Does the quotient ${\mathrm{Sp}(n) \over \mathrm{U}(n)}$ have group structure?

Question

Given the compact symplectic Lie group $\mathrm{Sp}(n)$ of $2n \times 2n$ matrices and the unitary Lie group $\mathrm{U}(n)$ of $n \times n$ matrices, is the quotient ${\mathrm{Sp}(n) \over\mathrm{U}(n)}$ a Lie group as well?

Answer 1

Not in a way that has anything to do with the group structure of $Sp(n)$; generally $G/H$ inherits a group structure from $G$ iff $H$ is normal, and $U(n)$ is very far from being normal in $Sp(n)$ (as I mentioned in the comments, conjugates of $U(n)$ can be constructed by picking different embeddings $\mathbb{C} \to \mathbb{H}$). In fact $Sp(n)$ is simple and so has no nontrivial connected normal subgroups.

When $n = 1$ the quotient $Sp(1)/U(1)$ is the $2$-sphere $S^2$ so does not admit any topological group structures whatsoever.