# Group of rational points on the unit circle

In mathematics, the rational points on the unit circle are those points (xy) such that both x and y are rational numbers ("fractions") and satisfy x2 + y2 = 1. The set of such points turns out to be closely related to primitive Pythagorean triples. Consider a primitive right triangle, that is, with integral side lengths a, b, c, with c the hypotenuse, such that the sides have no common factor larger than 1. Then on the unit circle there exists the rational point (a/cb/c), which, in the complex plane, is just a/c + ib/c, where i is the imaginary unit. Conversely, if (xy) is a rational point on the unit circle in the 1st quadrant of the coordinate system (i.e. x > 0, y > 0), then there exists a primitive right triangle with sides xcycc, with c being the least common multiple of x and y denominators. There is a correspondence between points (x,y) in the x-y plane and points x + iy in the complex plane which will be used below, with (ab) taken as equal to a + ib.

## Group operation

The set of rational points forms an infinite abelian group, which shall be called G in this article. The identity element is the point (1, 0) = 1 + i0 = 1. The group operation, or "product" is (xy) * (tu) = (xt − uyxu + yt). This product is angle addition since x = cosine(A) and y = sine(A), where A is the angle the radius vector (xy) makes with the radius vector (1,0), measured counter clockwise. So with (xy) and (tu) forming angles A and B, respectively, with (1, 0), their product (xt − uyxu + yt) is just the rational point on the unit circle with angle A + B. But we can do these group operations in a way that may be easier, with complex numbers: Write the point (xy) as x + iy and write (tu) as t + iu. Then the product above is just the ordinary multiplication (x + iy)(t + iu) = xt − yu + i(xu + yt), which corresponds to the (xt − uyxu + yt) above.

### Example

The points on the unit circle: 3/5 + i4/5 and 5/13 + i12/13 (corresponding to the two most famous Pythagorean right triangles:3,4,5 and 5,12,13) are elements of G, and their group product is (−33/65 + i56/65), which corresponds to a 33,56,65 Pythagorean right triangle. The sum of the squares of the numerators 33 and 56 is 1089 + 3136 = 4225, which is the square of the denominator 65.

### Other ways to describe the group

${\displaystyle G\cong {\mathrm {SO} }(2,{\mathbb {Q} }).}$

The set of all 2×2 rotation matrices with rational entries coincides with G.This follows from the fact that the circle group ${\displaystyle S^{1}}$ is isomorphic to ${\displaystyle {\mathrm {SO} }(2,{\mathbb {R} })}$, and the fact that their rational points coincide.

## Group structure

The structure of G is an infinite sum of cyclic groups. Let G2 denote the subgroup of G generated by the point 0 + 1i. G2 is a cyclic subgroup of order 4. For a prime p of form 4k + 1, let Gp denote the subgroup of elements with denominator pn, n a nonnegative integer. Gp is an infinite cyclic group. The point (a2 − b2)/p + (2ab/p)i is a generator of Gp. Furthermore, by factoring the denominators of an element of G, it can be shown that G is a direct sum of G2 and the Gp. That is:

${\displaystyle G\cong G_{2}\oplus \bigoplus _{p\equiv 1{\bmod {4}}}G_{p}.}$

Since it is a direct sum rather than direct product, only finitely many of the values in the Gps differ from zero.

### Example

Suppose we take the element in G corresponding to ({0};2,0,1,0,0,...0,...) where the first coordinate 0 is in C4 and the other coordinates give the powers of (a2 − b2)/p(r) + i2ab/p(r) where p(r) is the rth prime of form 4k + 1. Then this corresponds to, in G, the rational point (3/5 + i4/5)2 · (8/17 + i15/17)1 = −416/425 + i87/425). The denominator 425 is the product of the denominator 5 twice, and the denominator 17 once, and as in the previous example, the square of the numerator −416 plus the square of the numerator 87 is equal to the square of the denominator 425. It should also be noted, as a connection to help retain understanding, that the denominator 5 = p(1) is the 1st prime of form 4k + 1, and the denominator 17 = p(3) is the 3rd prime of form 4k + 1.

## The unit hyperbola's group of rational points

There is a close connection between this group on the unit hyperbola and the group discussed above. If ${\displaystyle {\frac {a+ib}{c}}}$ is a rational point on the unit circle, where a/c and b/c are reduced fractions, then (c/ab/a) is a rational point on the unit hyperbola, since ${\displaystyle (c/a)^{2}-(b/a)^{2}=1,}$ satisfying the equation for the unit hyperbola. The group operation here is ${\displaystyle (x,y)\times (u,v)=(xu+yv,xv+yu),}$ and the group identity is the same point (1,0) as above. In this group there is a close connection with hyperbolic cosine and hyperbolic sine, which parallels the connection with cosine and sine in the unit circle group above.

### Copies inside a larger group

There are isomorphic copies of both groups, as subgroups,(and as geometric objects) of the group of the rational points on the abelian variety in four-dimensional space given by ${\displaystyle w^{2}+x^{2}-y^{2}+z^{2}=0.}$ Note that this variety is the set of points with Minkowski metric, relative to the origin, equal to 0. The identity in this larger group is (1,0,1,0), and the group operation is ${\displaystyle (a,b,c,d)\times (w,x,y,z)=(aw-bx,ax+bw,cy+dz,cz+dy).}$
For the group on the unit circle, the appropriate subgroup is points of form (w,x,1,0), with ${\displaystyle w^{2}+x^{2}=1,}$ and its identity element is (1,0,1,0). The unit hyperbola group corresponds to points of form (1,0,y,z), with ${\displaystyle y^{2}-z^{2}=1,}$ and the identity is again (1,0,1,0). (Of course, since they are subgroups of the larger group, they both must have the same identity element.)