Motivating the Casimir element Weyl's theorem states that any finite-dimensional representation of a finite-dimensional semisimple Lie algebra is completely reducible.  In my mind, the "natural" way to prove this result is by way of Lie groups.  However, as a student, I first encountered Weyl's theorem in the textbook by Humphreys, in which he gives a purely algebraic proof.  I remember finding this proof very mysterious, and in particular it seemed that Humphreys pulled the Casimir element out of a hat.  Looking at the proof again now, I still find it somewhat mysterious, and if I had to present the proof in a graduate class, I don't think I would be able to motivate it very well.
How would you motivate this purely algebraic proof of Weyl's theorem?
 A: This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem. 
General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is a nontrivial irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel.
The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. Every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action via matrix coefficients. 
On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casimir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.
A: Maybe I should try to defend myself, or at least the self I was four decades ago when I improvised my graduate text.   But first I should disclaim any originality in the proof of Weyl's theorem, which I drew from Bourbaki (who didn't invent it either).    I was at the time far from being an anti-Bourbaki person and in fact actually associated with some of them.    But in my limited experience teaching real-life US graduate students, I was well aware that it was unrealistic to build up Lie algebras abstractly in the Bourbaki style.   In particular, I didn't want to talk about the universal enveloping algebra (and its center) until that was really necessary.    At the same time, I really felt that it was useful to talk about Weyl's theorem earlier than usual since I was emphasizing representation theory rather than general structure theory.   
Indeed, as a later addition to my book indicated, Victor Kac demonstrated in the mid-1970s the value of working with just a single Casimir-type operator in his approach to affine Lie algebras and what came to be known as the Weyl-Kac character formula.   
Anyway, it's clear that the approach of Weyl imitating the classical treatment of complete reducibility for finite groups is the most natural, but for this you have to be in the framework of compact Lie groups and invarioant integration.   There are advantages to working with Lie algebras directly in a purely algebraic framework, though of course some of the ideas lose their original group-theoretic motivation.    
For me the algebraic proof was a good illustration of the power of slightly abstract algebraic thinking, especially for students not previously exposed to such proofs.   But the main motivation for introducing the Casimir element would be to have an "invariant" commuting operator (essentially living in the undefined universal enveloping algebra) relative to the given finite dimensional representation.   Where this operator comes from is another question, since Casimir's own work had a physics origin.  Whether or not it helps to provide an intrinsic definition for the operator is another question, but for that there is an earlier MO question as well: 52587 along with my earlier historical question 41150.
A: I think that there are two things to be motivated: one is the Casimir, and the other is the proof of semi-simplicity.
First, for the Casimir, it might help to note that there is a ``formula-free" construction. A symmetric bilinear form $\kappa$ defines $\mathfrak{g}\simeq \mathfrak{g}^{\vee}$, and the Casimir is the image in $U(\mathfrak{g})$ of the element corresponding to the identity under the isomorphism $\mathfrak{g}\otimes\mathfrak{g}\simeq \mathfrak{g}^{\vee}\otimes\mathfrak{g}$.
The idea of the proof is actually very simple: you construct an element of the center of the algebra which ``detects" the trivial representation. I.e., it acts by zero on the trivial representation and by a non-zero scalar on all simple (finite-dimensional) representations. (This is in exact analogy to what happens for finite groups in characteristic zero: then $1-\frac{1}{|G|}\sum_{g\in{G}}\delta_g$ has the same property).
Once you have such a central element, let's call it $C$, the proof that finite-dimensional representations are semi-simple is easy. For all $V,W$, note that:
$$\operatorname{Ext}^i(V,W)=
\operatorname{Ext}^i(\mathbb{C},\underline{\operatorname{Hom}}(V,W))$$
(where $\underline{\operatorname{Hom}}$ is internal $\operatorname{Hom}$ relative to the usual tensor product of $\mathfrak{g}$-modules, and both $\operatorname{Ext}$s are of $\mathfrak{g}$-modules) because formation of internal $Hom$ is exact in both variables. Therefore, it's enough to show that $\operatorname{Ext}^1(\mathbb{C},V)=0$ for all finite-dimensional $\mathfrak{g}$-modules $V$ (substitute $\underline{\operatorname{Hom}}(V,W)$ for $V$).
Clearly any such $V$ has finite length, so by devissage, it's enough to prove for simple modules. Either $V$ is trivial or it is not. If $V$ is non-trivial, then $C$ acts on $\operatorname{Ext}^i(\mathbb{C},V)$ by two different scalars: the one by which it acts on $V$ and by $0$ (by which it acts on $\mathbb{C}$). Therefore, this vector space must be zero. If $V=\mathbb{C}$, then $\operatorname{Ext}^1(\mathbb{C},\mathbb{C})=0$ since for any extension $E$, the homomorphism $\mathfrak{g}\to\operatorname{End}(E)$ maps to the 1-dimensional subspace sending $E$ to $\mathbb{C}$ and sending $\mathbb{C}$ to $0$, but since $\mathfrak{g}$ has no codimension 1 ideals this must be the trivial homomorphism.
Of course, basically the same proof goes through for finite groups and is essentially the same as the usual proof.
A: I'm surprised that no one has mentioned the differential-geometric motivation for the Casimir element.
Suppose $\mathfrak{g}$ is the Lie algebra of a simple compact Lie group $G$.  Then $U(\mathfrak{g})$ is the algebra of left-invariant differential operators on $G$. 
$G$ has a natural bi-invariant metric, given by the Killing form.  But now there is an obvious central element of $U(\mathfrak{g})$: the Laplacian associated to this metric!  This is precisely the Casimir element.  It's centrality follows "by pure thought" from the bi-invariance of the Killing form.  To obtain Casimir elements in general and prove their centrality, simply copy the formulas from the case of the Lie algebra associated to a Lie group.
I had always assumed this was the original motivation for the construction.
A: I will toot my own horn a bit by mentioning the lecture notes for a summer course I taught on this very subject last year.  The relevant lecture is this one and in it, I decided (based on answers given to my question Why the Killing form?) to present Casimir as an analogue of the averaging operation for proving "Weyl's theorem" (i.e. Maschke's theorem) for finite groups.  I never did find a good way of actually drawing an exact parallel, but I did try to structure the proof in such a way that it was clear this was how it functioned.  (All the actual math is copied from Humphreys.)
