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

130 Problems of Chapter 1
contain only the elements and being different from it thus coincides
with the subgroup of central symmetries
c) The elements 1 and –1 commute with all the others elements of the
group of quaternions. Hence if one of the elements coincides with
1 or –1‚ then If is an arbitrary element different from
1 and –1‚ then i.e.‚ Therefore
if are any two elements different from of 1 and –1‚
But the square of every element in the group of
the quaternions is equal to 1 or to –1. Hence the commutant can contain
only the elements 1 and –1; because the group is not commutative‚ it
must be different from 1. It follows that the commutant is {1‚ –1}.
119. By the same arguments used to solve Problems 118 (a)‚(b) we
obtain that the commutant in the group of symmetries of the regular
contains only the rotations.
FIGURE 41 FIGURE 42
Let be odd and let be the reflection of the with respect
to axis (Figure 41)‚ the counterclockwise rotation of the by
the angle (sending A onto B). Thus is the rotation
of the sending (verify) B on C‚ i.e.‚ the counterclockwise rotation
by Since the commutant is a subgroup we obtain that for odd
it contains the rotations by all the angles which are a multiple of
Since the commutant contains only rotations of the regular for
odd it coincides with the subgroup of all rotations of the regular
isomorphic to (see 31).
Now let Inscribe in the regular a joining the
even vertices. Joining the odd vertices‚ we obtain a second regular
If is an arbitrary vertex of the both transformations and
either exchange or fix the two Therefore the element