that Go
We will now investigatean important relationship
and Gn are monaxial and hencecyclic groups.Let i
that
existsbetweenthe pole points of any givenrotabe a generatorof Goand let n : #(c) and,m: #(G).
group
G. We define0(G) to be the setof all pole
tion
We want to show that m : r. As observedin the
points belongingto the rotationsof G, i.e.,0(G) -sectionon propermonaxialgroups,n is the smallest
thereexistsat leastoneg€ G suchthat g * I
positiveintegersuchthat h" : 1. Considerthe rota- {p e Sl
p|. For example,in thecaseof 322,0(322)
g(p):
and
tion ghg-t where g(p) : q (such a g existsby the
We now define a
lPrrrPrrrPrr,Pr2,p2s,psbps2rprrl.
hypothesis
of thetheorem).Srnceg-'(q): p andh(p)
for
eachp"p" € 0(G)
relation
on
0(G)
by
defining
: p, we observe Ihat ghg-t(q) : S(h(g-'(q))) :
(symbolized
p,
is
to
that
equivalent
h - pr) if and
h
dh(p)) : g(p): 4. Henceghg-' @): q demonstrat- only if thereexistsa rotation € G suchthat g(p') =
I
ing that ghg-' e Go. We next show that gftg-r is a
pr. Hence,if p' - pr, then from TheoremI we know.
generatorof Go.Letf be an arbitraryrotationof Go.
that Go, and Go, are cyclic groups of identical
Then g-'/ge Go becauseg-'fg(p): g-'W@))):
orders,i.e., #(Gpr) : #(Go).
S-'(^q)) : S-'@) : p. Therefore,sinceGo : (h),
We observethe following important propertiesof
g-'fg : it for someinteger I < t < n. Hencegh'g-'
with respectto the relation definedin the last
0(G)
: S@-yg)S-t : .f. Therefore
paragraph:
(l) Reflexiue.'For all p e 0(G), p - p. (This is
f: sh's-'
: Shh ... hg-, (wherel occurst times)
becauseG has an identity elementI and 1(p) = p.
= S h ( S - ' S ) h ( S - ' g .)-h.
:
g
f
1
;
@-'glhg-'(since
Hencep - p.)
: @hg-')@hy-r).. ' (ghg-') (by associative
law)
(2) Symmetric:For allpr, pr€9(G), if p, - p2,then
: (ghg-')'
if ge G suchthatg(pr) : p",
pz - pr.(This is because
: pr.
Consequentlyf is a power of ghg-'. Since/ was an theng has an inverseg-'€ G so that g-'(p")
arbitrarily chosen rotation of Go, we have shown that Hencepz pr.)
(3) Transitiue.'For all pr, pr, Pse 0(G), if p, - p,
Go : @hg-t). Since m : #(Go), as observed earlier
if g(pt) : p,
andp2 - p", thenp, - p3.(This is because
z is the smallest positive integer such that (ghg-'1^
:
g,
and h@r) p, where h e G, then hg e G and
1. However, since n : #(G), (ghg-') : gl2ng-r
= h(g(p,)) : h(p") : p'. Henceh - pg.)
glg-' : 1, but m is the smallestsuch integer hence hg(p')
m 1 n. But (ghg-r)- : gh*g-' : 1 and accordingly Any relation definedon a set satisfyingthesethree
relation.Let p E
h- : g-'g:
. 1 .H o w e v e r n : l ( G o ) i s t h e s m a l l e s t propertiesis called an equiualence
classdetermined
we
define
the
equiualence
G(G), then
positive integer such that h" : I. Hence n 1m.
:
qI.*
p
That
is, [p] is the
O(G)lp
by
to
be
[p] {qe
Therefore, since n < m and m 1 n, we conclude that
- p,
p.
pole
points
to
Sincep
G
equivalent
all
of
set
of
m : n and so #(Gr) : #(G").
appears
p
of
0(G)
every
element
we
have
and
so
e lpl
In our example 322, if we consider the cyclic
class;hence0(G) is the union**
group (322)^t : ll,II101,2)we see Ihat ttt0lz(p2r): pr". in someequivalence
classes.Furthermore,if [p]
its
equivalence
of
all
of
Hence, according to Theorem I we can conclude that
or
more
elementsin common(symand
one
[4] have
#((322)e) : f;\(322)o"). We observe that this is ing),*** then [p]
bV
bolized
t
tql.
[p]O lql
deedtrue because(322) ez2: glotol2T and henceboth
be
ints belo
to
the
rotation
sets contain two rotations. Similarl, Vml21prr7:
pr2 and therefore we can conclude that (322)o' and
(322)pn have the same order. Again an examination
of Figure 4 shows this to be the casesince(322)prr:
ll ,3,3-'l and (322)rr" : 1l,3,3-'l . It may also be pre-
* The subsetof S consistingof all g(p) as I'ranges over all G is
sometimescalledthe orbit of p.
** The union of a collectionof given setsis the set consisting
of all thoseelementsthat appearin at leastone of the given sets.
The union of two sets A and B is symbolizedA l-l B and the
symbolism1) A, -"un, Ar U A, U Ar.
i* Those familiar with group theory will recognize that this
theorem shows that Go is isomorphic to Go since they are both
cychc.
*** The expression"A O B" denotesthe intersectionof sets
A and B, i.e., the set of all elementsthat are in both A and B.
The symbol0 denotesthe empty set;i.e., the set containingno
elements.
i-l
t55
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
Accordingly, distinct classesare disjoint, i.e., they H ; y , t h e n H 1 1: l h i 1 , h 1 t , " ' , h ' i l. M o r e o v e r t h e a x i s
haveno elements
in common.A thoroughdiscussion common to the rotations of Hlyis a vrfold axis (this
of equivalence
classes
includingproofsof our obser- axis of course passes through pit and the origin,
vationsabovecan be found in mostmodernabstract Q). In the context of our example 322, we see
alegbratexts(e.g.,Larsen,1969;Shapiro,1975).
that Hr, : ll ,3,3-'l,Hr, : U,3,3-t), Hr, : ll ,Uoo)21.
In our exampleof 322,the elements
of the equiva- H r "
u,ulol2L H''-: l[,utol2\
$,loto\)|, an
lenceclassof p2,in 0(322) arelprrf : llapg(322)lp^ H", : ll,U00l2l, and Haa : 11 pt0l2\. In addition
-Ptl:
we observedthat #(H'') : #(Hr) : t,r : 3, #(Hrr) :
|n e 0(322)l thereexistsan elementge322suchthat #(H",) : #(Hr') : vz: 2, and l(H") : #(H'r) : #(H")
:vs:2,
g(pz,): Pul
We are about to present a lengthy derivation of
some important relationships between many of the
: (I(p"r), 3(pr,),,rr]-,,rour,
conceptsdiscussedabove.These relationshipswill be
to'2(prr),
used to develop facts about G which will ultimately
totot2(prr)l
"'ot 2(p"r),
enable us to determine all of the proper
: l 4 " t , P z z ,P z a P
, z r ,P z z ,P z s l
crystallographicpoint groups which we then list in
Table
6. In an attempt to presentthis discussionin a
: l \ r r , P z z ,P z a l '
more understandablefashion, each of the more imFollowing the same procedure one may calculate portant relationships will be set off in a box at the
each of the equivalenceclassesof 0(322). The results beginningof each argument where the relationshipis
are: lp"rl : IP"rl: [prr] : lPzbPzz,Pzs],
[pr'] : lP'") = established.
:
:
:
lprr,prrl.We
lprr,prr,p"rl and [p1l
lpr"]
[prr]
t
notice that it is indeed true that these equivalence
(l)
2(N-l):En;Qt-r)
d-l
classesare disjoint and that 0(322) is their union.
The pole point notation for 322 was deliberately
Recall that N : #(G), vi : #(Hit), nt : #(C{G)),
chosen at the outset so that points with the same
classesof pole
first subscript would be in the same equivalenceclass. and t is the number of equivalence
Equation
Let t denote the number of distinct equivalence points. The basic strategy for establishing
the
counting
ways
of
classesof 0(G)where C'(G), Cr(G),"',C'(G) denotes (l) will be to find two distinct
be
two
will
result
The
the equivalenceclassesof 0(G), i.e., C{G) : [p] for nonidentity rotations of G.
expressionseach equaling twice the number of nonsomep€G(G),hence
identity rotations of G. This will establishEquation
t
( I ) sincethesetwo equal expressionswill be precisely
:
g if i + j.
-U c,(G) @(c) and c,(G)a cr(G):
those appearingin the equation. We begin by taking
each pole pointpi.r in 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
set and we let nl denotethe number of points in C1(G);
nonidentity rotations in G becauseeach ofthese rotathat is, n1 : fr(C{G)). Therefore, we may list the
tions leavesexactly two pole points fixed and hence is
elementsof C,(G) asp1, ptz,. . . ,ptn.l n the caseof our
counted twice. The number of non-identity rotations
example, C{322) : {pr,ppl, C"(322) : \pzr,pzz,pzel
- 1. Thus, for the
leaving piifixed is f(H,7) - | : vt
and Cr(322) : lpu,psz,par|and as observed above,
pole points in C'(G) we have
: lg(o,)lSe 3zzl
eQ22) :
E
c{322) : c1322)U c,(322)U
cs(322).In addition, notice that n, : #(CLQ22)): 2,
nz: #(Cz(322)) : 3 and n" : #(cs(322)) : 3.
Let Hiy denote G' the group of all rotations in G
that leave pu fixed. Sincep;y andp1pare both in Cr(G),
#(H,t) : f(H1e) becausepu - p,n (see Theorem l).
H e n c e H r r , H r r ,. . . , H r n t a l l h a v e t h e s a m e o r d e r
which we signify as 2,. It follows that H1,is a cyclic
group of order v1 and if we let hqbe the generator for
rotationsleaving
fi(non-identity
p1 fixed) : vr - |
leaving
rotations
f(non-identity
p ' , f i x e d )= ' v r - |
nr equatlons
l(non-identity rotations leaving
pt", fixed) : vr - |
M
156
Tasls 5.
B BOISEN. "/R.. AND G. V. GIBBS
Display of the Derivation of Equation (1) for Group 322
Notation for
group G
Notation for group 322
0(322):
e(G)
l\rr, Pn' Pzt,Pzz,Pzz,Psr,Pzz,P,nl
c{322)
,(G)
Pn
Pti
caQ22)
c2Q22)
Pzz
Pzt
Pzz
Ptt
I'23
Pzs
Gpii
: {c € Gls@') (322),,, :
(322)p,, :
(7))\^
:
Ptil
[ 1,
\v;
l)
{r, J,J-'}
(3- 1)
n
"(u;
{r, J,J-'}
(3 - 1)
2 ( 3- r ) : 4
l)
f n n @- , r )
'too'2]
(322)e"" --
(322),"":
l I , ' o t o t2 \
l1, "to'2]' |1, "'o' 2l
(2- r)
(2- r)
J
2 ( # ( 3 2 2- ) 1 ) : 2 ( 6 -
Summing up thesenumberswe flnd that the contribution from C,(G) is nr(u, - l). Following the same
argument, we seethat the contribution from C;(G) is
ni(vi- 1) for each I < i < t. Adding thecontribution
from each of the I equivalenceclasses,we find that
t h e s u m i s I n 1 @ 1 - I ) . S i n c eN - I i s t h e n u m b e r o f
t=1
non-identity rotations in G, it follows that twice the
n u m b e r o f n o n - i d e n t i t yr o t a t i o n si n G i s 2 ( N - l ) a n d
s o w e h a v e e s t a b l i s h e dE q u a t i o n ( l ) , n a m e l y
2(N-
1):
t
nili-r).
(1)
Returning to our example 322, we observed earlier
that N : 6, h : 2, n, : 3, fra: 3, u, : 3, uz : 2, and
v t : 2 . T h e n o t a t i o n u s e di n t h e d e r i v a t i o no f E q u a t i o n ( l ) i s d i s p l a y e df o r 3 2 2i n T a b l e 5 . I t i s c l e a rf r o m
t h e t a b l e t h a t , a s p r e d i c t e db y E q u a t i o n ( l ) , 2 ( # ( 3 2 2 )
- l)+ nr(vz- l)tne(us- l):
l)'
l0:nlvr
N :
nivi for I < i < t
(2- r)
(2- r)
(2)
Since the followrng argument is quite tedious, the
r e a d e r m a y c h o o s et o s k i p t o E q u a t i o n ( 3 ) o n f i r s t
r e a d i n g .T h i s c a n b e d o n e w i t h o u t s a c r i f i c i n ga b a s i c
understandingof the development.An argument est a b l i s . h i n gE q u a t i o n ( 2 ) u s i n g t h e f o l l o w i n g p a r a graph and the footnote is presentedfor those who are
a l r e a d yf a m i l i a r w i t h g r o u p t h e o r y .T h e r o l e o f E q u a t i o n ( 2 ) i s t o f a c i l i t a t et h e a n a l y s i so f t h e i m p l i c a t i o n s
o f E q u a t i o n( l ) .
Let i be any positive integer such that | < i < t.
(322),""--
(322)""":
{ . 1, " o o t2 l
{/,'o'o'z}
(2- r)
(2
r)
3(2-t1:3
1/?-l\-?
4+
2(#(G)- r)
(322)e", :
+
3
:10
l):10
Then for this particular i we want to show that N :
n;21.Consider the pole points belongingto C;(G). For
eachptj, there existsat least one rotation Xf€ G such
t h a t S ( p i r ) : P i i . W e s i n g l eo u t o n e s u c h r o t a t i o n f o r
eachTand label it g;. Then for each | < i < nt, g/Pir)
: pii. By the way the gr's were chosen, the collection of pole points lg'(p,'), gr(Pi'),
, gni
(p^)l : lPi', Pn,' '',Pt,,l : c,(G)' we now form
the following list where as before, hiris ageneratorof
H i ,Q . e .H
, i , : l h i r , h i 1 , ' ' ' ,h k \ ) :
g r h i r , g r h i J , ' ' ' ,g r h ' , 1
grt"'g'hi"' "
g"h'r\
"
:::
g^h,,,g-h,"''''' gn'h'i!*
We now wish to show that every rotation in G
appearsin this listing exactlyonce so that the number
o f e n t r i e si n t h e l i s t e q u a l s# ( G ) : N . L e t g € G , t h e n ,
s i n c e g m a p s p i l t o a n e q u i v a l e n tp o l e p o i n t , g ( p ; 1 ) €
C;(G) and henceg(p1'): pii : g/p,,,) for some | < i <
n;. Therefore, g1'g(pn) : pit. Accordingly, gi-'g e
Hi, : {hiuh,?,"',hhland so gr'g : ftf for some
* T h o s e f a m i l i a r w i t h g r o u p t h e o r y w i l l r e c o g n i z et h a t e a c h r o w
in our list is actually an enumerationof a coset of Hit Furtherm o r e , t h e r o t a t i o n s i n G t h a t r o L a l ep i t l o p i h a r e a l l l i s t e d i n t h e
k ' b r o w . H e n c e e a c h r o w c o r r e s p o n d st o a p o l e p o i n t i n C , ( G )
and conversely Consequently there is a one-to-one corres p o n d e n c eb e t w e e nt h e n u m b e r o f p o l e p o i n t s i n C 1 ( G )a n d c o s e t s
o f H , , . H e n c e t h e r e a r e n i c o s e t so f H i t a n d s i n c e l ( H , r ) = v 1 w a
h a v e , b y L a g r a n g e ' st h e o r e m , N : n p 1 .
157
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
I < k < /,. Hence, g : gth,l which appears in our
list. Sinceg is an arbitrarily chosenrotation of G, we
may concludethat every rotation in G appearsin the
list at least once. Our next task is to show that no
rotation of G appearsmore than once in the list. Suppose that g € G appears in the list as g = g,hfi and as
g : g r h nw i t h I 1 k , m l v t a n d I ( r , s 3 n i . W e
wish to show under these circumstancesthat r : s
and k = n. Sincep1 is a pole point associatedwith
hn, wa have g,hfrQtir) : g"(p,r) : pi, and g"hTr@,r):
g"(pt') : pi,. Since g,hfr, : g"hTr,by the definition of
equal mappitrss,pir: pis and so r = s. With this result we can now write g,hft : g,hfi which implies by
the cancellation law that hrt : hts and so k : m since
| < k, m 1 vi. Therefore, r : s and k : m as required. Thus, we have shown that every rotation in G
appearsexactly once in the list. Sincethe list consists
of r; rows and v1 columns, there are n1u;symbols
listed, and since each rotation in G is listed exactly
once, we conclude that l(G) : N : np1.
Returning to our example group 322, we construct
the same list as in the general case using per ?S our
pole point, as6 lttql2 as th.e.generator'of the group
H r r . S i n c el ( p i : p r r : t t t u t 2 l q a r ) . 3 ( p r r: ; p s z :
Iqtql2(prr),and 3-'(p3.) : pss : unl2(p"r), we must
select 91 to be either I orLtt0l2, g to be either J
or [0t012,and 93 to be either J-' or t100]2.we will
choose 1 as gr, 3 as g", un6lt00\2as gr. Hence, our
l i s t f o r 3 2 2 u s i n gp a r o S o u r p o l e p o i n t a n d u t 1 l 2 a s
our generator of H., is as follows:
2 ( N- 1 ) : E N Q - r / v )
2 - 2/N:
Because
H;7is a ,uUgrolpof G, we canconclude
that l(H;;) : vi ! #G) : N. Also, since Hs7is the set
of all rotations in G that leavethe pole pointply fixed,
it must, by the definition of a pole point, contain at
least one rotation in addition to the identity. Hence
2 1 vi 1N. Using the fact that zi must be greaterthan
or equal to 2, the right member of Equation (3)
satisfiesthe inequaltiy
Itt
Irr - r/v)t Irr - U2): E
<r/zt: t/2.
i=l
Rt-,]o.,1n"" r/ = 2. ;" can write
t
2 > 2 - 2 / N: I t t - r / , ) > t / 2 .
Therefore, 2 > t/2and hence4 ) L Thus / can only
be equal to 1,2, or 3. We can thus concludethat there
are no rotation groups having more than 3
equivalenceclassesof pole points. We now examine
each of thesethree casesfor the value of r.
1 . I n t h i s c a s e ,E q u a t i o n ( 3 ) b e Casewhere t:
comes
I
- 7/,):
(r - t/v)
2 - 2/N:
| <t
or
trool
rtlrot 2
Lt1rot22
Strto\ 2,
311r0t22
2 I r t o t2 ,
tot
trool
2tt
22
After forming these compositions (seeTable 4), the
list becomes
'"ot
t'
1r- r/r).
|
L-2/N:
-l/v'.
The left member of this equation is always
nonnegativebecauseN > 2, but the right member is
always negative because z.
contradiction. Therefore, t cannot equal I, from
which we concludethat 0(G) must contain more than
one equivalenceclassof pole points; t must equal 2 or
6ase where I : 2. In this caseEquation (3) becomes
J-t,
2
Itool
2.
"ott"'l
It is apparent that each rotation of 322 is listed
exactly once and that v".ns : 2.3 : 6 : #(322).
,
li -11 r
2-2/N:
2-2/N:
Itt
-r/,n)
i -l
:
(1 -
l/r,) *
(l -
r/vz)
By a little algebraic manipulation we find that
-r/,0)
(3)
2: N/u'l N/vz.
From Equation (2) we have that N/vi = n1,and so the
SinceN : ftrur,it followsthatn,: N/u,. Replacing
above expressionsimplifiesto
n; in Equation(l) by N/vL we obtain
2(N - l):
I
( N / v ) ( v ,- t )
2:nrln"'
Becausen, and r, are positive integers,we conclude
M. B. BOISEN, JR., AND G. V. GIBBS
158
Tasrp 6.
SymbolforG:vp2v3
#(G): N
Possible Proper Crystallographic Point Groups
#(C'(G):
N/v'
#(c,(G) -- N/r, #(c,(G)): N/vz
222
322
^,,'
622
2
2
2
2
2
2
6
8
l2
J
J
?1 t +
432
72
24
4
6
A
A
A
6
6
4
8
6
L2
Dihedral groups
Tetrahedralgroup
Octahedralgroup
by 23.
t The group 332is usuallydesignated
that n, : flz: I is the only possiblesolution. Hence,
for a rotation group with t : 2, we have two
equivalenceclassesconsistingof one pole point each.
Altogether G has a total of two pole points, which
accordingly define one and only one rotation axis.
Therefore,those groups with two equivalenceclasses
of pole points must be the monaxial groups given in
Table 3. The number of elements in each of these
possiblemonaxial groups is equal to the order of the
rotation axis, l(G) : ut : ttz.
Casewhere t : 3. In this caseEquation (3) expands
to
2 - 2 / N = ( l - l / v r ) + ( l - l / v z )+ ( l - l / u s ) '
which yields ni :
Table 6.
* S e e K l e i n , W e y l , o r Z a s s e n h a u sf o r a d e r i v a t i o n o f t h e n o n crystallographic rotation groups based on these ideas.
We begin by consideringthe interaxial anglesthat
must exist for a polyaxial combination of the form
N/vi. These ate also recorded in
Determination of the interaxial angles
We have shown above that there are only I I possible proper crystallographicpoint groups, namely l,
2, 3, 4, 6, 222, 322, 422, 622, 332, and 432. The
monaxial groups I, 2, 3, 4, and 6 were established
earlier. However, we have not yet establishedthe fact
that each of the polyaxial combinations actually
yields a group. Also, the possibility renlains that
some of thesecombinations may lead to two or more
groups which may arise by assumingdifferent intersection anglesbetweenthe rotation axes(theseangles
Rewriting this resultwe seethat
are called interaxial angles.).We will seelater in this
( 4 ) section, however, that each of the possibilitiesleads
| + 2/N : l/vr I l/vzl l/us'
Since N > 2, it follows that I * 2/N > I and so to preciselyone point group. In the processwe will
actually determine the interaxial angles used in the
(5)
l/v,*l/y"tl/ur)l'
formation of these point groups. Inherent in the
strategy we will follow in determining these interFor convenience,we can assumethat v, ) vz ) vs.
In constructing Table 6 we have considered all axial angles is the observation that before a polypossiblecombinations 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
6 3 3 , 4 4 3 , 6 4 3 , 6 6 3 , 4 4 4 , 6 4 4 , 6 6 4 , a n d 6 6 6 . A q u i c k given pole point to another pole point in the same
check shows that none of thesesetssatisfyInequality equivalence class. We recall that because equiva( 5 ) . ( N o t e t h a t t h e d e r i v a t i o n o f I n e q u a l i t y ( 5 ) d i d lence classesare 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 observalarge number non-crystallographicpoint groups may tion will also be of considerableuse in the determinabe 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 orders of each of
helpful to observe that 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
CROUP THEORY DERIVATION OF THE 32 POINT GROUPS
vr22 (when ur : 2, 3, 4, 6) to be a group. Table 6
showsthat there are exactlytwo pole points which we
denote prt and p* in the equivalenceclass Cr(r'22).
Let lr be a non-identity rotation in G such thath@r)
: pr, i.e., ftC Hrr. Then hftrz) must equal pn ot pn
sinceevery rotation in vQ2 must permute the equivalenceclassC1(vr22): lpnepl. Becauseit was assumed
above that h(prr) : p' and becausel cannot map
two distinct pole points to the same pole point, we
conclude that h(prz) must not equal pll and hence
h(p'r) : p'r. Thereforepll and Pu both lie on the axis
of the rotation ft. Consequently,p, andp* are antipodal points and so there is exactly one z,-fold axis
associatedwith Cr(vr22).Also each of the equivalence
classesassociatedwith the cyclic groups of order 2
contain z, pole points. Accordingly, we write Cr(vr22)
: lpa.ezz,... euI and C"(vr22)-- lp$esr, .'. Aul .
Moreover, since there is a total of ff(CdvQ2)) +
fi(Cs(vr22)) : 2v, pole points associated with the
2-fold axes,there must be exactly v12-fold axesassociated with Cz@r22)and C"(u'22).Now that we know
the number of each kind of rotation axis present in
vr22, the task is to discover the appropriate interaxial anglesbetweentheseaxesso as to form a group.
If we place the Z axis along the v'-fold axis, then
p' and ptz ?te the points where the Z axis intersects
with the surface E of the unit sphere.The question
now is "Where can a 2-fold axis associatedwith say
pz, be placed so that the associatedhalf-turn g will
permute lprr,prrl?" If g(ptt) : /rr, then g(prr) : pr".
Sinceg has only two pole points, p21must be either
pt ot prr, which is a contradiction because the
equivalenceclassesof pole points are disjoint. Hence
g(pi : prr, andp' bisects the arc ft,r.
Therefore,
159
tor so that the arc length between any two adjacent
points is 360"/vt.
We now observethat there is only one rotation lt
associatedwith the pole points in C'(vr22)such that
h(p) : p* wherep1 and p,2 are adjacent pole points
in C"(ur22). If we let trool2 denote the half-turn
pzt, and
about the X axis then uNl2(prr)
property
thathltnl2(prr)
the
the rotation ltltoo)2 has
: pzz. However, lrltu)2 I fr becaus" i1 lruoa)2 :
ft, then \runl2 : 11 / which according to the
cancellation law would yield the contradiction that
It00l2 : 1. But h was the only rotation associated
with Cr(vr22)which maps p2r to prr. Hence htt0012
must be a rotation associatedwith a pole point in
either C2@r22)or Cs@'22).Therefore iltml2 must be
a half-turn which maps pz, to pz" and whose rotation
axis is perpendicularto the Z axis. Thus the axis of
iltm\2 must bisect the arcs along the equator determined by p^ and pzz. Let q denote the pole point
belonging to il|ul2 which bisects the shorter arc between p21 and pr". Sincepzr and p2 are adjacent pole
points in C2@r22),q cannot be in Cr(v'22) and hence
must be in Cs(vQ2).The images of 4 under the rotations about the z,-fold axis are the remaining pole
points in Ct(vr22).In summary, we have shown that,
if v22 (where ut : 2, 3, 4, 6) is to be a group, there is
only one possiblearrangementof rotation axes.The
interaxial angles for this arrangementare such that
the angle betweenthe z,-fold rotation axis and each
of the 2-fold rotation axes associatedwith the pole
points in Cz(v22) and Cs(u,22)is 90o and the angle
betweenany two adjacent2-fold axesis 180'/z'. Refgrring to Figure 4, we see that our example group
322 does indeed satisfly this interaxial criteria. The
:
;;;fid;": f,i,,' andsince
ft,, +"'fr," l8oo,it fact that vr22 actually forms a group when its interfollows that ftrr:
90o. Accordingly, it follows that axial anglesare as describedabove can be shown by
t h e 2 - f o l d a x i s a s s o c i a t e d w i t h p ' m u s t b e 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 axesassociatedwith the remain- theory since the matrix representationfor the rotaing pole points in Cr(vr22) and Cs(vr22)are also per- tions in vr22 is completely determined by the interpendicular to the z'-fold axis.Thus, the z' 2-fold axes axial angles).
in u'22 are all perpendicularto the v'-fold axis.
Interaxial anglesfor the tetrahedral group 332 ( i.e., 23)
The remaining question is "What must the interOur next task is to determine the interaxial angles
axial angles be betweenthese2-fold axes?" Without
for
332. It is clear from Table 6 that there are two
generality,
(the
p'
is on the equator
loss of
because
great circle perpendicularto Z), we can place the X equivalenceclassesassociatedwith the 3-fold axes
axis of our coordinate system so that it passes containing4 points each,and one equivalenceclassof
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, because of the nature of the rotations and three 2-fold axes in 332.
We begin by consideringthe placementof the three
about the zr-fold axis, we obtain z, poirtts (hence all
of the points in C2@r22))spacedequally 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
the half-turns in G and sincethe antipodal point of
prt is associatedwith the same half-turn associated
with pr,, it follows that the antipodal point of pr,,
which we denote 6y prr, is in Cr(332).Without loss of
generality,we place Ihe Z axisalong the line segment
pt'1ttr.Let prr be any one of the remaining pole points
in Cr(332).We will show that flp"" :90' by assuming that frp* + 90o and arriving at a contradiction.
tf"frp""?-90;,then
eitherfr,, i 90"orfr,n < 90'
where p3n is antipodal to pga. Without loss of
generality,we may assumethat f ,j"" < 90'. If we let
2 denote the half-turn along the Z axis (with pole
pointsp31,p32),
then 2(ps) : pq for somepsj€ Cs(332).
Hence ffpr1 : frp*
observation on the effect of a half-turn, p' bisects
+ 90' : 180'. Thereforepr;
fi)rr, hencei|,pr" 190'
'
=
pgr
since
180o.
Moreover, it is clear that py
I
Gpr,
Hence
rotation
axis associatedwith
the
.
€ \p"r,p"",p""|
pr; is different from those associatedwith lltrr,prrl and
. Therefore,psj must be one of the remaining
Lptt,ptnl
pole points in Cr(332),saypat : ps,. We also observe,
since p' bisects 6prr, that Tpru Qp* and @"u
are coplanar where O is the origin of R3. Hence
the three axes associated with half-turns are all
coplanar. We now make an obvious but important
o b s e r v a t i o na b o u t t h i r d - t u r n s .l f g i s a t h i r d - t u r n a n d
4 is a point not bn its rotation axis, then the points
lqS@),g'(q)l determine a plane perpendicularto the
axis of g. Hence given 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 hypothesis that 6br"
* 9O', we have shown that all the pole points of
Ca(332)are coplanar and henceall of the 3-fold axes
must be perpendicular to the plane determined by
thesepole points. This means that 332 can 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 sincepsewas selectedto be any
pole point exceptp3landpr", we seeby similar reasoning 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 placement of the four 3fold axes.Let pt be a pole point in C1(332).Then p,,
appearson E in one of the octants delineatedby the
three 2-fold axes.Without loss of generality,we can
assumethat p' is located somewherein the first octant (denotedby the shadedregion in Fig. 5). If we let
p"rplay the role of q in our observationin the previous
paragraph,then we know that the 3-fold axis through
p' is perpendicularto some plane containingp.' and
two other pole points in C'(332). By inspecting all
such planescontainingp' and rememberingthat the
3-fold axis cannot be coincident with any of the 2fold axes,it is clear that the plane determinedbypr',
pss,?ndpru (seeFig. 5) is the only possibility. Therefore the third-turn associatedwith p'r mapsp31to pas,
p$ to p"u, and pss to p' requiring that Pr be
equidistant from the points pu, pss,and pss(seeFig.
6a). Therefore 6i,., = Gp,, : 6h". Solving in the
standard way the sphericaltriangles that arise from
theserefationships,we conclude that fijr, : 54.74".
The imagesof p,i under the half-turns in 332 yield the
four pole points belongingto C'(332) : lpt, pn, pn,
prnlas pictured in Figure 6a. The antipodal points to
t h o s e a p p e a r i n gi n C 1 ( 3 3 2 )a r e n o t i n C ' ( 3 3 2 ) , b u t
they are the pole points that form Cr(332).The resulting set of all pole points belonging to 332 are shown
in Figure 6b. We have therefore shown that this
arrangementyields the only possibilityfor 332 to be a
group. If a multiplication table is prepared,it can be
shown that this arrangement does indeed yield a
group. A list of selectedinteraxial anglesbetweenthe
rotation axesin 332 is given in Figure 6c, from which
the remaining anglesmay be deduced.
I
Frc 5. The orientations of the 2-fold rotation axes for the
t e t r a h e d r a l g r o u p 3 3 2 = 2 3 w i t h t h e i r a s s o c i a t e dp o l e p o i n t s
shown as solid circles The surface area on the unit sphere where
pole point prr must lie is shaded
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
Pe
til'.
o
Pse
t3,'.'
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
: frh"" = fipru : flp,, -- fipr" = f?rp"o= 54.74";
pT,p,,
arclengths:
f1fo,,= 6A":
-70.32';
fiit,,: 6fu"0: 109.47";
i?,p"" 90";6i",: 125.26'.
Interaxial angles for the octahedral group 432
fip,.:
flp,, =
axes in 432 must also be arranged in the same manner. That is to say that332 is containedin432. Hence
The final casein our considerationof proper rota- we have placed all of the 4-fold and 3-fold axesprestion 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(seeTable 6). Implicit implicit in the horizontally placed 4-folds and since
in the three 4-fold axes are three 2-fold axes. Since these axes are perpendicularto the vertically placed
there are more than one 3-fold axespresent,the argu- four-fold axis, we can conclude that 422 is contained
ment given in our discussion of 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 same orientation for these4- axes. Hence the point ps1placed as shown in Figure
'7a
fold axes as for the 2-fold axes in 332 (see Fig. 5).
is a pole point belonging to Csg32). The
pole
points
Sincethe 6
belongingto the 4-fold axesin remaining elevenpole points in C"(432)can be found
432 are arrangedin exactly the sameway as the 6 pole by locating the images of p., under the rotations of
points belonging to 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 are shown in Figure
7b. The interaxial anglescan be deducedfrom those
given for 332. Also we note that by forming a multiplication table, one can verify that this arrangement
ofthe axesinvolved in 432 doesin fact yield a group.
In summary we have found interaxial anglesfor the
e l e v e np r o p e r c r y s t a l l o g r a p h ipc o i n t g r o u p s 1 , 2 , 3 , 4 ,
6 , 2 2 2 , 3 2 2 , 4 2 2 , 6 2 2 , 3 3 2 ,a n d 4 3 2 . I t c a n b e s h o w n
that for each ofthese groups there existsa lattice that
is left invariant under each of the rotations of the
group and so they are all bonafide crystallographic
point groups (Boisen and Gibbs, in preparation).
Furthermore, we have shown that there.areno other
proper crystallographicpoint groups.
Furthermore, all of the sets constructed from a
proper crystallographicgroup G as in (l) or (2) are
groups and hence improper crystallographic point
groups.
Before presentingthe proof of Theorem 2 we will
use the theorem to derive the collection of all improper crystallographicpoint groups, give some examples, and record in Table 7 all of the crystallographicpoint groups.
If H is a subgroup of G such that #(G)/#(H) : 2,
then we say that H is a haluing group. This theorem
shows that if we take each proper crystallographic
point group G one at a time and first form G U Gi
and then H U (G\H)t for each of its halving groups
H (note that in certain casesG does not have any
Improper crystallographic point groups
halving groups), then the resulting collection of
In the previous section we showed that there are
groups is preciselythe collection of all improper crysonly I I proper rotation groups, and we derived each
tallographic point groups.
of them. In this sectionwe will show how each of the
Note that the improper crystallographic point
improper crystallographicpoint groups can be congroups
of the form G U Gi have order 2f(G) and
structed from these proper crystallographic point
are
those
that contain the inversion (sinceI € G).
groups. This will be easily accomplished once we
we
have a total of eleven improper rotation
Hence
have establishedthe following important theorem.
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 impo
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 U 9i where f is the inversion.(Here
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
(G\H)t
where
H
is
a
O
su
o
f
G
U
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 11ProperCrystallographic
Point Groups
G
1
2
3
4
6
222
322
4))
622
332: 23
432
Halving
Groups*
The 2l Improper CrystallographicPoint Groups
Containing i (centrosymmetrical)
# (G)
H
GUG'
I
2
none
I
none
I
2/m
2
3
2
4/m
6/m
mmm
J
4
6
6
8
t2
t2
ai
J
A
222
6
322
none
23
#(c u Gr)
4f mmm
2
4
6
8
12
8
12
16
6f mmm
.A
L+
J
5z1m
2imT
4/m32/m
24
48
Not containing r'
H \J (G\H)t
none
m
none
6
mm2
3mm
4mk
42m
6mm
62m
none
4zm
#(H U (G\ H)r)
2
4
6
4
6
8
8
l2
l2
24
* The halving groups H of a given proper crystallographicpoint group G can be easily found by examining those
g r o u p s i n c o l u m n l t h a t h a v e o n e h a ltfh e n u m b e r o f e l e m e n t s a s GI f. L i s i n c o l u m n l s u c h t h a t# ( 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
betweentheserotations 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,
the rotation axes appearing in G U Gt and H U
(G\H)i are the same as those in G Therefore it
follows that the interaxial anglesobtained for G apply to both G U Gt and H l-J (G\H), as well. By the
way Table 7 is constructedand becauseof the relationship betweenthe proper and improper groups, all
the crystallographicpoint groups with the sameinteraxial anglesappear in the same row. We note again
that in this paper we have not demonstratedfor each
of these crystallographic groups the existenceof a
lattice I which is left invariant. However, this fact
can be shown and will be exploredin a future paper.
We will now do several examples of the
construction described in Theorem 2 followed by
Table 7 showing the results we obtain by applying
this constructionto all elevenproper crystallographic
point groups. We recall that the orientation symbo
attached to a given rotation or rotoinversion is
dependent on the choice of the basis. In each of
our examples the basis will be chosen to be consistent with the International Tables for X-ray
Crystallography (see Table 2.3.1 , Henry and Lonsdale, 1952). We then conclude the paper with the
proof of Theorem 2.
To illustratethe useof Theorem2 in the construcgroups,we will
tion of the impropercrystallographic
H U (G\Hy
Gi
and
G
firstconsidertheconstructions U
whereG is thepropermonaxialgroup6. For thiscase
G U G t b e c o m e6s U 6 , : U , 6 , 3 , 2 , 3 ' ' , 6 - ' ) U
u, 6, 3, 2, 3-',, 6-'li : ll, 6, 3, 2, 3-" 6-'l U
U, 6, 3 3-" 6-"
lti, 6i, 3i, 2i, 3-'.i, 6-'il
i, 6,3, m, i-',6--'l *hich is denotedby 6/m. To
constructthe group H U (G\H)i, G must contain
a subgroupH suchthat l(H) : #(G)/2. As notedin
Table 7, if such an H exists,it must appearin column 1 of the table.In the caseunderconsideration
:
#(6) = 6, hence#(H) 3. The only groupin column
I with order 3 that qualifiesas a possiblehalving
group in 6 is the propermonaxialgroup 3. Sinceall
the rotationsin 3 are containedin 6, we conclude
that 3 is the only halvinggroup in 6. Accordingly,it
is clear that H U (G\H), becomes3 U (6\3), :
t1,3,3-1 U 16,2,6-Lli : \1,3,3-" 6i,2i,
6 'i1 : 11,3, 3-',6, m,d-'l denotedby 6-.
In our nextillustration,we will constructG U Gd
and H U (G\H)t whereG is the proper polyaxial
group 322. In this caseG U Gi becomes(322) U
lol2, l0t|l ), i, 3, 1-', l1ffilm,
(322)i : ll, 3, 3 -', U001
2, 11
Ittllm,l0tllmldenotedUy32/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 are con- 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
' " , (sriXs-i)}. Sincethe inversion
noted by 3mm.
{(s'd)(s'i),(s'i)(srd),
Finally we considerthe impropercrystallographiccommutes with any proper rotation we have
point groupsthat 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 groups 4 and 222 in a set of z distinct (by the cancellationlaw) proper
Table7 qualifyas halvinggroupsin 422.Lettingtt :
rotations in I and so m 1 r. Hence m : n. Next we
4 , w e c o n s t r u c4t U ( 4 2 2 ) \ 4 ) i : U , 4 , 2 , 4 - ' ,
consider the set of proper rotations G : {ftr, ... ,
l100lm,
Ut}lm, l|t|lm, lft|lml 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 332 lackshalvinggroupsdespite since they are proper rotations belonging to l, the
the fact that groups322 and6 both haveordersone- composition ab is also a proper rotation in I and so
half that of 332.This is becauiea six-foldaxisis not aD is one of the /r's listed in G. Second,if a : kt for
containedin 332 and the interaxialanglesbetween s o m e 1 < j < n a n d b = s , f o r s o m e I I r 1 n ,
the 2-fold and the 3-fold rotation axesin 322 and 332 then ab : kis,. But since (kf,)i : kls,i) e l, ab =
are different.
k;s, is, by the cancellationlaw, some s; listed in G.
Proof of Theorem2
Let I denotean improper crystallographic
point
groupand let kr, kr, . ' . , k, be theproperrotationsin
l. We know.that I doeshavesomcproper rotations
since1 E l, which is a proper rotation. We now
(1) , e I and(2)d6 l.
consider
two cases;
Casewherei € l. In this case,each of isometries/r,,
k r , ' ' ' , k n , k r i , k r i , . . . , k n i a r e i n l . W e w i l l s h o wt h a t
this is a complete listing of the elementsof t. Let t 6
l, then we want to show that I appearsin the list. lf t
is a proper rotation, then t already appearsin the list
kr, kr, ' . . , k,. If t is not a proper rotation, then it
must be an improper rotation and hencecan be written t = li where/is a proper rotation. Sincel' E I we
know that (fr)i =fe t. Hence
<j <
"f- kforsome |
r and consequentlyli : k,i is in the list. Therefore I
\kr, k", ..., kn, kLi, k2i, ..., k"il. Sincethe
composition of two proper rotations is again a proper
rotation, we see that lky k2, . . . , knl forms a
subgroup of l. Let G denote the group consistingof
lkr, . . . , /c,). Then | : G U Gi and so lis of the form
describedin statement(l) of the theorem.
Case where i €! l: Again let kr, . . . , rt, denote the
proper rotations in l. The set of all improper rotations in I can be written in the form {Jri,Jrd, .. . J-i}
where .r; is a proper rotation for all | < j < ru. Since
i€ l, st is not the identity for all I <i < z. First we
want to show that m = n. That is, the number of
proper rotations in I equals the number of improper
rotations in L We begin by considering the set of
T h i r d , i f a : s ; f o r s o m eI < j < n a n d b : / r " f o r s o m e
| < r < n, then ab : s1k,.Since (sri)ft, = s1k,i El,
ab : sft, is somesl listed in G. Fourth, if a : sy for
: s , f o r s o m el 1 r < n , t h e n
somel <j<nandb
ab : sis, : sis7i2: (s;d)(s"t')E l. Hence ab is a proper
rotation in I and so appearsas one of the ft's. Therefore it has been shown that G is closed under composition and hence a group. Note that if we set H :
'
\kr, . . , knl, then H, being the set of all proper rotations in l, is a group and so H is a subgroup of G.
\ l s o f i ( G ) / f t ( H ): 2 n / n : 2 . S i n c e | : l k , , - . . , k n ,
r,i, "' , rni), | = H U(G\H), as required.
Now to establishthe final remark of the theorem
suppose that G is a proper rotation group. Then,
using the techniquesemployed in the above proof, it
is easy to show that G (_JGi is closed and hence a
group. Next we suppose that G has a subgroup H
such that #(G)/#(H) : 2. We wish to show that the
set H U (G\H)t is closedand hencea group.* Leta,b
e H U (G\H)t. lf a,b e H then clearlyab eH and so ab
€ H U ( G \ H ) i a s r e q u i r e d .I f a e H a n d b e ( G \ H ) i ,
then D = gi for some g € G\H. Now ag € H for if ag e
H , t h e n g : a - ' ( a g ) E H w h i c h i s n o t t h e c a s e .H e n c e
49€G\Handso ab:(ag)i e (G\H)tandso againabE
H U (G\H)t. Similarly, if a e (G\H)i and b e H,then ab
€ H U (G\H)t. Now supposethat a,b E(G\H)i.Then a
: gri and b: grifor somegl,g2€G\H andaD: (Sri)(S"i)
: ggz. Hencewe want to show that gB2 g H.We have
+ The reader who is familiar with factor groups may recognize
immediatelythat H U (G\H)i is closed.
GROUP THEORY DERIVATION OF THE 32 POINT GROUPS
proved above that the product of a rotation in (G\H)
with a rotationin H is in G\H. Henceg,H : lg,hlft e Hl
q G\H. By the cancellation law, rt G'H) : # (H)
a n d s o # G ' H ) : # ( G \ H ) . T h e r e f o r e g ' H: G \ H . C o n sequently, if gB, e G\H, then gg" : g1ft for some
h e H . B u t t h e n , b y t h e c a n c e l l a t i o nl a w , g , =
h e H , w h i c h i s a c o n t r a d i c t i o n .T h e r e f o r c , 9 4 , e H
and so ab : (S'i)(gri) : 9.8, € H. Hence H U
( G \ H ) t i s c l o s e da n d s o i s a g r o u P .
Acknowledgments
W e t h a n k A m y B o u s k a E r b a c h , C a r l F r a n c i s ,a n d J o h n H i g g i n s
of VPI&SU and Dr. Felix Chayes of the Geophysical Laborat o r y , C a r n e g i eI n s t i t u t ea t W a s h i n g t o n ,D C . , f o r r e a d i n ga n e a r l i e r
v e r s i o n o f t h e m a n u s c r i p t a n d f o r m a k i n g s e v e r a lh e l p f u l s u g g e s t i o n s P r o f e s s o rF . D . B l o s so f V P I & S U , P r o f e s s o rP B M o o r e o f
t h e U n i v e r s i t y o f C h i c a g o , P r o f e s s o rW B . W h i t e o f t h e P e n n s y l v a n i a S t a t e U n i v e r s i t y , a n d D r . G E . B r o w n , J r , t o g e t h e rw i t h h i s
s t u d e n t sM a r y K e s k i n e n , J a m e sO O ' B r i e n t , a n d M a r k T a y l o r o f
Stanford University, reviewedthe final manuscript and made a
number of helpiul comments We also wish to thank Ramonda
H a y c o c k sf o r t y p i n g t h e m a n u s c r i p t a n d S h a r o n C h i a n g f o r d r a f t i n g t h e f i g u r e s .G . V G i s g r a t e f u l t o t h e E a r t h S c i e n c e sS e c t i o no f
t h e N a t i o n a l S c i e n c eF o u n d a t i o n f o r s u p p o r t i n g t h i s s t u d y w i t h
N . S . F G r a n t D E 5 7 l - 0 0 4 8 6A 0 3
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Manuscript receiued,December 12, 1974: accepled
for Publication' October 14' 1975