TITLE: algebraic group G vs. algebraic stack BG
QUESTION [25 upvotes]: I've gathered that it's "common knowledge" (at least among people who think about such things) that studying a (smooth) algebraic group G, as an algebraic group, is in some sense the same as studying BG as an algebraic stack. Can somebody explain why this is true (and to what extent it is true)? I can get as far as seeing that quasi-coherent sheaves on BG are the same as representations of G, but it feels like there's more to it.
In particular, Scott Carnahan mentioned here that deformations of BG as an algebraic stack should correspond exactly to deformations of G as an algebraic group. I assume this means that any deformation of BG must be of the form BG', where G' is a deformation of G (as a group). It's clear to me that such a BG' is a deformation, but why should these be the only deformations?

REPLY [12 votes]: The stack $BG$ only recovers $G$ up to inner automorphisms, not canonically (as suggested by blah) - this can lead to serious issues in families or equivalently over a nonalgebraically closed field, as Shenghao's comment points out. One way to say this is the following: the loops in $BG$ are $G/G$, the adjoint quotient of $G$. On the other hand, if you give a map
$pt \to BG$ then the based loop space (fiber product of $pt$ with itself over $BG$) is $G$,
so you recover the group canonically.