So the question was from a result of Serre which basically says if $H$ is a closed subgroup of $Sp_{2n}(\mathbb{Z}_p)$ that maps surjectively onto $Sp_{2n}(\mathbb{Z}/p\mathbb{Z})$, then $H=Sp_{2n}(\mathbb{Z}_p)$.
It seems natural to believe a similar result would also hold for orthogonal group $O_n$, and this is essentially asking: If H is a subgroup of $O_{n}(\mathbb{Z}/p^2\mathbb{Z})$ that maps onto $O_{n}(\mathbb{Z}/p\mathbb{Z})$, would this imply $H=O_{n}(\mathbb{Z}/p^2\mathbb{Z})$? Can anyone point out if this claim is known to be true or false?
My thought was that if denote the kernel of $O_{n}(\mathbb{Z}/p^2\mathbb{Z}) \twoheadrightarrow O_n(\mathbb{Z}/p\mathbb{Z})$ by $\Gamma$, then $\Gamma=\{E+pM |M^\intercal+M=0\}$ is abelian, and it turns out that $\Gamma$ as $\mathbb{F}_p[O_n(\mathbb{Z}/p\mathbb{Z})]$ module is irreducible so $\Gamma \cap H=E$, which also implies $H\Gamma= O_n(\mathbb{Z}/p\mathbb{Z})\rtimes \Gamma=O_{n}(\mathbb{Z}/p^2\mathbb{Z})$. This looks pathological but I couldn't tell myself why.