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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?

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Define a map $$F:G\to M,\quad F(g)=g\cdot p.$$ Then, you can easily show that $F$ has constant rank. Hence, $F^{-1}(p)=G_p$ is a properly embedded submanifold of $G$.

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