Suppose that $M$ is a manifold with a metric $g$, and three independent (independent in the sense of Killing vector fields) Killing vector fields $K_1,K_2,K_3$ are given with commutation relations $$ [K_i,K_j]=\sum_k\epsilon_{ijk}K_k. $$ Here $\epsilon_{ijk}$ is the standard (algebraic) Levi-Civita symbol. This setup is standard when spherical symmetry is given. Usually when spherical spacetimes are given, additional assumptions are made.
I am trying to see if those additional assumptions are necessary. In particular, does this setup imply that the orbits of the isometry group generated by the $K_i$ is two-dimensional?
I tried to prove by contradiction and assumed that the $K_i$ are pointwise independent. Then by Frobenius' theorem, $M$ is foliated (at least locally) by three dimensional surfaces to which the $K_i$ are tangent, and the $K_i$ then form a frame, and are also Killing fields of the induced metric. I tried to use the fact that 1) the $K_i$ is a frame, 2) the $K_i$ are Killing, 3) the $K_i$ satisfy the above commutation relation to arrive at a contradiction. If such a contradiction had been obtained then it would have followed that the $K_i$ cannot be pointwise independent, and thus the distribution generated by them would be two or one dimensional, but a one dimensional manifold cannot accomodate three independent Killing fields, so then the distribution is two-dimensional. Unfortunately I have not been able to prove this.
So, does the above setup imply that the distribution generated by the $K_i$ is two-dimensional? If so how? References are also welcome if not a direct proof.