Decomposition of semi-simple Lie algebras

Background: Let $G$ be a finite-dimensional Lie group, with Lie algebra $L(G)$. A subspace $I\subset L(G)$ which satisfies $[L(G),I]\subset I$ is called an ideal of $L(G)$. A non-abelian Lie algebra whose only ideals are $0$ and itself (trivial ideals) is called a simple Lie algebra. A semi-simple Lie algebra is a direct sum of simple Lie algebras.

Statement: A semi-simple Lie algebra $L(G)$ has a decomposition into commuting simple factors $L(G_i)$ $$L(G)=L(G_1)\oplus L(G_2)\oplus\cdots\oplus L(G_k)$$ and $L(G)$ cannot be reduced further. The associated group has structure $$G=G_1\times G_2\times\cdots\times G_k /\left\{\text{discrete group}\right\}$$ where $G_i$ are simple Lie groups.

Question: I'm a physicist, so I'd usually be happy with broad sketches of proofs (for instance, one can turn the sum of simple Lie algebras into a product of Lie groups via the exponential map) and examples (eg. the Standard Model of particle physics). However, I'd like to see a mathematical proof of the above statement, especially the derivation of the group structure. Further insights as to how to find the discrete group to remove are also welcome.

• What you out wrote is not quite right, for this description of G you have to assume that each $G_i$ is simply connected. You can get rid of the discrete subgroup by assuming that G is simply connected. What book are you reading? The proofs are a bit complicated (if done from scratch) and best read in a book, I do not think MSE is the right forum for this. The key step is that each Lie algebra determines a unique simply connected Lie group. – Moishe Kohan May 26 '16 at 22:00