My question concerns two results in the neighborhood of the standard theorem of Myers-Steenrod that isometry groups of Riemannian manifolds are Lie groups. Both appear in the first chapter of Kobayashi's book "Transformation Groups in Differential Geometry", and the longer I think about them, the more I doubt whether Kobayashi's exposition can be completely correct. One of them is a theorem due to Kobayashi that can be stated in modern terms as follows:

Theorem 3.2: Suppose $M$ is a smooth manifold with a fixed global trivialization of its tangent bundle, and $G \subset \operatorname{Diff}(M)$ is the group of diffeomorphisms that preserve this trivialization. Then $G$ admits the structure of a Lie group acting smoothly on $M$ such that for any $p \in M$, the map $G \to M : g \mapsto g \cdot p$ sends $G$ diffeomorphically to a smooth submanifold of $M$.

I'm stuck on the final sentence of Kobayashi's proof, so in order to ask my question, I have to summarize the proof. It's based on a theorem of Palais (3.1 in Kobayashi's book), which for simplicity I will state here in the case of a closed manifold:

Theorem 3.1(closed case): On a smooth closed manifold $M$, suppose $G \subset \operatorname{Diff}(M)$ is a subgroup and $S$ is the set of vector fields on $M$ whose flows belong to $G$. If the Lie algebra generated by $S$ is finite dimensional, then $G$ admits the structure of a Lie group acting smoothly on $M$ such that $S$ is its Lie algebra.

This theorem confused me for a while before I realized there is some sleight-of-hand in the statement: we like to imagine $\operatorname{Diff}(M)$ as carrying a natural topology, but the statement of Palais's theorem doesn't mention any topology on this group, and indeed, the topology produced by the proof might be very different from what you expect. That's because the main step in the proof is to apply the following lemma:

Lemma: Suppose $G$ is a group and $H \subset G$ is a topological group contained in $G$ as a normal subgroup. If the map $H \to H : h \mapsto g h g^{-1}$ is continuous for every $g \in G$, then $G$ admits a unique topology for which $H \subset G$ is an open subset.

In Kobayashi's proof of Theorem 3.1, the subgroup $H \subset G$ is the connected Lie group of diffeomorphisms generated by the flows of vector fields in the set $S$, so he endows $G$ with the topology promised by the lemma, making $H$ the identity component of $G$, and then transfers the smooth structure from $H$ to the other components. I could buy that, except for the following detail:

**Question 1**: Could the "Lie group" $G$ resulting from Theorem 3.1 have uncountably many connected components?

For instance, one could apply the lemma above to the normal subgroup ${\mathbb R} \subset {\mathbb R}^2$, with ${\mathbb R}$ assumed to carry its usual Lie group structure. The result would be a *very* strange smooth structure on ${\mathbb R}^2$ in which it is diffeomorphic to a disjoint union of uncountably many copies of ${\mathbb R}$. That is almost certainly not what you want to do in any given situation, and according to what I regard as the standard definitions these days, the result is not a Lie group --- it is not even a manifold, because it is not second countable. But I don't see why such a scenario couldn't happen in Theorem 3.1, and am thus left unsure as to whether the theorem is even true according to standard definitions. (I don't have access to Palais's original proof of this theorem, but I have looked at the exposition in Postnikov's book "Geometry VI: Riemannian Geometry"; Postnikov states the lemma above on page 124 as Exercise 10.3 and does not discuss it any further.)

Now, here is a quick outline of Kobayashi's proof of Theorem 3.2, in which I will again make life slightly easier by assuming $M$ is closed. Recall that $M$ is endowed with a global trivialization of its tangent bundle, so there is a natural notion of "constant" vector fields on $M$.

**Step 1**: The map $G \to M : g \mapsto g \cdot p$ is injective.**Step 2**: The $G$-orbit of $p$ is a closed subset of $M$.**Step 3**: The subgroup $H \subset G$ consisting of flows of vector fields is generated by the Lie algebra of all vector fields that commute with the constant vector fields. Moreover, all vector fields in this Lie algebra are nowhere zero.**Last step**: Since the Lie algebra in step 3 is finite dimensional, Theorem 3.1 endows $G$ with a Lie group structure, and the map $G \to M : g \mapsto g \cdot p$ is now an injective immersion with closed image, thus its image is a submanifold of $M$.

Leaving my previous caveat about Theorem 3.1 aside, I was fine until the the very last part, but...

**Question 2**: It is certainly not true in general that the image of an injective immersion is a submanifold whenever it is closed; so how does Kobayashi's proof actually guarantee that the orbit is a submanifold?

I can imagine various horror scenarios: for instance, if the group in Theorem 3.1 really can end up having uncountably many components, then the map $G \to M : g \mapsto g \cdot p$ might immerse an $(n-1)$-dimensional "manifold" onto an $n$-dimensional neighborhood of $p$. That scenario is not actually so worrying since it then becomes obvious that the orbit is in any case a submanifold of $M$ (namely it is an open subset), but even if $G$ has only countably many components, I can imagine the orbit might be something like a sequence of disjointly embedded connected $(n-1)$-dimensional submanifolds converging to one that passes through $p$, and that would still be a closed image of an injective immersion.

I realize that Kobayashi's theorem also follows from the proof of Myers-Steenrod that isometry groups are Lie groups, and while I have not read the latter in detail, I see no reason to be skeptical of it (it certainly puts a lot more effort into proving that orbits are submanifolds). Of course, I was hoping I wouldn't have to read the rest of Myers-Steenrod because Kobayashi's proof looked (deceptively?) easier.