Mmm, I would say that looks a little odd. The Galilean algebra can be obtained by a contraction of the Poincare algebra. In this contraction you scale, in the Poincare algebra, the spatial P_i and K_i by a factor epsilon. Then you calculate again the Lie algebra and take the limit epsilon --> 0. This epsilon can be seen as 1/c, and the contraction can be regarded as sending c to infinity.

The J's constitute the SO(3) subgroup of Poincare which stays untouched, and the boosts will commute. I would say that K is of order 1/c, not 1/v.

Yes :) Would you ever find the tendency of wanting to know more about it in a pedagogical way: the notes of Robert Gilmore about Lie groups explain it quite well, in chapter 13.

Well, because in this particular contraction you leave the SO(3) untouched (because SO(3) is a subgroup of both the Poincare as the Galilei group, so you want to keep it), the generators J of SO(3) are not scaled by epsilon. I think that's what Weinberg means, but I'll check his statement.

But I assume you mean by J the [itex]J_i[/itex] which generate the rotations, not the [itex]J_{\mu\nu}[/itex], right?

Ok, I have checked. My guess is that the 1/v of K is a typo, and should be read as 1/c. That would mean that K is of order [itex]\epsilon^{1}[/itex]. The J's are indeed the generators of SO(3), and because of the particular contraction you take they scale like [itex]\epsilon^{0}[/itex].