What is the defining property of reductive groups and why are they important? Having read (skimmed more like) many surveys of the Langlands Program and similar, it seems the related ideas apply exclusively to groups that are "reductive".
But nowhere, either in these surveys or elsewhere, have I been able to find a simple and compelling definition of what it means for a group to be reductive and why this property is important and why Langlands was naturally led to frame his conjectures for them and not, say, for any group.
Any suggestions or links?
 A: I understand two reasons why reductive groups play such a central role nowadays.
First:
A maybe not so satisfying answer to "Why are reductive groups important in the Langlands program" is: $\operatorname{GL}_n$ is important.
It happens that many conjectures and motivating ideas in the Langlands philosophy for $\operatorname{GL}_n$ can be easily formulated in the more general setting of arbitrary reductive groups.  Proving things in this more general setting is another matter entirely.
I don't believe there is any definition of reductive group which motivated Langlands to formulate his conjectures that way.  Rather, the structure theory of arbitrary reductive groups is so similar to that of $\operatorname{GL}_n$ that the important ideas for the general linear group carry right over.  
An example is parabolic induction.  A reductive group $G$ has various Levi subgroups $M$, which are reductive groups in their own right, but generally less complicated than $G$.    Many representations of $G$ are parabolically induced from representations of Levi subgroups.  So the idea is that if we understand representations of "smaller" reductive groups $M$, we will know more about representations of the bigger group $G$.  This is especially evident in the case $G = \operatorname{GL}_n$, where the Levi subgroups look like products of smaller $\operatorname{GL}$s.
Second:
Number theoretic objects associated to arbitrary reductive groups are expected to be special cases of those associated to the general linear group.  I sketch this argument below.  Breaking up these objects associated to $\operatorname{GL}_n$ into subcases might be insightful.
If $^LG$ is the Langlands dual group of a reductive group $G$ over a field $k$, $r: \, ^LG \rightarrow \operatorname{GL}_n(\mathbb C)$ is a representation, and $\pi$ is a representation of $G$, there should be an associated L-function $L(s,\pi,r)$.  The Langlands correspondence should (roughly) associate to this a homomorphism $\rho: \operatorname{Gal}(k_s/k) \rightarrow \space ^LG$.  This correspondence should preserve the L-functions
$$L(s,\pi,r) = L(s, r \circ \rho)$$
The composition $r \circ \rho$ is an $n$ dimensional Galois representation which by the Langlands correspondence for $\operatorname{GL}_n$ ought to associate this to a representation $\Pi$ of $\operatorname{GL}_n(k)$.  That Langlands correspondence should also preserve L-functions:
$$L(s, r \circ \rho) = L(s,\Pi)$$
and therefore
$$L(s,\pi,r) = L(s,\Pi)$$
So the philosophy is that everything should eventually come back to $\operatorname{GL}_n$.
A: A linear algebraic group is unipotent if it consists entirely of unipotent linear transformations, i.e. $I+N$ with $N$ nilpotent.
A linear algebraic group is reductive if it has no connected normal unipotent abelian subgroup except $\{I\}$.
See A. Borel and J. Tits, Groupes reductifs, IHES Publ Math. 1965 p. 59 or Armand Borel, Linear Algebraic Groups, p. 158.
But the essential idea is that a reductive group (over the complex numbers) has all finite dimensional representations completely reducible, so behaves like a finite group. 
A motivating example: the matrices $\begin{pmatrix}1&x\\0&1\end{pmatrix}$ form a group of matrices acting on the plane, preserving the horizontal axis, but not any complementary line. So this is not reductive. A linear algebraic group is reductive if it has no normal subgroups like this. Clearly such a subgroup gets in the way of having all representations completely reducible.
Another: the group $GL(2)$ of all $2 \times 2$ complex matrices is reductive, because it contains the unitary group $U(2)$, which is compact (so behaves like a finite group), as a Zariski dense subgroup, i.e. any holomorphic function $f$ defined on an open set $W \subset GL(2)$ with $W \cap U(2) \ne \emptyset$, and vanishing on that intersect, vanishes. So holomorphic representations of $GL(2)$ determine and are determined by continuous representations of $U(2)$.
A: For algebraic groups over $\mathbb{C}$, we have the following deep theorem of Cartan, Chevalley and Mostow describing concretely the reductive groups:
For an algebraic subgroup $G$ of $\mathrm{GL}_n(\mathbb{C})$, the following conditions are equivalent:


*

*$G$ is reductive.

*$G$ has a subgroup $K$ which is compact for the usual topology, and Zariski-dense in $G$.

*$G$ is conjugate to a subgroup of $\mathrm{GL}_n(\mathbb{C})$ which is self-adjoint, that is stable under $g \mapsto g^* := \overline{g}^T$.


The basic examples to have in mind: $\mathrm{GL}_n(\mathbb{C})$ and $\mathrm{SL}_n(\mathbb{C})$ are reductive, while the subgroup of upper-triangular matrices is not reductive (when $n \geq 2$). 
As Ben McKay explained, a complex algebraic group is reductive if and only if all its algebraic representations are semi-simple.
