Polygons and Divisibility

In the last post we introduced cyclic groups. The cyclic group Cn is generated by S, where S represents clockwise rotation through 360°/n (or 1/n of a revolution). It includes S2, which is 2/n of a revolution, and S3, which is 3/n of a revolution, and so on, all the way up to Sn-1. Notice that Sn = I is a complete revolution, which gets us back where we started.

More generally, suppose p is a whole number. Divide p by n to obtain the quotient q and remainder r. We can write p ÷ n = q R r or p = q ⋅ n + r. So Sp = Sqn+r = (Sn)q ∘ Sr. In other words, Sp consists of q complete revolutions composed with r/n of a revolution. It follows that Sp is the same as Sr. So, if we want to see what Sp does, we really only need to pay attention to the remainder of p ÷ n.

Suppose n = 8 for example. The group C8 is the group of rotations of the regular octagon, generated by a 45°-rotation S.

octagon symmetry

What transformation is (say) S29? Well, 29 ÷ 8 = 3 R 5, so S29 amounts to 3 complete revolutions composed with 5/8 of a revolution, or rotation through 5 ⋅ 45° = 225°. In other words, S29 = S5.

Now, we’ve noted that, if m is a divisor of n, then Cm is contained within Cn. This can be seen in the following way. Write n = m ⋅ q. Then Sq represents 1/m of a revolution. For instance, if n = 8, then we can take m = 4. Writing 8 = 4 ⋅ 2, we see that S2 represents one quarter of a revolution, i.e., a rotation through 90°.

polygon inscribed 2

This observation amounts to the following fact: a regular polygon with m sides can be inscribed in a regular polygon with n sides if and only if m is a divisor of n. (We regard a line segment as a “polygon” with two sides.) To draw the polygon with m sides, connect every qth vertex, where n = m ⋅ q.

For instance, the divisors of 6 are 2 and 3, so we can inscribe an equilateral triangle and a bisecting line segment in a regular hexagon. The triangle is constructed by connecting every second vertex. Or again, the divisors of 12 are 2, 3, 4, and 6, so we can inscribe a hexagon, a square, an equilateral triangle, and a bisecting line segment in the regular dodecagon. The square is constructed by connecting every third vertex, since 12 = 4 ⋅ 3.

polygon inscribed

The divisors of 15 are 3 and 5, so we can inscribe a regular pentagon and an equilateral triangle in the regular pentadecagon. The pentagon is constructed by connecting every third vertex, since 15 = 5 ⋅ 3.

So, if m is a divisor of n with n = m ⋅ q, then 〈Sq〉 = Cm. We’d like to answer the more general question now: If p is any whole number, then what does the group 〈Sp〉 amount to? In other words, if we keep composing p/n of a revolution with itself, what cyclic group do we obtain?

Let’s consider the example of n = 8 again. Take p = 3, and write T = S3. Then T is a 135°-rotation, or 3/8 of a revolution. Let’s start composing T with itself. To begin with, T2 is 6/8 of a revolution, or S6. Then T3 is 9/8 of a revolution, which is the same as 1/8 of a revolution. So T3 = S. Next, T4 is 12/8 of a revolution, which is the same as 4/8 of a revolution, or S4. Continuing like this, we find that the group generated by T consists of I, S3, S6, S, S4, S7, S2, and S5, in that order. So 〈T〉 = C8.

Now let’s take p = 6. Write U = S6. The group generated by U consists of S6, S4, S2, and I. We don’t obtain the entire group in this case. In fact, since S6 is a 270°-rotation, S4 is a 180°-rotation, and S2 is a 90°-rotation, we see that 〈U〉 = C4.

If we check each of the elements of C8, what we’ll find is the following. The group can be generated by each of S, S3, S5, and S7. The transformations S2 and S6 only generate C4. The transformation S4 generates C2. And the transformation I generates the trivial group C1.

Now, if p and n are whole numbers not both zero, then their greatest common divisor d is the largest whole number that divides both p and n. For instance, the greatest common divisor of 6 and 8 is 2, while the greatest common divisor of 12 and 30 is 6. In general, the group generated by Sp is the same as the group generated by Sd where d is the greatest common divisor of n and p. So, writing n = m ⋅ d, we find that 〈Sp〉 = Cm. If p and n share no common divisors larger than 1, then they are said to be relatively prime; in this case, m = n, and Sp generates the whole cyclic group.

Consider the case n = 12. Then S, S5, S7, and S11 each generate C12 since 12 shares no common divisors larger than 1 with 1, 5, 7, or 11. The transformations S2 and S10 each generate C6 since the greatest common divisor of 12 and 2, or 12 and 10, is 2. The transformations S3 and S9 each generate C4 since the greatest common divisor of 12 and 3, or 12 and 9, is 3. The transformations S4 and S8 generate C3. The transformation S6 generates C2. And I generates C1.

Or again, consider n = 15. Then S, S2, S4, S7, S8, S11, S13, and S14 each generate C15. Next, S3, S6, S9, and S12 each generate C5. The transformations S5 and S10 generate C3. And I generates C1.

Pick a single vertex of the n-sided polygon. Imagine repeatedly applying a rotation Sp to it (where p is less than n) and connecting each pair of consecutive points with a line segment. This amounts to connecting every pth vertex. If p happens to be 1 or n – 1, then we obtain the polygon itself. But if p is between 1 and n – 1, then we obtain either an inscribed polygon or a star. Also, the figure produced by p is the same as that produced by n – p; the direction is just reversed. It should also be clear that an inscribed star has n points if and only if p and n are relatively prime.

Everyone knows that there’s one 5-pointed star; this corresponds to p = 2 (or p = 3), because we draw it by connecting every second (or third) vertex.

star 5

Next, there are no 6-pointed stars because each of 2, 3, and 4 share common divisors with 6. But there are two 7-pointed stars, corresponding to p = 2 (or p = 5) and p = 3 (or p = 4).

star 7

We draw the first by connecting every second vertex, and the second by connecting every third vertex. There is only one 8-pointed star, corresponding to p = 3 (or p = 5).

star 8

It’s drawn by connecting every third vertex. If we connect every second vertex, we wind up with a square; if we connect every fourth vertex, we obtain a bisecting line segment. Next, there are two 9-pointed stars, corresponding to p = 2 (or p = 7) and p = 4 (or p = 5).

star 9

In general, if n happens to be a prime number, hence has no divisors other than 1 and itself, then there are (n – 3)/2 different n-pointed stars. For instance, there are (11 – 3)/2 = 4 different 11-pointed stars.

star 11


Rotational Ornaments

An isometry is a transformation of the plane that takes each line segment to a congruent line segment and each angle to a congruent angle. Basic types of isometries include reflections, rotations, and translations.

In the last last post, we saw that a symmetry of a geometrical figure is an isometry that preserves the figure. A group is a collection of symmetries satisfying: (1) if a symmetry T is in the group, then its inverse is also in the group; and (2) if T and U are in the group, then the composition U ∘ T is also in the group.

We discussed the dihedral groups D3 of the equilateral triangle and D4 of the square. Now consider the set of symmetries that preserves the regular pentagon.

pentagon symmetry 1

In this case we have five reflections: R1, R2, R3, R4, and R5. In addition, we have clockwise rotation S through one fifth of a complete revolution, or 72°, and its compositions with itself: S2, S3, S4. So the dihedral group D5 consists of ten symmetries: I, S, S2, S3, S4, R1, R2, R3, R4, and R5. Once again, we can write D5 = 〈R1,R2〉, where R1 and R2 are reflections across adjacent lines making an angle of 36° with one another. For instance, R1 ∘ R2 is a 72°-rotation.

Now imagine extending the sides of the pentagon as shown.

pentagon symmetry 2

This “breaks” the symmetry. No longer can we reflect. However, the rotations still preserve the figure. So the symmetry group now consists of five symmetries: I, S, S2, S3, and S4. This is referred to as the cyclic group C5. It’s generated by S, so we can write C5 = 〈S〉. The cyclic group is a subgroup of the dihedral group.

In general, the cyclic group Cn is generated by a rotation through 360°/n, or one nth of a revolution, while the dihedral group Dn is generated by two reflections whose lines make an angle of 360°/2n. Cyclic and dihedral groups abound in ornamental symmetry; Owen Jones’ 1910 Grammar of Ornament contains a wealth of examples.

The group D1 contains a single reflection; it may be thought of as the set of symmetries of the letter V. This is bilateral symmetry, as exhibited in the following Egyptian ornament.


The group D2 contains two reflections across perpendicular lines; it may be thought of as the set of symmetries of the letter H. We see examples of it in the following Greek and Pompeian ornaments.


Notice that if we ignore the color in the Pompeian ornament, then we have a D4 symmetry instead. A true example of D4 symmetry is given in the following Persian ornament.


Larger dihedral groups are rarer but still occur. The following ornaments are, respectively, Assyrian, Chinese, and Byzantine; they have symmetry groups D5, D8, and D22.

Dihedral other

Ornaments that exhibit only a cyclic symmetry generally have some kind of overlapping knot pattern or swirl element which prevents reflection. The cyclic group C1 is the trivial group consisting of just the identity.


The cyclic group C2 contains a single rotation through 180°. This may be thought of as the set of symmetries of the letter N. The following Celtic ornament has C2 symmetry; the overlapping of the knot prevents its having D2 symmetry.


The cyclic group C4 is generated by a 90°-rotation. Examples include the following medieval European and Moorish ornaments.


The group C6 is generated by a 60°-rotation, while C8 is generated by a 45°-rotation; these are exhibited in the following medieval European and Arabian ornaments.

Cyclic other

In the latter case, we have to ignore the box design around the circular ornament. If we don’t, then the symmetry group is C4 rather than C8. In general, Cn contains Cm as a subgroup if and only if m is a divisor of n. So C8 contains C1, C2, and C4. Similarly, C6 contains C1, C2, and C3. We’ll return to this in the next post on the topic.

Symmetry Groups

Think of a spacial arrangement of geometrical elements. It could be a polygon, or a honeycomb lattice, or some other arrangement. A symmetry is a rigid transformation that preserves the arrangement.  Consider an equilateral triangle, for instance:

triangle symmetry 1

There are six “moves” that would preserve the triangle. To begin with, we could reflect it across any of the three axes of symmetry.

triangle symmetry 2

In addition, we could spin it through 120° around the center, clockwise or counterclockwise. And, finally, we could just leave it be.

Symmetries have the following property: if we perform two symmetries in succession, then the composition obtained is also a symmetry. For instance, if we reflect the triangle across two lines of symmetry, then this amounts to a 120° rotation. In general, if T and U are two symmetries, then we’ll use the notation U ∘ T to represent the symmetry obtained by performing first T and then U. We can think of this as a “multiplication” of symmetries. Be careful, though: the order matters. The symmetry U ∘ T is not necessarily the same as T ∘ U.

To make all this clearer, let’s consider another example. A square has four lines of symmetry:

square symmetry

Reflection across any of these lines preserves the shape. Label the four reflections as R1, R2, R3, and R4. We can also rotate the square clockwise around the center through 90°, or 180°, or 270°. If S is the symmetry that rotates the square through 90°, then S ∘ S = S2 is rotation through 180°, and S ∘ S ∘ S = S3 is rotation through 270°. And, finally, there’s the identity symmetry, which just leaves the square as it is. We’ll call that I. Notice that S4 = I, since S4 amounts to a 360° rotation.

So we have eight symmetries in all: I, S, S2, S3, R1, R2, R3, and R4. As mentioned above, we can view composition of symmetries as a kind of “multiplication” on this set. However, it isn’t commutative: the order matters. To see this, consider S and R1. Let’s see what S ∘ R1 does to the square:

square symmetry 2

This amounts to the transformation R4. Now let’s see what R1 ∘ S does:

square symmetry 3

This amounts to R2. So S ∘ R1 = R4 while R1 ∘ S = R2. It follows that S ∘ R1 ≠ R1 ∘ S.

What is true is that the composition of any two symmetries in our set is also in the set. The set of symmetries is “closed” under composition. In addition, each symmetry has its inverse, which carries the shape back to the original position. For instance, R1 is its own inverse, since R1 ∘ R1 = I. The same is true of all the reflections. On the other hand, the inverse of S is S3, since S ∘ S3 is a 360° rotation, which amounts to I. And, finally, the inverse of S2 is itself, since two 180° rotations amount to a 360° rotation. If we wanted to, we could make an 8 × 8 multiplication table for this set of symmetries. Each row and each column would contain each symmetry exactly once.

Another interesting thing to note is that we could use a pair of reflections to obtain the other symmetries. Take R1 and R2, for instance. Notice that S = R1 ∘ R2. We can compose S with itself to obtain the other rotations: S2 = R1 ∘ R2 ∘ R1 ∘ R2 and S3 = R1 ∘ R2 ∘ R1 ∘ R2 ∘ R1 ∘ R2. Finally, R3 = R2 ∘ R1 ∘ R2, while R4 = R1 ∘ R2 ∘ R1. So we’ve written every symmetry as a product of R1 and R2.

A group is a set of symmetry operations satisfying the following two requirements: (1) the composition of two operations in the group is also in the group, and (2) for each operation in the group, the inverse of the operation is also in the group. The group we’ve been discussing is the dihedral group D4. It has order 8, meaning that it contains 8 operations: I, S, S2, S3, R1, R2, R3, and R4. We say that D4 is generated by R1 and R2, and write D4 = 〈R1,R2〉.

In coming posts, we’ll see how the theory of groups can be used to classify ornamental symmetries.

Eadem Mutata Resurgo

A golden rectangle is a rectangle ABCD whose dimensions are such that, if a square ABPQ is inscribed as shown, then the smaller rectangle PCDQ cut off is similar to the original.

golden rectangle 1

Because the sides of the smaller rectangle are in the same ratio as those of the original, the process may be repeated, thereby cutting off yet a smaller golden rectangle. This could be repeated ad infinitum, thus:

golden rectangle 2

It’s easy to find the ratio of the sides. If we take x = BC and y = AB, then the ratio of x to y is the same as the ratio of y to x – y. So x/y = y/(x – y). Writing x = ϕy, we have ϕ = 1/(ϕ – 1), or ϕ2 – ϕ – 1 = 0. Solving for ϕ yields (1 + √5)/2, or about 1.618. We call ϕ the golden ratio.

Now, if we take our sequence of squares and construct a quarter-circle in each square, then we obtain a spiral shape.

golden rectangle 4

It’s often claimed that this models the shell of the chambered nautilus and other objects found in nature.


This isn’t really the case, however. Part of the reason is that the “spiral” is not a true spiral. It’s pieced together from a sequence of quarter-circles, each of which has a different radius. In other words, the radius of curvature jumps discontinuously as we move around the spiral. The shell of a nautilus, on the other hand, is a logarithmic or equiangular spiral. The radius varies continuously and by a geometric progression as we turn around the spiral.

We can still use the golden rectangle to construct a logarithmic spiral. If we draw two successive diagonals as shown

golden rectangle 3

then they intersect at the point that the sequence of squares converges upon. Call this point O. As we look closer and closer at O, we have an infinite sequence of rectangles, each of which is similar to the original. For instance, if we were to expand the picture to make SU as long as BC is now, then it would look exactly the same, complete with the infinite sequence of squares spiraling in to O. The picture is self-similar. Each of the four line segments emanating from O touches infinitely many corners in the sequence of golden rectangles. For instance, the segment from O to B touches B, S, and so on—the points are too close together to label—and each occupies the same corner in its respective golden rectangle.

The point O is the fixed center of our true spiral. The segment lengths OB, OC are in a golden ratio, as are OC, OD, and OD, OQ, and so on. Furthermore, these successive pairs all meet at right angles.

Imagine tracing out the spiral starting at A. The point O is the center of the spiral. As we turn, the spiral draws closer and closer to O, in such a way that its distance decreases by a factor of ϕ with each quarter-turn. Thus, if the distance from O to A is 1, then, as we turn clockwise from OA to OP, the distance of the spiral from O should decrease to ϕ-1. As we turn from OP to OR, the distance should decrease to ϕ-2. And so on. Conversely, if we turn in the counterclockwise direction, then the distance of the spiral from O increases by a factor of ϕ with each quarter-turn.

The golden spiral thus constructed is closely approximated by the artificial spiral we constructed above. The latter is sometimes incorrectly called the golden spiral, but is actually known as the Fibonacci spiral. The golden spiral is but one example of a logarithmic spiral; it is based on the factor ϕ, but any other factor s > 1 could be used instead. In terms of polar coordinates, this would be parametrized by r(θ) = α ⋅ s2θ/π, where α > 0 is an arbitrary scaling factor.

The logarithmic spiral was studied by the Swiss mathematician Jacob Bernoulli (1654 – 1705), who called it the Spira miribilis—the “wonderful spiral”—because of its property of self-similarity. He was so taken with this property that he requested a logarithmic spiral to be incised on his epitaph, together with the motto EADEM MUTATA RESURGO (“though changed, I remain the same”). The craftsmen misunderstood and put a simple Archimedean spiral instead.

eadem mutata resurgo