I'm reading a paper which has this line:


A direct computation shows that $P\Omega\_8$($\mathbb K$) has an elementary abelian subgroup $X = 2^2$ such that $C\_{P\Omega\_8(\mathbb K)}(X) = T\_4.2^{1+4}\_+$. Now the action of $2^{1+4}\_+$ on $T\_4$ is also explicitly determined.


$\mathbb K$ is algebraically closed with char not equal 2 and $P\Omega\_8$($\mathbb K$) is the adjoint algebraic group of type D.


What is the action exactly here? I suppose the four 2 invert the four $T\_1$? What about the center? Not sure...



action of the extra-special group
Take $T\_4$ as the diagonal torus, which is composed by 4-tuples of 2x2 rotation matrices $a,b,c,d$: $$\begin{bmatrix}a & & & \\ & b & &\\ & & c &\\ & & & d\end{bmatrix}$$.


Let $T$ denote the matrix $\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$ and $I$ the $2\times2$ identity matrix.


Then the extraspecial group $E=2^{1+4}\_+$ acts on $T\_4$ by conjugation of the following matrices:


$\begin{bmatrix}X & & & \\ & X & &\\ & & X &\\ & & & X\end{bmatrix}$, $\begin{bmatrix}& X & & \\ X & & &\\ & & & X\\ & & X & \end{bmatrix}$, $\begin{bmatrix}& & X & \\ & & & X\\ X & & &\\ & X & & \end{bmatrix}$ and$\begin{bmatrix}& & & X \\ & & X &\\ & X & &\\ X & & &\end{bmatrix}$.


where $X$ represents an element that is either $T$ or $I$, and the number of $T$s is always even.


This group has $32$ elements, because every matrix above represent $8$ elements. There is a matrix representation of $Q\_8 \circ Q\_8$ on [GroupNames](https://people.maths.bris.ac.uk/%7Ematyd/GroupNames/1/ES+(2,2).html). By checking the matrices on the GroupNames page and exchanging $T$ for $-1$ and $I$ for $1$, it is evident that $Q\_8 \circ Q\_8$ is a subgroup of $E$. Since $E$ and $Q\_8 \circ Q\_8$ both have $32$ elements, we have $E=Q\_8 \circ Q\_8$.


Bonus: The group $X$ is generated by the following elements of the maximal torus.


$\begin{bmatrix}-I & & & \\ & -I & &\\ & & I &\\ & & & I\end{bmatrix}$ and $\begin{bmatrix}-I & & & \\ & I & &\\ & & -I &\\ & & & I\end{bmatrix}$


Another way of seeing this group is via the projections from the universal cover to the orthogonal group and then the projective orthogonal group, i.e. $Spin\_8^+(K) \rightarrow \Omega\_8^+(K) \rightarrow P\Omega\_8^+(K)$.


Let $\hat{T}\_4$ be the Cartan subgroup of $Spin\_8^+(K)$ and $T'\_4$ its image in $\Omega\_8^+(K)$. Then the image of the exponent-2 subgroup of $\hat{T}\_4$ in $\Omega\_8^+(K)$ are composed of the elements in $T'\_4$ that have order $2$ and reversing an even number of indicies, i.e. the elements $(a, b, c, d) \rightarrow (s\_1 a,s\_2 b,s\_3 c, s\_4 d)$ where the $s\_n$ are either $+1$ or $-1$ and $s\_1s\_2s\_3s\_4=1$. There are $8$ such elements.


In the map $\Omega\_8^+(K) \rightarrow P\Omega\_8^+(K)$, the element $(a, b, c, d) \rightarrow (-a,-b,-c,-d)$ is identified with the identity element, so the image of the subgroup in $T'\_4$ in $T\_4$ has only 4 elements.


