CHAPTER 10

SYMMETRY AND GROUPS

10.1 More About Groups

Recall that a set together with a binary operation · is called a group if the following conditions are satisfied:

1. is closed under the binary operation; that is, if x and y are elements of , then so is x · y.

2. The associative law holds. If x, y, and z are elements of , then

3. There is an identity element e in . For every x in , e · x = x · e = x.

4. is closed with respect to inversion. For every member x in , there is another member x′ also in such that x · x′ = x′ · x = e.

When dealing with groups in general, the binary operation is typically called group multiplication or simply multiplication, and the symbol for the operation is omitted.

There are a few simple but useful facts about groups that can occasionally save us some work.

Theorem 10.1.1. A group has only one identity element.

Proof. Suppose that e and f are identity elements in a group (, ·). Since xe = x for every x in , we have fe = f. Since fx = x for every x in , we have fe = e. Therefore, f = fe = e.

Theorem 10.1.2. Each element of a group has only one inverse.

Proof. Let x be an element of and suppose that x′ and x″ are both inverses for x. Then,

and

However, by the associative law,

Since there is exactly one inverse for each x in , there is no ambiguity in denoting it by x⁻¹.

An important consequence of the previous theorem is that the cancellation laws hold:

If ya = ax, can we conclude that y = x? We certainly could if it were true that ax = xa or that ay = ya. If a group has the property that xy = yx for every pair of elements x and y in , we say that the commutative law holds, and the group is called a commutative group or an abelian group. The group of all isometries of the plane is not a commutative group, although it is true that TS = ST for some pairs of isometries.

Returning to the question of whether ya = ax implies that x = y, the answer is that this may not be the case. If ya = ax, then multiplying on the right by a⁻¹ gives us

This is a fairly important operation in group theory and is called conjugation. More specifically, we say that y is the conjugate of x by a if y = axa⁻¹.

Example 10.1.3. Let a = T_AB and let x = R_{Q, θ}. Find the conjugate of x by a.

Solution. We know that a⁻¹ is T_BA, and so axa⁻¹ is the product of three direct isometries and must therefore be a direct isometry. To figure out what it is, we know from previous exercises that we need only look at how it affects two convenient points.

Let Q′ be the point that is mapped to Q by T_BA. In other words, let Q′ = T_AB(Q). Then

so axa⁻¹ is either the identity or a rotation about Q′.

Now apply axa⁻¹ to the point Q. Let Q″ = T_BA(Q). Then Q is the midpoint of Q′ Q″. Let X = R_{Q, θ}.(Q″) and let Y = T_AB(X). The segments and are equal in length and in the same direction, so Q′QXY is a parallelogram, and so ∠Q″QX is congruent to ∠QQ′Y. It therefore follows that axa⁻¹ = R_{Q′, θ}.

The image of a straight line under any isometry A is another straight line. The following theorem says that any conjugate of a reflection is again a reflection, but possibly in a different line.

Theorem 10.1.4. Let A be any isometry in the plane and let m be any line. Then

Proof. If A is a direct isometry, so is A⁻¹, and if A is an opposite isometry, then so is A⁻¹. It follows that AR_mA⁻¹ is an opposite isometry.

Let X be a point on A(m); that is, let X = A(Y) for some point Y on m. Then

This says that AR_mA⁻¹ fixes every point on the line A(m), and since AR_mA⁻¹ is an opposite isometry, it must be the reflection RA_(m).

The image of a directed line segment under any isometry A is another directed line segment of the same length. The following theorem says that any conjugate of a translation is again a translation through the same distance, although possibly in a different direction.

Theorem 10.1.5. Let A be any isometry in the plane, and let be a directed line segment. Then

where the directed segment

Proof. We first show that AT_CD A⁻¹ must be a translation. The isometry A T_CD A⁻¹must be a direct isometry, and since the only direct isometries without fixed points are translations, it suffices to show that A T_CD A¹ has no fixed point.

Suppose to the contrary that X is a fixed point of the isometry, that is, that

Now, X = A(Y) for some point Y, so that

This says that Y = T_CD (Y), that is, Y is a fixed point of T_CD. However, this contradicts the fact that a translation has no fixed points. Thus, we must conclude that AT_CDA⁻¹ has no fixed points, and therefore it must be a translation.

To pin down the translation T_EF, let us consider the effect of AT_CDA⁻¹ upon E:

so we have a translation that maps E to F. In other words,

Theorem 10.1.6. Let Q be a point on the line m. Then R_l R_{Q, θ} R_l = R_{Q, −θ}.

We leave the proof as an exercise.

10.1.1 Cyclic and Dihedral Groups

If a is an element of a group and m is a positive integer, then a^m, a^−m, and a₀ are defined as follows:

With these definitions, it is not difficult to verify that the following two laws of exponents hold for all integers m and n:

In general, we can expect that (ab)^m ≠ a^m b^m, unless the group is commutative.

The order of an element a of a group is the smallest positive integer n such that a_n = e. If there is no positive integer n such that a_n = e, then the order of a is infinite.

For example, in the group of all isometries of the plane, the order of I is 1, the order of R_{Q, 90°} is 4, the order of R_{Q, 180°} 2, the order of R_{Q, 270°} is 4, and the order of T_AB is infinite.

The order of a group is the number of elements in that group.

A group is said to be generated by a subset of if every element of can be expressed as a product of elements of and inverses of such elements (by this we mea a finite product, not something that is the limit of some infinite process). In this case, we write

A group is called a cyclic group if it is generated by an element a that is, . Here, the order of could be finite or infinite.

• For example, the group of integers, with addition as the binary operation, is generated by the set { 1 } ; that is, (, +) = 1 .
  Here, the group multiplication is addition of integers, and the inverse of 1 is − 1. To see why is generated by { 1 }, note that, for an integer m,
  
  Thus, = (1), and the order of the group is infinite.
• On the other hand, if a and aⁿ = e for some integer n ≥ 1, then the group = a has n elements, namely
  
  and the order of the group is n. This group is denoted by _n and is called the cyclic group of order n.

A group is called a dihedral group if it is generated by two elements a and b for which

The elements of this group are

Consequently, the group is called the dihedral group of order 2n and is denoted by _2n.

Unlike the situation with cyclic groups, it is not immediately obvious that the 2n elements listed on the previous page are the only elements of _2n.

• For example, how do we know that a product like
  
  is actually one of the 10 elements that are said to comprise ₁₀ or that the inverses of such products belong to ₁₀
  Here, n = 5, and the 10 elements are
  
  Since b² = e, we can replace the factors b^k with b if k is odd or with e (that is, omit the factor) if k is even. The product a⁸b³ (a⁻¹)¹⁰ a⁴b⁵ reduces to
  
  Then, recalling that a⁵ = e, replace a⁸ with a⁵ a³ = a³ and replace a⁻⁶ with a⁻⁶ (a¹⁰) = a⁴ to get
  
  Now replace a⁴ by a (b⁻¹ b) a (b⁻¹ b) a (b⁻¹ b) to get
  
  Since bab⁻¹ = a⁻¹, this becomes
  
  which is one of the 10 elements, as claimed.
  It should be clear that by applying the same process to any finite product, we will end up with one of a^s or ba^s.

Example 10.1.7. What is the inverse of ba^s in _2n?

Solution. It is its own inverse. Here is the proof:

Expanding (bab⁻¹), several b and b⁻¹ cancel each other out, and we get ba^sb⁻¹, so that

as claimed.

Example 10.1.8. Assuming that 0 < k < n, what is a^k b in _2n?

Solution. We have

Here is the prototypical concrete example of a dihedral group.

Example 10.1.9. Show that the group of symmetries of a square is ₈.

Solution. Label the vertices of the square A, B, C, and D counterclockwise, and let Q be the center point of the square. Any isometry T that carries the square onto itself must map the vertex A to T(A), where T(A) is one of the four vertices. The vertex B is mapped to T(B), which must be one of the two vertices that are adjacent to T(A). Since T must also map Q to Q, the action of T on A and B completely determines the isometry.

Two particular symmetries of the square are R_{Q, 90°} = and R_m, where m is a diagonal of the square. Letting a = R_{Q, 90°} and b = R_m, we see that

The last equality follows by Theorem 10.1.6.

Consequently, the isometrics R_{Q, 90°} and R_m generate the dihedral group ₈, and since there are exactly eight different symmetries of the square, we are finished.

The preceding example has an obvious generalization whose proof is virtually identical to the proof for the square:

Theorem 10.1.10. The group of symmetries of the regular n-gon is the dihedral group _2n.

Remark. This result is often used as the definition of _2n.

10.2 Leonardo’s Theorem

Leonardo da Vinci apparently worked out all possible symmetries for the floor plans of many chapels. Because of this, the following theorem is known as Leonardo’s Theorem.

Theorem 10.2.1. (Leonardo’s Theorem)

Every finite group of isometries of the plane is either a cyclic group or a dihedral group.

The proof of this theorem is long but not difficult and amounts to checking what can happen. In order to prove it, we need some facts about the product

that were developed earlier. We marshall them here for convenience.

We also need the following facts about the interaction between rotations and reflections:

Note that points (2) and (5) imply the following:

Lemma 10.2.2. Suppose that G is a finite group of isometries.

Proof. The proof uses the fact that if a group contains a translation, then it must be infinite. To see why this is true, note that if T_AB is in the group , then so are (T_AB)², (T_AB)³, and so on. However, each of (T_AB)², (T_AB)³, …, is a translation, and all of them are different.

(1) We want to show that if R_{P, ϕ} and R_{Q, θ} are two different members of , then P = Q. To establish this, we proceed by contradiction and assume that P ≠ Q. We consider two cases.

(a) ϕ + θ ≡ 0 (mod 360):
If P ≠ Q, then the product R_{P, ϕ} R_{Q, θ} would be a translation, contradicting the fact that is finite.

(b) ϕ + θ (mod 360):
The fact that is a group means that the product

is a member of . However, we would then have

for some X and

for some point Y. Consequently,

This is either the identity, if X = Y, or a translation, if X ≠ Y. However, it cannot be the case that X = Y, for this would mean that R_{Y, −(ϕ+θ)} is the inverse of R_{X, ϕ+θ} or in other words, that

However,

and we would have

which contradicts the fact that rotations with different centers do not commute. Since the centers X and Y are different, it follows that the product

is a translation, again contradicting the fact that is finite.

(2) Suppose that R_P,ϕ and R_{Q, θ} both belong to , and suppose for a contradiction that P is not on m. Then (R_m R_{P, ϕ})² is a translation, meaning that would have to be infinite.

This completes the proof.

We next prove a theorem that is part of Leonardo’s Theorem but which is useful in its own right.

Theorem 10.2.3. If is a finite group of isometries that consists of exactly n rotations (counting the identity as a rotation through 360°), then is the cyclic group _n.

Proof. From Lemma 10.2.2, we know that consists of rotations through various angles with all rotations centered at a common point Q, We may also assume that all rotations belonging to are through a positive angle no greater than 360°. Since there are only a finite number of rotations, one of them has the smallest positive angle of rotation, say θ. We claim that every other rotation, including R_{Q, 360°}, must be a multiple of θ.

We again use a proof by contradiction. Supposing that this were not the case, there would be a rotation R_{Q, ϕ} in where ϕ > 0 and where ϕ is not a multiple of θ. Let α be the remainder when ϕ is divided by θ; that is,

where m is a positive integer and 0 < α < θ. Since is a group, then

is in , and therefore so is the product

However, this is impossible, since R_{Q, θ} is the rotation with the smallest positive angle of rotation.

This shows that all of the rotations in are multiples of R_{Q, θ} and, conversely, since is a group, all multiples of R_{Q, θ} must be in . Letting a denote R_{Q, θ}, this means that consists of

for some integer n, where a_n = R_{Q, 360°} = I. This completes the proof of Theorem 10.2.3.

Theorem 10.2.4. Suppose that is a finite group that contains a rotation (other than the identity) and a reflection. Then there is a rotation R_{Q, α} in the group that generates all of the rotations in the group. That is, the set of all rotations in is

where nα = 360°.

Furthermore, if R_m is any reflection in , then every reflection in must be one of the following:

In particular, this means that is the dihedral group _2n

Proof. Let be the set of all rotations that are in , including the identity. Then by the previous theorems, these rotations form a group, and all of the rotations have the same center Q. Hence, by Theorem 10.2.3, is a cyclic group and so the members of are

where nα = 360°.

To complete the remainder of the proof, we have to show that if R_l; is any reflection that is in , then there is some integer k such that

Now, consider the product R_m R_l. Since both reflections belong to , it follows that the lines l and m both contain Q, and so R_m R_l is a rotation with center Q.

Since R_m R_l is a member of , and since contains all of the rotations in , it follows that

for some integer k. Therefore, we have

Leonardo’s Theorem now follows from Theorem 10.2.3 and Theorem 10.2.4.

One of the consequences of Leonardo’s Theorem is:

Theorem 10.2.5. The group of symmetries of a polygon in the plane is either a cyclic group or a dihedral group.

Proof. Given a vertex A of a polygon, together with an adjacent vertex B, any symmetry of the polygon must map A onto one of the vertices of the polygon, in which case there are at most two possible adjacent vertices that can be the image of B. In other words, there are only a finite number of symmetries. Leonardo’s Theorem now tells us that the group of symmetries is either a cyclic group or a dihedral group.

10.3 Problems

1. Prove that a finite group of isometries cannot contain two halfturns about distinct points.

2. Prove that the set of all halfturns and all translations forms a group.

3. Prove that if a triangle is invariant under a reflection, then the triangle must be isosceles.

4. Which of the following sets of transformations form a group, and which do not form a group?

(a) All translations.

(b) All reflections.

(c) All glide reflections.

(d) All rotations.

(e) All direct isometries.

(f) All opposite isometries.

5. If H_O1 H_O2 = H_O2 H_O1 = T, prove that T = I, the identity transformation.

6. Find a plane figure P such that its group of symmetries equal

(a) the cyclic group C₂ of order 2,

(b) the cyclic group C₁ of order 1.

7. Find a plane figure P such that its group of symmetries equal

(a) the dihedral group D₂ of order 4,

(b) the dihedral group D₁ of order 2.

8. Find the group of symmetries of each of the following figures.

9. Let be a group of isometries whose subgroup of translations is generated by , where AB ≠ 0. Prove that if R_ℓ , then either is parallel to ℓ or is perpendicular to ℓ.

10. Find the group of isometries of an ellipse.

11. If a and b are elements of a group and

show that ab = ba.

12. Let a and b be elements of a group such that b has order 2 and ab = ba⁻¹.

(a) Show that aⁿb = baⁿ for all integers n.

Hint: Evaluate the product (bab) (bab) in two different ways to show that ba²b = a⁻², and then extend this method.

(b) Show that the set = {aⁿ, baⁿ|n } is closed under multiplication and in fact forms a group.

(c) Show that = a, b, the dihedral group with generators a and b.

13. A cuboid is a rectangular parallelepiped: that is. a parallelepiped where each plane face is orthogonal to four other faces and parallel to the fifth, as in the figure.

Find the group of symmetries of a cuboid with three unequal sides.

14. What is the symmetry group of a rhombus that is not a square? Find all the symmetries of the rhombus and construct the Cayley table or multiplication table for the group of symmetries.

Hint: The diagonals of a parallelogram bisect each other, and a parallelogram is a rhombus if and only if its diagonals are perpendicular.

15. Let be a (nonequilateral) isosceles triangle.

Find the group of symmetries of and construct the Cayley table for the group.

• In the plane, the discrete groups fixing a line are the groups of symmetries of ribbons or friezes. To study the frieze groups, we first find an isometry fixing a line and then compose it with an isometry that has a fixed point. The group generated by this isometry is called a frieze group.

16. For each of the seven patterns given in the figure below, assuming each extends to infinity both to the left and to the right, name the types of isometries in the symmetry group of each pattern.

17. Prove that if H_O0 , a frieze group with translation subgroup

then H_On/2 for every integer n. where

18. Find the group of symmetries of the following repeated pattern on an infinite horizontal strip, as shown below.

19. Find a different set of two generators for the frieze group

where ℓ || , and O₀ and O are two fixed points on , as in the figure below.

20. With repeated use of only the symbol

construct a repeated pattern on a horizontal strip whose frieze group is

where ℓ || , and O₀ and O are points on ℓ such that , as in the figure in Problem 10.19.

21. Find the frieze group of an infinite horizontal strip consisting of repeated I’s if the I’s lie above the midline of the strip as shown below.

22. Consider a particular vertex of an equilateral triangle. Under a symmetry, it can land on any of the three vertices. The remaining vertices must follow in order, either counterclockwise or clockwise. Thus, there are only 3 × 2 = 6 symmetries for the equilateral triangle. They are: the identity I, a 120° counterclockwise rotation J, a 120° clockwise rotation K, and three reflections X, Y, and Z, as shown in the figure on the following page.

These six symmetries form a group under the operation of composition, the dihedral group of the equilateral triangle. Construct the operation table of this group.

23. Consider a particular vertex of a square. Under a symmetry of the square, this vertex can land on any one of the four vertices. The remaining vertices must follow in order, either counterclockwise or clockwise. Thus, there are only 4×2 = 8 symmetries for the square. They are: the identity I, a 180° rotation or halfturn R, a 90° counterclockwise rotation A, a 90° clockwise rotation C, and four reflections H, V, D, and U, as shown in the figure below

These eight symmetries of the square form a group under the operation of composition, the dihedral group of the square. Construct the operation table of this group.

24. Consider the set of symmetries of a non-square rectangle. It has only four elements: I, R, H, and V, analogous to the corresponding symmetries for the square. These four symmetries form a group with respect to composition. Construct the multiplication table of this group.

25. The complex numbers 1, − 1, i, and −i form a group under multiplication. Construct the operation table of this group.

26. The Quaternion group consists of the eight elements

with the operation of multiplication defined by

Construct the operation table.

27. Let

These functions form a group with respect to the operation of composition. Construct the operation table.