The exercise is as follows:
Let $L$ be a Galois extension of $K$ whose Galois group is $G$. If $H \subset G$ is a subgroup, let $L'\subset L$ be the subfield that is fixed by $H$. Show that if $p \in O_K$ has a split factor in $L'$, i.e. if there is some prime ideal $P'$ lying above $p$ in $O_{L'}$ such that $e=f=1$ (both ramification and residual degree are $1$) then there is a conjugate of $D_P$ in $H$, where $D_P$ denotes the decomposition group of $P$ lying above $p$ in $O_L$.
I read a solution to this exercise which I don't quite understand, I would be grateful if somebody could explain it to me. The proof goes as follows:
Suppose $p$ has a split factor in $L'$. Then there is $P'$ above $p$ such that $D_{P'}=\{Id\}$. So, for all primes $P$ lying above $P'$ in $O_L$, if $\sigma \in D_{P}$, $\sigma\mid_{L'}\in D_{P'}=\{Id\}$, therefore $\sigma\mid_{L'}=Id$ and thus $\sigma\in H$ since it fixes $L'$
I don't understand why $L'$ is Galois, as far as I know a subextension of a Galois extension is not necessarily Galois, otherwise any finite extension would be Galois. So I don't know how Galois theory is relevant here, in particular to assert that $D_{P'}=\{Id\}$. This would be true if $L'$ were Galois, because the decomposition group is trivial iff the prime ideal is completely split. Now, what if $L'$ is not Galois? (i.e. if $H$ is not a normal subgroup). Any hint on how to go on? Am I right to believe that the solution is incomplete or flawed? Thank you for your time.