Originally Posted by

**Deveno** partial answer: the trivial group, and Z2 are both solvable. therefore a polynomial which has these groups as galois groups is solvable by radicals.

In this particular case yes, in general you have to be working in characteristic noit dividing the order of the Galois group (for

example, characeristic zero always fulfils this, and this is why in undergraduate courses this is the usual condition)

Tonio

this is part of the beauty of galois theory, questions about polynomials (which are generally hard), are turned into questions about groups (which are easier).

so, if you start with the "hard part solved" (you've found, or been given, the galois group of a polynomial), you can use the "easy part" to recover information

about the polynomial without much effort.

the classic example is p(x) = x^5 - x - 1. its galois group is S5, which is not solvable, because A5 is simple. we can therefore abandon any hope of solving

p(x) by radicals (S5/A5 is abelian, but A5/{e} is not, and there is no other normal subgroup of A5).