# Is this a group on the circle?

Is there a way the set of real points on any circle be made into a group? I was trying to form a group from the real points of a circle as below:

• The identity element will be the point $(0,1)$.

• To find the product of any two points $A$ and $B$, draw a straight line through them (so we have the line $\overline{AB}$), then draw a line parallel to $\overline{AB}$ that passes through $(0,1)$, the identity point. The point where it intersects the circle is the product of $A$ and $B$.

• The inverse of any point is the point directly opposite that point horizontally. Example the inverse of the point $(1,0)$ is the point $(-1,0)$. With this setting, we see the inverse of the identity $(0,1)$ is itself. Also, the inverse of the $(0,-1)$ is itself.

• Not sure of your construction, but the answer to your question is yes: t's the group of planar rotations. (rotate the vector $(0,1)$, or whatever, to the chosen vector.
– lulu
Jul 6 '17 at 12:33
• The product of an element and its inverse isn't the identity in your construction.. Jul 6 '17 at 12:38
• The product of an element and its inverse is the identity because the line connecting them is parallel to (0,1) if we made a line that passes through it. and since no other point in the circle that intersects the line that passes through (0,1) but (0,1) itself! Jul 6 '17 at 12:42
• In this construction, the product of A and B may have two possible values Jul 6 '17 at 13:51
• Is your product associative? That requires proof. What is the square of an element? What, for example, is $(1,0)\times (1,0)$? Is it $(0,-1)$? That seems the most natural. But then...Well, look at $A=(1,0),B=(-1,0)$. Clearly $A(AB)=Ae=A$, yes? But $(AA)B=(0,-1)\times (-1,0)$ which does not appear to be $A$. Have I blundered? If not, that would be a failure of associativity.
– lulu
Jul 6 '17 at 13:52

This is indeed a group!

Moreover, this same group structure can be put onto any conic -- not just a circle. The reason why this is a group and why this works with conics is due to where the proof for associativity comes from: Pascal's theorem. As you have already discovered, the rest of the group axioms come more-or-less for free from the definition of the operation.

You may be interested in two papers on the subject:

As you will see in Lemmermeyers' paper, you can endow elliptic curves with a similar structure. You can read more about the elliptic curve group here or here. The group is also mentioned in some algebraic geometry texts as well.

Edit: N. J. Wildberger mentions this group in a lecture of his on Algebraic Topology and talks some about the proof of associativity.

• The group can't be cyclic because it is uncountable. The group is just too large to be generated by a single element. Jul 8 '17 at 13:39
• but if we take proper subgroups, we can always have a cyclic subgroup. Jul 9 '17 at 1:07
• If you consider the subgroup of rational points on the circle, this is not a cyclic group. Jul 9 '17 at 4:07
• Not sure about that because I can always construct a cyclic group from an arbitrary starting point close to the identity point. Jul 9 '17 at 5:11
• You can create many cyclic subgroups, but not all subgroups are cyclic. That said, all finite subgroups are cyclic! Jul 9 '17 at 5:19

This will give a group and it will be the usual circle group. This is the group on the set $\{e^{i\phi}\mid \phi\in[0,2\pi)\}$ with complex multiplication as group operation. Alternatively it can be written as $\Bbb R/2\pi\Bbb Z$ as states in Evargalo's answer.

Proof.

Lets say the two points of the circle are given as $e^{i\phi}$ and $e^{i\psi}$. The corresponding line through the $1$ according to your description is given by

$$g(t)=1+t(e^{i\phi}-e^{i\psi}).$$

We are looking for the (if possible) non-zero value of $t$ for which $\|g(t)\|=1$. We can agree that there is at most one from the geometric intuition on this problem. So let me show you that

$$t^*=\frac{e^{i(\phi+\psi)}-1}{e^{i\phi}-e^{i\psi}}$$

is this value. This will give you $g(t^*)=e^{i(\phi+\psi)}=e^{i\phi}e^{i\psi}$. All we need to show is that $t^*$ is real. For this, multiply nominator and denominator with $e^{-1/2i\phi}e^{-1/2i\psi}$ to obtain

$$t^*=\frac {e^{1/2i\phi}e^{1/2i\psi}-e^{-1/2i\phi}e^{-1/2i\psi}} {e^{1/2i\phi}e^{-1/2i\psi}-e^{-1/2i\phi}e^{1/2i\psi}}.$$

Setting $z=e^{1/2i\phi}e^{1/2i\psi}$ and $w=e^{1/2i\phi}e^{-1/2i\psi}$ with their complex conjugates $\bar z$ and $\bar w$, we see that

$$t^*=\frac{z-\bar z}{w-\bar w} =\frac{2i\cdot\mathrm{Im}(z)}{2i\cdot\mathrm{Im}(w)} =\frac{\mathrm{Im}(z)}{\mathrm{Im}(w)}$$

which is real. $\;\square$

To be completely rigorous you will have to argue seperately for the case $t^*=\mathrm{Im}(z)=0$ and $\mathrm{Im}(w)=0$, but I hope this will do it for now.

I wrote a question motivated on this one, and you can see from the answer that no matter what you try, as soon as you develop a (topological) group structure on $S^1$, it will be the usual circle group structure.

• Thank you for the proof and the links! I had a suspicion the groups were isomorphic, but I couldn't see the map. Jul 7 '17 at 19:48

If you use the argument-modulus representation of the points of the circle (choosing the center of the circle as the origin and its radius as the unit), then:

$\mathcal{C}=\{e^{i\theta}, \theta\in[0,2\pi)\}$ is isomorphic to the additive group $\mathbb{R}/2\pi$