Editing Abel–Ruffini theorem (section)

====Explicit example====
{{see also|Galois theory#A non-solvable quintic example}}

The equation <math>x^5-x-1=0</math> is not solvable in radicals, as will be explained below.

Let {{mvar|q}} be <math>x^5-x-1</math>.
Let {{mvar|G}} be its Galois group, which acts faithfully on the set of complex roots of {{mvar|q}}.
Numbering the roots lets one identify {{mvar|G}} with a subgroup of the symmetric group <math>\mathcal S_5</math>.
Since <math>q \bmod 2</math> factors as <math>(x^2 + x + 1)(x^3 + x^2 + 1)</math> in <math>\mathbb{F}_2[x]</math>, the group {{mvar|G}} contains a permutation <math>g</math> that is a product of disjoint cycles of lengths 2 and 3 (in general, when a monic integer polynomial reduces modulo a prime to a product of distinct monic irreducible polynomials, the degrees of the factors give the lengths of the disjoint cycles in some permutation belonging to the Galois group); then {{mvar|G}} also contains <math>g^3</math>, which is a [[transposition (mathematics)|transposition]]. Since <math>q \bmod 3</math> is irreducible in <math>\mathbb{F}_3[x]</math>, the same principle shows that {{mvar|G}} contains a [[cyclic permutation|5-cycle]]. Because 5 is prime, any transposition and 5-cycle in <math>\mathcal S_5</math> generate the whole group; see {{slink|Symmetric group|Generators and relations}}. Thus <math>G = \mathcal S_5</math>. Since the group <math>\mathcal S_5</math> is not solvable, the equation <math>x^5-x-1=0</math> is not solvable in radicals.