Lie groups: Show $L_y \circ x_t = x_t \circ L_y$ where $x_t$ is the flow of a left invariant differentiable vector field

Question

My definition of a left invariant vector field is one for which the equation $dL_xX = X$ is satisfied, where $X$ is the differentiable vector field and $x \in G$ where G is a Lie group.

Consider the two sides of the equation:

$L_y \circ x_t (p) = L_y(x_t(p)) = yx_t(p)$

$x_t \circ L_y (p) = x_t(L_y(p)) = x_t(yp))$

Why are these equal?

Answer 1

Hint: Let $\theta^{(p)}(t) = yx_t(p)$ and $\psi^{(p)}(t) = x_t(yp)$. Compute both ${\theta^{(p)}}'(0)$ and ${\psi^{(p)}}'(0)$. Then use uniqueness of integral curves.