Let $G$ be a Lie group acting smoothly on a manifold $M$. Then, for each $p \in M$, it is obvious that the isotropy group $G_p = \{ g \in G : gp=p\}$ is a closed subgroup of $G$. A standard theorem (wikipedia link) about Lie groups tells you that closed subgroups are (embedded) submanifolds, so this is true in particular of the isotropy group $G_p$. I must confess, however, that for me this theorem is really more of a "fact". The proof looks a little complicated, and I have never gone through it. So, I would like to know
Question: Is there a way to see that $G_p$ is a submanifold more directly, i.e. can this be shown with less work than it takes to prove the more general statement about closed subgroups?
