Reduction of $\operatorname{SO}_{2n+1}(\mathbb{Z})$ modulo prime powers Let $p$ be a prime, and let $n \geq 0$ be an integer. Is the quotient map $\operatorname{SO}_{2n+1}(\mathbb{Z}) \to \operatorname{SO}_{2n+1}(\mathbb{Z}/p^k\mathbb{Z})$ surjective for odd primes $p$ and all integers $k \geq 1$? What happens when $p = 2$? (I'm sure this is written down somewhere, but I can't find a reference.)
What I know: something strange happens at the prime $2$, because the matrix
$$\left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
is an involution $\operatorname{SO}_{2n+1}(\mathbb{Z}/2\mathbb{Z})$, but it does not lift to an element of $\operatorname{SO}_{2n+1}(\mathbb{Z}/4\mathbb{Z})$, because its square fails to be orthogonal over $\mathbb{Z}/4\mathbb{Z}$.

Let $G$ be the group $\text{SO}_{2n+1}(\mathbb{Z})$. This means its columns (and its rows) form an orthonormal basis of $\mathbb{R}^{2n + 1}$. So it seems that each column (and each row) must consist of all $0$'s and one $\pm1$. So $G$ is a pretty small finite group. I doubt it could surject onto $\text{SO}_{2n + 1}(\mathbb{Z}/p^k\mathbb{Z})$ for many or most values of $p$ and $k$—just compare the sizes of the groups. Take a look at Dickson's book on linear groups, you can find a lot of information about $\text{SO}_{2n+1}(\mathbb{Z}/p\mathbb{Z})$ that could help you with this.
https://archive.org/details/lineargroupswith00dickuoft
I do not know what is going on for $p = 2$. The matrix you wrote down doesn't look to be orthogonal, because the top row dotted with itself is $0$ (mod $2$) not $1$. Maybe I have not understood you correctly?
