Activity 9.4.1.
Make a multiplication table for the dihedral group \(D_5\text{.}\) (Label the rotations \(R_i\text{;}\) the index \(i\) indicates how many fifth of a full turn for the rotation. Label the reflections \(F_i\text{;}\) the index indicates reflection over a line through the \(i\) vertex, counting counterclockwise. Then start calculating the compositions using vertex opertaions. One you have done a number of these, you can start looking for patterns in the multiplication table and potentially use those patterns to finish the table.)
Solution.
There are many multiplications to calculate. I recommend calculating matrices by vertex operations. If I number the vertices of the pentagon counterclockwise from 1 to 5, then \(R_2\) (rotation by \(\frac{2}{5}\) of a turn) can be represented by this movement of vertices.
\begin{align*}
1 \amp \mapsto 3\\
2 \amp \mapsto 4\\
3 \amp \mapsto 5\\
4 \amp \mapsto 1\\
5 \amp \mapsto 2
\end{align*}
Reflection \(F_3\) (reflection over the line through vertex 3 and the middle of the opposite edge)) can be represented by this action on the vertices.
\begin{align*}
1 \amp \mapsto 5\\
2 \amp \mapsto 4\\
3 \amp \mapsto 3\\
4 \amp \mapsto 2\\
5 \amp \mapsto 1
\end{align*}
To calculate \(R_2 \circ F_3\text{,}\) I can use these vertex changes. In composition, I do the right transformation first. Therefore, starting at vertex 1, \(F_3\) send it to vertex 5, and then \(R_2\) sends vertex 5 to vertex 2. Therefore, the whole composition sends vertex 1 to vertex 2. I calculate the movement of the other four vertices similarly.
\begin{align*}
1 \amp \mapsto 2\\
2 \amp \mapsto 1\\
3 \amp \mapsto 5\\
4 \amp \mapsto 4\\
5 \amp \mapsto 3
\end{align*}
Then I have to identify this transformation, since is still should be something in the group. It cannot be a rotation, since vertex 4 is fixed. The only rotation that fixes vertex 4 is \(F_4\text{.}\) Therefore, I have \(R_2 \circ F_3 =
F_4\text{.}\) Note that the multiplication is not commutative. If I did \(F_3 \circ R_2\text{,}\) the result would be these vertex transformations.
\begin{align*}
1 \amp \mapsto 3\\
2 \amp \mapsto 2\\
3 \amp \mapsto 1\\
4 \amp \mapsto 5\\
5 \amp \mapsto 4
\end{align*}
This is the reflection \(F_2\text{,}\) so \(F_3 \circ R_2 =
F_2\text{.}\) This is just a demonstration calculation: you’ll have to many of these to produce the whole table. Between calculating compositions and using the patterns of the table to fill in remaining entries, you should end up with this multiplication for \(D_5\text{.}\) Table 9.4.1.
| \(\circ\) | \(\Id\) | \(R_1\) | \(R_2\) | \(R_3\) | \(R_4\) | \(F_1\) | \(F_2\) | \(F_3\) | \(F_4\) | \(F_5\) |
| \(\Id\) | \(\Id\) | \(R_1\) | \(R_2\) | \(R_3\) | \(R_4\) | \(F_1\) | \(F_2\) | \(F_3\) | \(F_4\) | \(F_5\) |
| \(R_1\) | \(R_1\) | \(R_2\) | \(R_3\) | \(R_4\) | \(\Id\) | \(F_4\) | \(F_5\) | \(F_1\) | \(F_2\) | \(F_3\) |
| \(R_2\) | \(R_2\) | \(R_3\) | \(R_4\) | \(\Id\) | \(R_1\) | \(F_2\) | \(F_3\) | \(F_4\) | \(F_5\) | \(F_1\) |
| \(R_3\) | \(R_3\) | \(R_4\) | \(\Id\) | \(R_1\) | \(R_2\) | \(F_5\) | \(F_1\) | \(F_2\) | \(F_3\) | \(F_4\) |
| \(R_4\) | \(R_4\) | \(\Id\) | \(R_1\) | \(R_2\) | \(R_3\) | \(F_3\) | \(F_4\) | \(F_5\) | \(F_1\) | \(F_2\) |
| \(F_1\) | \(F_1\) | \(F_3\) | \(F_5\) | \(F_2\) | \(F_4\) | \(\Id\) | \(R_3\) | \(R_1\) | \(R_4\) | \(R_2\) |
| \(F_2\) | \(F_2\) | \(F_4\) | \(F_1\) | \(F_3\) | \(F_5\) | \(R_2\) | \(\Id\) | \(R_3\) | \(R_1\) | \(R_4\) |
| \(F_3\) | \(F_3\) | \(F_5\) | \(F_2\) | \(F_4\) | \(F_1\) | \(R_4\) | \(R_2\) | \(\Id\) | \(R_3\) | \(R_1\) |
| \(F_4\) | \(F_4\) | \(F_1\) | \(F_3\) | \(F_5\) | \(F_2\) | \(R_1\) | \(R_4\) | \(R_2\) | \(\Id\) | \(R_3\) |
| \(F_5\) | \(F_5\) | \(F_2\) | \(F_4\) | \(F_1\) | \(F_3\) | \(R_3\) | \(R_1\) | \(R_4\) | \(R_2\) | \(\Id\) |
