The general Frobenius Theorem stating that
Let $u_1,\dots,u_k$ be $k$ smooth linearly independent vector field on $M$. Let $$ W=\operatorname{Span}(u_1,\cdots,u_k) $$ Then $[u_i,u_j]\in W$ for any $i,j$ if and only if there exist foliation by $k$ dimension hypersurface tangent to $M$.
To my understanding,
there exist foliation by $k$ dimension hypersurface tangent to $M$.
means there is a cooridnate $(w_1,\cdots,w_{n-k},x_1,\cdots,x_k)$ such that $$ u_i=\frac{\partial}{\partial x_i} $$
I know the proof of the special case where $k=2$. To prove the general case, there is a hint saying using induction on $k$. However, I am not clear how to commit the induction. I tried to consider $[u_{k-1},u_k]$, but we only have $$ [u_{k-1},u_k]\in\operatorname{Span}(u_1,\cdots,u_k) $$ rather than $$ [u_{k-1},u_k]\in\operatorname{Span}(u_{k-1},u_k) $$ So I cannot perform the simpler case. Could anyone help?
Here is a way of thinking about Frobenius theorem that might help to understand the theorem and its proof.
The condition that one has a set of $k$ linearly independent complete vector fields on $M$ closed by the Lie bracket of vector fields means that one has a Lie subalgebra of the Lie algebra of vector fields of $M$.
One can, then, invoke Picard-Lindelöf's theorem (on existence and uniqueness of solutions for ODE's) and use the flow of each vector field to provide an action of the associated Lie group on $M$. This action induces a foliation on $M$, the orbits of this action are the leaves of the foliation.