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

Solutions 111
39. a) See Table 8; b) see Table 9; c) Tee table 10.
40. We prove that all properties of a group are satisfied: 1) for three
arbitrary remainders we have modulo
because in both terms of the equality we have the remainder of the
division by of the number 2) the unit element is 0 because
for every remainder 3) if then the inverse
element (the opposite element) of is because modulo one has
the inverse element of element 0 is 0
itself. This group is cyclic with generator 1, because the smallest integer
such that modulo is equal to
41. Since one has and
Hence (see 33) is divisible by i.e., and
are equal modulo
42. Answer. They are isomorphic: the groups (1) and (4) (consider
the mapping and the groups
(2) and (3) (consider the mapping
(see solutions 6 and 7)).
43. Since is a bijective mapping, exists and is bijective. Let
and be two arbitrary elements of the group In the group there
are two (unique) elements and such that and Since
is an isomorphism, (the products are taken in the
corresponding groups). Therefore Since
and are two arbitrary elements of the group is an isomorphism.
44. Since the mappings and are bijective,
is also bijective. Let and be two arbitrary elements
of the group Thus
and hence is an
isomorphism (the products are taken in the corresponding groups).