In Ballentine's book on quantum mechanics (in 3rd chapter), he introduces the symmetry transformation of Galilean group associated with Schrodinger equation.
Now the Galilean group as such has 10 generators (3 rotations - $L_i$ , 3 translations - $P_i$, 3 boosts - $G_i$ and time translation - $H$). Apart from this the Schrodinger solution (the probability amplitude) is arbitrary upto to a phase factor ($e^{i\phi}$). Hence we include one more generator induced by the transformation of phase. With this the general Unitary transformation,
$$ U = \sum_{i=1}^3\Big(\delta\theta_iL_i + \delta x_iP_i + \delta\lambda_iG_i +dtH\Big) + \delta\phi\mathbb{\hat1} = \sum_{i=1}^{10} \delta s_iK_i + \delta\phi\mathbb{\hat1}$$
The commutation relations of the group altogether can be given as,
$$ [K_i,K_j] = i\sum_{n}C_{ij}^{\;\;n}K_n + ib_{ij}\mathbb{\hat1} $$.
Now, this commutation relation does not have the structure of Lie algebra. Since with Lie Algebra the elements are closed under commutation. This one has an extra element sitting with $\delta\phi\mathbb{\hat1}$.
What is really going on here ? Is this really a 11-parameter Lie group ? If so how do we convince about the algebra of generators ?