# Symmetry

y(x) = x+y is also an isometry

(Figure12.11); however, T cannot be inO(n) since it is not a linear map.

We are mostly interested in isometries in R2. In fact, the only isometries in R2 are

rotations and reflections about the origin, translations, and combinations of the two. For example, aglide reflection is a translation followed by a reflection (Figure 12.12). In Rn

all isometries are given in the same manner. The proof is very easy to generalize. y x x y x T(x)

Figure 12.12: Glide reflections

Lemma 12.13. An isometry f that fixes the origin in R2 is a linear transformation. In

particular, f is given by an element inO(2).

Proof. Letf be an isometry inR2 fixing the origin. We will first show that f preserves

inner products. Since f(0) = 0,∥f(x)=x; therefore,

x22f(x), f(y)+y2 =f(x)22f(x), f(y)+f(y)2 =⟨f(x)−f(y), f(x)−f(y) =∥f(x)−f(y)2 =xy2 =xy,xy =x22x,y+y2. Consequently, ⟨f(x), f(y)=x,y⟩. Now let e1 and e2 be(1,0)t and (0,1)t, respectively. If

x= (x1, x2) =x1e1+x2e2,

then

f(x) =⟨f(x), f(e1)⟩f(e1) +⟨f(x), f(e2)⟩f(e2) =x1f(e1) +x2f(e2).

The linearity of f easily follows.

For any arbitrary isometry, f, Txf will fix the origin for some vector x in R2; hence,

Txf(y) =Ayfor some matrixA∈O(2). Consequently,f(y) =Ay+x. Given the isometries

f(y) =Ay+x1

their composition is

f(g(y)) =f(By+x2) =ABy+Ax2+x1.

This last computation allows us to identify the group of isometries on R2 withE(2).

Theorem 12.14. The group of isometries on R2 is the Euclidean group, E(2).

A symmetry group inRn is a subgroup of the group of isometries on Rn that fixes a

set of points X R2. It is important to realize that the symmetry group of X depends

both onRn and on X. For example, the symmetry group of the origin inR1 isZ

2, but the

symmetry group of the origin inR2 isO(2).

Theorem 12.15. The only finite symmetry groups in R2 are Z

n and Dn.

Proof. Let G={f1, f2, . . . , fn} be a finite symmetry group that fixes a set of points in

X R2. Choose a point x∈X. This point may not be a fixed point—it could be moved by Gto another point in X. Define a setS ={y1,y2, . . .yn}, where yi=fi(x). Now, let

z= 1 n ni=1 xi.

While the pointzis not necessarily in the setX, it is fixed by every element in the symetry group. Without loss of generality, we may now assume that z is the origin.

Any finite symmetry group G in R2 that fixes the origin must be a finite subgroup

of O(2), since translations and glide reflections have infinite order. By Example 12.10, elements inO(2)are either rotations of the form

=

(

cosθ sinθ sinθ cosθ

)

or reflections of the form = ( cosϕ sinϕ sinϕ cosϕ ) ( 1 0 0 1 ) = ( cosϕ sinϕ sinϕ cosϕ ) .

Notice that det() = 1, det() =1, and2 =I. We can divide the proof up into two

cases. In the first case, all of the elements in Ghave determinant one. In the second case, there exists at least one element in Gwith determinant 1.

Case 1. The determinant of every element inGis one. In this case every element inGmust be a rotation. SinceGis finite, there is a smallest angle, sayθ0, such that the corresponding

element0 is the smallest rotation in the positive direction. We claim that 0 generates

G. If not, then for some positive integernthere is an angle θ1 between0 and (n+ 1)θ0.

If so, then (n+ 1)θ0−θ1 corresponds to a rotation smaller than θ0, which contradicts the

minimality ofθ0.

Case 2. The group G contains a reflection T. The kernel of the homomorphism ϕ :G {−1,1} given by A 7→ det(A) consists of elements whose determinant is 1. Therefore, |G/kerϕ|= 2. We know that the kernel is cyclic by the first case and is a subgroup of G of, say, ordern. Hence,|G|= 2n. The elements ofGare

Rθ, . . . , Rθn−1, T Rθ, . . . , T Rθn−1.

These elements satisfy the relation

T RθT =R−θ1.

12.2. SYMMETRY 141

The Wallpaper Groups

Suppose that we wish to study wallpaper patterns in the plane or crystals in three dimen- sions. Wallpaper patterns are simply repeating patterns in the plane (Figure 12.16). The analogs of wallpaper patterns in R3 are crystals, which we can think of as repeating pat-

terns of molecules in three dimensions (Figure 12.17). The mathematical equivalent of a wallpaper or crystal pattern is called a lattice.

Figure 12.16: A wallpaper pattern inR2

Figure 12.17: A crystal structure in R3

Let us examine wallpaper patterns in the plane a little more closely. Suppose that x andyare linearly independent vectors inR2; that is, one vector cannot be a scalar multiple

of the other. Alattice of x and y is the set of all linear combinationsmx+ny, where m and nare integers. The vectors xand yare said to be a basis for the lattice.

Notice that a lattice can have several bases. For example, the vectors (1,1)t and (2,0)t

have the same lattice as the vectors (1,1)t and (1,−1)t (Figure 12.18). However, any lattice is completely determined by a basis. Given two bases for the same lattice, say {x1,x2}and {y1,y2}, we can write

y1=α1x1+α2x2

y2=β1x1+β2x2,

whereα1,α2,β1, and β2 are integers. The matrix corresponding to this transformation is

U = ( α1 α2 β1 β2 ) .

If we wish to givex1 and x2 in terms of y1 and y2, we need only calculateU−1; that is, U−1 ( y1 y2 ) = ( x1 x2 ) .

Since U has integer entries, U−1 must also have integer entries; hence the determinants of both U and U−1 must be integers. Because U U1 =I,

det(U U−1) =det(U)det(U−1) = 1;

consequently, det(U) = ±1. A matrix with determinant ±1 and integer entries is called unimodular. For example, the matrix

(

3 1

5 2

)

is unimodular. It should be clear that there is a minimum length for vectors in a lattice.

(2,0) (1,1) (1,1)

(1,−1)

Figure 12.18: A lattice in R2

We can classify lattices by studying their symmetry groups. The symmetry group of a lattice is the subgroup ofE(2)that maps the lattice to itself. We consider two lattices inR2

to be equivalent if they have the same symmetry group. Similarly, classification of crystals in R3 is accomplished by associating a symmetry group, called a space group, with each

type of crystal. Two lattices are considered different if their space groups are not the same. The natural question that now arises is how many space groups exist.

A space group is composed of two parts: a translation subgroup and a point. The translation subgroup is an infinite abelian subgroup of the space group made up of the translational symmetries of the crystal; the point group is a finite group consisting of ro- tations and reflections of the crystal about a point. More specifically, a space group is a subgroup ofG⊂E(2)whose translations are a set of the form{(I, t) :t∈L}, whereLis a lattice. Space groups are, of course, infinite. Using geometric arguments, we can prove the following theorem (see [5] or [6]).

Theorem 12.19. Every translation group in R2 is isomorphic to Z×Z.

The point group of G isG0 ={A: (A, b)∈Gfor someb}. In particular, G0 must be a

subgroup ofO(2). Suppose that xis a vector in a latticeLwith space groupG, translation groupH, and point groupG0. For any element (A,y)in G,

(A,y)(I,x)(A,y)1 = (A, Ax+y)(A−1,−A−1y) = (AA−1,−AA−1y+Ax+y) = (I, Ax);

hence,(I, Ax) is in the translation group ofG. More specifically,Axmust be in the lattice L. It is important to note thatG0 is not usually a subgroup of the space groupG; however,

ifT is the translation subgroup of G, thenG/T =G0. The proof of the following theorem

12.2. SYMMETRY 143 Theorem 12.20. The point group in the wallpaper groups is isomorphic to Zn or Dn,

where n= 1,2,3,4,6.

To answer the question of how the point groups and the translation groups can be combined, we must look at the different types of lattices. Lattices can be classified by the structure of a single lattice cell. The possible cell shapes are parallelogram, rectangular, square, rhombic, and hexagonal (Figure12.21). The wallpaper groups can now be classified according to the types of reflections that occur in each group: these are ordinarily reflections, glide reflections, both, or none.

Rectangular

Square Rhombic

Parallelogram

Hexagonal

Figure 12.21: Types of lattices in R2

Notation and Reflections or

Space Groups Point Group Lattice Type Glide Reflections?

p1 Z1 parallelogram none p2 Z2 parallelogram none p3 Z3 hexagonal none p4 Z4 square none p6 Z6 hexagonal none pm D1 rectangular reflections

pg D1 rectangular glide reflections

cm D1 rhombic both

pmm D2 rectangular reflections

pmg D2 rectangular glide reflections

pgg D2 rectangular both

c2mm D2 rhombic both

p3m1, p31m D3 hexagonal both

p4m, p4g D4 square both

p6m D6 hexagonal both

Table 12.22: The 17 wallpaper groups Theorem 12.23. There are exactly 17 wallpaper groups.

p4m p4g Figure 12.24: The wallpaper groups p4m and p4g

The 17 wallpaper groups are listed in Table 12.22. The groups p3m1 and p31m can be distinguished by whether or not all of their threefold centers lie on the reflection axes: those of p3m1 must, whereas those of p31m may not. Similarly, the fourfold centers of p4m must lie on the reflection axes whereas those of p4g need not (Figure12.24). The complete proof of this theorem can be found in several of the references at the end of this chapter, including [5], [6], [10], and [11].

Sage We have not yet included any Sage material related to this chapter. Historical Note

Symmetry groups have intrigued mathematicians for a long time. Leonardo da Vinci was probably the first person to know all of the point groups. At the International Congress of Mathematicians in 1900, David Hilbert gave a now-famous address outlining 23 problems to guide mathematics in the twentieth century. Hilbert’s eighteenth problem asked whether or not crystallographic groups in n dimensions were always finite. In 1910, L. Bieberbach proved that crystallographic groups are finite in every dimension. Finding out how many of these groups there are in each dimension is another matter. InR3 there are 230 different

space groups; in R4 there are 4783. No one has been able to compute the number of

space groups for R5 and beyond. It is interesting to note that the crystallographic groups

were found mathematically for R3 before the 230 different types of crystals were actually

discovered in nature.

### Exercises

1. Prove the identity

x,y= 1 2 [

x+y2− ∥x2− ∥y2].

2. Show thatO(n) is a group.

3. Prove that the following matrices are orthogonal. Are any of these matrices inSO(n)? (a) ( 1/2 1/2 1/2 1/2 ) (b) ( 1/5 2/5 2/5 1/5 )

12.3. EXERCISES 145 (c)   4/ 5 0 3/5 3/5 0 4/5 0 1 0   (d)  1/32/3 2/32/3 1/32/3 2/3 1/3 2/3  

4. Determine the symmetry group of each of the figures in Figure12.25.

(b) (a)

(c)

Figure 12.25

5. Let x,y, and wbe vectors in Rn and α R. Prove each of the following properties

of inner products. (a) x,y=y,x.

(b) x,y+w=x,y+x,w. (c) ⟨αx,y=x, αy=α⟨x,y.

(d) x,x⟩ ≥0with equality exactly when x= 0. (e) Ifx,y= 0 for allx inRn, theny= 0.

6. Verify that

E(n) ={(A,x) :A∈O(n) and xRn}

is a group.

7. Prove that{(2,1),(1,1)} and {(12,5),(7,3)} are bases for the same lattice.

8. Let G be a subgroup of E(2) and suppose that T is the translation subgroup of G. Prove that the point group of Gis isomorphic toG/T.

9. LetA∈SL2(R)and suppose that the vectorsxandyform two sides of a parallelogram

inR2. Prove that the area of this parallelogram is the same as the area of the parallelogram

with sides Axand Ay.

10. Prove that SO(n) is a normal subgroup ofO(n). 11. Show that any isometryf inRn is a one-to-one map.

12. Prove or disprove: an element in E(2) of the form(A,x), where x̸= 0, has infinite order.

13. Prove or disprove: There exists an infinite abelian subgroup ofO(n).

14. Letx= (x1, x2)be a point on the unit circle inR2; that is,x21+x22 = 1. IfA∈O(2),

show thatAx is also a point on the unit circle.

15. LetGbe a group with a subgroupH(not necessarily normal) and a normal subgroup N. ThenG is asemidirect product ofN byH if

H∩N ={id}; • HN =G.

Show that each of the following is true.

(a) S3 is the semidirect product ofA3 by H ={(1),(12)}.

(b) The quaternion group, Q8, cannot be written as a semidirect product.

(c) E(2) is the semidirect product ofO(2) by H, where H consists of all translations in

R2.

16. Determine which of the 17 wallpaper groups preserves the symmetry of the pattern in Figure 12.16.

17. Determine which of the 17 wallpaper groups preserves the symmetry of the pattern in Figure 12.26.

Figure 12.26 18. Find the rotation group of a dodecahedron.

19. For each of the 17 wallpaper groups, draw a wallpaper pattern having that group as a symmetry group.

[1] Coxeter, H. M. and Moser, W. O. J. Generators and Relations for Discrete Groups, 3rd ed. Springer-Verlag, New York, 1972.

[2] Grove, L. C. and Benson, C. T. Finite Reflection Groups. 2nd ed. Springer-Verlag, New York, 1985.

[3] Hiller, H. “Crystallography and Cohomology of Groups,” American Mathematical Monthly 93(1986), 765–79.

[4] Lockwood, E. H. and Macmillan, R. H.Geometric Symmetry. Cambridge University Press, Cambridge, 1978.

[5] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.

[6] Martin, G.Transformation Groups: An Introduction to Symmetry. Springer-Verlag, New York, 1982.

[7] Milnor, J. “Hilbert’s Problem 18: On Crystallographic Groups, Fundamental Do- mains, and Sphere Packing,” t Proceedings of Symposia in Pure Mathematics 18, American Mathematical Society, 1976.

12.3. EXERCISES 147 [8] Phillips, F. C.An Introduction to Crystallography. 4th ed. Wiley, New York, 1971. [9] Rose, B. I. and Stafford, R. D. “An Elementary Course in Mathematical Symmetry,”

American Mathematical Monthly 88(1980), 54–64.

[10] Schattschneider, D. “The Plane Symmetry Groups: Their Recognition and Their Notation,” American Mathematical Monthly 85(1978), 439–50.

[11] Schwarzenberger, R. L. “The 17 Plane Symmetry Groups,” Mathematical Gazette 58(1974), 123–31.

## The Structure of Groups

The ultimate goal of group theory is to classify all groups up to isomorphism; that is, given a particular group, we should be able to match it up with a known group via an isomorphism. For example, we have already proved that any finite cyclic group of order n is isomorphic to Zn; hence, we “know” all finite cyclic groups. It is probably not reasonable to expect

that we will ever know all groups; however, we can often classify certain types of groups or distinguish between groups in special cases.

In this chapter we will characterize all finite abelian groups. We shall also investigate groups with sequences of subgroups. If a group has a sequence of subgroups, say

G=Hn⊃Hn−1 ⊃ · · · ⊃H1⊃H0 ={e},

where each subgroupHiis normal inHi+1and each of the factor groupsHi+1/Hi is abelian,

thenGis a solvable group. In addition to allowing us to distinguish between certain classes of groups, solvable groups turn out to be central to the study of solutions to polynomial equations.

### Finite Abelian Groups

In our investigation of cyclic groups we found that every group of prime order was isomorphic to Zp, where p was a prime number. We also determined that Zmn = Zm ×Zn when

gcd(m, n) = 1. In fact, much more is true. Every finite abelian group is isomorphic to a direct product of cyclic groups of prime power order; that is, every finite abelian group is isomorphic to a group of the type

Z1

1 × · · · ×Zp

αn n ,

where eachpk is prime (not necessarily distinct).

First, let us examine a slight generalization of finite abelian groups. Suppose that G is a group and let {gi} be a set of elements in G, where i is in some index set I (not

necessarily finite). The smallest subgroup ofG containing all of thegi’s is the subgroup of

G generated by thegi’s. If this subgroup of G is in fact all of G, then G is generated by

the set {gi :i∈ I}. In this case the gi’s are said to be thegenerators of G. If there is a

finite set{gi :i∈I} that generatesG, thenG isfinitely generated.

Example 13.1. Obviously, all finite groups are finitely generated. For example, the group S3 is generated by the permutations (12)and(123). The groupZ×Zn is an infinite group

but is finitely generated by {(1,0),(0,1)}.

Example 13.2. Not all groups are finitely generated. Consider the rational numbers Q under the operation of addition. Suppose that Q is finitely generated with generators

13.1. FINITE ABELIAN GROUPS 149 p1/q1, . . . , pn/qn, where eachpi/qi is a fraction expressed in its lowest terms. Letpbe some

prime that does not divide any of the denominators q1, . . . , qn. We claim that 1/p cannot

be in the subgroup of Qthat is generated by p1/q1, . . . , pn/qn, since p does not divide the

denominator of any element in this subgroup. This fact is easy to see since the sum of any two generators is

pi/qi+pj/qj = (piqj+pjqi)/(qiqj).

Proposition 13.3. LetHbe the subgroup of a groupGthat is generated by{gi ∈G:i∈I}.

Then h∈H exactly when it is a product of the form h=1

i1 · · ·g

αn

in ,

where thegiks are not necessarily distinct.

Proof. Let K be the set of all products of the form 1

i1 · · ·g

αn

in , where the giks are not

necessarily distinct. CertainlyKis a subset ofH. We need only show thatK is a subgroup of G. If this is the case, then K =H, since H is the smallest subgroup containing all the

Outline

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