Recall that a set together with a binary operation · is called a group if the following conditions are satisfied:
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 = TAB and let x = RQ, θ. Find the conjugate of x by a.
Solution. We know that a−1 is TBA, 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 TBA. In other words, let Q′ = TAB(Q). Then
so axa−1 is either the identity or a rotation about Q′.
Now apply axa−1 to the point Q. Let Q″ = TBA(Q). Then Q is the midpoint of Q′ Q″. Let X = RQ, θ.(Q″) and let Y = TAB(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 = RQ′, 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 ARmA−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 ARmA−1 fixes every point on the line A(m), and since ARmA−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 ATCD A−1 must be a translation. The isometry A TCD A−1must be a direct isometry, and since the only direct isometries without fixed points are translations, it suffices to show that A TCD 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 = TCD (Y), that is, Y is a fixed point of TCD. However, this contradicts the fact that a translation has no fixed points. Thus, we must conclude that ATCDA−1 has no fixed points, and therefore it must be a translation.
To pin down the translation TEF, let us consider the effect of ATCDA−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 Rl RQ, θ Rl = RQ, −θ.
We leave the proof as an exercise.
If a is an element of a group and m is a positive integer, then am, 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 ≠ am bm, unless the group is commutative.
The order of an element a of a group is the smallest positive integer n such that an = e. If there is no positive integer n such that an = 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 RQ, 90° is 4, the order of RQ, 180° 2, the order of RQ, 270° is 4, and the order of TAB 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.
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.
Example 10.1.7. What is the inverse of bas 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 basb−1, so that
as claimed.
Example 10.1.8. Assuming that 0 < k < n, what is ak 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 RQ, 90° = and Rm, where m is a diagonal of the square. Letting a = RQ, 90° and b = Rm, we see that
The last equality follows by Theorem 10.1.6.
Consequently, the isometrics RQ, 90° and Rm 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.
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 TAB is in the group , then so are (TAB)2, (TAB)3, and so on. However, each of (TAB)2, (TAB)3, …, is a translation, and all of them are different.
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 RQ, 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 RQ, 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 RQ, θ is the rotation with the smallest positive angle of rotation.
This shows that all of the rotations in are multiples of RQ, θ and, conversely, since is a group, all multiples of RQ, θ must be in . Letting a denote RQ, θ, this means that consists of
for some integer n, where an = RQ, 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 RQ, 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 Rm 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 Rl; is any reflection that is in , then there is some integer k such that
Now, consider the product Rm Rl. Since both reflections belong to , it follows that the lines l and m both contain Q, and so Rm Rl is a rotation with center Q.
Since Rm Rl 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.