I am reading Arnold's ODE but I cannot solve this problem.
This is on page 79.
$\mathbf{Problem.}$ Suppose a one-parameter group of symmetries of a direction field in an $n$-dimensional domain is known. Reduce the problem of integrating the corresponding differential equation to finding the integral curves of a direction field on a domain of dimension $n-1$.
This problem comes with a hint, which I don't know how to prove either.
Hint: The space of orbits of the symmetry group has dimension $n-1$
$\mathbf{Q1.}$ How to prove the hint ?
My thought on Q1.
The symmetry $\mathbf{g}$ is a differmorphism such that $\mathbf{g'(x)v(x)}=\mathbf{v(g(x))}$.
It's a PDE of $n$-dimension but solving $\mathbf{g(x)}$ explicitly does not look promising to prove the hint.
$\mathbf{Q2.}$ If the statement in the hint is true, then how to reduce the ODE by one dimension?
My thought on Q2.
Locally, $\mathbf{g(x)}$ acts as translation so $\mathbf{g'(x)v(x)}=\mathbf{v(g(x))}$ will go like $\mathbf{v(x)}=\mathbf{v(x+dx)}$.
If we have $n-1$ dimension local translation, the direction field will only depend on one coordinate. The original problem will reduce to integrate the curve only on one coordinate and not on $n-1$ dimension locally.
Please help.