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

?)