= Solution
There are two necessary qualifications to the literal formulation. The whole <group> $G$ is itself a closed <normal subgroup>, so the claim about containment in the centre must concern \b[proper closed <normal subgroups>]. Also, the homomorphism conclusion needs \b[a nontrivial target $H$]: for example, $SU(2)\to\{1\}$ has <surjective> differential onto the zero <Lie algebra>, but its <group kernel> is all of $SU(2)$ and no quotient by a finite subgroup is a one-point <group>. We prove the precise assertions, including these exceptional cases.
Write $\mathfrak g=\operatorname{Lie}(G)$. The centre $\mathfrak z(\mathfrak g)$ is invariant under the <Adjoint representation>, so irreducibility makes it either zero or all of $\mathfrak g$. In the latter case $\mathfrak g$ is <Abelian>. For <connected> $G$, the <adjoint action> is then trivial: the <group> is generated by exponentials, and $\operatorname{Ad}_{\exp X}=e^{\operatorname{ad}X}=I$. A nonzero <trivial representation> is <irreducible> only in <dimension> one. A <connected> <compact> one-dimensional <Lie group> is a circle, so $G\cong U(1)$.
For every other $G$, the <Lie algebra> of its centre is zero. The centre is a closed <compact> <Lie subgroup>; a zero-dimensional <compact Lie group> is finite. Let $N$ be a closed <normal subgroup>. The <closed-subgroup theorem> identifies $\mathfrak n=\operatorname{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak g$, hence $\mathfrak n=0$ or $\mathfrak n=\mathfrak g$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero <Lie algebra> and is finite. For any $a\in N$, the <continuous map> $g\mapsto gag^{-1}$ takes <connected> $G$ into the <finite set> $N$, so it is constant and equal to $a$. Thus
$$
\boxed{G\not\cong U(1)\Longrightarrow Z(G)\text{ is finite};\qquad N\lhd G, N\text{ closed and proper}\Longrightarrow N\subseteq Z(G).}
$$
Proper <closed subgroups> of the circle are finite as well, and of course central.
Now let $d\Phi$ be <surjective>. The <submersion theorem> makes $\Phi(G)$ contain an identity neighbourhood in $H$, so it is an <open subgroup>. Since $H$ is <connected>, $\Phi(G)=H$. Its <group kernel> $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central <group kernel>; this also holds for $G\cong U(1)$, whose proper <closed subgroups> are finite. The Lie-group <first isomorphism theorem> gives
$$
\boxed{H\cong G/F,\qquad F\subseteq Z(G)\text{ finite},\quad H\ne\{1\}.}
$$
If $H$ is trivial, the correct statement is instead $H=G/G$.
For the <Spin groups>, use the real <Clifford algebra> with $v^2=-\|v\|^2$. The <group> $\operatorname{Spin}(m)$ consists of products of an <even number> of <unit vectors> inside its invertible even part. <Conjugation> on the embedded $\mathbb R^m$ defines the standard double covering $\operatorname{Spin}(m)\to SO(m)$ with <group kernel> $\{\pm1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted <conjugation>, is a hyperplane reflection; the even products give orientation-preserving transformations. The <Cartan–Dieudonné theorem> states that every <orthogonal transformation> is a product of reflections, proving <surjectivity>. For $m\geq3$ this <connected> covering <group> is <simply connected>; $\operatorname{Spin}(2)$ is a circle.
Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E=\mathbb C^4$ with its standard <Hermitian form> and <volume> $e_1\wedge e_2\wedge e_3\wedge e_4$. On $\Lambda^2E$, write $e_{ij}=e_i\wedge e_j$ and define an <antilinear map> $J$ by
$$
Je_{12}=e_{34},\quad Je_{13}=-e_{24},\quad Je_{14}=e_{23},\quad
Je_{34}=e_{12},\quad Je_{24}=-e_{13},\quad Je_{23}=e_{14}.
$$
It satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\,\mathrm{vol}$ and the <Hermitian inner product> $h$ is <linear> in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the <real structure of the SU4 exterior square>.
The fixed space $W$ has the following real <orthonormal basis>, with all vectors divided by $\sqrt2$:
$$
e_{12}+e_{34},\quad i(e_{12}-e_{34}),\quad
e_{13}-e_{24},\quad i(e_{13}+e_{24}),\quad
e_{14}+e_{23},\quad i(e_{14}-e_{23}).
$$
Thus $W$ is six-dimensional over $\mathbb R$, its complexification is $\Lambda^2E$, and the exterior-square action restricts to an <orthogonal representation> on $W$. Its <determinant> is one because $SU(4)$ is <connected>. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.
To compute the <group kernel>, diagonalize $g\in SU(4)$ with <eigenvalues> $z_1,\ldots,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i z_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A <unitary matrix> with that single <eigenvalue> is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two <scalar matrices> plainly act trivially on the <exterior square>.
The differential of $R$ is <injective> because its <group kernel> is finite. Both <Lie algebras> have <dimension> $15$, so the differential is an <isomorphism>. Its image is consequently open in <connected> $SO(6)$, hence all of $SO(6)$. Therefore
$$
\boxed{SU(4)/\{\pm I_4\}\cong SO(6).}
$$
This also realizes $SU(4)$ as $\operatorname{Spin}(6)$, since $SU(4)$ is <simply connected>, by uniqueness of the <connected> <simply connected> covering <group>.
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