In the last post we introduced cyclic groups. The cyclic group C_{n} is generated by S, where S represents clockwise rotation through 360°/n (or 1/n of a revolution). It includes S^{2}, which is 2/n of a revolution, and S^{3}, which is 3/n of a revolution, and so on, all the way up to S^{n-1}. Notice that S^{n} = 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 S^{p} = S^{qn+r} = (S^{n})^{q} ∘ S^{r}. In other words, S^{p} consists of q complete revolutions composed with r/n of a revolution. It follows that S^{p} is the same as S^{r}. So, if we want to see what S^{p} does, we really only need to pay attention to the remainder of p ÷ n.

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

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

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

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 q^{th} 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.

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 〈S^{q}〉 = C_{m}. We’d like to answer the more general question now: If p is any whole number, then what does the group 〈S^{p}〉 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 = S^{3}. Then T is a 135°-rotation, or 3/8 of a revolution. Let’s start composing T with itself. To begin with, T^{2} is 6/8 of a revolution, or S^{6}. Then T^{3} is 9/8 of a revolution, which is the same as 1/8 of a revolution. So T^{3} = S. Next, T^{4} is 12/8 of a revolution, which is the same as 4/8 of a revolution, or S^{4}. Continuing like this, we find that the group generated by T consists of I, S^{3}, S^{6}, S, S^{4}, S^{7}, S^{2}, and S^{5}, in that order. So 〈T〉 = C_{8}.

Now let’s take p = 6. Write U = S^{6}. The group generated by U consists of S^{6}, S^{4}, S^{2}, and I. We don’t obtain the entire group in this case. In fact, since S^{6} is a 270°-rotation, S^{4} is a 180°-rotation, and S^{2} is a 90°-rotation, we see that 〈U〉 = C_{4}.

If we check each of the elements of C_{8}, what we’ll find is the following. The group can be generated by each of S, S^{3}, S^{5}, and S^{7}. The transformations S^{2} and S^{6} only generate C_{4}. The transformation S^{4} generates C_{2}. And the transformation I generates the trivial group C_{1}.

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 S^{p} is the same as the group generated by S^{d} where d is the greatest common divisor of n and p. So, writing n = m ⋅ d, we find that 〈S^{p}〉 = C_{m}. 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 S^{p} generates the whole cyclic group.

Consider the case n = 12. Then S, S^{5}, S^{7}, and S^{11} each generate C_{12} since 12 shares no common divisors larger than 1 with 1, 5, 7, or 11. The transformations S^{2} and S^{10} each generate C_{6} since the greatest common divisor of 12 and 2, or 12 and 10, is 2. The transformations S^{3} and S^{9} each generate C_{4} since the greatest common divisor of 12 and 3, or 12 and 9, is 3. The transformations S^{4} and S^{8} generate C_{3}. The transformation S^{6} generates C_{2}. And I generates C_{1}.

Or again, consider n = 15. Then S, S^{2}, S^{4}, S^{7}, S^{8}, S^{11}, S^{13}, and S^{14} each generate C_{15}. Next, S^{3}, S^{6}, S^{9}, and S^{12} each generate C_{5}. The transformations S^{5} and S^{10} generate C_{3}. And I generates C_{1}.

Pick a single vertex of the n-sided polygon. Imagine repeatedly applying a rotation S^{p} to it (where p is less than n) and connecting each pair of consecutive points with a line segment. This amounts to connecting every p^{th} 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.

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).

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).

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).

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.