Chapter 9
Symmetry
In this chapter we explore a connection between algebra and geometry. One of the main
topics of plane geometry is that of congruence; roughly, two geometric gures are said to be
congruent if one can be moved to coincide exactly with the other. We will be more precise
below in our description of congruence, and investigating this notion will lead us to new
examples of groups.
9.1
Isometries
Consider the following pair of triangles.
They are congruent since the rst can be moved to coincide with the second. However, the
next pair of triangles are not congruent.
In order to move the rst to coincide with the second, one would also have to reshape the
triangle. One way of seeing that they are not congruence is that the distances between the
vertices of the rst triangle are not the same as the distances between the vertices of the
second. We use the idea of distance to make more precise the idea of congruence.
113
114
Chapter 9. Symmetry
De nition 9.1. An isometry of the plane is a 1-1 and onto map of the plane to itself that
preserves distances. That is, an isometry f is a 1-1 onto function of the plane such that,
for every pair P; Q of points, the distance between P and Q is equal to the distance between
f (P ) and f (Q).
Two geometric gures are then congruent if there is an isometry that maps one gure
exactly to the other. By representing points as ordered pairs of real numbers, we can give
algebraic formulas for distance between points. First, let kP k be the distance from P to the
origin; this is the length of P . The distance between P = (a; b) and Q = (c; d) is then given
by the distance formula
p
kP Qk = (a c)2 + (b d)2 :
In terms of groups, what is important about isometries that the composition of two
isometries is again an isometry, as we will prove shortly. Note that since an isometry is a
1-1 and onto function, it has an inverse function.
Lemma 9.2. (1) The composition of two isometries of the plane is again an isometry. (2)
The inverse of an isometry is an isometry.
Proof. To prove (1), let f and g be isometries of the plane and let P; Q be points. Then
k(f
g)(P )
(f
g)(Q)k = kg(P )
g(Q)k = kP
Qk
since f and g are each isometries. Therefore, f g preserves distance. Moreover, f g is
the composition of two 1-1 and onto functions, so it is also 1-1 and onto. Therefore, f g is
an isometry. For part (2), let f be an isometry, and let f 1 be its inverse function. Let P
and Q be points. We need to prove that kf 1 (P ) f 1 (Q)k = kP Qk. Let P 0 = f 1 (P )
and Q0 = f 1 (Q). Then P = f (P 0 ) and Q = f (Q0 ) by de nition of f 1 . Since f is an
isometry, kP 0 Q0 k = kf (P 0 ) f (Q0 )k. In other words, kf 1 (P ) f 1 (Q)k = kP Qk.
This is exactly what we need to see that f 1 preserves distance. Since f 1 is also 1-1 and
onto, it is an isometry.
Let Isom(R2 ) be the set of all isometries of the plane R2 . The lemma above shows that
composition of functions is a binary operation on Isom(R2 ). Associativity of composition
always holds among functions, the identify function is clearly an isometry, and the lemma
also shows that every element of Isom(R2 ) has an inverse. Therefore, Isom(R2 ) is a group.
9.2
Symmetry Groups
Let X be a subset of the plane. We associate a group to X, called the symmetry group of
X. This notion makes perfect sense for subsets of Rn for any n; but, we will restrict our
attention to plane gures.
9.2. Symmetry Groups
115
De nition 9.3. If X is a subset of the plane, then the symmetry group of X is the set of
all isometries f of the plane for which f (X) = X. This group is denoted by Sym(X).
To better understand the de nition, we look at it more carefully. If f is an isometry and
X is a subset of the plane, then f (X) = ff (P ) : P 2 Xg. Therefore, f 2 Sym(X) if for every
point P 2 X, we have f (P ) 2 X and, for every Q 2 X there is a P 2 X with f (P ) = Q. We
do not require that f (P ) = P for P 2 X. For example, if X = f(0; 1); (1; 0); ( 1; 0); (0; 1)g,
then a rotation of 90 about the origin sends this set of four points to itself. Thus, this
rotation is a symmetry of X. To see, in general, that Sym(X) is a subgroup of Isom(R2 ),
we only need to show that if f; g 2 Sym(X), then f g 1 2 Sym(X). However, if f; g 2
Sym(X), then f (X) = g(X) = X. Therefore, g 1 (X) = X, and so f (g 1 (X)) = f (X) = X.
We start by giving some examples of symmetries.
Translations. Let v be a xed element of R2 . The function f (x) = x + v is the translation
by the vector v. It is trivial to see that f is an isometry. Its inverse is translation by v.
We will denote the translation by a vector v by v .
v
Re ections. Let ` be a line in R2 . Then the map that re ects points across ` is an
isometry.
116
Chapter 9. Symmetry
If ` is the line through the origin parallel to a vector w, then the re ection across ` is given
by the formula
x w
f (x) = 2
w x:
w w
This formula comes from the formula for projection of one vector onto another that one sees
in multivariable calculus. From it one can see that f is an isometry. A re ection f satis es
f 2 = id.
Rotations. If be an angle, then the rotation (counterclockwise) by an angle
origin is given in coordinates by
r(x; y) =
cos
sin
sin
cos
x
y
about the
T
= (x cos
y sin ; x sin + y cos ):
From this formula one can see that a rotation about the origin is an isometry. We can use
this to describe a rotation about any point. If r0 is the rotation by about a point P 2 R2 ,
and if t is the translation by P , then r0 = t r t 1 . As a consequence, this shows that any
rotation is an isometry. Note that r 1 is rotation by
about the origin. If = 2 =n for
n
some integer n, then r = id.
Glide Re ections. We can produce new isometries by composition. One can check that
the composition of a rotation and a translation is again a rotation, and that the composition
of a rotation and a re ection is either a rotation or a translation. The composition of a
re ection and a translation may be a re ection, although it may also be a new type of
isometry. We will call such a composition a glide re ection if it is not a re ection.
9.3. Examples of Symmetry Groups
117
We will show later that any isometry of R2 is a composition of a translation with either
a rotation or a re ection. Therefore, we have accounted for all types of isometries of the
plane.
Exercises
1. Prove that the composition of a rotation and a translation is again a rotation.
2. Let f1 and f2 be re ections about the lines `1 and `2 , respectively. Suppose that `1 and
`2 are parallel and that b is the vector perpendicular to these lines such that translation
by b sends `1 to `2 . Show that f2 f1 is translation by 2b.
3. Let r be a rotation about a point P and let f be a re ection. Prove that f r is a
rotation about P if f (P ) = P , but that f r is a translation if f (P ) 6= P .
4. Let f be a re ection about a line ` and let be a translation by a vector b. If b is
parallel to `, show that f is a re ection about the line ` + b=2.
5. Let f be a re ection about a line ` and let be a translation by a vector b. If b is not
parallel to `, show that f xes no point, so is not a re ection. Morevoer, if b = b1 + b2
with b1 parallel to ` and b2 perpendicular to `, show that f is the composition 0 f 0 of
the translation 0 by b2 and f 0 is the re ection about the line ` + b1 =2.
(By this problem, it follows that any glide re ection is the composition of a re ection
followed by a translation by a vector perpendicular to the re ection line.)
9.3
Examples of Symmetry Groups
The examples we give in this section will help us to understand the de nition of symmetry
group and to introduce as symmetry groups two important classes of groups, the cyclic and
dihedral groups.
Example 9.4. We calculate the symmetry group of an equilateral triangle T . For convenience, we view the origin as the center of the triangle.
118
Chapter 9. Symmetry
A
B
C
The identity function is always a symmetry of any gure. Also, the rotations about the origin
by an angle of 120 or 240 are elements of Sym(T ). Furthermore, the three re ections across
the three dotted lines of the following picture are also isometries.
A
B
C
We have so far found six isometries of the triangle. We claim that Sym(T ) consists precisely
of these six isometries. To see this, rst note that any isometry must send the set fA; B; Cg
of vertices to itself. There are only six 1-1 functions that send fA; B; Cg to itself, and any
isometry is determined by what it does to three non-collinear points, by Proposition 9.14,
so we have found all symmetries of the triangle.
If r is the rotation by 120 and f is any of the re ections, then an exercise will show that
f; rf; r2 f are the three re ections. Moreover, r2 is the 240 rotation. Therefore, Sym(T ) =
fe; r; r2 ; f; rf; r2 f g. The elements r and f satisfy r3 = e and f 2 = e. Another exercise will
show that r and f are related via the relation f r = r2 f . Alternatively, this equation may
be rewritten as f rf = r 1 , since f 1 = f and r2 = r 1 .
Example 9.5. Now let us determine the symmetry group of a square, centered at the origin
for convenience. As with the triangle, we see that any isometry that preserves the square
must permute the four vertices. There are 24 permutations of the vertices. However, not
all come from isometries. First, the rotations of 0 , 90 , 180 , and 270 about the origin are
symmetries of the square. Also, the re ections about the four dotted lines in the picture
below are also symmetries.
9.3. Examples of Symmetry Groups
119
A
B
D
C
We now have eight symmetries of the square and we claim that this is all. To see this, we
give a counting argument. There are four choices for where vertex A can be sent since it
must go to one of the vertices. Once a choice has been made, there are just two choices
for where B is sent since it must go to a vertex adjacent to the image of A; this is forced
upon us since isometries preserve distance. After images for A and B have been chosen, the
images of the other two vertices are then xed; C is sent to the vertex across from the image
of A, and D is sent to the vertex across from the image of B. So, there is a total of 4 2 = 8
possible isometries. Since we have found eight, we have them all.
If r is the rotation by 90 and f is any re ection, then the four re ections are f; rf; r2 f; r3 f ;
this can be seen by an exercise similar to that needed in the previous example. The four rotations are r; r2 ; r3 ; r4 = e. Thus, the symmetry group of the square is fe; r; r2 ; r3 ; f; rf; r2 f; r3 f g.
We have r4 = e and f 2 = e, and an exercise shows that f r = r3 f , or f rf = r 1 . This is the
same relation that holds for the corresponding elements in the previous example.
Example 9.6. If we generalize the previous two examples by considering the symmetry
group of a regular n-gon, then we would nd that the symmetry group has 2n elements. If
r is a rotation by 360 =n and if f is a re ection, then the rotations in the group are the
powers r; r2 ; : : : ; rn 1 ; rn = e of r, and the re ections are f; rf; : : : ; rn 1 f . Thus, half of the
elements are rotations and half are re ections. The elements r and f satisfy the relations
rn = e, f 2 = e, and f rf = r 1 . This group is commonly called the Dihedral group, and
is denoted Dn . We also consider the subgroup of all rotations in Dn . This is the cyclic
subgroup hri = fe; r; : : : ; rn 1 g. We denote this group by Cn . This group also arises as a
symmetry group. We show how C4 arises in the following example.
Example 9.7. The following gure has only rotations in its symmetry group; because it has
rotations only of 0 , 90 , 180 , and 270 , its symmetry group is C4 .
120
Chapter 9. Symmetry
By drawing similar but more complicated pictures, we can represent Cn as a symmetry group
of some plane gure.
Example 9.8. The Zia symbol has symmetry group D4 ; we see that besides rotations by
0 , 90 , 180 , and 270 , it has horizontal, vertical, and diagonal re ections in its symmetry
group, and no other symmetry.
Example 9.9. Consider a circle of radius 1 centered at the origin 0. Any isometry of
the circle must map the center to itself. Thus, any such isometry is linear, by Proposition
9.12. Conversely, if ' is any linear isometry, and if P is any point with d(P; 0) = 1, then
d('(P ); '(0)) = d('(P ); 0) = 1. Thus, ' sends the circle to itself. Therefore, the symmetry
group of the circle is O2 (R).
Exercises
1. Let G be the symmetry group of an equilateral triangle. If r is a 120 rotation and f
is a re ection, show that f , rf , and r2 f are the three re ections sof G.
2. Show that there are exactly six 1-1 functions from a set of 3 elements to itself.
3. Let G be the symmetry group of an equilateral triangle. If r is a 120 rotation and f
is a re ection, so that f r = r2 f . Use this to show that f rf = r 1 .
4. Let G be the symmetry group of a square. If r is a 90 rotation and f a re ection,
show that the four re ections of G are f , rf , r2 f , and r3 f .
5. Let G be the symmetry group of a square. If r is a 90 rotation and f a re ection,
show that f r = r3 f . Use this to show that f rf = r 1 .
9.4
The Structure of Isom(R2)
In this section we give a group-theoretic decomposition of the group Isom(R2 ). We will
use the description of isometries obtained in this section to aid us in our classi cation of
9.4. The Structure of Isom(R2 )
121
wallpaper groups. There are two special subgroups we will consider. One is the group T
of all translations of R2 . It is clear that this is a subgroup of Isom(R2 ). The second is the
orthogonal group O2 (R), the set of all isometries that are also linear transformations of the
vector space R2 . An alternative description, which we will prove below, is that this is the
group of all isometries that x the origin. Recall that a linear transformation T : R2 ! R2
is a function satisfying T (u + v) = T (u) + T (v) for all u; v 2 R2 , and T (au) = aT (u) for all
u 2 R2 and scalars a 2 R.
By coordinatizing the plane, we may consider points in the plane as vectors. Following
usual notation, we will write kuk for the length of a vector u; recall that if u = (a; b), then
p
kuk = a2 + b2 . The distance between two vectors u and v is then ku vk. With this
notation, we see that a map f of the plane is an isometry if kf (u) f (v)k = ku vk for
any vectors u and v.
Let g be an isometry with g(0) = 0. From this condition we see for any u 2 R2 that
kg(u)k = kg(u)
g(0)k = ku
0k = kuk :
In other words, g preserves the length of a vector. Recall that if u and v are vectors, then
there is a unique angle with 0
such that
ku
vk2 = kuk2 + kvk2
2 kuk kvk cos :
This fact is a consequence of the Cauchy-Schwartz inequality, which says that kuk kvk
ju vj. A consequence of this inequality is that the dot product is given by the formula
u v = kuk kvk cos .
Lemma 9.10. If g is an isometry of R2 with g(0) = 0, then g preserves angles. That is, for
any u; v 2 R2 , the angle between g(u) and g(v) is the same as the angle between u and v.
Proof. Let g be an isometry. If
ku
If
0
is the angle between two vectors u and v, then
vk2 = kuk2 + kvk2
2 kuk kvk cos :
is the angle between g(u) and g(v), then
kg(u)
g(v)k2 = kg(u)k2 + kg(v)k2
2 kg(u)k kg(v)k cos 0 :
However, since kg(u)k = kuk and kg(v)k = kvk, we get
ku
This forces cos
0
vk2 = kg(u)
= cos . Since 0
g(v)k2 = kuk2 + kvk2
;
0
2 kuk kvk cos 0 :
, we conclude that
0
= .
Lemma 9.11. If g is an isometry of R2 with g(0) = 0, then g preserves dot products. In
other words, g(u) g(v) = u v for all u; v 2 R2 .
122
Chapter 9. Symmetry
Proof. If is the angle between u and v, then u v = kuk kvk cos . By Lemma 9.10, is
also the angle between g(u) and g(v). Therefore, g(u) g(v) = kg(u)k kg(v)k cos . Since
kg(u)k = kuk and kg(v)k = kvk, this yields g(u) g(v) = u v.
Proposition 9.12. If g is an isometry of R2 with g(0) = 0, then g is a linear transformation.
Proof. Let [v1 ; v2 ] be an orthonormal basis of R2 . First of all, kwi k = kg(vi )k = kvi k = 1
since g preserves length. Therefore each wi is a unit vector. Next, since g preserves angles,
the angle between w1 and w2 is equal to the angle between v1 and v2 , which is =2. This
means that fw1 ; ; w2 g is another orthonormal basis of R2 . Recall that if u = 1 v1 + 2 v2 , then
the coe cients i are determined by the formula i = u vi . So, we have i = g(u) g(vi ) =
g(u) wi since g is dot product preserving. However, g(u) = 1 w1 + 2 w2 since [w1 ; w2 ] is an
orthonormal basis. We conclude that g(u) = (g(u) v1 ) w1 + (g(u) v2 )w2 . From this formula
we can see that g is a linear transformation. For, let u; v 2 R2 . Then
g(u + v) = ((u + v) w1 ) w1 + ((u + v) w2 ) w2
= ((u w1 )w1 + (v w2 )w2 ) + ((v w1 )w1 + (v w2 )w2 )
= g(u) + g(v);
and if
is any scalar, then
g( u) = ( u w1 )w1 + ( u w2 )w2 = (u w1 )w1 + (u w2 )w20
=
((u wi )w1 + (u w2 )w2 ) = g(u):
This proves that g is a linear transformation.
We will refer to any isometry that is a linear transformation as a linear isometry. If g
is a linear isometry of R2 , by viewing the elements of R2 as column matrices we can write
g(u) = Au for some 2 2 matrix A. The matrix for g is not arbitrary. We can get a restriction
on A by knowing that g preserves the dot product. If g(x) = Ax, then the (i; j)-entry of AT A
is g(ei ) g(ej ) = ei ej , which is 1 for i = j and 0 otherwise. This shows us that AT A = I,
the 2 2 identity matrix. Conversely, if A is a matrix with AT A = I, then the linear map g
de ned by g(x) = Ax is an isometry, since
g(u) g(v) = (Au) (Av) = (Au)T (Av)
= uT (AT A)v = uT v = u v:
The set of matrices that satisfy the condition AT A = I is called the orthogonal group,
and is denoted O2 (R).
Corollary 9.13. If f is an isometry of R2 , then f (x) = Ax + b for some b 2 R2 and some
2 2 matrix A with AT A = I.
Corollary 9.14. An isometry is determined by its action on any three non-collinear points.
9.4. The Structure of Isom(R2 )
123
Proof. Suppose that f and g are isometries that agree on non-collinear points P; Q; R. There
are matrices A; B and vectors b; c such that f (X) = AX + b and g(X) = BX + c for any
point X. Then
AP + b = BP + c;
AQ + b = BQ + c;
AR + b = BR + c;
From these we obtain the two equations A(P Q) = B(P Q) and A(Q R) = B(Q R).
Now, since P; Q; R are not collinear, the vectors P Q and Q R are not parallel; thus,
they form a basis for R2 . Thus, AX = BX for any X 2 R2 ; it follows that A = B. From
AP + b = BP + c and A = B, we also conclude that b = c. Therefore, f = g.
Because of the connection between linear transformations and matrices, we get the following connection between O2 (R) and Isom(R2 ).
Proposition 9.15. Let H be the subgroup of isometries of R2 that preserve the origin. Then
H = O2 (R).1
Proof. We de ne a map
(A)(x) = Ax. We have
: O2 (R) ! H by (A) is the isometry x 7! Ax. In other words,
(AB)(x) = (AB)x = A(Bx) = (A)(Bx)
= (A)( (B)(x))
= ( (A) (B))(x):
Therefore, (AB) = (A) (B). So, is a group homomorphism. If (A) is the identity
function, then (A)(x) = x for all x. Then Ax = x for all x. But then the matrix A de nes
the identity linear transformation, so A = In . Therefore, is injective. Finally, if g 2 H,
then g(x) = Ax for some matrix A by Corollary 9.13; the element b in that corollary is 0
since b = g(0) = 0. Corollary 9.13 shows that AT A = I, so A 2 O2 (R). This proves that
g = (A), so is surjective. Therefore, is a group isomorphism.
We now show how Isom(R2 ) can be constructed from the group T of all translations
and O2 (R). We view O2 (R) as both the group of linear isometries and the group of all
orthogonal matrices. We write G = Isom(R2 ) and H = O2 (R) for ease of notation. We
proved above that every isometry is a composition of a linear isometry and a translation.
Therefore, G = T H, meaning that every element of G has the form tf for some t 2 T and
some f 2 H. We next note that T is a normal subgroup of G. To see this, the equation
G = T H shows that it is enough to prove that if f is a linear isometry and t is a translation,
1
one page somewhere about isomorphism.
124
then f
So,
Chapter 9. Symmetry
t f
1
is a translation. Suppose that t(x) = x + a. Then t(f
f tf
1
(x) = f (f
1
1
(x)) = f
1
(x) + a.
(x) + a) = x + f (a)
since f is linear. Therefore, f t f 1 is translation by f (a). It is clear that T \ H = fidg
since any nontrivial translation xes no point. We therefore have written G = T H with T a
normal subgroup, H a subgroup, and T \ H = fidg.
Exercises
1. Prove that O2 (R) is a subgroup of the group Gl2 (R) of all invertible matrices over R.
2. If SO2 (R) = fA 2 O2 (R) : det(A) = 1g, prove that SO2 (R) is a subgroup of Gl2 (R).
Structure of O2 (R)
As we saw in the previous section, the group Isom(R2 ) is built from the subgroup of translations and the subgroup O2 (R) of linear isometries. We point out some facts about O2 (R).
If we use the standard basis for R2 , then an element of O2 (R) can be represented by a 2 2
matrix. Furthermore, any such matrix A satis es the condition AT A = I2 . Taking determinants, we obtain det(A) = 1. The subset of O2 (R) consisting of matrices with determinant
1 is a subgroup, and is called the special orthogonal group, and is denoted SO2 (R). Thus,
SO2 (R) = fA 2 O2 (R) : det(A) = 1g. Note that [O2 (R) : SO2 (R)] = 2
Let A be the matrix with respect to the standard basis for an element in O2 (R). If
a b
c d
A=
;
then the condition AT A = I2 is
a b
c d
T
a b
c d
=
a2 + c2 ab + cd
ab + cd b2 + d2
=
1 0
0 1
:
This yields a2 + c2 = 1 and b2 + d2 = 1. Therefore, there is an angle with a = cos and
c = sin . Furthermore, the condition ab + cd = 0 says that the vector (b; d) is orthogonal
to (a; c). Since b2 + d2 = 1, this forces (b; d) = ( sin ; cos ) or (b; d) = (sin ; cos ).
The rst choice gives a matrix with determinant 1 and the second choice gives a matrix of
determinant 1. From this we see that if A 2 SO2 (R), then we may write
A=
cos
sin
sin
cos
for some angle . In other words, A is a rotation about the origin by an angle . On the
other hand, if A 2
= SO2 (R), then
A=
cos
sin
sin
cos
:
9.4. The Structure of Isom(R2 )
125
From the formula given earlier for a re ection across a line through the origin, we can see
that A is the re ection across the line y = (tan =2)x.
To summarize, elements of SO2 (R) are rotations and elements of O2 (R)n SO2 (R) are
re ections. In particular, if r 2 SO2 (R) and f 2
= SO2 (R), then rf 2
= SO2 (R), so rf is a
2
1
re ection. Thus, (rf ) = 1, or f rf = r . The dihedral group Dn is the group of symmetries
of a regular n-gon. As we will see below, if r is a rotation by 2 =n and f is any re ection,
then Dn can be described as
Dn = id; r; : : : ; rn 1 ; f; rf; : : : ; rn 1 f
and the isometries r and f satisfy the relations
rn = f 2 = 1;
f rf = r 1 :
Thus, we can identify Dn as a subgroup of O2 (R) by setting r to be the rotation by an
angle of 2 =n and f any re ection. It is a proposition of group theory, which we will not
prove here, that if G is a group of order 2n containing elements x; y with xn = y 2 = e and
yxy = x 1 , then G is isomorphic to Dn .
Lemma 9.16. Let G be a subgroup of O2 (R) and let R be the subset of rotations in G. Then
R is a subgroup of G and [G : R] 2.
Proof. We know that the composition of rotations about the origin is again a rotation about
the origin, and the inverse of a rotation (by ) about the origin is a rotation (by
) about
1
the origin. Thus, if r; s 2 R, then r s 2 R. Therefore, R is a subgroup of G. If R = G,
then [G : R] = 1. Suppose that R 6= G. Let f 2 G R. Then f is a re ection. If g 2 G is
another re ection, then g f is a rotation, so g f 2 R. Therefore, since f 1 = f , we see
that gR = f R. Thus, G consists of two cosets, R and f R. This proves that [G : R] = 2.
Thus, in either case, we have [G : R] 2.
We have the following property of nite subgroups of O2 (R).
Proposition 9.17. Let G be a nite subgroup of O2 (R). Then G is isomorphic to either a
cyclic group of order n or a dihedral group of order 2n, for some integer n.
Proof. Let N = G \ SO2 (R), a normal subgroup of G. Since [O2 (R) : SO2 (R)] = 2 and
G=N is isomorphic to a subgroup of O2 (R)= SO2 (R), we get [G : N ] 2. If N = f1g, then
either G = f1g is cyclic, or G = hf i for some re ection f , so G = D1 . Therefore, assume
that N 6= f1g. The group N consists of rotations. Since it is nite, there is a nontrivial
rotation r 2 N of minimal possible angle . If r0 is any other nontrivial rotation in N , and
if r0 is a rotation by , then there is an integer m with m
< (m + 1) . The rotation
0 m
r(r )
is a rotation by the angle 0
m < . Minimality of then forces m = . In
0
other words, r 2 hri. This proves that N = hri is cyclic. If G = N , then G is cyclic. If
126
Chapter 9. Symmetry
G 6= N , then [G : N ] = 2. If f 2 GnN , then as pointed out before, f rf = r 1 . If n = jN j,
then jGj = 2n, and G is generated by r and f , and satis es the relations rn = f 2 = 1 and
f rf = r 1 . Therefore, G = Dn .
Chapter 10
Symmetry Groups of Bounded
Figures and of Strip Patterns
In this chapter we classify the symmetry groups of two classes of gures, bounded gures
and strip patterns. As we will see, we do not need a great deal of group theory to do this.
However, in the next chapter, we classify symmetry groups of wallpaper patterns. This will
require a substantial amount of group theory.
10.1
Symmetry Groups of Bounded Figures
In this section we classify the symmetry groups of a bounded plane gure. We will see that
any such symmetry group is isomorphic to a subgroup of O2 (Z). Moreover, if the gure has
a \smallest" rotation, then we will see that its symmetry group is either Cn or Dn .
To start, we will be speci c about what type of gures we consider. A subset S of R2 is
said to be bounded if there is a positive number r such that S is a subset of the closed disk
fP 2 R2 : jP j rg of radius r. All examples in the previous section were bounded gures.
We begin with a simple but important lemma restricting the type of group that can arise as
the symmetry group of a bounded gure.
Lemma 10.1. Let S be a bounded gure, and set G = Sym(S). Then G does not contain a
nontrivial translation.
Proof. Suppose that 2 G is a translation, and suppose that is translation by the vector
b. Then, since G is a group, n 2 G for every positive integer n. Thus, translation by nb is
a symmetry of S for every n. Since S is bounded, there is an r such that jP j r for every
P 2 S. But then n (P ) = P + nb 2 S for every n. The triangle inequality implies that
jnbj jP j + jP + nbj 2r. This yields n jbj 2r. However, since this is true for every n,
we must have jbj = 0; thus, b = 0. Therefore, is the identity map.
To prove the main result about symmetry groups of bounded gures, we need some facts
about compositions of isometries, whose proofs we leave for exercises.
127
128
Chapter 10. Symmetry Groups of Bounded Figures and of Strip Patterns
1. The composition of two re ections about lines intersecting at a point P is a rotation
about P .
2. The composition of two rotations about a common point P is another rotation about
P.
3. If r and s are rotations about di erent centers, then rsr 1 s
1
is a nontrivial translation.
4. If r is a rotation about P and f is a re ection not xing P , then rf is a glide re ection.
5. Let r be a rotation and f a re ection in O2 (R). Prove that f rf = r 1 .
Proposition 10.2. Let S be a bounded gure, and set G = Sym(S). Then G is a subgroup
of O2 (R). In particular, every isometry of S is linear.
Proof. By the lemma there are no nontrivial translations in G. As a consequence, there are
no nontrivial glide re ections, since the square of a nontrivial glide is a nontrivial translation.
Thus, G must consist solely of rotations and/or re ections. If G is either the trivial group
or is generated by a single re ection, then the result is clear. We may then assume that
G contains a nontrivial rotation or two re ections. In either case G contains a nontrivial
rotation since the composition of two re ections is a rotation. If G contains two rotations
r and s, then rs is a rotation if r and s share a common rotation center. However, if they
have di erent centers, then rsr 1 s 1 is a nontrivial translation. By the lemma, this cannot
happen. Therefore, every rotation in G has the same center, which we will view as the
origin. If r 2 G and if f 2 G is a re ection, then f r is a glide re ection if f does preserve
the rotation center of r. Since G does not contain a glide re ection, f must preserve the
origin. Therefore, all symmetries of S preserve the origin, and so G O2 (R) is comprised
of linear isometries.
Let S be a bounded gure. By the proposition, G = Sym(S) is a subgroup of O2 (R). We
can then consider the subgroup R of rotations of S. We consider the case where R contains
a nontrivial rotation of smallest possible angle.
Corollary 10.3. If S is a bounded gure such that Sym(S) contains a rotation of smallest
possible angle, then Sym(G) is either isomorphic to Cn or to Dn for some n.
Proof. Let r be a rotation in G = Sym(S) of smallest possible nonzero angle . Recall that
is only unique up to multiples of 2 . We may assume that 0 <
, since if G contains
a rotation by , then it contains a rotation by
, and if < < 2 , then 0 < 2
< ,
1
and r is rotation by 2
. If s is another rotation in G, and if s is rotation by , then
there is an integer m with m
< (m + 1) . The symmetry sr m has rotation angle
m , which by the inequality is smaller than . By choice of , we must have
m = 0,
m
so = m . Thus, s = r . Therefore, the group R of rotations of S is the cyclic group
generated by r. Moreover, we claim that = 2 =n for some n. To prove this, let n be the
10.2. Symmetry Groups of Strip Patterns
129
smallest integer with 2 =
rewriting a little gives
and
n. Then n
2
1 < 2 = . Multiplying these inequalities by
n <2+ ;
so 0
n
2 < . However, the rotation angle in the interval [0; 2 ] for rn is n
2 .
n
Since r 2 G, minimality of forces n
2 = 0, giving = 2 =n, as desired. Since we
n
have shown that R = hri, and r = id but rm 6= id for 0 < m < n, we see that jRj = n,
and so jRj < 1. If G = R, then G = Cn . If not, then G contains a re ection f . Therefore,
[G : R] = 2 by Lemma 9.16. In any case, [G : R] < 1. Since jGj = [G : R] jRj, the group
G is nite. By Proposition 9.17, we see that G is Cn or Dn for some n.
Exercises
1. Prove that the composition of two re ections about lines intersecting at a point P is
a rotation about P . Moreover, if is the angle between the two re ection lines, prove
that the composition is a rotation by =2.
2. Prove that the composition of two rotations about a common point P is another
rotation about P . Moreover, if the rotations are by angles and , prove that the
composition is rotation by + .
3. If r and s are rotations about di erent centers, prove that rsr 1 s
translation.
1
is a nontrivial
4. If r is a rotation about P and f is a re ection not xing P , prove that rf is a glide
re ection.
10.2
Symmetry Groups of Strip Patterns