Decomposition of semisimple Lie groups (lie algebras) into direct products (direct sums) and the correspondence. Let $G$ be a semisimple Lie group and $Lie(G)$ be its Lie algebra. I guess the following proposition should be true by the theory of smooth manifolds (Let me know if not):
(1) If $G \cong G_1\times \cdots \times G_n$ ($\cong$ means isomorphic as Lie groups), then
$$Lie(G)=Lie(G_1)\oplus \cdots \oplus Lie(G_n).$$
But I am not sure if the following converse is also true:
(2) If $Lie(G)=\mathfrak{g_1} \oplus \cdots \oplus \mathfrak{g_n}$, then there exists Lie groups $G_1,\cdots G_n$, with $Lie(G_i)=\mathfrak{g_i},\forall i$, such that (as isomorphism of Lie groups)
$$G \cong G_1\times \cdots \times G_n.$$
I have noticed that
If the Lie algebra is a direct sum, then the Lie group is a direct product?
But here I am not assuming $G$ to be simply connected
 A: No, in the non-simply connected case this is not true, and the fundamental group is exactly the obstruction. If $\text{Lie}(G) = \mathfrak{g}_1 \times \dots \times \mathfrak{g}_n$ and for each index $i$ we take $G_i$ to be the simply connected Lie group with Lie algebra $\mathfrak{g}_i$, then $G_1 \times \dots \times G_n$ is necessarily the universal cover of $G$, so there's a covering map
$$G_1 \times \dots \times G_n \to G$$
and it's an isomorphism iff $G$ is simply connected.
In the non-simply connected case here is the kind of thing that can go wrong. Let $G = SO(4)$. Its Lie algebra satisfies $\mathfrak{so}(4) \cong \mathfrak{su}(2) \times \mathfrak{su}(2)$ so its universal cover is $SU(2) \times SU(2)$, but the covering map
$$SU(2) \times SU(2) \to SO(4)$$
is nontrivial, and in fact has diagonal kernel $(-1, -1)$. The Lie groups with Lie algebra $\mathfrak{so}(4)$ which can be expressed as a nontrivial product must be expressible as a nontrivial product of two Lie groups with Lie algebra $\mathfrak{su}(2)$, so the possibilities are $SU(2) \times SU(2), SO(3) \times SU(2), SU(2) \times SO(3), SO(3) \times SO(3)$ and $SO(4)$ is none of these (this can be seen by inspecting the kernels of the corresponding covering maps).
