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under matrix multiplication. This group is called the special orthogonal group
and is isomorphic to the circle group W . SO(2) acts on R2 as follows. The matrix
M ∈ SO(2) takes the vector x ∈ R2 to the vector Mx. The orbit of any element
x ∈ R2 is the circle through x with center at the origin. Since the origin is the
only ¬xed point for any of the nonidentity transformations, the stabilizer of the
origin is the whole group, whereas the stabilizer of any other element is the
subgroup consisting of the identity matrix only.
The orbits of the cyclic group C2 acting on the hexagon in Figure 4.4 are
{1, 5}, {2, 4}, {3}, and {6}.
There is an important connection between the number of elements in the orbit
of a point x and the stabilizer of that point.

Lemma 4.39. If G acts on X, then for each x ∈ X

|G: Stab x| = |Orb x|.

Proof. Let H = Stab x and de¬ne the function

ξ : G/H ’ Orb x

by ξ(Hg) = g ’1 (x). This is well de¬ned on cosets because if Hg = H k, then
k = hg for some h ∈ H , so k ’1 (x) = (hg)’1 (x) = g ’1 h’1 (x) = g ’1 (x), since
h’1 ∈ H = Stab x.
The function ξ is surjective by the de¬nition of the orbit of x. It is also injec-
’1 ’1 ’1
tive, because ξ(Hg1 ) = ξ(Hg2 ) implies that g1 (x) = g2 (x), so g2 g1 (x) = x
’1
and g2 g1 ∈ Stab x = H . Therefore, ξ is a bijection, and the result follows.

Note that ξ is not a morphism. G/Stab x is just a set of cosets because Stab
x is not necessarily normal. Furthermore, we have placed no group structure on
Orb x.

Theorem 4.40. If the ¬nite group G acts on a set X, then for each x ∈ X,

|G| = |Stab x||Orb x|.

Proof. This follows from Lemma 4.39 and Corollary 4.8.

Example 4.41. Find the number of proper rotations of a cube.

Solution. Let G be the group of proper rotations of a cube; that is, rotations
that can be carried out in three dimensions. The stabilizer of the vertex 1 in
Figure 4.6 is Stab 1 = {(1), (245) Ž (386), (254) Ž (368)}. The orbit of 1 is the set
of all the vertices, because there is an element of G that will take 1 to any other
vertex. Therefore, by Theorem 4.40,

|G| = |Stab 1| |Orb 1| = 3 · 8 = 24.
ACTION OF A GROUP ON A SET 99




2 1

3 4

6 5


7 8

Figure 4.6. Cube. Figure 4.7. Re¬‚ection in a plane.


The full symmetry group of the cube would include improper rotations such as
the re¬‚ection in the plane as shown in Figure 4.7. This induces the permutation
(24) Ž (68) on the vertices, and it cannot be obtained by physically rotating the
cube in three dimensions. Under this group

Stab 1 = {(1), (245) Ž (368), (254) Ž (386), (24) Ž (68), (25) Ž (38), (45) Ž (36)},

so the order of the full symmetry group of the cube is

|Stab 1||Orb 1| = 6.8 = 48.

Therefore, there are 24 proper and 24 improper rotations of the cube.
The article by Shapiro [32] contains many applications, mainly to group the-
ory, of the actions of a group on a set.
We conclude by mentioning the action of the symmetric group Sn on the set
of polynomials in n variables x1 , x2 , . . . , xn . A permutation σ ∈ Sn acts on a
polynomial f = f (x1 , x2 , . . . , xn ) by permuting the variables:

σ (f ) = f (xσ 1 , xσ 2 , . . . , xσ n ).

This was the action we used in Chapter 3 to de¬ne the parity of a permutation
and prove the parity theorem. It has historical interest as well. It is the context in
which Lagrange proved Lagrange™s theorem”in essence, what he actually did
is prove Theorem 4.40 for this action. Moreover, this is the group action that
launched Galois theory, about which we say more in Chapter 11.
100 4 QUOTIENT GROUPS


EXERCISES

In Exercises 4.1 to 4.4, which of the relations are equivalence relations? Describe
the equivalence classes of those relations which are equivalence relations.
4.1. The relation ∼ on P — P de¬ned by (a, b) ∼ (c, d) if and only if a + d =
b + c.
4.2. The relation T on the set of continuous functions from R to R, where fTg
if and only if f (3) = g(3).
4.3. The inclusion relation on the power set P (X).
4.4. The relation C on a group G, where aCb if and only if ab = ba.
4.5. Find the left and right cosets of H = {(1), (12), (34), (12) Ž (34)} in S4 .
4.6. Let H be the subgroup of A4 that ¬xes 1. Find the left and right cosets of
H in A4 . Is H normal? Describe the left cosets in terms of their effect on
the element 1. Can you ¬nd a similar description for the right cosets?
In Exercises 4.7 to 4.12, verify that each of the functions is well de¬ned. Determine
which are group morphisms, and ¬nd the kernels and images of all the morphisms.
The element of Zn containing x is denoted by [x]n .
f : Z12 ’ Z12 , where f ([x]12 ) = [x + 1]12 .
4.7.
f : C12 ’ C12 , where f (g) = g 3 .
4.8.
f : Z ’ Z2 — Z4 , where f (x) = ([x]2 , [x]4 ).
4.9.
f : Z8 ’ Z2 , where, f ([x]8 ) = [x]2 .
4.10.
f : C2 — C3 ’ C3 , where f (hr , k s ) = (12)r Ž (123)s .
4.11.
f : Sn ’ Sn+1 , where f (π) is the permutation on {1, 2, . . . , n + 1} de¬ned
4.12.
by f (π)(i) = π(i) if i n, and f (π)(n + 1) = n + 1.
4.13. If H is a subgroup of an abelian group G, prove that the quotient group
G/H is abelian.
If H is a subgroup of G, show that g ’1 Hg = {g ’1 hg|h ∈ H } is a subgroup
4.14.
for each g ∈ G.
Prove that the subgroup H is normal in G if and only if g ’1 Hg = H for
4.15.
all g ∈ G.
4.16. If H is the only subgroup of a given order in a group G, prove that H is
normal in G.
4.17. Let H be any subgroup of a group G. Prove that there is a one-to-one
correspondence between the set of left cosets of H in G and the set of
right cosets of H in G.
Is the cyclic subgroup {(1), (123), (132)} normal in S4 ?
4.18.
Is the cyclic subgroup {(1), (123), (132)} normal in A4 ?
4.19.
Is {(1), (1234), (13) Ž (24), (1432), (13), (24), (14) Ž (23), (12) Ž (34)} nor-
4.20.
mal in S4 ?
4.21. Find all the group morphisms from C3 to C4 .
Find all the group morphisms from Z to Z4 .
4.22.
EXERCISES 101


4.23. Find all the group morphisms from C6 to C6 .
4.24. Find all the group morphisms from Z to D4 .
In Exercises 4.25 to 4.29, which of the pairs of groups are isomorphic? Give
reasons.
4.25. C60 and C10 — C6 .
4.26. (P {a, b, c}, ) and C2 — C2 — C2 .
4.27. Dn and Cn — C2 . 4.28. D6 and A4 .
√ √
4.29. Z4 — Z2 and ({±1, ±i, ±(1 + i)/ 2, ±(1 ’ i)/ 2}, ·).
4.30. If G — H is cyclic, prove that G and H are cyclic.
4.31. If π is an r-cycle in Sn , prove that ρ ’1 Ž π Ž ρ is also an r-cycle for each
ρ ∈ Sn .
4.32. Find four different subgroups of S4 that are isomorphic to S3 .
4.33. Find all the isomorphism classes of groups of order 10.
4.34. Find all the ten subgroups of A4 and draw the poset diagram under inclu-
sion. Which of the subgroups are normal?
4.35. For any groups G and H , prove that (G — H )/G ∼ H and (G — H )/H ∼
= =
G, where G = {(g, e) ∈ G — H |g ∈ G} and H = {(e, h) ∈ G — H |h ∈
H }.
4.36. Show that Q/Z is an in¬nite group but that every element has ¬nite order.
4.37. If G is a subgroup of Sn and G contains an odd permutation, prove that
G contains a normal subgroup of index 2.
4.38. In any group (G, ·) the element a ’1 b’1 ab is called the commutator of
a and b. Let G be the subset of G consisting of all ¬nite products of
commutators. Show that G is a normal subgroup of G. This is called the
commutator subgroup. Also prove that G/G is abelian.
4.39. Let C— be the group of nonzero complex numbers under multiplication and
let W be the multiplicative group of complex numbers of unit modulus.
Describe C— /W .
4.40. Show that K = {(1), (12) Ž (34), (13) Ž (24), (14) Ž (23)} is a subgroup of S4
isomorphic to the Klein 4-group. Prove that K is normal and that S4 /K ∼ =
S3 .
4.41. If K is the group given in Exercise 4.40, prove that K is normal in A4 and
that A4 /K ∼ C3 . This shows that A4 is not a simple group.
=
4.42. The cross-ratio of the four distinct real numbers x1 , x2 , x3 , x4 in that order
is the ratio » = (x2 ’ x4 )(x3 ’ x1 )/(x2 ’ x1 )(x3 ’ x4 ). Find the subgroup
K of S4 , of all those permutations of the four numbers that preserve the
value of the cross-ratio. Show that if » is the cross-ratio of four numbers
taken in a certain order, the cross-ratio of these numbers in any other order
must belong to the set

1 1 1 »
», 1 ’ », , 1 ’ , , .
» 1’» »’1
»
102 4 QUOTIENT GROUPS


Furthermore, show that all permutations in the same coset of K in S4 give
rise to the same cross-ratio. In other words, prove that the quotient group
S4 /K is isomorphic to the group of functions given in Exercise 3.42. The
cross-ratio is very useful in projective geometry because it is preserved
under projective transformations.
4.43. (Second Isomorphism Theorem) Let N be a normal subgroup of G, and
let H be any subgroup of G. Show that H N = {hn|h ∈ H, n ∈ N } is a
subgroup of G and that H © N is a normal subgroup of H . Also prove that

H /(H © N ) ∼ H N/N.
=

4.44. (Third isomorphism theorem) Let M and N be normal subgroups of
G, and N be a normal subgroup of M. Show that φ: G/N ’ G/M is a
well-de¬ned morphism if φ(Ng) = Mg, and prove that

(G/N )/(M/N ) ∼ G/M.
=

4.45. If a ¬nite group contains no nontrivial subgroups, prove that it is either
trivial or cyclic of prime order.
4.46. If d is a divisor of the order of a ¬nite cyclic group G, prove that G
contains a subgroup of order d.
4.47. If G is a ¬nite abelian group and p is a prime such that g p = e, for all
g ∈ G, prove that G is isomorphic to Zn , for some integer n.
p
4.48. What is the symmetry group of a rectangular box with sides of length 2,
3, and 4 cm?
4.49. Let
ab
Gp = ∈ M2 (Zp )|ad ’ bc = 1 in Zp .
cd

If p is prime, show that (Gp , ·) is a group of order p(p 2 ’ 1), and ¬nd a
group isomorphic to G2 .
Show that (R— , ·) acts on Rn+1 by scalar multiplication. What are the
4.50.
orbits under this action? The set of orbits, excluding the origin, form the
n-dimensional real projective space.
Let G be a group of order n, and let gcd(n, m) = 1. Show that every
4.51.
element h in G has an mth root; that is, h = g m for some g ∈ G.
4.52. Let G denote the commutator subgroup of a group G (see Exercise 4.38).
If K is a subgroup of G, show that G ⊆ H if and only if K is normal in
G and G/K is abelian.
4.53. Call a group G metabelian if it has a normal subgroup K such that both
K and G/K are abelian.
(a) Show that every subgroup and factor group of a metabelian group is
metabelian. (Exercises 4.43 and 4.44 are useful.)
(b) Show that G is metabelian if and only if the commutator group G is
abelian (see Exercise 4.38).
EXERCISES 103


4.54. Recall (Exercise 3.76) that the center Z(G) of a group G is de¬ned by
Z(G) = {z ∈ G|zg = gz for all g ∈ G}. Let K ⊆ Z(G) be a subgroup.
(a) Show that K is normal in G.
(b) If G/K is cyclic, show that G is abelian.
For Exercises 4.55 to 4.61, let Zm — = {[x] ∈ Zm |gcd(x, m) = 1 }. The number of
elements in this set is denoted by φ(m) and is called the Euler φ-function. For
example, φ(4 ) = 2 , φ(6 ) = 2 , and φ(8 ) = 4 .
Show that φ(p r ) = p r ’ p r’1 if p is a prime.
4.55.
Show that φ(mn) = φ(m)φ(n) if gcd(m, n) = 1.
4.56.
Prove that (Zm — , ·) is an abelian group.
4.57.
Write out the multiplication table for (Z8 — , ·).
4.58.
Prove that (Z6 — , ·) and (Z17 — , ·) are cyclic and ¬nd generators.
4.59.
Find groups in Table 4.6 that are isomorphic to (Z8 — , ·), (Z9 — , ·), (Z10 — , ·),
4.60.
and (Z15 — , ·) and describe the isomorphisms.
Prove that if gcd(a, m) = 1, then a φ(m) ≡ 1 mod m. [This result was known
4.61.
to Leonhard Euler (1707“1783).]
Prove that if p is a prime, then for any integer a, a p ≡ a modp. [This
4.62.
result was known to Pierre de Fermat (1601“1665).]
4.63. If G is a group of order 35 acting on a set with 13 elements, show that G
must have a ¬xed point, that is, a point x ∈ S such that g(x) = x for all
g ∈ G.
If G is a group of order p r acting on a set with m elements, show that G
4.64.
has a ¬xed point if p does not divide m.
5
SYMMETRY GROUPS
IN THREE DIMENSIONS


In this chapter we determine the symmetry groups that can be realized in two-
and three-dimensional space. We rely heavily on geometric intuition, not only to
simplify arguments but also to give geometric ¬‚avor to the group theory. Because
we live in a three-dimensional world, these symmetry groups play a crucial role
in the application of modern algebra to physics and chemistry.
We ¬rst show how the group of isometries of Rn can be broken down into
translations and orthogonal transformations ¬xing the origin. Since the orthog-
onal transformations can be represented as a group of matrices, we look at the
properties of matrix groups. We then use these matrix groups to determine all the
¬nite rotation groups in two and three dimensions, and we ¬nd polyhedra that
realize these symmetry groups.


TRANSLATIONS AND THE EUCLIDEAN GROUP

Euclidean geometry in n dimensions is concerned with those properties that are
preserved under isometries (rigid motions) of euclidean n-space, that is, bijections
±: Rn ’ Rn that preserve distance. The group of all isometries of Rn is called the
euclidean group in n dimensions and is denoted E(n). Given w ∈ Rn , the map
Rn ’ Rn with v ’ v + w is called translation by w, and we begin by showing
that the group T (n) of all translations is a normal subgroup of E(n), and that
the factor group is isomorphic to the group of all orthogonal n — n matrices (that
is, matrices A such that A’1 = AT , the transpose of A”re¬‚ection of A in its
main diagonal).
Recall that a function »: Rn ’ Rn is called a linear transformation if »(av +
bw) = a»(v) + b»(w) for all a, b ∈ R and all (column) vectors v, w ∈ Rn . Let
{e1 , e2 , . . . , en } denote the standard basis of Rn , that is, the columns of the

Modern Algebra with Applications, Second Edition, by William J. Gilbert and W. Keith Nicholson
ISBN 0-471-41451-4 Copyright ™ 2004 John Wiley & Sons, Inc.

104
TRANSLATIONS AND THE EUCLIDEAN GROUP 105


n — n identity matrix. Then the action of » is matrix multiplication »(v) =
Av for all v in Rn , where the matrix A is given in terms of its columns by
A = [»(e1 )»(e2 ) · · · »(en )] and is called the standard matrix of ±. Moreover,
the correspondence » ” A is a bijection that preserves addition, multiplica-
tion, and the identity. So we may (and sometimes shall) identify » with the
matrix A.
If v √ w are vectors in Rn , let vžw = vT w denote their inner product. Then
and
||v|| = vžv is the length of v, and ||v ’ w|| is the distance between v and w.
Thus a function ±: Rn ’ Rn is an isometry if

(— )
||±(v) ’ ±(w)|| = ||v ’ w|| for all v, w ∈ Rn .

Since ||v ’ w||2 = ||v||2 + 2(vžw) + ||w||2 for any v, w ∈ Rn , it follows from (— )
that every isometry ± preserves inner products in the sense that

(—— )
±(v)ž±(w) = vžw for all v, w ∈ Rn .

Lemma 5.1. If ±: Rn ’ Rn is an isometry such that ±(0) = 0, then ± is linear.
Proof. It follows from (—— ) that {±(e1 ), ±(e2 ), . . . , ±(en )} is an orthonormal
basis of Rn . If a ∈ R and v ∈ Rn , then (—— ) implies that

[±(av) ’ a±(v)]ž±(ei ) = (av)žei ’ a(vžei ) = 0 for each i.

Hence ±(av) = a±(v), and ±(v + w) = ±(v) + ±(w) follows in the same way
for all v, w ∈ Rn .

Hence the isometries of Rn that ¬x the origin are precisely the linear isometries.
An n — n matrix A is called orthogonal if it is invertible and A’1 = AT ,
equivalently if the columns of A are an orthonormal basis of Rn . These matrices
form a subgroup of the group of all invertible matrices, called the orthogonal
group and denoted O(n).

Proposition 5.2. Let »: Rn ’ Rn be a linear transformation with standard
matrix A.

(i) » is an isometry if and only if A is an orthogonal matrix.
(ii) The group of linear isometries of Rn is isomorphic to O(n).
Proof. If A is orthogonal, then for all v, w ∈ Rn ,

||»(v) ’ »(w)||2 = A(v ’ w)žA(v ’ w) = (v ’ w)T AT A(v ’ w) = ||v ’ w||2

and it follows that » is an isometry. Conversely, if » is an isometry and
{e1 , e2 , . . . , en } is the standard basis of Rn , then (—— ) gives

ei žej = »(ei )ž»(ej ) = Aei žAej = eT (AT A)ej = the (i, j )-entry of A
i
106 5 SYMMETRY GROUPS IN THREE DIMENSIONS


for all i and j . It follows that AT A = I , so A is orthogonal, proving (i). But then
the correspondence » ” A between the linear transformation » and its standard
matrix A induces a group isomorphism between the (linear) isometries ¬xing the
origin and the orthogonal matrices. This proves (ii).

Given a vector w ∈ Rn , de¬ne „w : Rn ’ Rn by „w (v) = v ’ w for all v ∈ Rn .
Thus „w is the unique translation that carries w to 0. Because

„w Ž „w = „w+w for all w, w ∈ Rn ,

the correspondence w ” „w is a group isomorphism (Rn , +) ∼ T (n). In partic-
=
ular, T (n) is an abelian group.

Theorem 5.3. For each n 1, T (n) is an abelian normal subgroup of E(n) and
E(n)/T (n) ∼ O(n). In fact, the map E(n) ’ O(n) given by
=

± ’ the standard matrix of „±(0) Ž ±

is a surjective group morphism E(n) ’ O(n) with kernel T (n).

Proof. Write G(n) for the group of all linear isometries of Rn . Observe that
if ± ∈ E(n), then „±(0) Ž ± is linear (it is an isometry that ¬xes 0), so we have
a map

φ: E(n) ’ G(n) φ(±) = „±(0) Ž ± for all ± ∈ E(n).
given by

By Proposition 5.2 it suf¬ces to show that φ is a surjective group morphism with
kernel T (n). To see that φ is a group morphism, observe ¬rst that

(——— )
± Ž „w = „±(w) Ž ± for all w ∈ Rn .

[Indeed, (± Ž „w )(v) = ±(v ’ w) = ±(v) ’ ±(w) = („±(w) Ž ±)(v) for all v.]
Hence, given ± and β in E(n), we have

φ(±) Ž φ(β) = „±(0) Ž ± Ž „β(0) Ž β = „±(0) Ž (± Ž „β(0) Ž ± ’1 ) Ž (± Ž β),

so it suf¬ces to show that „±(0) Ž (± Ž „β(0) Ž ± ’1 ) = „(± Ž β)(0) . But this follows
because ± Ž „β(0) Ž ± ’1 is a translation by (——— ), so „±(0) Ž (± Ž „β(0) Ž ± ’1 ) is the
unique translation that carries (± Ž β)(0) to 0. Hence φ is a group morphism.
Moreover, φ is surjective because φ(») = » for every » ∈ G(n), and Ker(φ) =
T (n) because φ(±) = 1Rn if and only if ±(v) = v + ±(0) for all v ∈ Rn , that is,
if and only if ± = „’±(0) .

If ± ∈ E(n) and ±(0) = w, the proof of Theorem 5.3 shows that „w Ž ± ∈ G(n).
’1
Hence every isometry ± of Rn is the composition of a linear isometry „w Ž ±
followed by a translation „w .
MATRIX GROUPS 107


Proposition 5.4. Every ¬nite subgroup of isometries of n-dimensional space
¬xes at least one point.

Proof. Let G be a ¬nite subgroup of isometries, and let x be any point of
n-dimensional space. The orbit of x consists of a ¬nite number of points that are
permuted among themselves by any element of G. Since all the elements of G
are rigid motions, the centroid of Orb x must always be sent to itself. Therefore,
the centroid is a ¬xed point under G.

If the ¬xed point of any ¬nite subgroup G of isometries is taken as the origin,
then G is a subgroup of O(n), and all its elements can be written as orthogonal
matrices. We now look at the structure of groups whose elements can be written
as matrices.


MATRIX GROUPS

In physical sciences and in mathematical theory, we frequently encounter mul-
tiplicative group structures whose elements are n — n complex matrices. Such a
group is called a matrix group if its identity element is the n — n identity matrix
I . To investigate these groups, we have at our disposal, and shall freely apply,
the machinery of linear algebra.
For example, if

’ sin(2πk/m)
cos(2πk/m)
Ak = ,
sin(2πk/m) cos(2πk/m)

then ({A0 , A1 , . . . , Am’1 }, ·) is a real matrix group of order m isomorphic to
Cm . The matrix Ak represents a counterclockwise rotation of the plane about the
origin through an angle (2πk/m).
The matrices
’1 ’i 0
10 0 0 01
i
, , , , ,
0 ’1 0 ’i ’1 0
01 0i
0 ’1 0 ’i 0i
and
, ,
’i
1 0 0 i0

form a group under matrix multiplication. This is a complex matrix group of
order 8 that is, in fact, isomorphic to the quaternion group Q of Exercise 3.47.
Since the identity of any matrix group is the identity matrix I and every
element of a matrix group must have an inverse, every element must be a non-
singular matrix. All the nonsingular n — n matrices over a ¬eld F form a group
(GL(n, F ), ·) called the general linear group of dimension n over F . Any
matrix group over the ¬eld F must be a subgroup of GL(n, F ).

Proposition 5.5. The determinant of any element of a ¬nite matrix group must
be an integral root of unity.
108 5 SYMMETRY GROUPS IN THREE DIMENSIONS


Proof. Let A be an element of a matrix group of order m. Then, by Corol-
lary 4.10, Am = I . Hence (detA)m = detAm = detI = 1.

Hence, if G is a real matrix group, the determinant of any element of G is
either +1 or ’1. If G is a complex matrix group, the determinant of any element
is of the form e2πik/m .
The orthogonal group O(n) is a real matrix group, and therefore any element
must have determinant +1 or ’1. The determinant function

det: O(n) ’ {1, ’1}

is a group morphism from (O(n), ·) to ({1, ’1}, ·). The kernel, consisting of
orthogonal matrices with determinant +1, is called the special orthogonal group
of dimension n and is denoted by

SO(n) = {A ∈ O(n)|detA = +1}.

This is a normal subgroup of O(n) of index 2. The elements of SO(n) are
called proper rotations, whereas the elements in the other coset of O(n)
by SO(n), consisting of orthogonal matrices with determinant ’1, are called
improper rotations.
An n — n complex matrix A is called unitary if it is invertible and A’1 is
the conjugate transpose of A. Thus the real unitary matrices are precisely the
orthogonal matrices. Indeed, if x, y = xT y denotes the inner product in Cn ,
the matrix A is unitary if and only if it preserves inner products in Cn (that
is, Ax, Ay = x, y for all x, y ∈ Cn ), if and only if the columns of A are
orthonormal.
The unitary group of dimension n, U(n), consists of all n — n complex uni-
tary matrices under multiplication. The special unitary group, SU(n), is the
subgroup of U(n) consisting of those matrices with determinant +1. The group
SU(3) received some publicity in 1964 when the Brookhaven National Labo-
ratory discovered the fundamental particle called the omega-minus baryon. The
existence and properties of this particle had been predicted by a theory that used
SU(3) as a symmetry group of elementary particles.

Proposition 5.6. If » ∈ C is an eigenvalue of any unitary matrix, then |»| = 1.

Proof. Let » be an eigenvalue and x a corresponding nonzero eigenvector of
the unitary matrix A. Then Ax = »x, and since A preserves distances, ||x|| =
||Ax|| = ||»x|| = |»|||x||. Since x is nonzero, it follows that |»| = 1.

The group {Ak |k = 0, 1, . . . , m ’ 1} of rotations of the plane is a subgroup of
SO(2), and the eigenvalues that occur are e±(2πik/m) . The matrix group isomorphic
to the quaternion group Q is a subgroup of SU(2), and the eigenvalues that occur
are ±1 and ±i.
FINITE GROUPS IN TWO DIMENSIONS 109


Cayley™s theorem (Theorem 3.38) showed that any group could be represented
by a group of permutations. Another way to represent groups is by means of
matrices. A matrix representation of a group G is a group morphism φ: G ’
GL(n, F ). This is equivalent to an action of G on an n-dimensional vector
space over the ¬eld F , by means of linear transformations. The representation
is called faithful if the kernel of φ is the identity. In this case, φ is injective
and G is isomorphic to Imφ, a subgroup of the general linear group. Matrix
representations provide powerful tools for studying groups because they lend
themselves readily to calculation. As a result, most physical applications of group
theory use representations.
It is possible to prove that any representation of a ¬nite group over the real or
complex ¬eld may be changed by a similarity transformation into a representation
that uses only orthogonal or unitary matrices, respectively. Therefore, a real or
complex faithful representation allows us to view a group as a subgroup of O(n)
or U(n), respectively.


FINITE GROUPS IN TWO DIMENSIONS

We determine all the ¬nite subgroups of rotations (proper and improper) of the
plane R2 . That is, we ¬nd all the ¬nite matrix subgroups of SO(2) and O(2).
This was essentially done by Leonardo da Vinci when he determined the possible
symmetries of a central building with chapels attached. See Field and Golubit-
sky [29], where they construct interesting symmetric patterns in the plane using
chaotic maps.

Proposition 5.7

(i) The set of proper rotations in two dimensions is

’ sin θ
cos θ
SO(2) = θ ∈R .
sin θ cos θ

(ii) The set of improper rotations in two dimensions is

cos θ sin θ
θ ∈R .
’ cos θ
sin θ

’ sin θ
cos θ
are e±iθ and
(iii) The eigenvalues of the proper rotation
sin θ cos θ
those of any improper rotation are ±1.

1 0
p q p
Proof. (i) Let A = ∈ SO(2), so that A = =
and A
0 1
r s r
q
. Since A preserves distances, p 2 + r 2 = 1, and q 2 + s 2 = 1; thus there
s
110 5 SYMMETRY GROUPS IN THREE DIMENSIONS


exists angles θ and φ such that p = cos θ, r = sin θ, q = sin φ, and s =
cos φ. Therefore,

detA = ps ’ qr = cos θ cos φ ’ sin θ sin φ = cos(θ + φ).

If A is proper, detA = 1, so θ + φ = 2nπ. Hence φ = 2nπ ’ θ , and A is of the
cos θ ’ sin θ
. Conversely, if A is of this form, then AAT = I and
form
sin θ cos θ
A ∈ O(2). Since detA = +1, A is a proper rotation, and A ∈ SO(2).
1 0
(ii) One improper rotation in R2 is , so the coset of improper rota-
0 ’1
tions is
1 0 cos θ sin θ
= θ ∈R .
SO(2)
0 ’1 sin θ ’ cos θ

(iii) If » is an eigenvalue of

’ sin θ (cos θ ) ’ » ’ sin θ
cos θ
= 0.
then det
,
(cos θ ) ’ »
sin θ cos θ sin θ

Therefore, »2 ’ 2» cos θ + 1 = 0 and » = cos θ ± i sin θ = e±iθ .
cos θ sin θ
If » is an eigenvalue of the improper rotation , then
sin θ ’ cos θ
(cos θ ) ’ » sin θ
= 0. Hence »2 ’ 1 = 0 and » = ±1.
det
’(cos θ ) ’ »
sin θ

cos θ sin θ
The improper rotation B = always has an eigenvalue 1
sin θ ’ cos θ
and hence leaves an axis through the origin invariant because, for any corre-
sponding eigenvector x, Bx = x. It can be veri¬ed that this axis of eigenvectors,
corresponding to the eigenvalue 1, is a line through the origin making an angle
θ/2 with the ¬rst coordinate axis. The matrix B corresponds to a re¬‚ection of
the plane about this axis.
Hence we see that an improper rotation is a re¬‚ection about a line through
the origin, and conversely, it is easy to see that a re¬‚ection about a line through
the origin is an improper rotation.

Theorem 5.8. If G is a ¬nite subgroup of SO(2), then G is cyclic, and so is
isomorphic to Cn for some n ∈ P.
Proof. By Proposition 5.6, every element A ∈ G ‚ SO(2) is a counter-
clockwise rotation through an angle θ (A), where 0 θ (A) < 2π. Since G is
¬nite, we can choose an element B ∈ G so that θ (B) is the smallest positive
angle. For any A ∈ G, there exists an integer r 0 such that rθ (B) θ (A) <
(r + 1)θ (B). Since θ (AB ’r ) = θ (A) ’ rθ (B), it follows that 0 θ (AB ’r ) <
θ (B). Therefore, θ (AB ’r ) = 0, AB ’r = I , and A = B r .
PROPER ROTATIONS OF REGULAR SOLIDS 111


Hence G = {I, B, B 2 , . . . , B r , . . . , B n’1 }, and G is a ¬nite cyclic group that
must be isomorphic to Cn for some integer n.

Theorem 5.9. If G is a ¬nite subgroup of O(2), then G is isomorphic to either
Cn or Dn for some n ∈ P.

Proof. The kernel of the morphism det: G ’ {1, ’1} is a normal subgroup,
H , of index 1 or 2 consisting of the proper rotations in G. By the previous
theorem, H is a cyclic group of order n, generated by B, for example.
If G contains no improper rotations, then G = H ∼ Cn . If G does contain an
=
improper rotation A, then

G = H ∪ H A = {I, B, B 2 , . . . , B n’1 , A, BA, B 2 A, . . . , B n’1 A}.

Since A and B k A are re¬‚ections, A = A’1 and B k A = (B k A)’1 = A’1 B ’k =
AB n’k . These relations completely determine the multiplication in G, and it is
now clear that G is isomorphic to the dihedral group Dn by an isomorphism that
takes B to a rotation through 2π/n and A to a re¬‚ection.

Theorem 5.9 shows that the only possible types of ¬nite symmetries, ¬x-
ing one point, of any geometric ¬gure in the plane are the cyclic and dihedral
groups. Examples of such symmetries abound in nature; the symmetry group of
a snow¬‚ake is usually D6 , and many ¬‚owers have ¬ve petals with symmetry
group C5 .
We have found all the possible ¬nite symmetries in the plane that ¬x one point.
However, there are ¬gures in the plane that have in¬nite symmetry groups that
¬x one point; one example is the circular disk. The group of proper symmetries
of this disk is the group SO(2), whereas the group of all symmetries is the whole
of O(2).


PROPER ROTATIONS OF REGULAR SOLIDS

One class of symmetries that we know occurs in three dimensions is the class
of symmetry groups of the regular solids: the tetrahedron, cube, octahedron,
dodecahedron, and icosahedron. In this section, we determine the proper rotation
groups of these solids. These will all be subgroups of SO(3). We restrict our
consideration to proper rotations because these are the only ones that can be
physically performed on models in three dimensions; to physically perform an
improper symmetry on a solid, we would require four dimensions!

Theorem 5.10. Every element A ∈ SO(3) has a ¬xed axis, and A is a rotation
about that axis.
Proof. Let »1 , »2 , and »3 be the eigenvalues of A. These are the roots of the
cubic characteristic polynomial with real coef¬cients. Hence, at least one eigen-
value is real and if a second one is complex, the third is its complex conjugate.
112 5 SYMMETRY GROUPS IN THREE DIMENSIONS


By Proposition 5.6, |»1 | = |»2 | = |»3 | = 1. Since detA = »1 »2 »3 = 1, we can
relabel the eigenvalues, if necessary, so that one of the following cases occurs:

(i) »1 = »2 = »3 = 1.
(ii) »1 = 1, »2 = »3 = ’1.
(iii) »1 = 1, »2 = »3 = eiθ (where θ = nπ).

In all cases there is an eigenvalue equal to 1. If x is a corresponding eigenvec-
tor, then Ax = x, and A ¬xes the axis along the vector x. We can change the
coordinate axes so that A can be written in one of the following three forms:
®  ®  ® 
100 1 0 0 1 0 0
(i) ° 0 1 0 » (ii) ° 0 ’1 0 » (iii) ° 0 cos θ ’ sin θ » .
0 ’1
001 0 0 sin θ cos θ

The ¬rst matrix is the identity, the second is a rotation through π, and the third
is a rotation through θ about the ¬xed axis.

A regular solid is a polyhedron in which all faces are congruent regular
polygons and all vertices are incident with the same number of faces. There are
¬ve such solids, and they are illustrated in Figure 5.1; their structure is given
in Table 5.1. The reader interested in making models of these polyhedra should
consult Cundy and Rollett [28].
Given any polyhedron, we can construct its dual polyhedron in the following
way. The vertices of the dual are the centers of the faces of the original poly-
hedron. Two centers are joined by an edge if the corresponding faces meet in




Tetrahedron Cube Octahedron Dodecahedron Icosahedron

Figure 5.1. Regular solids.


TABLE 5.1. Regular Solids

Number of Number of Number of Number of Faces at
Polyhedron Vertices Edges Faces Faces Each Vertex

Tetrahedron 4 6 4 Triangles 3
Cube 8 12 6 Squares 3
Octahedron 6 12 8 Triangles 4
Dodecahedron 20 30 12 Pentagons 3
Icosahedron 12 30 20 Triangles 5
PROPER ROTATIONS OF REGULAR SOLIDS 113


an edge. The dual of a regular tetrahedron is another regular tetrahedron. The
dual of a cube is an octahedron, and the dual of an octahedron is a cube. The
dodecahedron and icosahedron are also duals of each other. Any symmetry of
a polyhedron will induce a symmetry on its dual and vice versa. Hence dual
polyhedra will have the same rotation group.

Theorem 5.11. The group of proper rotations of a regular tetrahedron is isomor-
phic to A4 .
Proof. Label the vertices of the tetrahedron 1, 2, 3, and 4. Then any rotation
of the tetrahedron will permute these vertices. So if G is the rotation group of the
tetrahedron, we have a group morphism f : G ’ S4 whose kernel contains only
the identity element. Hence, by the morphism theorem, G is isomorphic to Imf .
We can use Theorem 4.40 to count the number of elements of G. The stabilizer
of the vertex 1 is the set of elements ¬xing 1 and is {(1), (234), (243)}. The vertex
1 can be taken to any of the four vertices under G, so the orbit of 1 is the set of
four vertices. Hence |G| = |Stab 1| |Orb 1| = 3.4 = 12.
There are two types of nontrivial elements in G that are illustrated in
Figures 5.2 and 5.3. There are rotations of order 3 about axes, each of which
joins a vertex to the center of the opposite face. These rotations perform an even
permutation of the vertices because each ¬xes one vertex and permutes the other
three cyclically. There are also rotations of order 2 about axes, each of which joins
the midpoints of a pair of opposite edges. (Two edges in a tetrahedron are said
to be opposite if they do not meet.) The corresponding permutations interchange
two pairs of vertices and, being products of two transpositions, are even.
Hence Imf consists of 12 permutations, all of which are even, and
Imf = A4 .

The alternating group A4 is sometimes called the tetrahedral group.
There are many different ways of counting the number of elements of the
rotation group G of the tetrahedron. One other way is as follows. Consider
the tetrahedron sitting on a table, and shade in an equilateral triangle on the
table where the bottom face rests, as in Figure 5.4. Any symmetry in G can
be performed by picking up the tetrahedron, turning it, and replacing it on the
table so that one face of the tetrahedron lies on top of the shaded equilateral
triangle. Any of the four faces of the tetrahedron can be placed on the table, and
each face can be placed on top of the shaded triangle in three different ways.


2
2
4 4
1
1
3
3

Element (1 2) Ž (3 4).
Figure 5.2. Element (2 3 4). Figure 5.3.
114 5 SYMMETRY GROUPS IN THREE DIMENSIONS




Figure 5.4



Hence |G| = 4 · 3 = 12. This really corresponds to applying Theorem 4.40 to the
stabilizer and orbit of a face of the tetrahedron.

Theorem 5.12. The group of proper rotations of a regular octahedron and cube
is isomorphic to S4 .
Proof. The regular octahedron is dual to the cube, so it has the same rotation
group. There are four diagonals in a cube that join opposite vertices. Label these
diagonals 1, 2, 3, and 4 as in Figure 5.5. Any rotation of the cube will permute
these diagonals, and this de¬nes a group morphism f : G ’ S4 , where G is the
rotation group of the cube.
The stabilizer of any vertex of the cube is a cyclic group of order 3 that
permutes the three adjacent vertices. The orbit of any vertex is the set of eight
vertices. Hence, by Theorem 4.40, |G| = 3 · 8 = 24.
Consider the rotation of order 2 about the line joining A to A in Figure 5.5.
The corresponding permutation is the transposition (12). Similarly, any other
transposition is in Imf . Therefore, by Proposition 3.35, Imf = S4 .
By the morphism theorem, G/Kerf ∼ S4 and |G|/|Kerf | = |S4 |. Since |G| =
=
|S4 | = 24, it follows that |Kerf | = 1, and f is an isomorphism.

The symmetric group S4 is sometimes called the octahedral group.

Theorem 5.13. The group of proper rotations of a regular dodecahedron and a
regular icosahedron is isomorphic to A5 .
Proof. A regular dodecahedron is dual to the icosahedron, so it has the same
rotation group.


2 1
A

3 4

4 3

A′
1 2

Figure 5.5. Diagonals of the cube.
PROPER ROTATIONS OF REGULAR SOLIDS 115


There are 30 edges of an icosahedron, and there are 15 lines through the center
joining the midpoints of opposite edges. (The re¬‚ection of each edge in the center
of the icosahedron is a parallel edge, called the opposite edge.) Given any one
of these 15 lines, there are exactly two others that are perpendicular both to the
¬rst line and to each other. We call three such mutually perpendicular lines a
triad. The 15 lines fall into ¬ve sets of triads. Label these triads 1, 2, 3, 4, and
5. Figure 5.6 shows the top half of an icosahedron, where we have labeled the
endpoints of each triad. (The existence of mutually perpendicular triads and the
labeling of the diagram can best be seen by actually handling a model of the
icosahedron.)
A rotation of the icosahedron permutes the ¬ve triads among themselves, and
this de¬nes a group morphism f : G ’ S5 , where G is the rotation group of the
icosahedron.
The stabilizer of any vertex of the icosahedron is a group of order 5 that
cyclically permutes the ¬ve adjacent vertices. The orbit of any vertex is the set
of all 12 vertices. Hence, by Theorem 4.40, |G| = 5 · 12 = 60.
There are three types of nontrivial elements in G. There are rotations of order
5 about axes through a vertex. The rotations about the vertex A in Figure 5.6
correspond to multiples of the cyclic permutation (12345), all of which are even.
There are rotations of order 3 about axes through the center of a face. The rota-
tions about an axis through the point B, in Figure 5.6, are multiples of (142)
and are therefore even permutations. Finally, there are rotations of order 2 about
the 15 lines joining midpoints of opposite edges. The permutation correspond-
ing to a rotation about an axis through C, in Figure 5.6, is (23) Ž (45), which
is even.
Every 3-cycle occurs in the image of f so, by Proposition 3.37, Imf = A5 .
Since G and A5 both have 60 elements, the morphism theorem implies that G is
isomorphic to A5 .

The alternating group A5 is sometimes called the icosahedral group.



3
1

5
4
4 5 2 B
C
1
3 2
2 A
3
1
4 5
4
5 2

3 1

Figure 5.6. Ends of the triads of the icosahedron.
116 5 SYMMETRY GROUPS IN THREE DIMENSIONS


FINITE ROTATION GROUPS IN THREE DIMENSIONS

We now proceed to show that the only ¬nite proper rotation groups in three
dimensions are the three symmetry groups of the regular solids, A4 , S4 , and A5
together with the cyclic and dihedral groups, Cn and Dn .
The unit sphere S 2 = {x ∈ R3 ||x|| = 1} is mapped to itself by every element
of O(3). Every rotation group ¬xing the origin is determined by its action on the
unit sphere S 2 . By Theorem 5.10, every nonidentity element A ∈ SO(3) leaves
precisely two antipodal points on S 2 ¬xed. That is, there exists x ∈ S 2 such that
A(x) = x and A(’x) = ’x. The points x and ’x are called the poles of A. Let
P be the set of poles of the nonidentity elements of a ¬nite subgroup G of SO(3).

Proposition 5.14. G acts on the set, P , of poles of its nonidentity elements.

Proof. We show that G permutes the poles among themselves. Let A, B, be
nonidentity elements of G, and let x be a pole of A. Then (BAB ’1 )B(x) =
BA(x) = B(x), so that B(x) is a pole of BAB ’1 . Therefore, the image of any
pole is another pole, and G acts on the set of poles.

We classify the rotation groups by considering the number of elements in the
stabilizers and orbits of the poles. Recall that the stabilizer of a pole x, Stab x =
{A ∈ G|A(x) = x}, is a subgroup of G, and that the orbit of x, Orb x = {B(x)|B ∈
G}, is a subset of the set P of poles. In Table 5.2 we look at the stabilizers and
orbits of the poles of the rotation groups we have already discussed.

TABLE 5.2. Poles of the Finite Rotation Groups
Group G = Cn Dn A4
|G| = n 2n 12
Symmetries of n-agonal cone n-agonal cylinder tetrahedron
Looking down
on the pole, x
|Stab x| = n n n
2 2 2 3 3
|Orb x| = n n
1 1 2 6 4 4


Group G = S4 A5
|G| = 24 60
Symmetries of cube or octahedron dodecahedron or icosahedron
Looking down
on the pole, x

or or or or or or




|Stab x| = 2 3 4 2 3 5
|Orb x| = 12 8 6 30 20 12
FINITE ROTATION GROUPS IN THREE DIMENSIONS 117


x x x




Rotation:



2p 4p
p p
or or p
3 3 2



Looking down on x
x x
the pole x:

Order of the rotation: 2 3 4 or 2

Figure 5.7. Rotations of the cube.


We take Cn to be the rotation group of a regular n-agonal cone whose base
is a regular n-gon. (The sloping edges of the cone must not be equal to the base
edges if n = 3.) Dn is the rotation group of a regular n-agonal cylinder whose
base is a regular n-gon. (The vertical edges must not be equal to the base edges
if n = 4.)
Each stabilizer group, Stab x, is a cyclic subgroup of rotations of the solid
about the axis through x. The orbit of x, Orb x, is the set of poles of the same
type as x. As a check on the number of elements in the stabilizers and orbits, we
have |G| = |Stab x||Orb x| for each pole x.
For example, the cube has three types of poles and four types of nontrivial
elements in its rotation group; these are illustrated in Figure 5.7.

Theorem 5.15. Any ¬nite subgroup of SO(3) is isomorphic to one of

1), Dn (n 2), A4 , S4 or A5 .
Cn (n

Proof. Let G be a ¬nite subgroup of SO(3). Choose a set of poles x1 , . . . , xr ,
one from each orbit. Let pi = |Stab xi | and qi = |Orb xi |, so that pi qi = n = |G|.
Each nonidentity element of G has two poles; thus the total number of poles,
counting repetitions, is 2(n ’ 1). The pole xi occurs as a pole of a nonidentity
element pi ’ 1 times. There are qi poles of the same type as xi . Therefore, the
total number of poles, counting repetitions, is
r r
2(n ’ 1) = qi (pi ’ 1) = (n ’ qi ),
i=1 i=1

so r
2 1
2’ = 1’ . (—)
n pi
i=1
118 5 SYMMETRY GROUPS IN THREE DIMENSIONS


2
2’
If G is not the trivial group, n 2 and 1 < 2. Since xi is a pole of
n
some nonidentity element, Stab xi contains a nonidentity element, and pi 2.
1 1
1’
Therefore, < 1. It follows from (*) that the number of orbits, r, must
2 pi
be 2 or 3.
If there are just two orbits, it follows that

2 1 1
2’ =1’ +1’
n p1 p2
n n
and 2 = + = q1 + q2 . Hence q1 = q2 = 1, and p1 = p2 = n. This means
p1 p2
that x1 = x2 , and there is just one axis of rotation. Therefore, G is isomorphic
to the cyclic group Cn .
If there are three orbits, it follows that
2 1 1 1
2’ =1’ +1’ +1’
n p1 p2 p3
2 1 1 1
1+ = + +. (**)
n p1 p2 p3

Suppose that p1 p3 . If p1 3, we would have
p2

1 1 1 111
+ + + + = 1,
333
p1 p2 p3

2 n
which is a contradiction, since > 0. Hence p1 = 2 and q1 = .
2
n
2 1 1 1 12 1 1
Now 1 + = + + , so + = + . If p2 4, we would
2 p2 2n
n p3 p2 p3
1 1 11 1 2
+ + = , which is a contradiction, since > 0. Hence p2
have
44 2
p2 p3 n
is 2 or 3. The only possibilities are the following.

n n n
Case (i). p1 = 2, p2 = 2, p3 = , n is even and n 4, q1 = , q2 = , and
2 2 2
q3 = 2.
Case (ii). p1 = 2, p2 = 3, p3 = 3, n = 12, q1 = 6, q2 = 4, and q3 = 4.
Case (iii). p1 = 2, p2 = 3, p3 = 4, n = 24, q1 = 12, q2 = 8, and q3 = 6.
Case (iv). p1 = 2, p2 = 3, p3 = 5, n = 60, q1 = 30, q2 = 20, and q3 = 12.

1 1 11 1
If p2 = 2 and p3 + + = , which contradicts (**), since
6,
36 2
p2 p3
2
> 0.
n
FINITE ROTATION GROUPS IN THREE DIMENSIONS 119


Case (i). Let H = Stab x3 . This is a group of rotations about one axis, and it
is a cyclic group of order n/2. Any other element A that is not in H is of
order 2 and is a half turn. Therefore, G = H ∪ H A, and G is isomorphic
to Dn/2 of order n.
Case (ii). Let y1 , y2 , y3 , and y4 be the four poles in Orb x2 . Now p2 =
|Stab yi | = 3; thus Stab y1 permutes y2 , y3 , and y4 as in Figure 5.8. There-
fore, ||y2 ’ y1 || = ||y3 ’ y1 || = ||y4 ’ y1 ||. We have similar results for Stab
y2 and Stab y3 . Hence y1 , y2 , y3 , and y4 are the vertices of a regular tetra-
hedron, and G is a subgroup of the symmetries of this tetrahedron. Since
|G| = 12, G must be the whole rotation group, A4 .
Case (iii). Let y1 , y2 , . . . , y6 be the six poles in Orb x3 . Since p3 = 4, a rota-
tion in Stab yi must ¬x two of the poles and rotate the other four cyclically.
Hence y1 , y2 , . . . , y6 must lie at the vertices of a regular octahedron. Again,
since |G| = 24, G must be the whole rotation group, S4 , of this octahedron.
Case (iv). Let y1 , y2 , . . . , y12 be the 12 poles in Orb x3 . Any element of
order 5 in G must permute these poles and hence must ¬x two poles
and permute the others, as in Figure 5.9, in two disjoint 5-cycles, say
2 3 4 5 6 Ž 7 8 9 10 11 , where we denote the pole yi
by i. The points y2 , y3 , y4 , y5 , and y6 form a regular pentagon and their
distances from y1 are all equal. Using similar results for rotations of order
5 about the other poles, we see that the poles are the vertices of an
icosahedron, and the group G is the proper rotation group, A5 , of this
icosahedron.

Throughout this section we have considered only proper rotations. However,
if we allow improper rotations as well, it can be shown that a ¬nite subgroup


y2 y1
y3



y4
Stab y1

Figure 5.8



8
2
7 3
1
Stab y1
6
9
11 4
12
5
10

Figure 5.9
120 5 SYMMETRY GROUPS IN THREE DIMENSIONS


of O(3) is isomorphic to one of the groups in Theorem 5.15 or contains one of
these groups as a normal subgroup of index 2. See Coxeter [27, Sec. 15.5] for a
more complete description of these improper rotation groups.


CRYSTALLOGRAPHIC GROUPS

This classi¬cation of ¬nite symmetries in R3 has important applications in crys-
tallography. Many chemical substances form crystals and their structures take
the forms of crystalline lattices. A crystal lattice is always ¬nite, but in order to
study its symmetries, we create a mathematical model by extending this crystal
lattice to in¬nity. We de¬ne an ideal crystalline lattice to be an in¬nite set of
points in R3 of the form
n1 a1 + n2 a2 + n3 a3 ,

where a1 , a2 , and a3 form a basis of R3 and n1 , n2 , n3 ∈ Z. Common salt forms
a cubic crystalline lattice in which a1 , a2 , and a3 are orthogonal vectors of the
same length. Figure 5.10 illustrates a crystalline lattice.
This use of the term lattice is not the same as that in Chapter 2, where a
lattice referred to a special kind of partially ordered set. To avoid confusion, we
always use the term crystalline lattice here.
A subgroup of O(3) that leaves a crystalline lattice invariant is called a crys-
tallographic point group. This is a ¬nite subgroup of O(3) because there are
only a ¬nite number of crystalline lattice points that can be the images of a1 , a2 ,
and a3 when the origin is ¬xed.
However, not all ¬nite subgroups of O(3) are crystallographic point groups.
Suppose that A ∈ SO (3) leaves a crystalline lattice L invariant. Then, by The-
orem 5.10, A is a rotation through an angle θ , and the trace of A is 1 + 2 cos θ .
If we choose a basis for R3 consisting of the vectors a1 , a2 , a3 of the crystalline
lattice L, the matrix representing A will have integer entries. The trace is invari-
ant under change of basis, so the trace of A must be an integer. Hence 2 cos θ




a3




a2
a1

Figure 5.10. Crystalline lattice.
EXERCISES 121


must be integral, and θ must be either kπ/2 or kπ/3, where k ∈ Z. It follows
that every element of a crystallographic point group in SO(3) can only contain
elements of order 1, 2, 3, 4, or 6.
It can be shown that every crystallographic point group in SO(3) is isomorphic
to one of C1 , C2 , C3 , C4 , C6 , D2 , D3 , D4 , D6 , A4 , or S4 .
If we allow re¬‚ections, the only other such groups in O(3) must contain one
of these groups as a normal subgroup of index 2. Every one of these groups
occurs in nature as the point group of at least one chemical crystal. See Coxeter
[27, Sec. 15.6] or Lomont [31, Chap. 4, Sec. 4].


EXERCISES

Find the group of proper rotations and the group of all rotations of the ¬gures in
Exercises 5.1 to 5.7.
5.1. 5.2. 5.3. 5.4.


5.5. 5.6. 5.7.



5.8. Let G be the subgroup of O(2) isomorphic to Dn . Find two matrices A and
B so that any element of G can be written as a product of A™s and B™s.
5.9. What is the group of proper rotations of a rectangular box of length 3 cm,
depth 2 cm, and height 2 cm?
Find the proper rotation group of the 13 Archimedean solids in Exercises 5.10
to 5.22. All the faces of these solids are regular polygons and all the vertices are
similar. (See Cundy and Rollet [28] for methods on how to construct these solids.)
5.10. 5.11. 5.12. 5.13.




5.14. 5.15. 5.16.




5.17. 5.18. 5.19.
122 5 SYMMETRY GROUPS IN THREE DIMENSIONS


5.20. 5.21. 5.22.




5.23. It is possible to inscribe ¬ve cubes in a regular dodecahedron. One such
cube is shown in Figure 5.11. Use these cubes to show that the rotation
group of the dodecahedron is A5 .




Figure 5.11. Cube inside a dodecahedron.



5.24. What is the proper symmetry group of a cube in which three faces, coming
together at one vertex, are painted green and the other faces are red?
5.25. Find the group of all rotations (both proper and improper) of a regular
tetrahedron.
5.26. Let G be the full symmetry group of the cube. De¬ne f : G ’ S4 as in
Theorem 5.12. Find the kernel of f and the order of G.
5.27. Let the vertices of a tetrahedron be (1, 1, 1), (’1, ’1, 1), (’1, 1, ’1),
and (1, ’1, ’1). Find matrices in SO(3) of orders 2 and 3 that leave the
tetrahedron invariant.
5.28. Let the vertices of a cube be (±1, ±1, ±1). Find matrices in SO(3) of
orders 2, 3, and 4 that leave the cube invariant.
Find the symmetry groups of the chemical molecules in Exercises 5.29 to 5.31.
(Assume that all of the C-C bonds are equivalent.)
5.29. 5.30. 5.31. H NO2
H H CH3 CH3
C C
C C C C
NO2 C C CH3
H C C H H C C H
C C
C C C C
H NO2
H H H H

Trinitrotoluene (TNT)
Benzene Xylene
EXERCISES 123


Find matrices in SO(3) that preserve the crystalline lattices described in Exercises
5.32 to 5.34, and ¬nd their crystallographic point groups. The points of the crys-
talline lattice are n1 a1 + n2 a2 + n3 a3 , where ni ∈ Z and the basis vectors ai
are given below.
a1 = (1, 1, 0), a2 = (’1, 1, 0), a3 = (0, 1, 1).
5.32.
a1 = (1, 0, 0), a2 = (0, 1, 0), a3 = (0, 0, 2).
5.33.
√ √
a1 = (1, 0, 0), a2 = (0, ’3 3, 3), a3 = (0, 3 3, 3).
5.34.
Let G(n) denote the group of linear isometries of Rn .
5.35.
(a) Show that E(n) = T (n)G(n) = G(n)T (n), where the product of sub-
groups is as de¬ned in Exercise 4.43.
(b) Show that T (n) © G(n) = {1Rn }.
(c) Use parts (a) and (b) to prove that E(n)/T (n) ∼ G(n).
=
6
´
POLYA“BURNSIDE METHOD
OF ENUMERATION


This chapter provides an introduction to the P´ lya“Burnside method of counting
o
the number of orbits of a set under the action of a symmetry group. If a group
G acts on a set X and we know the number of elements of X, this method will
enable us to count the number of different types of elements of X under the
action of G.
For example, how many different chemical compounds can be obtained by
attaching a CH3 or H radical to each carbon atom in the benzene ring of
Figure 6.3? There are 26 different ways of attaching a CH3 or H radical on paper,
but these do not all give rise to different compounds because many are equivalent
under a symmetry. There are six different ways of attaching one CH3 radical and
¬ve H radicals, but they all give rise to the same compound. The dihedral group
D6 acts on the 26 ways of attaching the radicals, and the number of different
compounds is the number of orbits under the action of D6 , that is, the number of
formulas that cannot be obtained from each other by any rotation or re¬‚ection.
We have seen that the number of different switching circuits that can be
n
obtained with n switches is 22 . This number grows very quickly as n becomes
large. Table 2.11 gives the 16 switching functions of two variables; when n = 3,
there are 256 different circuits, and when n = 4, there are 65,536 different cir-
cuits. However, many of these circuits are equivalent if we change the labels of
n
the switches. That is, the symmetric group, Sn , acts on the 22 different circuits
by permuting the labels of the switches. The number of nonequivalent circuits is
the number of orbits under the action of Sn .


BURNSIDE™S THEOREM

Let G be a ¬nite group that acts on a ¬nite set X. The following theorem
describes the number of orbits in terms of the number of elements left ¬xed

Modern Algebra with Applications, Second Edition, by William J. Gilbert and W. Keith Nicholson
ISBN 0-471-41451-4 Copyright ™ 2004 John Wiley & Sons, Inc.

124
BURNSIDE™S THEOREM 125


by each element of G. It was ¬rst proved by W. Burnside in 1911 and was
called Burnside™s lemma; it was not until 1937 that its applicability to many
combinatorial problems was discovered by G. P´ lya.
o
Theorem 6.1. Burnside™s Theorem. Let G be a ¬nite group that acts on the
elements of a ¬nite set X. For each g ∈ G, let Fix g = {x ∈ X|g(x) = x}, the
set of elements of X left ¬xed by g. If N is the number of orbits of X under
G, then
1
N= | Fix g|.
|G| g∈G

Proof. We count the set S = {(x, g) ∈ X — G|g(x) = x} in two different ways.
Consider Table 6.1, whose columns are indexed by the elements of X and whose
rows are indexed by the elements of G. Put a value of 1 in the (x, g) position if
g(x) = x; otherwise, let the entry be 0.
The sum of the entries in row g is the number |Fix g| of elements left ¬xed
by g. The sum of the entries in column x is |Stab x|, the number of elements of
G that ¬x x.
We can count the number of elements of S either by totaling the row sums or
by totaling the column sums. Hence

|S| = |Fix g| = |Stab x|.
g∈G x∈X

Choose a set of representatives, x1 , x2 , . . . , xN , one from each orbit of X under
G. If x is in the same orbit as xi , then Orb x = Orb xi , and by Theorem 4.40,
|Stab x| = |Stab xi |. Hence
·
N N
|Fix g| = |Stab x| = |Orb xi ||Stab xi | = N · |G|
g∈G i=1 x∈Orb xi i=1

by Theorem 4.40. The theorem now follows.

TABLE 6.1. Elements of S Correspond to the 1™s in
This Table
Elements of X Row
Sums
x

.
.
.
1
|Fix g|
Elements 0 1 1 1 0. . .
g . . .1

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