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

Solutions 109
on the left and by on the right, one obtains Since
and one obtains i.e., Since and are
two arbitrary elements of the group G, this group is commutative.
26. We need to prove that We
have
Exactly in the same way one proves
that (For a more rigourous proof use the induction
method).
27. We will consider some cases: a) if then
b) if then
c)if then
d) if
then
the case is treated in the same
way as the cases (c) and (d). The cases or are easily verified.
28. We will consider some cases: a) if then
b) if then
are easily verified.
then
The cases or
29. Answer. In the group of symmetries of the triangle (see 3) and
are of order 3, and of order 2; for the square (see 5) and are of
order 4, of order 2; for the rhombus (see solution 6) all elements
(except ) have order 2.
30. 1) Let where and If we multiply on
the right the two members of the equality by we obtain
and (see 27) Since we obtain a contradiction
of the hypothesis that the element has order
if