$\operatorname{U}(2n) \supset \frac{\operatorname{Sp}(n) \times \operatorname{Sp}(m)}{\mathbb{Z}_2}$ for what maximum of $m$? I know that the Sp($n$) group is a real Lie group which is compact, connected, and simply connected
with $n(2n+1)$real Lie algebra generators. It can be constructed out of the intersections between
a non-compact, simply connected, simple Lie group $\operatorname{Sp}(2n;\mathbf C)$ and the unitary group $\operatorname{U}(2n)$ as
related by
$$
\operatorname{Sp}(n):=\operatorname{Sp}(2n;\mathbf C)\cap\operatorname{U}(2n)=\operatorname{Sp}(2n;\mathbf C)\cap\operatorname {SU} (2n) \tag{1}.
$$
I also know that:
$$
\operatorname{U}(2n) \supset \operatorname{SU}(2n) 
\supset\operatorname{Sp}(n) \supset \operatorname{U}(n) \tag{2}.
$$

Now can we show the following:
$$
\operatorname{U}(2n) \supset 
\frac{\operatorname{Sp}(n) \times \operatorname{Sp}(1)}{\mathbb{Z}_2}? \text{ for some large enough $n$}? \tag{Q1}.
$$
$$
\operatorname{U}(2n) \supset 
\frac{\operatorname{Sp}(n) \times \operatorname{Sp}(m)}{\mathbb{Z}_2} \text{ for some large enough $n$ and for what maximum of $m$}? \tag{Q2}.
$$

Q1 and Q2 are my questions,
for what maximum of $m$?
Lie group experts, please illuminate! Thanks!
p.s. For $n=1$, Q1 is wrong since
$\operatorname{U}(2n) \supset 
\frac{\operatorname{Sp}(n) \times \operatorname{Sp}(1)}{\mathbb{Z}_2}$ but it is wrong $\operatorname{U}(2) \not\supset 
\frac{\operatorname{Sp}(1) \times \operatorname{Sp}(1)}{\mathbb{Z}_2}$.
 A: This never occurs.  That is, the is no group which is locally isomorphic to $Sp(n)\times Sp(1)$ which is a subgroup of $U(2n)$.
First, a general fact:  If $H$ is a connected semisimple group (i.e., no positive dimensional normal abelian subgroups) and $H\subseteq U(n)$, then $H\subseteq SU(n)$.  The idea of this proof is to consider the restriction of $\det:U(n)\rightarrow S^1$ to $H$.  If $\det|_{H}$ is non-trivial, the image is all of $S^1$ (being a connected subgroup), so the kernel is a codim $1$ normal subgroup of $H$.  This can't happen for semisimple groups, so $\det|_{H}$ is trivial.  That is, $H\subseteq SU(n)$.
So, your question is equivalent to asking if $Sp(n)\times Sp(1)/\mathbb{Z}/2\mathbb{Z}$ embeds into $SU(2n)$.  In fact, we will show that every homomorphism $f:Sp(n)\times Sp(1)\rightarrow SU(2n)$ is trivial on at least one of the factors.
Using the dimension formula for representations of $Sp(n)$ (e.g., Fulton and Harris, Rep Theory, pg. 406), it follows easily that the smallest non-trivial representation is of dimension $2n$ - all others are larger.  Thus, up to conjugacy, there is a unique non-trivial homomorphism $Sp(n)\rightarrow SU(2n)$.
Now, the homogeneous space $SU(2n)/Sp(n)$ is actually an irreducible symmetric space.  In particular, it is isotropy irreducible: the action of $Sp(n)$ on the orthogonal complement (with respect to a bi-invariant metric) to $\mathfrak{sp}(n)\subseteq\mathfrak{su}(2n)$ is an irreducible representation.
It follows that $Sp(n)\subseteq SU(2n)$ is maximal among connected groups.  For if we have $Sp(n)\subseteq K\subseteq SU(2n)$, the, the fact that $Sp(n)\subseteq K$ implies the isotropy action preserves $\mathfrak{sp}(n)^\bot \cap \mathfrak{k}$.  Irreducibility now forces $\mathfrak{k} = \mathfrak{sp}(n)$ or $\mathfrak{k} = \mathfrak{su}(2n)$.  Since $K$ is connected, $K = Sp(n)$ or $K = SU(2n)$.
Now, given any homomorphism $f:Sp(n)\times Sp(1)\rightarrow SU(2n)$, if $f$ cannot be non-trivial on both factors.  To see this, first note that $f|_{I\times Sp(1)}$ must commute with $f|_{Sp(n)\times 1}$.  Since $Sp(n)$ has discrete center, this means the images $f|_{I\times Sp(1)}$ and $f_{Sp(n)\times 1}$ must be distinct.  In particular, $f(Sp(n)\times Sp(1))$ stictly contains $Sp(n)$.  By the previous paragraph, this means that $f(Sp(n)\times Sp(1)) = SU(2n)$.  But this is absurd, just by counting dimensions.
Edit  Here is a proof that a connected compact semisimple Lie group cannot have a codimension one normal subgroup.  As a by product of the proof, it also cannot have a codimension two normal subgroup.
As a consequence of the well known fact that every Lie group has a cover of the form $\Pi_{i=1}^m G_i\times T^k$ with the $G_i$ simply connected and simple, and $T^k$ a $k$-dimensional torus, it follows that every Lie algebra of a compact Lie group splits as a direct sum $\mathfrak{g} = \bigoplus_{i=1}^m \mathfrak{g}_i \oplus \mathbb{R}^k$ with each $\mathfrak{g}_i$ simple and non-abelian and where $\mathbb{R}^k$ has trivial Lie bracket.  (Here, direct sum means brackets between the factors are $0$).  Ideals in $\mathfrak{g}$ correspond to connected normal subgroups of $G$, so the semisimplicity assumption means $k = 0$.
We now claim that every ideal in $\mathfrak{g}$ splits as a sum of factors.  That is, it is of the form $\bigoplus_{i=1}^m \mathfrak{h}_i$ where each $\mathfrak{h}_i$ is either trivial or equal to $\mathfrak{g}_i$.  Believing this for the moment, since the minimal dimension of simple non-abelian Lie algebra is $3$, the result will follow.
The proof of the claim is by induction, with the base case (one factor) simply being the definition of simple.  So, assume the result is true for the sum of any $m$ simple non-abelian Lie algebras, and let $\mathfrak{k}$ be an ideal in $\bigoplus_{i=1}^{m+1} \mathfrak{g}_i$ with each $\mathfrak{g}_i$ non-abelian and simple.
Consider $[\mathfrak{g}_1, \mathfrak{k}]$.  This is an ideal in $\mathfrak{g}_1$, which is simple, so $[\mathfrak{g}_1, \mathfrak{k}]$ is either trivial or equal to $\mathfrak{g}_1$.
Assume we are in the first case:  $[\mathfrak{g}_1,\mathfrak{k}] = 0$.  Since $\mathfrak{g}_1$ is centerless (being simple and non-abelian), this then implies that the projection of $\mathfrak{k}$ to $\mathfrak{g}_1$ is trivial, so $\mathfrak{k}$ is  naturally an ideal of the sum $\bigoplus_{i=2}^{m+1} \mathfrak{g}_i$, so the inductive hypothesis gives the result.
So, assume we are in the second case:  $\mathfrak{g}_1 = [\mathfrak{g}_1,\mathfrak{k}]$.  Because $\mathfrak{k}$ is an ideal, we have $\mathfrak{g}_1= [\mathfrak{g}_1,\mathfrak{k}] \subseteq \mathfrak{k}$.  Since $\mathfrak{k}$ contains the entire $\mathfrak{g}_1$ factor, it now follows that $\mathfrak{k}$ splits as $\mathfrak{g}_1\oplus (\mathfrak{k} \cap \bigoplus_{i=2}^{m+1}\mathfrak{g}_i)$.  Noting that $(\mathfrak{k} \cap \bigoplus_{i=2}^{m+1}\mathfrak{g}_i)$ is an ideal in $\bigoplus_{i=2}^{m+1} \mathfrak{g}_i$, the inductive hypothesis finishes the proof.
