I have been reading John Stillwell's 'Naive Lie Theory' and in Section 3.7, the author tries to prove that $Z(SO(2m)) = {\pm 1}$. In the proof, a matrix $I^\star$ is introduced with the form:
$$ I^\star = \left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right) $$
and it's mentioned that a diagonal arrangement of $I^\star$ is in $SO(2m)$. However, when $m$ is odd, wouldn't the determinant of the matrix $I^\star$ be -1? By definition, matrices in $SO(2m)$ should have a determinant of 1. Is there an error in this part of the proof, or am I misunderstanding something? Any clarification would be greatly appreciated. Thank you!
Although I agree that there is a problem here, it is easy to go around it. The final goal is to prove that if $R_{\theta_1,\theta_2,\ldots,\theta_n}$ belongs to the center of $SO(2m)$, then each $\sin\theta_k$ is equal to $0$, right?! So, consider a block matrix $J$ whose blocs consist of $m-2$ copies of the identity matrix and two copies $I^*$, one of which is located at the same position as $\left(\begin{smallmatrix}\cos\theta_k&-\sin\theta_k\\\sin\theta_k&\cos\theta_k\end{smallmatrix}\right)$. Then $J\in SO(2m)$ and, since $R_{\theta_1,\theta_2,\ldots,\theta_n}$ and $J$ commute, $\sin\theta_k=0$.