# Making finite groups out of neat collections of functions.

This is a piece of folklore I found on the Internet:

$$\begin{array}{ll} \varepsilon = x & f = 1 - x \\ g = \frac{1}{x} & h = \frac{1}{1 - x} \\ m = \frac{x - 1}{x} & n = \frac{x}{x - 1} \end{array}$$

There are 6 functions, and it is claimed that they form the dihedral group $$D_3$$ under composition. I computed some part of the corresponding (hypothetical yet) Cayley table, and it does look like a group. However, the derivations, trivial as they may be, take a lot of paper and attention to actually perform, and a slight mistake will lead one to believe that this is not a group. What kind of tricks can I use to find out if this is a group, short of carefully computing the whole table?

1. It appears that all the functions here are distinct, but technically I should provide for every pair of functions a value that differentiates them. For example, $$\varepsilon(2) = 2, h(2) = -1$$, and so on. This is really so boring. As a shortcut, I may compare limits and observe:

$$\begin{array}{ll} \varepsilon \to +\infty & f \to -\infty \\ g \to 0^+ & h \to 0^- \\ m \to 1^- & n \to 1^+ \end{array}$$

So far so good.

2. $$\varepsilon$$ is the identity.

3. $$g$$ is the multiplicative inverse, so in the group $$\langle\mathbb{R}, \times\rangle$$ $$g \circ g$$ would be $$(a^{-1})^{-1} = a$$, so $$g \circ g$$ = $$\varepsilon$$ and we have our first subgroup $$Z_2$$.

This is about as far as I could get, until I decided to explore the compositional structure of the given functions. Turns out that I can express them all in terms of these trivial functions:

$$\begin{array}{rl} \alpha & = & x + 1 \\ \beta & = & -x \\ \gamma & = & \frac{1}{x} \end{array}$$

$$\begin{array}{rl} & f = \alpha\beta \\ g = \gamma & h = \gamma\alpha\beta \\ m = \alpha\beta\gamma & n = \alpha\beta\gamma\alpha\beta \end{array}$$

Only a handful of simple theorems are enough to infer the whole Cayley table, giving a witness to group structure:

$$\beta\beta = \gamma\gamma = \varepsilon \tag{involution}$$ $$\beta\alpha\beta = \alpha^- \tag{inverse}$$

It took only a page to get there. (Identity omitted for clarity.)

$$\begin{array}{c|cccccc} & f & g & h & m & n \\ \hline f & & m & n & g & h \\ g & h & & f & n & m \\ h & g & n & m & & f \\ m & n & f & & h & g \\ n & m & h & g & f & \\ \end{array}$$

I wonder though if there is a yet simpler way. The given functions are so neat, they are obviously specially prepared. But I am not seeing how. For example, would it be difficult to offer a collection of 8 functions that form the group $$D_4$$ of the symmetries of the square under composition? Is there some generality I am not seeing?

To any nonzero point $$(x, y)$$ in $$\mathbb{R}^2$$, we can associate a (possibly infinite) slope $$s = \frac{y}{x}$$. Suppose that under a linear transformation $$T$$, the point $$(x, y)$$ is sent to the new point $$(ax + by, cx + dy)$$. Then the slope of this new point is equal to $$\frac{cx + dy}{ax + by} = \frac{ds + c}{bs + a},$$ so we see that linear transformations on $$\mathbb{R}^2$$ induce so-called "fractional linear transformations" on the slopes of points. The map $$\varphi$$ taking a linear transformation to its induced fractional linear transformation is a group homomorphism from the group of invertible linear transformations on $$\mathbb{R}^2$$ (aka GL$$_2(\mathbb{R})$$) to the group of invertible fractional linear transformations (for both groups the group multiplication is composition). You can straightforwardly compute that the kernel of $$\varphi$$ is the group of dilations of the plane, so if $$G$$ is a subgroup of GL$$_2(\mathbb{R})$$ that doesn't contain any dilations, then $$\varphi(G)$$ is a collection of fractional linear transformations that is isomorphic to $$G$$.
For your particular collection of fractional linear transformations, consider the three points $$(0,1)$$, $$(1,1)$$, and $$(1,0)$$ in the plane. You can compute that the subgroup $$G$$ of GL$$_2(\mathbb{R})$$ that stabilizes these points is isomorphic to $$D_3$$, that $$G$$ doesn't contain any dilations, and that $$\varphi(G)$$ is exactly the collection of fractional linear transformations that your post is about. To construct $$D_n$$ for any odd $$n$$ out of fractional linear transformations, you can instead start with the stabilizer in GL$$_2(\mathbb{R})$$ of the vertices of a regular $$n$$-gon, and apply the same procedure. This won't work to construct $$D_n$$ for even $$n$$, since one of the symmetries of an even $$n$$-gon is 180-degree rotation, which is the same thing as a dilation by a factor of -1. Perhaps there's a more clever geometric construction that one can do.
• Awesome answer! For the future reader: this group is called $\mathrm{PGL}_1(\mathbb{R})$ elsewhere. – Ignat Insarov Dec 24 '19 at 11:29