CHAPTER 8

THE ALGEBRA OF ISOMETRIES

8.1 Basic Algebraic Properties

Consider a large iron grate (G) in the shape of a right triangle, as shown in the figure below, The grate has to move from its current position to cover the hole (H). It must fit exactly within the dotted lines.

The grate is far too heavy to be lifted by hand, so a machine has been rigged that can lift the grate and flip it around any desired axis, thereby placing it on the other side of the turning axis. The action of the machine is shown in the figure below. If necessary, the axis can pass through the grate.

Find a sequence of flips that will move the grate to cover the hole. What is the minimum number of flips it will take to cover the hole?

The machine is a “reflecting” device—after the machine does its work, the new position of the grate is the reflection through the axis of the original position of the grate. The solution to the problem is that the grate can be moved in a step-by-step manner to cover the hole in three flips, and the minimum number of flips necessary is three. A sequence of flips is depicted in the figure below.

The first step uses a reflection through the line l, which is the right bisector of the segment joining the right angle vertex B of the grate with the corresponding vertex of the hole. With this reflection, the grate ABC is moved into position A′B′C′.

We now have one vertex in the correct position. Vertices A′ and C′ are not in the correct position, and the next step is to do another flip to get one of them into the correct place.

In (2), vertex C″ of the hole corresponds to vertex C′ of the moved grate, and by relecting A′B′C′ about the right bisector of C′C″ we can move C to the correct positon.

Note that the right bisector of C′C″ passes through B′ because B′C′ = B′C″. After the second step, the grate is in position A″B′C″. The third and final step is to reflect A″B′C″ about the line B′C″.

The Composition of Transformations

Earlier we defined the composition of two isometries. This terminology is used for any collection of transformations. If T and S are two transformations, the composition or product of T and S is denoted by either S T or ST. In terms of the individual points in the plane, ST acts as follows. First, the point X is mapped by T to T(X), then T(X) is mapped by S to S (T(X)). In other words, for each point X in the plane:

To move the grate to its desired position, we applied the product R_n R_m R_l of three reflections. Again, we emphasize that we follow the “right-to-left” rule. When evaluating R_n R_m R_l(X), first find R_l(X), then R_m (R_l(X)), and finally R_n (R_m (Rl (X))).

Equal Transformations

In the iron grate problem, we used the product of three reflections to shift the grate to its desired position. If we had a better machine—one that could perform rotations as well as reflections—we could accomplish the same thing by performing just two operations, as shown in the figure below.

In this case, the grate is moved by applying the product R_s R_O,α. In the first solution, we used R_n R_m R_l. The individual transformations that make up the two products are quite different, but the net effect is the same. So we can write:

Two transformations are said to be equal if they have the same effect on every point in the plane. In other words, saying that T = S means that T(X) = S(X) for every point X in the plane.

As the example shows, T = R_s R_O,α and S = R_n R_m R_l does not mean that T and S necessarily have the same description. The situation is similar to equality of functions in trigonometry: if

then the functions f and g are equal (since we have pointwise equality), although their descriptions are quite different.

Other examples of equal transformations that are defined differently are the halfturn and a reflection in a point.

The halfturn about a point O, denoted by H_O, is the transformation R_{O, 180°}. The reflection in the point O is the transformation that takes each point P to the point P′ so that O is the midpoint of PP′. The point O is called the center of the halfturn or the center of reflection. In the plane, since a reflection through a point is identical to a halfturn, there is no further need to talk about reflections through a point, and there is really no need for a special symbol to denote it.

Closure

In Chapter 7, we mentioned that the composition of two isometries results in a transformation that is also an isometry. In algebra, we would describe the situation by saying that the set of isometries of the plane is closed under the operation of composition.

More generally, if is a set of elements and if is a binary operation on , we say that is closed under the operation if for every pair of elements a and b in the product a b is also in . For example, the set of positive integers is closed under addition.

Associativity

If R, S, and T are three transformations, the product TSR means first apply R, then S, and then T. For a point X, the notation TSR(X) means T (S (R(X))).

We can overrule this by using parentheses since the operations inside parentheses are carried out first. The notation (TS) R is interpreted as follows: first, determine what (TS) is. It will be some transformation, call it H, and H is usually different than either T or S. The notation (TS) R means HR; that is, first apply the transformation R, then apply the transformation H.

For example, let us suppose that S and T are reflections about two different parallel Lines and R is a rotation. Then, as we will see later, TS is some translation H. So we interpret (TS) R as meaning first do the rotation R, then follow it by the translation H.

In a similar way, the notation T (SR) means that we should first determine what SR is, namely, some different isometry L, and then take the product of L and T: T (SR) = TL. Note, however, that the parentheses in this case could be omitted, because in the absence of parentheses, the expression is evaluated from right to left.

The following theorem shows that (TS) R and T (SR) are equal.

Theorem 8.1.1. (Associative Law)

The associative law holds for the product of transformations; that is, given three transformations T, S, and R in the plane,

Proof. Let X be a point in the plane. We will evaluate T (SR) (X) and (TS) R(X).

The notation T (SR) (X) means “first evaluate SR(X), then evaluate T (SR(X)).” By definition, SR(X) = S (R(X)), so

The notation (TS) R(X) tells us to evaluate TS (R(X)). Now, TS(Z) = T (S(Z)) for all Z in the plane. In particular, when Z = R(X), we get

Thus,

Since (TS) R and T (SR) have the same effect on every point X in the plane, we conclude that they are equal transformations.

8.2 Groups of Isometries

In algebra, a set of elements together with a binary operation · is called a group and is denoted by (, ·) if it possesses the following properties:

1. The set is closed under the binary operation.

2. The associative law holds: (a · b) · c = a · (b · c) for all a, b, and c in .

3. G has an identity element: there is some element e in G such that e · a = a · e = a for every a in .

4. For every a in , there is an inverse element a′: there is an element a′ in such that a · a′ = a′ · a = e.

If it is also true that the commutative law holds (that is, if a · b = b · a for all a and b in ), then is called an Abelian group.

The notation for the binary operation · is usually omitted, so that we write ab instead of a · b.

We learned in Chapter 7 that every isometry has an inverse. This, along with the results of the previous section, show that the family of isometries in the plane, together with the composition operation, satisfy all four of the conditions listed above. We can summarize this with the following theorem:

Theorem 8.2.1. The set of all isometries of the plane, together with the operation of composition, forms a group.

Given an isometry T, we denote its inverse by T⁻¹. Theorem 7.2.1 stated that

Example 8.2.2. What are the inverses of G_{l, AB} and H_O?

Solution. We have

In any group, an element a other than the identity e such that a² = a · a = e is called an involution. In other words, an involution is an element other than the identity that is its own inverse.

Any transformation that is an involution in a group of transformations is said to be an involutory or involutoric transformation; that is, T is involutory if and only if TT = T T = I. Among the transformations

the involutions are R_l, H_P, and R_{P, θ} when θ is a multiple of 180°.

8.2.1 Direct and Opposite Isometries

In the figure above, triangle ABC has been carried into A′B′C′ by a reflection in the line m. The orientation of the triangle has changed—the path ABC proceeds in a clockwise (or negative) direction around the triangle, while the image path A′B′C′ goes in a counterclockwise (or positive) direction. This reversal of orientation will obviously happen to every triangle to which R_m is applied, and R_m is therefore called an opposite isometry. In general, any isometry T that reverses the orientation of every triangle is called an opposite isometry. On the other hand, an isometry T that preserves the orientation of every triangle is said to be a direct isometry.

There is a strong analogy between the products of direct and opposite isometries and the products of positive and negative numbers: direct isometries are like positive numbers and opposite isometries are like negative numbers. Some people prefer to draw an analogy to the addition of even and odd numbers, with direct isometries corresponding to the even numbers, so that the product of a direct and opposite isometry is like the sum of an even and an odd number. Some texts use the terms even and odd isometries instead of direct and opposite isometries.

Theorem 8.2.3. The product of two direct isometries is a direct isometry. The product of two opposite isometries is a direct isometry. The product of a direct isometry and an opposite isometry is an opposite isometry.

Theorem 8.2.4. The direct isometries of the plane form a group .

Proof. It is clear that is closed (under multiplication). Since the associative law holds for all transformations, it also holds for . The identity map is a direct isometry; that is, I is in . Every isometry has an inverse, and an isometry and its inverse are either both direct or both indirect. Therefore, whenever T is in , so is T⁻¹. This completes the proof.

It is relatively easy to invent a transformation of the plane that is neither direct nor opposite. For example, let B and C be two different points in the plane, and let T be the transformation that interchanges B and C but leaves every other point where it is. The transformation T is almost the identity map, and there are many triangles PQR that remain unchanged under T. However, if A is a point that forms a triangle with B and C, the transformation maps A, B, and C to A, C, and B, respectively, thereby reversing the orientation. In other words, T preserves the orientation of some triangles while reversing the orientation of others. Thus, T is neither direct nor opposite.

The situation where T is an isometry is quite different as a consequence of two fundamental theorems about isometries in the plane, namely, that a plane isometry is completely determined by its action on three noncollinear points and that every plane isometry is the product of at most three reflections.

We begin the proofs of these facts with a theorem that characterizes the identity transformation.

Theorem 8.2.5. If T is an isometry that fixes each of three noncollinear points, then T is the identity.

Proof. Let A, B, and C be three collinear points that are fixed by T, so that

We want to show that T(X) = X for every point X in the plane. Suppose that this is not the case. Then there is a point P such that

Since T is an isometry with T(A) = A and T(P) = P′, we have

This means that A is on the right bisector of PP′. Similarly, B and C must also be on the right bisector of PP′, contradicting the fact that A, B, and C are noncollinear. This contradiction shows that our supposition that T is not the identity must be false and completes the proof.

When we say that a transformation T fixes a point X we mean that T(X) = X, and the point X is called a fixed point of T. In Martin’s text [50], he uses the phrase “T fixes a set ” to mean that under T the image of is . He uses the phrase “Tfixes S pointwise” to means that T(X) = X for each point X of . Another way of saying the same thing is to say that the set is invariant under the transformation T.

As an example of the difference in the meaning of this language, consider the mapping of an equilateral triangle ABC onto itself by a rotation of 120° around the centroid. The triangle ABC is fixed by the transformation, but it is not fixed pointwise.

Question 8.2.6. Suppose that an isometry fixes each of two different points. Can we assume that the isometry is the identity?

One of the virtues of group theory is that once you have proven a theorem about groups, it is valid for every group and does not have to be proven separately for each different group. Here is a very simple yet useful result:

Theorem 8.2.7. Let a and b be elements of a group (, ·) such that a · b = e, where e is the identity. Then a = b⁻¹ and b = a⁻¹.

Proof. Since G is a group, the element a has an inverse a⁻¹. Multiplying each side of the equation

by a⁻¹ on the left, we have

so that

and so

and thus, b = a⁻¹.

The proof that a = b ⁻¹ can be obtained by multiplying the equation a · b = e on the right by b⁻¹.

The following theorem can be proven in much the same way as Theorem 8.2.5, but we can give a more satisfying proof by using the fact that the isometries of the plane form a group. This proof illustrates how nicely algebra fits with geometry.

Theorem 8.2.8. An isometry of the plane is completely determined by its action on three noncollinear points.

Proof. Let A, B, and C be three noncollinear points in the plane, and let S and T be two isometries such that

We want to show that S = T. Now, since the isometries of the plane form a group, we know that T has an inverse T⁻¹ so that

Thus, the isometry T⁻¹S fixes each of the points A, B, and C, and by Theorem 8.2.5, this means that T⁻¹,S must be the identity. By the previous theorem, S must be equal to the inverse of T⁻¹; that is,

Theorem 8.2.9. Every isometry of the plane that is not the identity can be decomposed into the product of at most three reflections.

Proof. Let T be a given isometry, and let A, B, and C be three noncollinear points.

From the iron grate example, we know how to map A, B, and C to T(A), T(B), and T(C), respectively, by a sequence of at most three reflections.

Suppose that it actually took three reflections, say,

in that order. Then

Thus, R_n R_m R_l and T both have exactly the same effect on A, B, and C, so R_n R_m R_l = T, by Theorem 8.2.8.

Theorem 8.2.9, together with the fact that a reflection in a line is an opposite isometry, now allows us to show that every isometry in the plane is either a direct isometry or an opposite isometry.

Theorem 8.2.10. Every isometry in the plane is either a direct isometry or an opposite isometry.

Proof. If the isometry T is the identity or the product of two reflections, it is a direct isometry. If T is a reflection or the product of three reflections, it is an opposite isometry.

8.3 The Product of Reflections

In this section, we use reflections to show that there are only four different types of plane isometries: reflections (in a line), rotations, translations, and glide reflections. We begin by examining the product of two reflections.

Example 8.3.1. Let l and m be two distinct parallel lines. Show that R_mR_l = T_XY, where is a directed segment perpendicular to l and m and twice the distance from l to m.

Solution. Let h be a line parallel to l and m such that l is midway between h and m, as in the figure above.

Let A and B be points on h and let C be a point on l R_l maps A and B to points A′ and B′ on m and leaves C where it is.

Note that the two directed segments and are parallel and twice the distance from l to m.

The reflection R_m leaves A′ and B′ where they are and maps C to C′, where is parallel to and equal in length and direction to .

Thus, the effect of R_m R_l is to map A, B, and C to A′, B′, and C′, respectively.

However, the translation T_AA′ also does the same thing, and by Theorem 8.2.8,

which completes the proof.

Example 8.3.2. Let l and m be two different lines intersecting at a point P. Show that R_m R_l = R_P,θ, where θ is twice the directed angle α from l to m.

Solution. Let h be the line through such that the directed angle from h to l is α, and let n be the line through such that the directed angle from m to n is α, as in the figure above.

Let A be a point on h and let C be a point on l. Note that A, C, and P cannot be collinear. If they were collinear, then h = l, and it would follow that l, h, and m are all the same line, which is a contradiction.

Consider the effect of R_m R_l on the points A, P, and C. It is clear that R_P,θ has exactly the same effect, so by Theorem 8.2.8, R_m R_l = R_{P, θ}.

If l and m are the same line, then R_m R_l is the identity. This fact, together with the previous two examples, proves the following theorem:

Theorem 8.3.3. The only direct isometries of the plane are the identity, the translations, and the rotations.

We next turn our attention to the product of three reflections.

Theorem 8.3.4. Let l, m, and n be three lines with a common point P. Then R_n R_m R_l is a reflection.

Proof. If l and m coincide, then

If n and m coincide, then

If neither of these two cases occurs, we proceed as follows: let α be the directed angle from l to m. Now, there is a unique line h through P such that the angle from h to n is α

Thus,

and, therefore,

To handle the combination R_n R_m R_l when the lines l, m, and n are all different and have no point in common, we break it down into separate cases. First, we consider the cases where l and m are both perpendicular to n, then the cases where m and n are both perpendicular to l.

We note first that if all three lines are parallel, then R_n R_m R_l is a reflection. We leave the proof of this as an exercise.

Theorem 8.3.5. Let l, m, and n be three different lines.

Proof. For (1), we have R_m R_l = T_AB by Example 8.3.1, and so R_nn R_m R_l is the glide reflection R_n T_AB = G_{n, AB} as claimed.

For (2), Let A = l ∩ m, let B = l ∩ n, and let C be any point on m other than A. We leave it as an exercise to show that

for each of the points X = A, B, C so that

by the fundamental Theorem 8.2.8, and R_l R_n R_m is the glide reflection G_{l, CD} by (1).

The cases not covered by the previous theorem are depicted in the figures below, which show the possible arrangements of the reflecting lines.

Although arrangements (1), (2), and (3) appear to be identical, they are not the same because the order of the reflections matters. It may appear that these cases are complicated, but the associative law comes to the rescue in a rather magnificent way.

Theorem 8.3.6. Let l, m, and n be three nonconcurrent lines. Then R_n R_m R_l is a glide reflection.

Proof. The proof of each of the four cases is very similar. Here is how it is done for case (4).

The combination R_m R_l is a rotation about A through the angle 2α, where α is the directed angle from l to m, as in (1) below. Let m′ be a line through A perpendicular to n, intersecting n at C′. Let l′ be the line through A such that the angle from l′ to m′ is α, as in (2) below. Then

Now apply the associative law:

The transformation R_n R_m′ is the rotation H_C′ Let n″ be the line through C′ perpendicular to l′, and let m″ be the line through C′ perpendicular to n″, as in (3) below. Then

Therefore,

and since m″ and l′ are parallel and both perpendicular to n″, we have a glide reflection.

8.4 Problems

1. Let S and T be two involutive transformations of the plane.

(a) Prove that ST is involutive if and only if ST = TS.

(b) Assume that S, T, and I are distinct transformations, where I is the identity, such that

Let Γ = {I, S, T, X}. Prove that Γ is a commutative subgroup of G, the group of all transformations on the plane, by constructing the multiplication table.

2. Let P, Q, and R be three points in the plane, and let P′, Q′, and R′, respectively, be their images under an isometry T. Show that the points P, Q, and R are collinear, with Q between P and R, if and only if the points P′, Q′, and R′ are collinear, with Q′ between P′ and R′.

Hint: When does equality hold in the Triangle Inequality?

3. Let T be an isometry of the plane. Show that if P and Q are fixed points of T, then every point X on the line through P and Q is a fixed point of T.

4. Let T be an isometry of the plane. Show that if T has three fixed points that are not collinear, then T = I, the identity.

5. Let S and T be isometries and let A, B, and C be three noncollinear points for which

Show that S = T.

6. Let H_A be a halfturn about a point A so that

where A, P, and P′ are collinear and d (A, P) = d (A, P′).

(a) Show that H_A is an isometry.

(b) Show that H_A is an involution; that is, H_A = .

(c) Show that if l is a line in the plane, then H_A (ℓ) is a line parallel to ℓ.

7. Let D, E. and F be the midpoints of the sides BC, AC, and AB, respectively, of ΔABC and let T be the transformation of the plane that is the product of the three halfturns

Show that the vertex A of ΔABC is a fixed point of T; that is, that T(A) = A.

Hint. Draw the picture.

8. Let T be an isometry of the plane that admits an invariant line ℓ (that is, T(ℓ) = ℓ) and a fixed point P. Prove that there is a point Q ℓ such that T(Q) = Q and a line ℓ′ through P such that T(ℓ′) = ℓ′.

9. Show that if a circle is invariant under the isometry T, then its center is a fixed point of T.

10. Let T ≠ I be an involutive isometry. Show that T has at least one fixed point.

11. Let T be an isometry that is an involution and has exactly one fixed point O in the plane. Show that T is the halfturn H_O about the point O.

12. If the isometry T is an involution, show that for any point P in the plane the midpoint of the line segment joining P and T(P) is a fixed point of T.

13. Let T be an isometry of the plane and let ℓ be the perpendicular bisector of the segment . Prove that T(ℓ) is the perpendicular bisector of the segment .

14. Let R_l be a reflection in the line l so that

where either ℓ is the perpendicular bisector of the segment PP′ for each P or P and P′ coincide on the line ℓ for each P.

(a) Show that R_ℓ is an isometry.

(b) Show that R_ℓ is an involution; that is, that R_ℓ = R⁻¹_ℓ.

(c) Show that if m is a line distinct from ℓ, then R_ℓ(m) is distinct from ℓ.

15. Show that if m and n are perpendicular lines that intersect at a point P in the plane, then

16. Given a point O and a directed segment .

(a) Find the point Q such that

(b) What is the product H_O T_AB?

17. Let A and C be distinct points in the plane. Show that B is the midpoint of the segment if and only if

18. In the triangle ΔABC, show that G is the centroid if and only if

where I is the identity.

19. Using halfturns, prove that the diagonals of a parallelogram bisect each other.

Hint. Show that if N is the midpoint of the diagonal AC of parallelogram ABCD so that H_A H_N = H_N H_C, then N is the midpoint of BD also, that is, H_D H_N = H_N H_B.

20. Using halfturns, prove that if ABCD and EBFD are parallelograms, EAFC is also a parallelogram.

21. Find all triangles such that three given noncollinear points are the midpoints of the sides of the triangle.

Hint. Given P, Q, and R, H_R H_Q H_P fixes a vertex of a unique triangle ΔP′Q′R′, as in the figure below.

22. Given ∠ABC, construct a point P on AB and a point Q on BC such that PQ = AB, and the line PQ intersects the line BC at an angle of 60°.

Hint. Take a point D such that AB = BD and BD intersects BC at an angle of 60°.

23. Prove that if R_n R_m fixes the point P and m ≠ n, then the point P is on both lines m and n.

24. Show that if m and n are distinct lines in the plane, then

if and only if m and n are perpendicular.

25. Let m be a line with equation 2x + y = 1. Find the equations of the transformation R_m.

26. Suppose that the lines ℓ and m have equations x + y = 0 and x − y = 1, respectively. Find the equations for the transformation R_ℓ R_m.

27. Given triangles ABC and DEF, where ΔABC ≡ ΔDEF and where

find the equations of the lines such that the product of reflections in the lines maps ΔABC to ΔDEF.

28. Let A₀ be a given point and ℓ₁, ℓ₂, …, ℓ_n be given lines. For 1 ≤ k ≤ n, let A_k be obtained from A_k−1 by a reflection across _ℓk, and let A_{n + k} be obtained from A_{n + k −1} by a reflection across ℓ_k.

(a) Prove that A_2n will coincide with A₀ if n is odd.

(b) Can the same conclusion be drawn if n is even?

29. Let A₀ = B₀ be a given point and ℓ₁, ℓ₂, …, ℓ_n be given lines. For 1 ≤ k ≤ n, let A_k be obtained from A_k−1 by a reflection across ℓ_k, and let B_k be obtained from B_k−1 by a reflection across ℓ_n-k+1. For what values of n will A_n coincide with B_n?