Why is the symmetric group of letters not solvable for ?

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- Jul 12th 2008, 07:26 PMparticlejohnsymmetric group
Why is the symmetric group of letters not solvable for ?

- Jul 12th 2008, 08:18 PMThePerfectHacker
The group is not solvable for . This is an interesting result which is not so easy to prove.

We will restate the conditions of a group being solvable with a criterion. Let be a finite group. The group is solvable if and only if for some positive integer . The notational meaning of is the -th__derived subgroup__. It is defined as follows. An element of is a__commutator__iff it has the form . The group generated by the set of all commutators is the__commutator subgroup__. We denote it by . We define . Thus, the equivalent definition of a finite solvable group is a group such that taking the commutator subgroup enough times will produce the trivial subgroup. The other important property of commutator subgroups is that if is a normal subgroup and is abelian then .

We can now prove that is not solvable for . Our first step will be to show where is the alternating group. Let be a -cycle in . Then . Thus, any -cycle is a commutator. But any element of can be expressed as a product of -cycles (we will prove this in the end). Thus, this shows . Since is a normal subgroup with which is abelian it means (by above) . But since . Thus, . This means for all . Thus, is not solvable.

It just remains to show any element in is a product of -cycles. Remember if then is a product of an even # of transpositions i.e. -cycles (by definition). We can therefore pair each transposition with another transposition. The product of two transpositions must take one of three forms: (i) (ii) (iii) . In (i) we can write . In (ii) we can write . In (iii) we can write . Thus, we have expressed everything in terms of -cycles.

There is a more elegant proof that is not solvable but it uses a fact which is also takes work proving. That is is a simple group for . This time we prove non-solvability directly by definition. Write . This is a composition series because the factors and are simple. It follows by the Jordan Holder theorem (one of my favorite theorems) that this composition series is unique. Since is not an abelian group it means it is impossible to get a composition series with abelian groups. Thus, is not solvable.

(Do you see where we use the fact ?) - Jul 12th 2008, 09:00 PMparticlejohn
- Jul 12th 2008, 09:04 PMThePerfectHacker
A polynomial in Q[x] is solvable by radicals if and only if its Galois group is a solvable group.

If the degree of f(x) is at most four then its Galois group is embedded in which is solvable. However if f(x) has degree 5 then for it to be unsolvable we required that its Galois group is . We need to actually construct such a polynomial (because just because it has degree 5 does not mean it automatically is unsolvable). - Jul 12th 2008, 11:48 PMNonCommAlg
- Jul 13th 2008, 10:55 AMThePerfectHacker
- Oct 31st 2008, 11:34 PMparticlejohn
yes