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

Solutions
135
143. Let be the three axes through the middle points of opposite
edges of the tetrahedron. By every symmetry of the tetrahedron
these axes are permuted in some way‚ i.e.‚ we obtain a mapping of the
group of symmetries of the tetrahedron in the group of permutations of
the three axes This mapping is a mapping onto the entire group of
these permutations‚ because (verify) every permutation can be obtained
by a suitable symmetry of the tetrahedron. It is easy to see that for
any two symmetries and the permutation of the axes corresponding
to the symmetry is the composition of the permutations
corresponding to symmetries and i.e.‚ It
follows that is an isomorphism.
We put every symmetry of the equilateral triangle KLM in correspondence
with every permutation of the axes in a natural way. We
obtain an isomorphism of the group of the permutations of the axes
into the group of symmetries of the triangle KLM.
The mapping is a surjective homomorphism (see 135) of the
group of symmetries of the tetrahedron onto the group of symmetries
of the triangle KLM. The kernel of this homomorphism contains all
symmetries of the tetrahedron that send each of the axes into itself.
Such symmetries are the identity and the rotations by 180° about the axes
From the result of Problem 138 and Theorem 3 we obtain that
these symmetries form a normal subgroup in the group of symmetries of
the tetrahedron and that the corresponding quotient group is isomorphic
to the group of symmetries of the triangle.
144. Let be the axes through the centres of opposite faces of
the cube. As in the solution of Problem 143 we define an isomorphism
of the group of rotations of the cube in the group of symmetries of the
equilateral triangle KLM. The kernel of this homomorphism contains all
rotations of the cube that send each of the axes into itself. Such
rotations are only the identical transformation and the rotations by 180°
about the axes From the result of Problem 138 and Theorem 3 we
obtain that these four rotations form a normal subgroup in the group of
rotations of the cube and the corresponding quotient group is isomorphic
to the group of symmetries of the triangle.
145. Denote by the counterclockwise rotation of the plane around
the point O by an angle The mapping is a homomorphism
of the group R into itself‚ because