Q- Let G be the group of the symmetries of the regular hexagon. List the elements of G (there are 12 of them), then write the table of G.
So for the listing the elements of G, they want it like this:
$R_0$=\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \end{smallmatrix}
$R_1$=\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 5 & 6 & 1 \end{smallmatrix}
all the way until $R_{12}$. Hopefully, you see the pattern.
Now for constructing the table, I have started as you can see below:
However, I realize a couple of things:
$R_0$=$R_6$=$R_{12}$
$R_1$=$R_7$
$R_2$=$R_8$
So let's say for the row $R_0$ and the column $R_6$ instead of writing $R_6$, would it be correct to write $R_0$. Same for row $R_0$ and the column $R_7$, would be it correct to leave it as $R_1$ or $R_7$.
This idea developed because I was looking at this link: http://www.cs.uleth.ca/~holzmann/notes/dihedral/dihedral.html
and the writer only replaces for the identity which they call id but in our case its $R_0$ but nothing else gets replace. So that kind of leaves me pretty confused. If anyone can clarify, I'd appreciate it a lot!
It seems that you want to represent the symmetries of the hexagon as permutations of its edges, assuming that these edges are numbered from 1 to 6 (counter-clockwise, lets say). Your phrase "hopefully you see the pattern" seems to indicate, that you assume that the permutations appearing are obtained by cyclically shifting the numbers in the second row. Then indeed $R_0=R_6$ etc. But this procedure does not give you all the different symmetries: you are missing the reflections. The symmetries $R_0$ to $R_5$ you obtain that way are the different rotations of the hexagon. If you proceed with the procedure $R_6$ to $R_{12}$ are the same rotations that you already got.
Example of a permutation describing a reflection: take the reflection at the axis passing through the edges 1 and 4 (these two are opposite to each other in the numbering I mentioned). The permutation describing this reflection is
$ \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2 \end{array} $