Homotopy group of the quotient $O(2n)/(U(p) \times U(n-p))$ The question is about the homotopy group of $O(2n)/(U(p) \times U(n-p))$:
$$
\pi_k(O(2n)/(U(p) \times U(n-p))
$$
Let $O(2n)$ be the orthogonal group represented by the rank-$2n$ matrix $V$ with real coefficients and with $V^T V =1$ where $T$ is trh transpose, and with $\det(V)=\pm 1$.
Let $U(n)$ be the unitary group represented by the rank-$nn$ matrix $V$ with complex coefficients and with $V^T V =1$ and with $|\det(V)|=1$ and $\det(V) \in \mathbb{C}$.
We can show that
$$
O(2n) \supset SO(2n) \supset U(n) \supset U(p) \times U(n-p)
$$
with some positive integer $n$ and $p$, and $1 \leq p<n$.
Question: How to determine
$$
\pi_k(O(2n)/(U(p) \times U(n-p)))=?
$$
In particular, I am correct to claim $\pi_2(O(10)/(U(2) \times U(3)))=\mathbb{Z}\times \mathbb{Z}?
$
 A: The fibration $U(p)\times U(n-p) \to O(2n) \to O(2n)/(U(p)\times U(n-p))$ induces a long exact sequence in homotopy groups
$$\dots \to \pi_{k+1}(O(2n)/(U(p)\times U(n-p))) \to \pi_k(U(p)\times U(n-p)) \to \pi_k(O(2n)) \to \pi_k(O(2n)/(U(p)\times U(n-p)) \to \pi_{k-1}(U(p)\times U(n-p)) \to \dots$$
The relevant section for the group you ask about is
$$\dots \to \pi_2(O(2n)) \to \pi_2(O(2n)/(U(p)\times U(n-p))) \to \pi_1(U(p)\times U(n-p)) \to \pi_1(O(2n)) \to \dots$$
For any Lie group $G$ we have $\pi_2(G) = 0$, so $\pi_2(O(2n)/(U(p)\times U(n-p)))$ is isomorphic to the kernel of the map $\pi_1(U(p)\times U(n-p)) \to \pi_1(O(2n))$. We have $\pi_1(U(p)\times U(n-p)) \cong \mathbb{Z}\times\mathbb{Z}$ and $\pi_1(O(2n)) \cong \mathbb{Z}_2$ (using the fact that $n > 1$). As the loop $\operatorname{diag}(e^{i\theta}, 1, \dots, 1)$ generates $\pi_1(U(a))$ for every $a$, and $\operatorname{diag}\left(\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}, I_2, \dots, I_2\right)$ generates $\pi_1(O(b))$ for every $b$, the induced map $\mathbb{Z}\times\mathbb{Z} \to \mathbb{Z}_2$ is given by $(1, 0) \mapsto 1$, $(0, 1) \mapsto 1$. The kernel is $\langle (1, -1), (0, 2)\rangle \cong \mathbb{Z}\times\mathbb{Z}$ and therefore
$$\pi_2(O(2n)/(U(p)\times U(n-p))) \cong \mathbb{Z}\times\mathbb{Z}.$$
For general $k$, you will need to understand the maps $\pi_*(U(p)\times U(n-p)) \to \pi_*(O(2n))$.
