Find the solution to $x$ in $D_4$ to $x^2 = r_0$

Question

I attempt to solve $x^2 = r_0$.

Things I know

$r_i ∘ r_j =$ $r_{i+_nj}$

$r_i ∘ s_j =$ $s_{i+_nj}$

$s_i ∘ r_j =$ $s_{i-_nj}$

$s_i ∘ s_j =$ $r_{i-_nj}$

I know how to solve the linear type, I am not sure how to proceed with quadratics.

Answer 1

Note that the $D_4$ group is the group of symmetries on a square and the symmetries are reflection wrt axis of symmetry and rotations. Obviously every reflection performed twice gives the initial configuration. Hence every reflection satisfies the equation $x^2 = r_0$. On the other side the only rotations which gives the initial configuration when applied twice are the identity and the rotation by $\pi$. Therefore every element of $D_4$ except the rotations by $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ satisfy the equation $x^2 = r_0$

Another way to solve this problem is to use the equations you have written. For $s_i$ we have that $s_i^2 = r_{i-_ni} = r_0$. Therefore every $s_i$ satisfies the equation. For $r_i$ we have $r_i^2 = r_{i+_ni}$, which is $0$ if and only $2i \equiv 0 \pmod 4 \iff i \equiv 0 \pmod 2$. As $i < 4$ we have that only $r_0$ and $r_2$ satisfy the equation.