Using the modern term, the Galois group originally is defined as follows:
Let $\Bbb K\subset\Bbb C$. Given any polynomial $f\in\Bbb K[x]$ with roots $r_1,...,r_n\in\Bbb C$, a Galois resolvent of $f$ is some $r\in \Bbb K[r_1,...,r_n]$ such that $\Bbb K[r]=\Bbb K[r_1,...,r_n]$. Choose $p_r$ the minimal polynomial of $r$ over $\Bbb K$ and let $t_1,t_2,...,t_m$ the roots of $p_r$ over $\Bbb C$. Express $r_k=f_k(r)$ for $f_k\in\Bbb K[x]$ then since $f(f_k(r))=p_r(r)=0$ for each $1\leq k\leq m$ while $p_r$ is irreducible, $p_r$ is factor of $f\cdot f_k$ for each $k$. Thus for any root $t_j$ of $p_r$, $f(f_k(t_j))=0$. Given additionally that $f$ is separable, it is not hard to see that for each $j$ the tuple $(f_1(t_j),...,f_n(t_j))$ is a permutation of $(r_1,...,r_n)$. Galois proved that the resolvent always exists for separable polynomials and these $m$ permutations form a group which is independent of the choice of the resolvent $r$.
Let $L$ be the splitting field of $f$ over $\Bbb K$. The current version of Galois group is defined as the group of all automorphisms $L\to L$ which fixes $\Bbb K$ pointwise. I have two questions:
(1) How does Galois himself come up with the definition of the first form? I cannot see the connection between "solvability of the solvent" and "the group defined with the conjugates of the solvent in the minimal polynomial" from the first glance. My book does not specify about this.
(2) Are the two definitions equivalent at all?