I do not understand one line in the proof of the following theorem. Within the proof of d) it's stated that, "if $y \in \mathfrak{h}$, then y commutes with x and hence also with s and n." I do not get why the fact that y commutes with x=s+n already implies that y commutes with both s and n. Does anybody know how the to things are related? I feel like the reason must be that n is the nilpotent and s the semisimple component, but what is the concrete connection?
Theorem: Let $ \mathfrak{h} $ be a Cartan subalgebra of a semisimple Lie algebra $ \mathfrak{g} $. Then:
a) The restriction of the Killing form of $ \mathfrak{g} $ to $\mathfrak{h} $ is nondegenerate.
b) $ \mathfrak{h} $ is abelian.
c) The centralizer of $ \mathfrak{h} $ is $ \mathfrak{h} $.
d) Every element of $ \mathfrak{h} $ is semisimple.
Proof for d:
Let $x \in \mathfrak{h}$ and let s ( resp. n) be its semisimple (resp. nilpotent) component. If $y \in \mathfrak{h}$, then y commutes with x and hence also with s and n. We therefore have $s ,n \in \mathfrak{c(h)}=\mathfrak{h}$. However, since y and n commute and $ad(n)$ is nilpotent, $ad(y) \circ ad(n)$ is also nilpotent and its trace $B(y,n)$ is zero. Thus n is orthogonal to every element of $\mathfrak{h}$. Since it belongs to $\mathfrak{h}$, n is zero by (a). Thus x = s which shows that x is indeed semisimple.
To turn Callum's and my comments into an answer, "standard" facts about Jordan decomposition in Lie algebras are that
a. for an endomorphism $e$ of a finite-dim. vector space $V$, there exist polynomials without constant term $p(T), q(T)$ such that $e_s := p(e)$ and $e_n := q(e)$ (making use of the associative structure of $End(V)$) are the semisimple and nilpotent components of $e$, respectively;
b. if $\mathfrak g$ is any semisimple Lie algebra, then there is a well-defined notion of decomposition of each element $x$ into a semisimple and nilpotent part $x_s+x_n$, commuting with each other, and such that when taking the adjoint representation, $ad(x_s)$ and $ad(x_n)$ are the semisimple and nilpotent parts of $ad(x)$ viewed as element of $End(\mathfrak g)$, respectively.
See e.g. Humphreys Introduction to Lie Algebras and Representation Theory 4.2 or Bourbaki's volume on Lie theory, vol. I §6 no. 3.
Now to apply that to the given situation, one assumes that $ad(x) (y) =0$ and wants to show e.g. $ad(x_s) (y) =0$. But from a and b it follows that $ad(x_s)$ can be written (as element of $End(\mathfrak g)$) as a polynomial in $ad(x)$, so ...