Galois Theory

1998/99 GCW

Solvabilty

A finite group with no nontrivial proper subgroups is cyclic of prime order. A group is called simple if it has no nontrivial proper quotients, or, equivalently, if it has no nontrivial proper normal subgroups. An abelian simple group is cyclic of prime order.

We say that a group is obtained by extending a group by a group if there is a surjective homomorphism with kernel isomorphic to . If is a maximal proper normal subgroup of a group then is simple. So every finite group can by obtained by extending a finite simple group by a smaller group, and that in turn can be got by extending a finite simple group, and so on, until we reach a simple group. It is a theorem of Zassenhaus that the finite simple groups obtained this way from a finite group are always the same; only the order in which they appear depends upon the choices of maximal proper normal subgroups. They are called the simple composition factors of . Just as positive integers are products of prime numbers, so finite groups are obtained by extending finite simple groups. However, nonisomorphic groups may have the same composition factors. For example, the groups both have as their simple composition factors.

A finite group is called solvable if its simple composition factors are all abelian (and hence cyclic of prime order).

Proposition: A finite group is solvable if and only if either of the following hold

• it is abelian,
• it is obtained by extending a solvable group by a solvable group.

Proposition: Subgroups and quotient groups of solvable groups are solvable.

We recall that denotes the group of permutations on symbols, and denotes the subgroup of even permutations.

Proposition: For the group is solvable.

Proof: We have , cyclic of order 2.

We have .

We have , where is the subgroup

For any group the commutator subgroup is defined to be the subgroup generated by all the elements of expressible in the form . It is a normal subgroup, and the quotient is abelian. Every abelian quotient of is a quotient of . A group is called perfect if its commutator subgroup is the whole group.

Theorem: For the group is perfect.

Proof: is generated by transpositions, i.e. 2-cycles. Since and we see that is generated by 3-cycles when . But since

we see that is perfect for .

A nontrivial perfect group cannot be solvable, hence is not solvable for .

We say that an extension of the form where has minimal polynomial over of the form is an adjunction of an -th root. We write it as . A radical extension is a composite of extensions of this kind. To any polynomial over we can associate the splitting extension of . That is, we adjoin all the roots of . We say that its Galois group is the Galois group of . To say that this is a radical extension is equivalent to saying that the roots of can be expressed in terms of the coefficients of by means of radical expressions, i.e. expressions involving -th roots, addition, subtraction, multiplication and division.

Theorem: If a Galois extension is radical, the Galois group is solvable.

Proof: The Galois group of is abelian.

Corollary: If the Galois group of a polynomial is not solvable there is no formula for its roots as radical expressions of its coefficients.

Proposition: Let be a prime number. Any subgroup of containing a 2-cycle and a -cycle

is the whole of .

Proof: Without loss of generality we can suppose that the subgroup contains and .

Let . Conjugation of by yields all the 2-cycles with .

Conjugating yields , so we get all the 2-cycles with . Induction yields all the 2-cycles with . Since and are coprime we get all the 2-cycles and hence all of .

Corollary: An irreducible polynomial over , of prime degree > 4, having just one pair of complex conjugate roots, has roots that cannot be expressed by radicals.

Proof: Let be the degree of the polynomial. Its Galois group is a subgroup of , as we may identify its elements with permutations of the roots. Since complex conjugation is evidently an automorphism of the splitting field, the Galois group contains a 2-cycle. Adjunction of a single root gives an intermediate extension of degree , so by the multiplication theorem the degree of the splitting field contains a factor . Hence the Galois group contains a -cycle.

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