A question regarding the dihedral group $D_4$.

Question

I know that the dihedral group $D_4$ is generated by two elements $\sigma$ and $\tau$ such that $\sigma^4=\tau^2=e$, and $\tau\sigma\tau^{-1}=\sigma^3$. This is essentially the group of eight elements $\{e,\sigma,\sigma^2,\sigma^3,\tau,\tau\sigma,\tau\sigma^2,\tau\sigma^3\}$.

Here, I am representing by $\sigma$ a rotation by $\pi/2$, and by $\tau$ a reflection about the horizontal axis, whence $\tau\sigma$ is a reflection about the vertical axis. Now, my question is the following: How are symmetries of the square defined by reflections along the two diagonals covered in the dihedral group? For $\tau=\tau\sigma^2$, and $\tau\sigma=\tau\sigma^3$ geometrically, and that exhausts all eight elements without covering the diagonal symmetries.

Any clarification is much appreciated.

Answer 1

Let's consider the following square with a $90^\circ$-rotation followed by a horizontal reflection : $$ \begin{array}{ccc} 1 & - & 2 \\ | & & | \\ 4 & - & 3 \end{array} \quad \overset{\sigma}\longrightarrow \quad \begin{array}{ccc} 2 & - & 3 \\ | & & | \\ 1 & - & 4 \end{array} \quad \overset{\tau}\longrightarrow \quad \begin{array}{ccc} 1 & - & 4 \\ | & & | \\ 2 & - & 3 \end{array} $$ One gets indeed a reflection through the left diagonal, which exchanges 2 and 4 but keeps 1 and 3 invariant; it is thus represented by the element $\tau\sigma$.

Answer 2

Hint: Every reflection has a line of symmetry (fixed vectors, i.e. $\lambda = 1$ eigenspace). Take any reflection, e.g $\tau\sigma$, and find the invariant axis.

Try it!

Assuming that we read these symmetries as functions acting on the left (composition goes from right to left as usual), $\sigma$ is a counterclockwise rotation by a quarter turn, and $\tau$ is a reflection across the horizontal axis, then the line through the origin at heading $3\pi/4$ (or equivalently opposite $-\pi/4$) is fixed. This is pretty easy to see by just drawing a square centered at the origin and following the NW or SE corner through the rotation followed by the reflection.

Here's what it looks like explicitly with matrices:

$$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} =\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$$