I was working on the following problem:
Let $K$ be a field of characteristic $0,$ $f \in K[X]$ an irreducible polynomial and $L/K$ a splitting field for $f.$ Let $\alpha_1, \dots, \alpha_n \in L$ be the roots of $f.$ Show that $Gal(L/K)$ is abelian if and only if for all $\sigma \in Gal(L/K)$ there exists a $p_\sigma \in K[X]$ such that $\sigma(\alpha_i) = p_{\sigma}(\alpha_i)$ for all $i.$
The backward direction is simple and the forward direction depends on one step which is the following: If $\sigma(\alpha_i) = \alpha_j$, $H_i$ is the subgroup that corresponds to $K(\alpha_i)$ under the Galois Correspondence, and likewise for $H_j$, can we say that $\sigma H_i \sigma^{-1} = H_j$?
Someone told me that this is always the case. I don't know if they meant in this situation specifically, because I know not all subgroups of Galois groups are conjugate. Were they right in any sense? From here we can use the fact of abelian group to say $ H_i = H_j$.
Thanks
