Why Isn't $L/\text{rad}(L),$ Commutative?

Question

My question is asking for help understanding the Lie algebra analogue of the following process:

Given a group, one of the most complicated things about that group is it's lack of commutativity, and we can measure that with commutators because $$ab = [a,b]ba = (aba^{-1}b^{-1})ba,$$ so we package all commutators into a subgroup, $$G' = [G,G],$$ and then throw away all the non-commutativity by forming the quotient $$G/G' = G/[G,G],$$ which is the largest Abelian group we can form from a given group, great - now for finite groups we can use fundamental theorem of finite abelian groups to classify $G/G'$, and so all that's left to do to fully classify a group $G$ is to analyze it's commutator subgroup $G'$, but with this we can repeat the process, deriving $$G'' = [[G,G],[G,G]]$$ and then analyzing $$G'/G'' = [G,G]/[[G,G],[G,G]],$$ iterating if necessary... A group is solvable if this process terminates in the identity, equivalently $G^{(n)} = 0$, and we can fully classify a finite group in terms of a sequence of Abelian groups if it's solvable, great... This followed page 15 here.

Now, for a Lie Algebra $L$, Erdmann shows that if $I$ is an ideal of $L$ then $L/I$ is abelian iff $I$ contains the derived algebra $L' = [L,L]$, defines a lie algebra to be solvable if $L^{(n)} = 0$ (e.g. $L^2 = [[L,L],[L,L]]$), then defines the radical $$\text{rad}(L)$$ to be the largest solvable ideal of $L$. Seems like you do this to ensure you get the largest Abelian quotient, thus so far exactly copying the above process, that is - it seems like the radical is the analogue of $G' = [G,G]$ so that $G/G'$, I mean $$L/\text{rad}(L),$$ is Abelian. In the context of Lie Algebras, $L/\text{rad}(L)$ is defined to be a semi-simple Lie algebra, having no non-trivial ideals, and will be shown to be a direct sum of simple lie algebras, but Erdmann defines a simple lie algebra to be non-Abelian. It's like you copied a process that generates something Abelian, $G/G'$, $L/\text{rad}(L)$, but then you just define it to not be Abelian, how does this make sense? Why isn't $L/\text{rad}(L)$ Abelian when you copied a process to generate Abelian groups? This is apparently the Levi decomposition, page 28 here.

Answer 1

The answer to the question is, that $L/rad(L)$ is semisimple, because we divide out the largest solvable ideal of $L$, and a Lie algebra without non-zero solvable ideals is semisimple, see here. But a semisimple Lie algebra is not abelian. The first example here is the Lie algebra $\mathfrak{sl}_2$ with brackets $[e,f]=h,\;[h,e]=2e,\;[h,f]=-2f$. Since the brackets are not all zero, the Lie algebra is not abelian.
In contrast, $L'=[L,L]$ is the commutator ideal, and as in the group case, $L/L'$ is abelian.
In other words, the interpretation of radical and derived ideal is quite different.

Answer 2

There are different ways of approaching the study of groups and of Lie algebras. It is true that we can study a group by looking at the sequence of commutators, but we then also have to figure out how these quotients fit together to give the structure of the whole group. This is not straightforward: finite abelian groups are classified but it is not easy to deduce from that a classification of all finite non-abelian groups. Another important sequence of normal subgroups is a composition series, and the quotient groups in that case are simple groups, so we can also think of groups as being made up from simple groups. Lie algebras (at least in case of characteristic zero) are easier because of Levi's Theorem which says that every finite-dimensional Lie algebra can be written as a semi-direct sum of the radical and a semisimple subalgebra. This splits the study of such algebras into three problems: study the semisimple Lie algebras, study the solvable Lie algebras and study how they fit together via this sum.