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

Solutions 133
ement of the commutant of the group of rotations of the cube there corresponds
a rotation of the tetrahedron
Let be a rotation of the cube by 90° about the axis through the
centres of the faces ABCD and and such that the vertex B
is sent onto A. Let be the rotation of the cube by 120° about the axis
through the vertices and and such that the vertex A is sent onto
Thus the rotation sends the vertex A onto itself (verify) and
the vertex onto D‚ i.e.‚ it is a non-trivial rotation of the cube about
the axis through the vertices A and This rotation is also a rotation
of the tetrahedron about the axis through the vertex A. Now it
is easy to show (see 121) that the commutant in the group of rotations
of the cube contains all the rotations fixing the tetrahedron. But since it
contains only these rotations‚ the commutant in the group of rotations of
the cube is isomorphic to the group of rotations of the tetrahedron.
129. Let A and B be two arbitrary cosets and their representant.
Since the element belongs to the commutant‚
and therefore AB = BA.
130. Let be two arbitrary elements of the group and let A and B
be the cosets to which they belong. Since AB = BA then
E. The commutator therefore belongs to a normal subgroup
N. In this way all the commutators belong to N and hence N is the
commutant.
131. Let be two arbitrary elements of N and an arbitrary
element of the group G. Since N is a normal subgroup the elements
and belong to N. Thus
is a commutator in the normal subgroup N‚ i.e.‚ it belongs
to K(N). Hence K(N) is a normal subgroup of the group G.
132. Let and be two arbitrary elements of the group F. Since
is a surjective homomorphism of the group G onto the group F there exist
two elements and of the group G such that and
Thus one has
This means that the group F is commutative.
The converse proposition is not true: see Example 12 in §1.13.
133. Let Thus
Therefore and