I've been trying to compute the cohomology of the group $G:=\text{SL}(2, p^2)$ for an odd prime $p$ (with coefficients in $\mathbb{F}_p$. To clarify, $G$ is the group of $2\times2$-matrices with entries in $\mathbb{F}_{p^2}$ and $1$ as their determinant. The outcome is gigantic, so I would like to know if my reasoning is correct. Thanks in advance :)
(1) The order of $G$. I know that $\text{GL}(2, p^2) \cong G \rtimes \mathbb{F}_{p^2}^\times$, where $\mathbb{F}_{p^2}^\times$ is the multiplicative group of units in $\mathbb{F}_{p^2}$. Since $|\text{GL}(2, p^2)| = |G\rtimes\mathbb{F}_{p^2}^\times| = |G|\cdot|\mathbb{F}_{p^2}^\times|$, we obtain $|G|=\frac{|\text{GL}(2, p^2)|}{|\mathbb{F}_{p^2}^\times|}$. Furthermore, the orders of those groups are $|\text{GL}(2, p^2)| = (p^4-1)(p^4-p^2)=p^2(p^4-1)(p^2-1)$ and $|\mathbb{F}_{p^2}^\times|=p^2-1$, so $|G| = p^2(p^4-1)$.
(2) Finding a Sylow $p$-subgroup of $G$. We know that the order of a Sylow $p$-subgroup $P$ will have to be $p^2$, and so it is quite easy to check that if we let $P$ consist of those matrices with $1$'s on the diagonal, any element of $\mathbb{F}_{p^2}$ in the top right corner, and $0$ in the bottom left corner, we obtain a subgroup of order $p^2$, and so we have our Sylow. In particular, it is an abelian group because of its order, and so it is isomorphic to either $(\mathbb{Z}/p\mathbb{Z})^2$ or $\mathbb{Z}/p^2\mathbb{Z}$. The characteristic of $\mathbb{F}_{p^2}$ is $p$, so we can show that $P\cong(\mathbb{Z}/p\mathbb{Z})^2$.
(3) Finding the normaliser of $P$. If we let $N = \text{N}_G(P)$, it follows that $N$ consists of those matrices with the following properties: (i) The diagonal has as its entries $a$ and $a^{-1}$ for some non-zero $a\in\mathbb{F}_{p^2}$, (ii) the bottom left corner is $0$, (iii) the top right corner can be any element in $\mathbb{F}_{p^2}$. It is easy to calculate the order of $N$, and so we obtain $|N|=p^2(p^2-1)$. We let $N$ act on $P$ by conjugation.
(4) Since $P$ is an abelian subgroup of $G$, there is an isomorphism $H^*(G; \mathbb{F}_p) \cong H^*(N; \mathbb{F}_p)$. Furthermore, the decomposition of $H^*(N; \mathbb{Z})$ into primary components yields an isomorphism $H^*(N; \mathbb{F}_p)\cong H^*(P; \mathbb{F}_p)^N$, the elements of $H^*(P; \mathbb{F}_p)$ invariant under the action of $N$. Since $P$ is normal in $N$, we have $H^*(P; \mathbb{F}_p)^N\cong H^*(P; \mathbb{F}_p)^{N/P}$. As $P\cong(\mathbb{Z}/p\mathbb{Z})^2$, we have $H^(P; \mathbb{F}_p)\cong \mathbb{F}_p[x_1, x_2]\otimes_{\mathbb{F}_p}\Lambda(y_1, y_2)$, where $|x_i|=2, |y_i|=1$ and $x_i$ is the Bockstein of $y_i$, for $i=1, 2$.
(5) Given $A = \begin{pmatrix}a & b \\ 0 & a^{-1} \end{pmatrix}\in N, B = \begin{pmatrix}1 & x \\ 0 & 1 \end{pmatrix}\in P$, we see that $A^{-1}BA = \begin{pmatrix} 1 & a^{-2}x\\ 0 & 1\end{pmatrix}$. Since $p$ is odd, $p^2-1$ is even, and so the action of $A^{\frac{p^2-1}{2}}$ on $P$ is trivial on $P$ regardless of $a$ and $b$. In fact, using $N/P$ instead of $N$, we may assume $b=0$. This suggests that the lowest degree invariants would be scalar multiples of elements of $x_1^\alpha x_2^\beta y_1^\gamma y_2^\delta$, where $\alpha, \beta\ge0$, $0\le\gamma, \delta\le1$, and $\alpha+\beta+\gamma+\delta=\frac{1}{2}(p^2-1)$. The products of these elements will also be invariants, and so we have a reasonable structure here.
It would seem that $$H^*(G;\mathbb{F}_p) \cong P[x_1^\alpha x_2^\beta\ |\ \alpha+\beta = \frac{1}{2}(p^2-1)]\otimes_{\mathbb{F}_p}\Lambda(x_1^\alpha x_2^\beta y_i\ |\ \alpha+\beta=\frac{1}{2}(p^2-1)-1, i=1, 2)\otimes_{\mathbb{F}_p}\Lambda(x_1^\alpha x_2^\beta y_1y_2\ |\ \alpha+\beta = \frac{1}{2}(p^2-1)-2).$$ I am just wondering if this can be correct. Structurally, it makes sense, but it gets so horribly big...