Thus far in the symmetric series we have found all the symmetries of regular polygons in Symmetry of regular polygons and shown that, together with composition operation, the set of symmetries of a triangle forms a group in Symmetric group of an equilateral triangle. Today, we will continue to work by looking at the set of symmetries on any polygon together with the composition operation. We will show this will be a group for any polygon and that this group will look like a cyclic group or one of the dihedral groups. As I alluded to in Symmetry, this result is in fact due to Leonardo da Vinci.
Symmetries form a group
When we constructed the symmetry group for the equilateral triangle we saw the under compositions our symmetries were closed under composition, associative, had an identity, and had inverses. In order to show that the set of symmetries of a polygon is a group, we make the following observations.
When composing symmetries, we note that if we start with a polygon and perform a symmetry, the polygon will end up in the same position (though the orientation of points may have changed). Since the second symmetry will take something in this position and place it back in the same position, we see that the composition of symmetries is again a symmetry.
If we consider symmetries as mappings, then we note that the composition of mapping is always associative, so composition satisfies associativity.
We next note that not moving the polygon is always a symmetry, so the identity is in the set of symmetries for any polygon.
Furthermore, if we have a symmetry, then doing exactly the opposite of what we did will also move the polygon from its position back to the same place.
Therefore the set of symmetries of any polygon does indeed form a group. As a final note, the symmetry of a polygon must send the vertices to other vertices and edges to other edges. Because polygons have a finite number of edges, this also implies that the set of symmetries of a polygon is also finite.
In the previous post I had mentioned that we would have symmetries that consisting of rotating our regular polygons and reflecting the polygons over some line. We will therefore be looking at what types of reflections and rotations will work for different polygons. Often we will do this by discussing which of these would not lead be a symmetry of a polygon. Because of this, we look at the mapping of a shift (or translation).
In order to shift a polygon, you would take your polygon and shift it in one direction. For example, take a square and move it to the right an inch. You can do this in any direction and you will maintain the shape of the polygon and the distance between any points within. As a new question, is there any distance you can move the square so that it lines up with its initial position?
If you imagine of infinite plane and moving the square in the same direction, there is no distance you can move it that would cause it to end up in its original position (aside from not moving it at all). This is because the plane will go off in every direction infinitely far without repeating. Therefore, any shift (other than not shifting) cannot be a symmetry of a polygon.
We note here that if you said you could move the square the entire circumference of the Earth and end up where you started, this is an amazing insight. However, since the Earth is curved, it does not behave like the Euclidean plane, so we don’t have this as a possibility with Euclidean geometry. Instead, the surface of the Earth would form a spherical geometry where there are no parallel lines. If you’d like to learn more about situations like this, I would refer to you to a Modern Geometry book. I prefer Modern Geometries; however, since it is out of print if you can’t get a used copy, I would look at College Geometry.
We will now consider what rotations are possible. To do this, we will have to find the composition of rotations. Therefore, suppose that we are looking at the plane. If we have a rotation around the point A by the angle α, R(A,α), we will move all points so that they stay the same distance from A and are rotated the appropriate angle. In the figure below, D’ would be the image of D under R(A,α).
If we now imaging doing the rotation R(A,β), we would then continue to rotate D’ at the same distance by an angle of β. This would result in moving D a total angle of α+β, that is R(A,β)oR(A,α)=R(A,α+β). Therefore, if the set of symmetries includes any rotation, R(A,α), it must also include the rotations about A by all multiples of α. If there is some other rotation, R(A,β), the set must also include R(A,nα+mβ) since we have shown the set is closed under composition. Recall that we must have a finite set of symmetries of a polygon, therefore such a set of rotations must be finite. This would mean that the rotations listed are not unique for every n and m, that is, there is repetition of rotations as defined above. This occurs because we consider rotating by any multiple of 360 the same as doing nothing.
Now, since we have a finite number of rotations, we must have a rotation with smallest angle. If we call this angle γ, we note that every other angle of rotation must be a multiple of γ. This is the case because if we had some angle that was not a multiple of γ, say δ, then we could show that there exists an angle of rotation smaller the γ, which would be a contradiction. Therefore, we have that the set of rotations about the point A, viewed as a subset of symmetries would look like R(A,nγ) for some γ. That is, the set would be generated by the rotation R(A,γ).
Next, suppose that we had two rotations about different points, that is R(A,α) and R(E,β) were symmetries of a polygon where A and B where not equal. Then, we must have the composition of these two within our set. To find this composition, we refer to the following diagram. Here we have that α is angle CAB and β is angle DEF (note the direction matters). We have drawn these so that AE is the angle bisector of both of these angles (if this was not the case you can move C,B and D,F to make it so). We then note that if we perform R(A,α) the point O will go to the point of intersection of lines AB and ED, call this O’. If we then perform R(E,β), this point will then be sent back to O. Hence O is a fixed point.
If we look at the image of A under this composition, we note that the first rotation does not move A, so we end up with the image of A being A’.
If we follow the through corresponding angles and sides, we find that AO=A’O and angle AOA’ is α+β. That is, R(E,β)oR(A,α)=R(O,α+β). Furthermore, R(A,-α), R(E,-β), and R(E,-β)oR(A,-α) are also in the set of symmetries. We note that, by the same argument, R(E,-β)oR(A,-α)=R(O’,-(α+β)). Therefore, R(O,α+β)oR(O’,-(α+β)) is in the set of symmetries.
At this point, we should note that our picture above doesn’t work in every situation. In particular, we assumed that lines AC and EF intersected and AB and ED intersected. As long as the sum of the corresponding angles are not a multiple of 360, this will indeed be the case. We shall look at an example of this case. Suppose that α=β=180. Then, if we look at the image of A, it will move nowhere under R(A,α). Under R(E,β) it will then rotate around E 180 degrees and end up in the direction of line AE with a distance of 2(AE) from its original position. If we then check other points, we will note that these as well will move in the direction of AE a distance of 2(AE). Therefore, we will have a shift. Now, note that for R(O,α+β)oR(O’,-(α+β)), the sum of angles is 0, so we have shown that there is a shift in our set of symmetries. However, we already know this is not possible, so we cannot have rotations about two distinct centers within the symmetric group of any polygon.
We therefore have that the set of rotations will form a group generated by the element R(A,α). A group generated by a single element is called a cyclic group, therefore, the set of rotational symmetries of a polygon forms a cyclic group.
In addition to rotations, we can also have reflections over a line be symmetries of a polygon. As was the case with rotations, we will look at what we can determine if know that certain rotations are symmetries of the polygon we are looking at. Therefore, we will need find the composition of reflections.
To begin with, let Rl be the reflection over the line l. Note that RloRl would be the process of flipping over the same line twice. The second flip will undo the first flip, so we get that the composition of a flip with itself is the identity. It is therefore possible to have just one reflection in the set of symmetries (this is the case with an isosceles triangle).
If we let l and m be lines such that l and m are parallel. If we will then have the following situation.
Here, we note that A moved on the line perpendicular to both l and m by a distance of 2 times the perpendicular distance between l and m. Since this will be the case for every point, we have that RloRm is a shift, so we cannot have both be symmetries of the same polygon.
On the other, hand, suppose that l and m intersect. Then we have the following figure.
Note that the image of A under Rl is A’ and the image of A’ under Rm is A”. If we again follow through the process, we will note that A was rotated about the intersection of the lines l and m by and angle which is double the angle from l to m. Therefore, if we have two reflections over intersecting lines, we must also have the rotation about the intersection of the lines by twice the angle between the lines.
Suppose now that we had 3 reflections. If any two were parallel, we would have a shift, so this cannot be the case. If the line met at three distinct points, then the composition of any two would form a rotation about the point of intersection. Choosing a different pair than the first, we would have rotations about two separate centers, which we have already shown cannot be the case. Hence, if we have reflections of more than 2 distinct lines, they must all be concurrent at a point which would also be the center of any rotations in the set of symmetries.
Rotations and Reflections
Suppose that R(A,α) and Rl are symmetries of the polygon we have. We then note that RloR(A,α)=RloRloRm=Rm where m is the line that intersects l at A and the angle from m to l is α/2. Similarly, R(A,α)oRl=Rn where is the line that intersects l at A and the angle from l to n is α/2. Therefore, the composition of a rotation and a reflection is again a reflection, so we have not added any new symmetries.
Now Let R be the set of all rotations and F the set of all reflections. Note that the set of reflections may well be empty. This would be the case if there was some distinction between the front and the back of the polygon (such as if your sheet of paper was one color in the front and a different color in the back) or if you have a triangle with 3 distinct side lengths. If this is the case, the set of symmetries would just be the set of rotations and would therefore be a cyclic group as we saw before.
On the other hand, suppose that F is nonempty. Then there exists a reflection Rl. Since the identity mapping is a rotation by an angle of 0, the set of rotational symmetries must also be nonempty. Now, let g be the mapping from R to F defined as R(A,α) is mapped to RloR(A,α). We note that the image under this mapping is the reflection defined above and this reflection must be a symmetry of the polygon. Hence g maps R to F. If we then define h from F to R by RloRm, we get that h will send reflections to rotations as defined in our reflection section that must be symmetries of the polygon. Now, since g and h are inverse well-defined mappings, and R and F are finite, we must have that the number of elements in each is the same. That is, the number of reflections is the same as the number of rotations.
We then define the dihedral groups as D2 is the set consisting of the identity mapping and one reflection over any line, D4 is the set consisting of a R(O,0), R(O,180),Rl, Rm, where O is the origin and l is x-axis and m is the line through the origin at an angle of 90 above the x-axis, and D2n is the set of rotations about the origin by angles of 360/n and reflections of the lines through the origin at angles above the x-axis of (360/2n)*m for m between 0 and n/2.
We have seen in Symmetry of regular polygons that the elements of the dihedral group, D2n are precisely the symmetries of the regular polygon with n sides. According to the mapping we gave above, if we treat the point of intersection of an lines of reflection, which is also the center of any rotation as the origin, then the symmetries of any polygon look precisely like one of the dihedral group for some n. That is, the group of symmetries of any polygon is either equivalent to a cyclic group consisting of rotations, or a dihedral group, that is the group of symmetries of a regular polygon.
We have now seen if we are looking for symmetries of polygons, we need only look at the corresponding symmetries for regular polygons. Therefore, we need only find the rotational symmetries and if we have any reflections as symmetries. The set will then be the same as the set of symmetries for some regular polygon (with a special case defined to cover for regular polygons of size 1 or 2).
Here, I tried to provide a justification more than a proof that this was actually true in order to keep the post from getting too technical (it is already rather in-depth). If you want to know more details of how I made certain jumps, let me know and I’ll be happy to send the details along. On the other hand, if you don’t want to follow all the details I have provided, I have italicized the conclusions of each section so that you can quickly refer to the results that we obtained.