Given a field $F$ and $f(x) \in F[x]$ a separable polynomial of degree $n$, and denote the roots of $f$ by $\alpha_1,…,\alpha_n$. We can consider the Galois group Gal$(f)$ as a subgroup, $H$, of $S_n$. Keith Conrad states:
"Two different choices for indexing the roots of f(T) can lead to different subgroups of Sn, but they will be conjugate subgroups"
(Link: https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisaspermgp.pdf)
My question is how does one prove the above statement – or in other words; that permutation of $\alpha_1,…,\alpha_n$ is that same as conjugation of $H$ inside $S_n$?
I know that two groups, say $H$, $T$ (where $T$ is a group in $S_n$ which corresponds to permutation of roots of Gal$(f)$), are conjugate inside $S_n$ iff there exist a $\pi\in S_n$ s.t.:
\begin{equation}
H=\pi T \pi^{-1}
\end{equation}
But how does one prove that this equality holds?
Best Answer
The proof of this question is to be found in:
https://kconrad.math.uconn.edu/blurbs/grouptheory/conjclass.pdf
(another paper from Keith Conrad)
More precisely, it follows straight from thr 5.1. in the paper.