Galois (see EQUATION) showed that, corresponding to every irreducible equation of the nth degree, there exists a transitive substitution-group of degree n, such that every function of the roots, the numerical value of which is unaltered by all the substitutions of the group can be expressed rationally in terms of the coefficients, while conversely every function of the roots which is expressible rationally in terms of the coefficients is unaltered by the substitutions of the group. This group is called the group of the equation. In general, if the equation is given arbitrarily, the group will be the symmetric group. The necessary and sufficient condition that the equation may be soluble by radicals is that its group should be a soluble group. When the coefficients in an equation are rational integers, the determination of its group may be made by a finite number of processes each of which involves only rational arithmetical operations. These processes consist in forming resolvents of the equation corresponding to each distinct type of subgroup of the symmetric group whose degree is that of the equation. Each of the resolvents so formed is then examined to find whether it has rational roots. The group corresponding to any resolvent which has a rational root contains the group of the equation; and the least of the groups so found is the group of the equation. Thus, for an equation of the fifth degree the various transitive subgroups of the symmetric group of degree five have to be considered. These are (i.) the alternating group; (ii.) a soluble group of order 20; (iii.) a group of order 10, self-conjugate in the preceding; (iv.) a cyclical group of order 5, self-conjugate in both the preceding. If x0, x1, x2, x3, x4 are the roots of the equation, the corresponding resolvents may be taken to be those which have for roots (i.) the square root of the discriminant; (ii.) the function (x0x1 + x1x2 + x2x3 + x3x4 + x4x0)(x0x2 + x2x4 + x4x1 + x1x3 + x3x0); (iii.) the function x0x1 + x1x2+ x2x3 + x3x4 + x4x0; and (iv.) the function x0^2x1 + x1^2x2 + x2^2x3 + x3^2x4 + x4^2x0. Since the groups for which (iii.) and (iv.) are invariant are contained in that for which (ii.) is invariant, and since these are the only soluble groups of the set, the equation will be soluble by radicals only when the function (ii.) can be expressed rationally in terms of the coefficients. If (x0x1 + x1x2 + x2x3 + x3x4 + x4x0)(x0x2 + x2x4 + x4x1 + x1x3 + x3x0)