Oh, good Lord! I was trying to find a way to prove that was associative, instead of answering if that was associative or not! I'm such an idiot.
By the way, I came with a property of an algebra A on vectors over a basis ##x_i## such that
##x_i*x_j = l_{ijk}*x_k##
that would make it associative; without taking much space, for three vectors u, v, w in Einstein's notation:
##u = u_i*x_i; v = v_i*x_i; w = w_i*x_i; ##
then ##u*(v*w) = (u*v)*w## makes ##u_a*x_a*(v_b*x_b*w_c*x_c) = (u_d*x_d*v_e*x_e*w_e)*w_f*x_f##
as ##u_?, v_?, w_?## are all free variables, then ##d=a, e=b, f=c##, and ## x_a*(x_b*x_c)=(x_a*x_b)*x_c##, that is the algebra is associative if the multiplication of the basis is associative; then, replacing with the expression for multiplication of the basis:
##x_a*(x_b*x_c) = x_a*(l_{bcg}*x_g) = l_{bcg}*x_a*x_g = l_{bcg}*l_{agh}*x_h##
##(x_a*x_b)*x_c = (l_{abi}*x_i)*x_c = l_{abi}*x_i*x_c = l_{abi}*l_{icj}*x_j##
as ##x_h## and ##x_j## are independent (that is, I assume they are independent), then ##h=j##; then, eliminating ##x_h## and renaming ##i=g## and ##d=h## to make it more readable,
##l_{abi}*l_{icd} = l_{bci} * l_{aid}## for any ##a, b, c, d##
And that's where my skill ends; now, question on that... this seems like a tensor operation of some kind, is that true? That is, is ##l_{ijk}## a 3-tensor, and the expression above is a tensor operation of some kind?