I'm working on a lemma concerning some Galois theory and arithmetics. Let $p$ be an odd prime and $K/F$ be a finite Galois extension of number fields of order prime to $p$ with Galois group $H$. Let $S$ be a finite set of places of $K$ and denote by $K_S(p)$ the maximal $p$-extension of $K$ that is unramified outside the places of $S$ (compositum of all Galois extensions of $K$ that are of $p$-power-degree). I know that $H$ acts transitively on the set of all places of $K$, but my claim is now
Claim: If the set $S$ is invariant under the action of $H$, then $K_S(p)/F$ is normal.
Attempt of a proof: Let $\sigma \in H$ be an F-automorphism of $K$. This gives a F-homomorphismus $\tilde\sigma: K_S(p) \longrightarrow \bar K$, where $\bar K$ is a fixed algebraic closure (this is an algebraic closure for $K_S(p)$ as well). If the image of $K_S(p)$ under $\tilde\sigma$, $M$, was not $K_S(p)$, then the field compositum $M \cdot K_S(p)$ would have the property of being unramified outside of $S$ again, because $\sigma$ and hence $\tilde\sigma$ leaves the places of $S$ invariant. That contradicts the fact that $K_S(p)$ is maximale. Therefore $M$ has to be $K_S(p)$ and $\tilde\sigma$ is in fact an $F$-automorphism of $K_S(p)$.
Is that proof correct?
Thank you for sharing your experience and knowledge with me..