In studying representations of a reductive group G, a standard technique is to use parabolic induction. The idea is that one studies such groups as a family (or perhaps in smaller families, like say just taking (products of) GL_n's) and then trying to understand representations of larger groups via representations of their smaller subgroups in the same family. Induction is the natural functor for doing this, but inducing directly from a Levi subgroup L (a natural class of reductive subgroups) gives you very large representations — the homogenous space G/L is not projective and so bundles on it will have lots of sections. The remedy is to use parabolic induction: take the bigger subgroup P which contains L as it's reductive part, and extend representations of L to representations of P by letting the unipotent radical U of P act trivially. Then you induce the result from P to G and (since G/P is projective) you get a much smaller representation, which you can feasibly study via Hecke algebras etc.
Now in all of the above, it seems to me to make more sense to think of L not as a subgroup of G at all, but rather the reductive quotient of the parabolic group P — that is, as a subquotient of G only. Now however, comes my question. One of the theorems you prove when studying parabolic induction (I'm begin sloppy about the context deliberately, but if you like, when studying finite group of Lie type say) is that it does not depend of the choice of a parabolic containing L. Since the same Levi can lie in two non-conjugate parabolic subgroups, this draws you back to the conclusion that it was better to think of parabolic induction as an operation on representations of subgroups of G after all.
How do people think about this? "Why" is parabolic induction independent of the choice of parabolic? The proof of this in the context of finite groups of Lie type goes as follows: prove a formula for the composition of a parabolic induction followed by a parabolic restriction (using two arbitrary pairs of Levi inside parabolic) which expresses the composition as a sum of (parabolic) inductions and restrictions for smaller groups (as for the normal Mackey identity in finite group). Then consider the Hom space between two parabolic induction functors for the same Levi, and notice that Frobenius reciprocity lets you write this in terms of the Mackey identity you have, and then induction on rank finishes you off.
This argument doesn't feel very enlightening me (though that might be because I don't really understand it), and I know another slick geometric proof in the context of character sheaves on a Lie algebra, which I'm not sure I understand the representation-theoretic content of. The result also is not just a curiosity — the Harish-Chandra strategy of classifying representations of G via cuspidal data wants to associate to an irreducible representation of G a unique "cuspidal datum" $(L,\rho)$ consisting of a Levi subgroup L and a cuspidal representation of L (up to conjugation in G), and this doesn't make sense without the independence of parabolic.
Also I don't remember the situation for p-adic groups: the "Mackey formula" argument I sketched should show that the parabolic induction functors you get don't depend on the parabolic at least in at the level of the K-group (which would be all you need to get a cuspidal theory running) but the functors themselves are perhaps not isomorphic (because the identity becomes some sort of filtration on the composition of the induction and restrictions, which is probably already is really in the finite groups type, so perhaps the same thing happens already for finite groups of Lie type when you study modular representations?
Best Answer
Let me contribute some confusion. In the situation I am familiar with (algebraic groups), one may first restrict to a Borel subgroup $B$ with the property that $B\cap L$ is a Borel subgroup of $L$. Recall that if $P$ contains both $L$ and $B$, then restricting from $P$ to $B$ and next inducing back up to $P$ does nothing to $P$-modules. So instead of inducing up from $P$ one might as well first restrict to $B$ and then induce up from $B$ to $G$. And all Borel subgroups are conjugate.