Maximal Closed Subgroups of $ SU_4 $ What are the maximal closed subgroups of $ SU_4 $?
 A: I claim that $N:=N_{SU(4)}(Sp(2))$ is a maximal subgroup of $Sp(2)$, and that $N = Sp(2) \cup iI Sp(2)$.
To see this, consider the double covering $\pi:SU(4)\rightarrow SU(4)/\{\pm I\}\cong SO(6)$.  Note that $-I\in Sp(2)\subseteq SU(4)$, so $\pi|_{Sp(2)}$ is the double covering $Sp(2)\rightarrow SO(5)$.  Up to conjugacy, there is an essentially unique $SO(5)\subseteq SO(6)$, the usual block form.
So, instead of studying $Sp(2)\subseteq SU(4)$, we'll study $SO(5)\subseteq SO(6)$ and pull the information back via $\pi$.
Proposition:  The only proper subgroup of $SO(6)$ which properly contains $SO(5)$ is $O(5) = \{ \operatorname{diag}(A,\det(A)):A\in O(5)\}$.
Proof:  The isotropy action of $SO(5)$ on $S^5 = SO(6)/SO(5)$ is transitive on the unit sphere in $T_{I SO(5)} S^5$, so is, in particular, irreducible.  This, then, implies that $SO(5)$ is maximal among connected groups:  if $SO(5)\subseteq K\subseteq SO(6)$, then on the Lie algebra level, the adjoint action of $\mathfrak{so}(5)$ would preserve both $\mathfrak{k}$ and $\mathfrak{k}^\bot$, contradicting irreducibility.  (Here, I'm using the fact that we can naturally identify the isotropy action of $H$ on $T_{I SO(5)} SO(6)/SO(5)$ with $\mathfrak{so}(5)^\bot\subseteq \mathfrak{so}(6)$.)
Since we now know that $SO(5)$ is maximal among connected subgroups of $SO(6)$, and the identity component of a Lie group is always a normal subgroup, it now follows that $N_{SO(6)}(SO(5))$ is a maximal subgroup of $SO(6)$.
Of course, $O(5)\subseteq N_{SO(6)}(SO(5))$, but why is the reverse inclusion true?  Well, every matrix $B\in SO(5)$ fixes the basis vector $e_6$ of $\mathbb{R}^6$, and $\operatorname{span}\{e_6\}$ is the unique subspace of $\mathbb{R}^6$ fixed by all of $SO(5)$.  A simple computation reveals that for any $C\in N_{SO(6)}(SO(5))$, that $CSO(5)C^{-1} = SO(5)$ fixes $Ce_6$.  It follows that $Ce_6 \in \operatorname{span}\{e_6\}$.  Moreover, since $C\in SO(6)$, we must in fact that $Ce_6 = \pm e_6$.  In either case, the fact that $CC^t = I$ now implies that $C\in O(5)$.  $\square$
Now, let's pull that information back to to better understand $N = N_{SU(4)}(Sp(2))$.  It's not too hard to see that $\pi|_N:N\rightarrow N_{SO(6)}(SO(5)$ is a double covering, with $\pi$ mapping $Sp(2)$ and $iI Sp(2)$ to the two different components of $N_{SO(6)}(SO(5))$.
Now, let $g\in N$ be arbitrary.  Since $\pi|_{Sp(2)\cup iI Sp(2)}$ is surjective onto $N_{SO(6)}(SO(5))$, there is an $h\in Sp(2)\cup iI Sp(2)$ with $\pi(g) = \pi(h)$.  Then $gh^{-1}\in \ker \pi = \pm I$, so $g = \pm I h$.  Since both $h, \pm I\in Sp(2)\cup iI Sp(2)$, it follows that $g\in Sp(2)\cup iI Sp(2)$ as well.
