$\operatorname{U}(4n) \supset \frac{\operatorname{Sp}(n) \times \operatorname{Sp}(m)}{\mathbb{Z}_2}$ for some 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:
$$
\operatorname{U}(4n) \supset 
\frac{\operatorname{Sp}(n) \times \operatorname{Sp}(1)}{\mathbb{Z}_2}? \tag{Q1}.
$$
$$
\operatorname{U}(4n) \supset 
\frac{\operatorname{Sp}(n) \times \operatorname{Sp}(m)}{\mathbb{Z}_2} \text{ for some $n$, and for what maximum of $m$}? \tag{Q2}.
$$

Q1 and Q2 are my questions,
for what maximum of $m$?
Lie group experts, please illuminate! Thanks!
 A: The answer to Q1 is yes, since there is a standard homomorphism $Sp(n)\times Sp(1)\to SO(4n)$ with kernel $\mathbb Z_2$ and $SO(4n)\subset SU(4n)\subset U(4n)$. To see thist, you view $\mathbb R^{4n}$ as quaternionic $n$-space $\mathbb H^n$ and the standard inner product on $\mathbb R^{4n}$ as the real part of the quaternionic Hermitian inner product. THen $Sp(n)$ acts on $\mathbb H^n$ by quaternionically linear Hermitian maps. On the other $Sp(1)$ acts by quaternionic scalar multiplications, which preserve the real part of the quaternioncally Hermitian form, but are not quaternionically linear maps since the quaternions are non-commutative.
As you can see, this is very specific for $Sp(1)$, so I have no idea about Q2.
A: I claim that for $m\geq 2$, there is no embedding of $(Sp(n)\times Sp(m))/ \mathbb{Z}/2\mathbb{Z}$ into $U(4n)$.  As Andreas shows, there is such an embedding when $m=1$.
As in my answer to a previous question of yours, we will study homomorphisms $Sp(n)\times Sp(m)\rightarrow U(4n)$.  We'll find that as long as $m\leq n$, there is a non-trivial homomorphism, but that no such homomorphisms have the right kernel.
To that end, we recall that a representation of a compact Lie group always decomposes as a direct sum of irreducible representations, and that an irreducible representation of a product of Lie groups is always given as an outer tensor product of irreducible representations.  That is, given irreducible representations $V_i$ of $G_i$, the representation of $G_1\times G_2$ on $V_1\otimes V_2$ given by $(g_1,g_2)\ast v_1\otimes v_2 = g_1 v_1\otimes g_2v_2$ is irreducible, and all such irreducible representations of a product arise in this fashion.
Now, the smallest (in terms of dimension) representation of $Sp(n)$ is the usual $2n$ dimensional one obtained from the inclusion $Sp(n)\rightarrow SU(2n)$, arising from the identification of $\mathbb{H}^n$ with $\mathbb{C}^{2n}$.
Suppose we have a homomorphism $f:Sp(n)\times Sp(m)\rightarrow U(4n)$.  Thought of as a representation, it decomposes as a sum $\sum \phi_i\otimes \psi_i$ where the $\phi_i$ are representations of $Sp(n)$ and the $\psi_j$ are representations of $Sp(m)$.
If $m\geq 2$, then we claim that each factor in this sum either has $\phi_i$ trivial or $\psi_i$ trivial.  That is, they cannot both be non-trivial.  For if $\phi_i$ and $\psi_i$ are both non-trivial, then $$\dim f \geq \dim (\phi_i\otimes \psi_i) = \dim \phi_i \cdot \dim \psi_i = 2n\cdot 2m = 4nm > 4m,$$  i.e., the image of $f$ lands in $U(4nm)$ and nothing smaller.  This contradicts that fact that $f$ has image in $U(4n)$.
(If $m=1$, we get no contradiction as long as $\phi_i\otimes \psi_i$ is the tensor product of the smallest representations and $f$ has no other summands.  This representation is precisely the one Andreas found.)
So, we now know that $f$ decomposes as $\sum \phi_i\otimes 1 + \sum 1\otimes \psi_j$ where $1$ denotes the trivial $1$-dimensional representation.  And note that there are certainly examples which are trivial on individual factors, at least if $m\leq n$.  Specifically, if $\phi$ and $\psi$ are the standard representations, then $\phi\otimes 1 + 1\otimes \psi$ is one such example.  This corresponds to the block embedding of $Sp(n)\times Sp(m)$ into $U(2n)\times U(2m)\subseteq U(4n)$.  If $m$ is much smaller than $n$, there will typically be many more options for embeddings.
However, because $f$ decomposes in that nice way, it is easy to see that $\ker f = (\cap_i \ker \phi_i)\times (\cap_j \ker \psi_j)\subseteq Sp(n)\times Sp(m)$.  In particular, $\ker f$ is not the diagonal $\mathbb{Z}/2\mathbb{Z}$.  So none of these homomorphisms have the right kernel.
