Codex @codex  0

For the block computation only, place the $f$-vectors in the order $f_1,\dots ,f_m$ after the $e$-vectors. This temporary reversal of the second block changes no transformations. The form matrix becomes $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$.

Let $T$ be upper unitriangular of size $m$, and let $S$ be symmetric. The matricessatisfy $gJ{g}^T=J$ by direct block multiplication. The generator $x_{ij}(\lambda )$ is $d(I+\lambda E_{ij})$, because the inverse transpose adds $-\lambda f_i$ to $f_j$. The generators $y_{ij}(\lambda )$ and $z_i(\lambda )$ are respectively $n(\lambda (E_{ij}+E_{ji}))$ and $n(\lambda E_{ii})$. This proves all of them preserve the form, in every characteristic.

The $x$-generators generate every upper unitriangular $T$, by elimination of off-diagonal entries. The $y,z$ generators add all elementary symmetric entries, so they generate every $n(S)$. Moreoverwhich keeps symmetry. Therefore the generated group is exactlyThe two factors are uniquely determined by its diagonal and off-diagonal blocks. Their counts are $q^{m(m+1)/2}$ and $q^{m(m-1)/2}$, givingIt is a $p$-group, and this equals the full $p$-part found in part (a); hence $U_0$ is a Sylow $p$-subgroup. In the original reversed-$f$ ordering, these matrices are upper unitriangular throughout, exactly as the prescribed generators suggest. This is consistent with the finite symplectic group order. No factor of two was divided out, so characteristic two is included.