Abel's theorem in problems and solutions - School of Mathematics

140 Problems of Chapter 1
166. Let K(G) be the commutant in the group G and
Consider the natural homomorphism (see §1.13) of the group G into the
quotient group G/N. Thus (see 155)
and in general Since by hypothesis the
group G/N is soluble‚ for a certain the subgroup is the unit
group‚ i.e.‚ Since the
subgroup is contained in the normal subgroup N. The group N
being soluble‚ for a certain the subgroup is the unit group. Since
is contained in N the subgroup is contained in
and it is thus the unit group. It follows that the group G is soluble.
167. From Problem 146 we obtain
Since the groups and are isomorphic to the groups G
and F‚ respectively‚ and they are therefore soluble‚ the group G × F (see
166) is also soluble.
168. Since the group G is soluble‚ in the sequence of commutants
... we will have‚ for a certain Consider
the sequence of groups G‚ K(G)‚ This sequence of
groups is the sequence we seek‚ because every group (after the first one) is
the commutant of the preceding group‚ and is therefore a normal subgroup
(see 116) of it. Moreover‚ all the quotient groups as well
as G/K(G)‚ are commutative (see 129); the group too‚ is
commutative.
169. Since by hypothesis the quotient group is commutative‚
the commutant is contained in (see 130). It follows that the
subgroup is contained in and in general the subgroup
is contained in for every and
The subgroup is contained in the subgroup which is
contained in etc.‚ up to It follows that the subgroup
is contained in But because the group
is by hypothesis commutative. Hence i.e.‚ the group
G is soluble.
We can obtain the result of Problem 169 also by induction‚ going from
to later to etc. and using the result of Problem 166.
170. By hypothesis the group G is soluble. This means that for a certain
the subgroup is the unit group and therefore the subgroup