Determinant of group representation p4m The group p4m consists of all compositions of translations,mirror reflections, and rotations by 90 degrees about any center of rotation in the grid.
$g(m, r, u, v)=\begin{bmatrix}
(-1)^{m}\cos \frac{r \pi }{2} & -(-1)^{m}\sin \frac{r \pi }{2} & u\\ 
\sin \frac{r \pi }{2} & \cos \frac{r \pi }{2} & v\\ 
0 & 0 & 1
\end{bmatrix}$
In the paper Group Equivariant Convolutional Networks (Section 4.3) they have mentioned that:
$m = \frac{1}{2}(1-\det(g)) $
That means, $\det(g) = 1 - 2m$
But when I tried to find the determinant manually I got the following:
$\det(g) = \begin{vmatrix}
(-1)^{m}\cos \frac{r \pi }{2} & -(-1)^{m}\sin \frac{r \pi }{2} & u\\ 
\sin \frac{r \pi }{2} & \cos \frac{r \pi }{2} & v\\ 
0 & 0 & 1
\end{vmatrix}\\
\Rightarrow \det(g) = 1.\begin{vmatrix}
(-1)^{m}\cos \frac{r \pi }{2} & -(-1)^{m}\sin \frac{r \pi }{2}\\ 
\sin \frac{r \pi }{2} & \cos \frac{r \pi }{2}\\
\end{vmatrix} \\
\Rightarrow \det(g) = (-1)^{m}\cos^{2} \frac{r \pi }{2} + (-1)^{m}\sin^{2} \frac{r \pi }{2}\\
\Rightarrow \det(g) = (-1)^{m}(\cos^{2} \frac{r \pi }{2} + \sin^{2} \frac{r \pi }{2}) \\
\Rightarrow \det(g) = (-1)^{m}$
Clearly, $1-2m \neq (-1)^{m}$
Am I doing anything wrong? If yes please correct me. If no then how can I find the value of m?

$1-2m$ is indeed equal to $(-1)^m$. In that paper it is mentioned that $m\in \{0,1\}$. $1-2\cdot 0=1=(-1)^0$, and $1-2\cdot 1=-1=(-1)^1$.
