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
?)