In John Lee's book Riemannian Manifolds on page 33, Lee writes "Clearly a homogeneous Riemannian manifold that is isotropic at one point is isotropic at every point".
It seems that he means that he concluded this from the following argument: Suppose a Riemannian manifold $(M, g)$ is isotropic at $p \in M$ and homogeneous, and take $q \in M$, $v, w \in T_q M$ with $g_q(v, v) = g_q(w, w) = 1$. Because $M$ is homogeneous, there is a Lie group $G$ that acts smoothly and transitively on $M$ by isometries, so we can find $x \in G$ such that $x \cdot q = p$, hence $x_* v, x_*w \in T_qM$. Because $M$ is isotropic at $p$, there exists another Lie Group $H$ acting on $M$ smoothly by isometries such that the isotropy subgroup $H_p \subseteq H$ acts transitively on $T_pM$, so there exists $y \in H$ such that $y_*(x_* v) = x_* w$. We would then have $(x^{-1}yx)_*v = w$ as desired. However, Lee does not specify in the definitions of homogeneity and isotropy that $G, H$ have to be the same Lie group, so I am not sure why there is a Lie group acting smoothly by isometries on $M$ that contains all elements of the form $x^{-1}yx$, where $x \in G$, $y \in H$.
I do not want to use the fact that the isometry group of any Riemannian manifold can be made into a Lie group, since this fact is beyond the scope of this chapter, and is supposedly pretty difficult to prove. One solution I was thinking of was maybe using a coproduct (free product) of the Lie groups $G, H$, but I was not sure how to give this a topology or smooth structure. Is there a better way of finishing the above argument?
