i'm trying to understand how to build a group H that contains complex functions and its operation is function composition.
I really don´t undestand how to build that group for a hexagon in the complex plane, however I found the matrix 2 by 2 representation in $\mathbb{R}^2$ (there are 12 matrixes and I found them).
So, the question is:
Determine the group of symmetries of the hexagon as a group H of complex functions $\in \mathbb{C}$
Write out the multiplication table of H
I need 12 functions ( I think ) because I need to make a Cayley table and then compare with matrixes Caley table.
I hope you can help me, thanks.
It is the dihedral group with 12 elements. Its group table can be found [here].
Let $\omega:=e^{i\pi/3}$, a primive $6$-th root of unity.
The group of isometries of hexagon
$$1,\omega, \ \omega^2, \ \omega^3, \ \omega^4, \ \omega^5 $$
can be generated by the two complex functions
$$f_1(z)=\omega z \ \ \text{and} \ \ f_2(z)=\bar z$$
(conjugate of $z$)
by combining $f_2$ (rotation by $\pi/3$) and $f_3$ (symmetry with respect to $x$-axis), you generate all transformations keeping the hexagon invariant.
Edit :
Here are some of the constituents of this group, where $c=\cos(\pi/3)$ and $s=\sin(\pi/3)$ :
$$\begin{matrix}f_1(z)=z & \leftrightarrow & \binom{1 \ 0}{0 \ 1} \\ f_2(z)=\omega z & \leftrightarrow & \binom{c \ -s}{s \ \ \ \ c} \\ f_3(z)=\bar z & \leftrightarrow & \binom{0 \ 1}{1 \ 0} \\ f_4(z)=f_2(f_3(z))=\omega \bar z & \leftrightarrow & \ \ \ \ \ \ \ \ \binom{c \ -s}{s \ \ \ \ c}\binom{0 \ 1}{1 \ 0}=...\\ f_5(z)=f_2(f_2(z))=\omega^2 z & \leftrightarrow & \binom{-c \ -s}{s \ \ \ -c}\\ ... &...&...\end{matrix}$$