I am noticing that to explain a topic simply, it helps to understand it, but I will persist. Be advised however to read an expert, I suggest e.g. Artin, Van der Waerden, Jacobson, Milne, Brauer, Lang, as well as Dummit and Foote, Stewart.
And check this article out, for a discussion of the history of Galois theory in textbooks, with many specific recommendations:
https://ncm.gu.se/pdf/normat/Juliusz_textbooks.pdf
A few words about symmetry
Solving a polynomial equation f with coefficients in F, means writing the roots of f in terms of the coefficients of f, and elements of F. This is inverse to the problem of writing the coefficients in terms of the roots, something easy to do, and also enlightening. Namely if r1,…,rn are the roots of a (monic) polynomial f, then f(X) factors completely into linear factors, as f(X) = (X-r1)(X-r2)….)(X-rn). Multiplying out thus gives the coefficients of f in terms of the roots:
f(X) = X^n -(r1+r2+…+rn)X^(n-1) + (r1r2+r1r3+…+rn-1rn)X^(n-2) -…±(r1r2r3…rn).
E.g. for a quadratic f with roots r,s, we have f(X) = (X-r)(X-s) = X^2 - (r+s)X + rs.
These coefficients are symmetric combination of the roots, in the sense that they do not change value under any reordering of the roots. In fact these particular symmetric functions of the roots, can be used to express any symmetric polynomial function of the roots. For this reason they are called the “elementary symmetric functions”. I.e. every symmetric polynomial function of the roots, that is one which does not change values under reordering the roots, can be written as a polynomial in the symmetric functions. This can be used to help solve polynomial equations.
E.g. if r,s are the roots of a quadratic polynomial f(X) = X^2 -bX +c, where thus b = r+s and c=rs, then the expression D = (r-s)^2 does not change value after reordering the roots, hence D must be writable as a polynomial in b and c. It is well known that D, the “discriminant” of f, equals D = b^2-4c = (r+s)^2 - 4rs = (r-s)^2. Since we know that r+s = b, we can solve for r and s by taking the two square roots of D, to get r-s, and s-r, and then add to b to get 2r and 2s.
I.e. b ±sqrt(D) = (r+s) ± (r-s) = 2r, or 2s. hence r,s = (1/2)(b ± sqrt(b^2-4c)), the “quadratic formula.
The discriminant exists in general, and always tells us a little about how far away the solution field of a polynomial is from the coefficient field. I.e. since the roots are themselves in the root field, any polynomial in the roots also lies there, and any fully symmetric polynomial in the roots lies in the coefficient field. But if some only partially symmetric polynomial in the roots also lies in the coefficient field, then the root field is not as far away from the coefficient field as it might be.
E.g. if r,s,t are the roots of a cubic f(X), then ∆ = (r-s)(r-t)(s-t) is a square root of the discriminant, hence changes sign under any transposition of two roots, but remains unchanged under any even permutation, i.e. any composition of an even number of transpositions, (such even permutations form a normal subgroup An of Sn called the "alternating group"). If ∆ lies in the coefficient field, it implies that the solution field has only dimension 3 over the coefficient field, not dimension 6, but if not, the dimension is 6. For the cubic X^3 - 2, the discriminant D = ∆^2 = -27(4), is negative, so its square root ∆ does not lie in the coefficient field Q, so this gives another argument for why the solution field in this case has dimension 6 = 3! over Q.
Galois groups of splitting fields
Galois in his genius, after studying the works of LaGrange, realized that understanding the root field of a polynomial, and its relation to the coefficient field, rested on understanding which permutations of the roots can be realized by automorphisms of the root field that leave fixed all elements of the coefficient field. (To simplify matters, note that if the coefficient field is Q the rational numbers, then all automorphisms of the root field leave Q fixed.)
E.g in the case of a quadratic polynomial, the roots are already rational if and only if the only permutation of the roots that fixes the coefficient field is the identity, if and only if the discriminant has a rational square root. In general, for an irreducible polynomial over Q of degree n, the discriminant has a rational square root if and only if the root field has dimension at most (1/2)n! over Q, whereas in general it could be n!.
An automorphism of K over F, is a bijection *:K—>K that equals the identity on the subfield F, and preserves both addition and multiplication in K.
Definition: If F is a field and f(X) is a polynomial with coefficients in F, and if K is the field extension of F obtained by adjoining a full complement of roots of f, i.e. if K is a splitting field of f over F, then the set of field automorphisms of K which restrict to the identity on F, is a group called the
Galois group of K over F, [and also the Galois group of f]. [K is uniquely determined by f and F, up to isomorphism.]
E.g. complex conjugation, sending a+bi to (a+bi)* = a-bi is an automorphism of C over R, hence an element of the Galois group of X^2+1 over R.
Basic theorem: The Galois group of f is naturally isomorphic to a permutation group of the roots of f, i.e. if degree f = n, to a subgroup of Sn.
proof: One only needs to show that an automorphisms of K over F carries a root of f to a (not necessarily different) root of f. Recall an automorphism of K over F, is a bijection *:K—>K that equals the identity on the subfield F, and preserves both addition and multiplication in K. Hence if ß in K is a root of f (with coefficients in F), then ß* is also a root of f, since * fixes the coefficients of f, including 0, so f=f*(= the polynomial obtained by applying * to the coefficients of f), hence 0 = f(ß) =(f(ß))* = f*(ß*) = f(ß*).
This shows every automorphism of K over F restricts to a permutation of the roots of f, and since K is generated over F by these roots, * is determined by what it does to the roots, so the restriction map is an injection from the Galois group to Sn. QED.
Remark: The Galois group of f, over F, is actually a subgroup of An if and only if the discriminant of f has a square root in F.
Galois’ insight was that this group has a special form in the case of root fields K obtained from polynomials that are solvable by radicals, i.e. that are contained in a union of field extensions of F, formed as root fields of polynomials of the special form X^r-c. so we look at those next, and their Galois groups.
Simplest case: K = root field of f(X) = X^n - 1. Then we claim the Galois group is abelian, in fact it is isomorphic to (Zn)* = the multiplicative group of units in the ring Zn = Z/nZ.
proof: The roots of X^n-1 themselves form a finite multiplicative subgroup of the complex numbers, the powers of e^(2πi/n) on the unit circle in the complex plane. Hence any field automorphism of the root field of f restricts to a multiplicative group automorphism of this group of roots, which is also injective since the elements of this group generate the root field. Each such automorphism sends a primitive root, such as e^(2πi/n), a generator of the group, to another primitive root, hence the map is an isomorphism from the Galois group to a subgroup of the group of units of (Z/nZ). The fact the map is onto that group follows from the (non trivial) algebraic result that the polynomial satisfied by the primitive roots of 1 is irreducible over Q, and the fact that the degree of this polynomial equals the order of the Galois group. In particular the Galois group in this case is abelian.
Next simplest case: K = root field of X^n - c.
Since the quotient of two nth roots of c is an nth root of 1, this root field contains the previous root field, so this one may be obtained in two stages, first adjoining all nth roots of 1, and then adjoining one root of X^n-c. Then an automorphism of this field, fixing the root field of X^n-1, sends each root a of X^n-c to some multiple ua, where u is a root of X^n-1, and u is independent of a. The map sending such an automorphism to u is an embedding from the Galois group of K over F = root field of X^-1, into the group of roots of X^-1, i.e. into the cyclic group Z/nZ. In particular, this group is abelian.
Remark: A key concept in this topic is that of a "normal" subgroup, which is a subgroup H of G such that one can form a quotient group G/H in a way such that the natural surjection G-->G/H is a homomorphism taking everything in H to the identity of G/H.
Corollary: If K is a field extension of Q, obtained in stages as the root fields of equations of form X^n-c, then the Galois group of K/Q is the union of a sequence of subgroups, each normal in the next larger one, with all successive quotient groups abelian.
proof: If F is contained in L contained in K, all root fields, then G(K/L) is a normal subgroup in G(K/F), and G(L/F) ≈ G(K/F)/G(K,L). QED.
Example: If F = Q, the rationals, and L is the root field of X^3-1, while K is the root field of X^3-2, then the Galois group G(K/Q) ≈ S3 has normal subgroup G(K/L) ≈ A3 ≈ Z/3Z. Both the subgroup G(K/L) and the quotient G(K/Q)/G(K/L) ≈ G(L/Q) ≈ Z/2Z, are abelian, (even though G(K/Q) ≈ S3 itself is not). Hence the Galois group of X^3-2 satisfies the necessary criterion for solvability over Q.
Here the sequence of subgroups in increasing order, is G(K/K) = {id}, G(K/L)≈Z/3Z, G(K/Q)≈S3, and the corresponding sequence of successive abelian quotients is G(K/L)/G(K/K) ≈ Z/3Z, and G(K/Q)/G(K/L) ≈ G(L/Q) ≈ Z/2Z.
In general, if the increasing succession of subfields of K, each formed by adding one more rjth root, is ....Fj, Fj+1,..., then the increasing sequence of subgroups of G(K/Q) is ....G(K/Fj+1), G(K/Fj),...., and the sequence of abelian quotient groups is......,G(K/Fj)/G(K/Fj+1) ≈ G(Fj+1/Fj),.....
Corollary: If f is a polynomial whose Galois group does not admit such a sequence of subgroups as described in the previous corollary, then f is not solvable by radicals.
Remark: If the splitting field H of f is merely contained in a field K of the form just described, rather than equaling it, then the Galois group G(H/Q) is a quotient of G(K/Q), i.e. then G(H/Q) ≈ G(K/Q)/G(K/H), and hence can be shown to have the same "abelian subquotient" property.
Corollary: The polynomial f(X) = X^5-80X+2 is not solvable by radicals.
proof: By calculus f has 3 real roots, hence 2 non real ones, and complex conjugation acts as a transposition on these 2 roots. Since f is irreducible, the Galois group G acts “transitively” on the 5 roots (every root can be sent to every other), hence G contains a 5-cycle and a transposition, hence equals all of S5. But S5 has A5 as index 2 subgroup; and A5 ≈ the Icosahedral group, which has no normal subgroups, i.e. A5 is “simple”. QED.