A

group *G* is

solvable if there exist groups

*G*_0,

*G*_1, . . .,

*G*_

*k* such that:

*G*_0 is the trivial group.
*G*_*k* is *G*.
*G*_*i* is a normal subgroup of *G*_{*i*+1}, for 0 <= *i* < *k*.
*G*_{*i*+1} / *G*_*i* is abelian for 0 <= *i* < *k*.

An important theorem of

Galois theory is that a

polynomial can be

solved by radicals iff its

Galois group is solvable. One can construct a

quintic polynomial with Galois group

S_5---said group being unsolvable

^{[1]}---so there can be no general

solution by radicals for polynomials of degree five or higher. That is, there is no

analogue of the

quadratic formula for polynomials of

degree >= 5.

^{[1]}: S_5's only proper normal subgroups are *e* (the trivial group) and A_5. S_5 / *e* = S_5 is not abelian, so that doesn't work. S _5 / A_5 = **Z**_2 *is* abelian, but A_5, being simple and nonabelian, is not itself solvable. Hence S_5 cannot be solvable.