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

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Solutions 131<br />

sends every <strong>in</strong>to itself. Consequently for the commutant<br />

can conta<strong>in</strong> only rotations by angles which are a multiple <strong>of</strong> Let<br />

be the rotation <strong>of</strong> the regular by the reflection with respect<br />

to the axis (Figure 42).<br />

Thus is the rotation send<strong>in</strong>g (verify) the vertex C onto B‚<br />

i.e.‚ the counterclockwise rotation by Hence the commutant<br />

conta<strong>in</strong>s all rotations by angles which are a multiple <strong>of</strong> <strong>and</strong><br />

only them. It is the subgroup <strong>of</strong> rotations <strong>of</strong> the plane send<strong>in</strong>g every<br />

regular <strong>in</strong>to itself. This group is isomorphic to<br />

120. Let be the axes through the middle po<strong>in</strong>ts (K‚L‚M‚<br />

respectively) <strong>of</strong> the opposite edges <strong>of</strong> the tetrahedron. Jo<strong>in</strong> the po<strong>in</strong>ts<br />

K‚ L‚ M so obta<strong>in</strong><strong>in</strong>g the regular triangle KLM. For an arbitrary rotation<br />

<strong>of</strong> the tetrahedron‚ either each one or none <strong>of</strong> the three axes is<br />

fixed (verify). If we put the permutation <strong>of</strong> the axes <strong>in</strong> correspondence<br />

with the permutation <strong>of</strong> the vertices K‚ M‚ L <strong>of</strong> the triangle KLM‚<br />

we obta<strong>in</strong> that to every rotation <strong>of</strong> the tetrahedron there corresponds a<br />

transformation <strong>of</strong> the triangle KLM‚ which is <strong>in</strong> fact a rotation <strong>of</strong> the<br />

triangle. Hence to every commutator <strong>in</strong> the group <strong>of</strong> rotations <strong>of</strong> the<br />

tetrahedron there corresponds a commutator <strong>in</strong> the group <strong>of</strong> rotations <strong>of</strong><br />

the triangle KLM. S<strong>in</strong>ce the group <strong>of</strong> rotations <strong>of</strong> the triangle is commutative‚<br />

every commutator <strong>in</strong> it is equal to Thus every commutator<br />

<strong>in</strong> the group <strong>of</strong> rotations <strong>of</strong> the tetrahedron must fix every one <strong>of</strong> the<br />

three axes Therefore the commutant <strong>in</strong> the group <strong>of</strong> rotations <strong>of</strong><br />

the tetrahedron can conta<strong>in</strong> only the identity <strong>and</strong> the rotations by 180°<br />

about the axes through the middle po<strong>in</strong>ts <strong>of</strong> the opposite edges. S<strong>in</strong>ce the<br />

group <strong>of</strong> rotations <strong>of</strong> the tetrahedron is not commutative‚ its commutant<br />

is different from the commutant be<strong>in</strong>g a normal subgroup‚ by 113 it<br />

co<strong>in</strong>cides with the subgroup conta<strong>in</strong><strong>in</strong>g the identity <strong>and</strong> all the rotations<br />

by 180° about the axes through the middle po<strong>in</strong>ts <strong>of</strong> the opposite edges<br />

<strong>of</strong> the tetrahedron.<br />

121. See solution 113.<br />

122. Both symmetries <strong>of</strong> the tetrahedron <strong>and</strong> either change or<br />

fix the orientation <strong>of</strong> the tetrahedron (see solution 67). Thus every commutator<br />

preserves the orientation <strong>of</strong> the tetrahedron. Consequently<br />

the commutant <strong>in</strong> the group <strong>of</strong> symmetries <strong>of</strong> the tetrahedron<br />

conta<strong>in</strong>s only the rotations <strong>of</strong> the tetrahedron. If

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