Ok, going down same rabbit hole as Edwards, trying to explain Galois in modern terms, in particular, why the table at the bottom of page 916 of the linked AMS Notices article, which is on page 116-117, of Neumann, does display the Galois group of the equation g whose roots are a,b,c….
Permutation groups as “substitutions”
We will describe the Galois group of an equation as a certain group of permutations, or substitutions, of the roots. To clarify, a permutation of say the first 5 integers, is a list of those integers, in some order, say: (2, 1, 4, 5, 3). One may consider this as defining a substitution, by taking as the original permutation the usual ordering (1, 2, 3, 4, 5).
Then the permutation (2, 1, 4, 5, 3), defines the substitution taking (1, 2, 3, 4, 5) to (2, 1, 4, 5, 3), i.e. taking 1—>2, 2—>1, 3—>4, 4—>5, 5—>3.
If we had begun with a different permutation as original, say (2, 1, 4, 5, 3) rather than the usual one, then this same permutation (2, 1, 4, 5, 3) would define a different substitution, namely the identity substitution taking (2, 1, 4, 5, 3) to itself.
But once an initial arrangement has been chosen, then each arrangement, i.e. each permutation, defines a unique substitution. Thus when an initial permutation is chosen, we use the terms permutation and substitution interchangeably. Thus if we are given the permutations (1, 2, 3), (2, 3, 1), (3, 1, 2), we think of the first one as the initial arrangement, and then the three given permutations define three substitutions, namely those taking the first to itself, the first to the second, and the first to the third. These three substitutions define a cyclic group of order three. The sequence (2, 3, 1), (3, 1, 2), (1, 2, 3) define exactly the same substitutions, where now (2, 3, 1) is the initial arrangement. It is the group of substitutions that matter; it can be described using any initial ordering of the elements.
We are used to discussing permutations as reorderings of the usual ordering of integers, i.e. we always take the usual order as initial, so to us a permutation is the same as a substitution, but roots of a polynomial do not have a usual ordering, and Galois was trying to emphasize that his group does not depend on the choice of ordering.
Automorphisms of a simple field extension:
Let F(V) = K be a simple algebraic field extension of degree n of F. Then V satisfies a unique irreducible monic polynomial h of degree n over F, and we have an isomorphism of fields K = F(V) ≈ F[X]/(h), i.e. V corresponds to X, mod h. Assume K contains all n roots of h, say V, V’, V’’,……,V^(n-1).
Then K is generated over F also by every root of h, i.e. K = F(V) = F(V’) = F(V’’) = ……. Thus the isomorphism of K with F[X]/(h) can be given so that Xmodh, corresponds to any one of the n roots V, V’, V’’,…. In particular, by composing these isomorphisms K—>F[X]/(h)—>K, we get exactly n automorphisms of K over F, and these take an element p(V) to itself, or to p(V’), or to p(V’’),…..
The Galois group of an equation
Now let F be the coefficient field of g, and let a,b,c…. be all roots of g in a suitable extension field, e.g. an algebraic closure of F. Then let K = F(a,b,c…) be the field generated over F by the full set of roots of g, i.e. K = the splitting field of g over F. Suppose m = degree(g), so there are m roots a,b,c…., and let n = the degree of K over F, i.e. the dimension as vector space.
Then by the theorem of the “primitive element”, proved earlier by Galois, the splitting field K is actually a simple extension of F, i.e. there is some element V such that K = F(a,b,c…) = F(V). This means that every element of the splitting field, in particular every root a,b,c…, is expressible as a rational function, in fact actually as a polynomial, in V, with coefficients in F. E.g. let a = ƒ(V), b = ƒ1(V), c = ƒ2(V),…, ƒm-1(V). (don't know why these alternate f's look different)
Now since the degree of K over F is n, the degree of the irreducible polynomial of V over F is n, so let V, V’, V’’,…,V^(n-1) be the full set of roots of this polynomial.
Then the splitting field K has exactly n automorphisms over F defined by sending an element p(V) to itself, or to p(V’), or to p(V’’),…., p(V^(n-1)). Moreover these automorphisms form a group. Since the roots a,b,c,…of g, generate K as a field over F, each automorphism of K is determined by its restriction to the set of roots. Moreover, each root of g is mapped by any F-automorphism of K to another root of g. Hence the group of automorphisms of K over F restricts isomorphically to a group of “permutations” (substitutions) of the set of roots of g.
Choosing some initial ordering of these m roots, we write them as polynomials in V, i.e. choose as above a = ƒ(V), b = ƒ1(V), c = ƒ2(V),…, ƒm-1(V). then we get the group of the equation g from the following sequence of n permutations of these m roots:
ƒ(V), ƒ1(V), ƒ2(V),…, ƒm-1(V).
ƒ(V’), ƒ1(V’), ƒ2(V’),…,ƒm-1(V’).
…….
ƒ(V^(n-1)), ƒ1(V^(n-1)),…….,ƒm-1(V^(n-1)).
Compare this to the table at bottom of page 916 of the AMS article of Edwards, and to that at top of page 117, of Neumann. Galois explains this in about two pages, (115,117, Neumann), which I recommend.
Remark: Edwards objects, on page 917, to how Galois describes the actions of the permutations on the roots in this table, but in my mind, his revised version of Galois' Prop.1 is no different from what I understood Galois to mean originally. I.e. I think the objection is due to a different interpretation of the term "rational function of the roots". It is true the action on such a function, of a permutation of the roots, is not well defined if you mean by that term "the value of the function", but it is well defined if you mean by that term "the form of the function", i.e. if you mean a rational expression in the roots, rather than the numerical value of that expression. I believe Galois made it clear he considered both meanings when he emphasized that "invariance" of a function, under a substitution, referred not just to invariance of the form, but also of the value of the function.
I.e. let p(x1,...,xm) be a rational function of m variables, with coefficients in the base field F, and given a permutation s of the variables, let (sp)(x1,...,xm) = p(s(x1),...,s(xm)).
If a1,...,am are the m roots of a polynomial g, in a splitting field K, there is no reason for the value p(a1,...,am), an element of K, to equal the value (sp)(a1,...,am) = p(sa1,...,sam), usually another element of K. If they are equal, i.e. if p(a1,...,am) = (sp)(a1,...am), Galois says "the rational function of the roots" p(a1,...,am), is invariant under the permutation s.
If the values are equal for all p such that p(a1,...,am) belongs to F, then Galois says s belongs to the group G of the polynomial g. Moreover, when G is so defined, he then says that conversely, if (the values) p(a1,...,am) = (sp)(a1,...,am) are equal for all s in G, then (the value) p(a1,...,am) belongs to F. This is Galois' definition of the Galois group of g.
It follows that if p(x1,...,xm) is any polynomial relation among the aj, i.e. such that p(a1,...,am) = 0, then also (sp)(a1,...,am)= 0, for any s in G, (since 0 is in F), which implies that each s in G acts, not just on polynomials p(x1,...,xm), but on the field K of their values p(a1,.,..,am). I.e. since the field K is a quotient of the polynomial ring F[x1,...,xm], by the maximal ideal of all such relations, and since s induces a ring automorphism on the polynomial ring, it then also induces a field automorphism of K, fixing F.
Thus to me, this rephrasing of Galois' Prop.1 by Edwards, although perhaps more precise, is not essentially different, nor is Galois' version "flawed". But these are matters of individual taste. For me, I never realized until reading Galois, that one can actually write down the Galois group in such an explicit elementary way. Of course "explicit" is generous, since upon examination, although one can prove that, in Galois' table, the entry ƒ(V') say, is one of the roots, one does not know which one it is!