is the group of transformations of of the form , .

is all the translations .

and , .

is a group of transformation in containing the group of translations as normal subgroup. Show that is a subgroup of , this means is the normalizer of .

Suppose and let , has the form . Hence . If , then .

Any solvable transitive subgroup of , a prime, is equivalent to a subgroup of containing the group of translations.

Suppose is this group. Then has a composition series

whose factors are all cyclic groups of prime order.

By lemma 4, we know that is also a solvable transitive subgroup of . Since is cyclic of prime order, then it must be generated by some p-cycle, otherwise it can not be transitive.

Apply some conjugation to in , we can assume is generated by . So is a translation group. We can identify the element of as the transformation of . By lemma 2, we know that is equivalent to some subgroup of .

Let be a normal nontrivial subgroup of a transitive subgroup of of transformations of . Show that all orbits have the same cardinality. Hence show that if is a prime, then is transitive.

Let be irreducible of prime degree over of characteristic 0, a splitting field over of . Show that is solvable by radicals over if and only if for any two roots of .

Since has characteristic 0, is solvable by radicals if and only if is a solvable subgroup of symmetric group . is irreducible if and only if is transitive.

(1) if is solvable by radicals, then is a solvable transitive subgroup of .

By lemma 3, we know is equivalent to a subgroup of containing . WLOG, assume . The action of on roots looks like , .

Consider , . , unique such that and . Then . Since , we get and , which means . So .

(2) If , then . This Galois group contains a normal Sylow p-subgorup which is isomorphic to . By lemma 2, we know is equivalent to a subgroup of which is solvable. So is solvable, is solvable by radicals.

Jacobson p262.