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

Solutions 113
of the real numbers into the real positive numbers. For any two positive
numbers and we have and it
follows that is an isomorphism of the group of the real numbers under
addition in the group of the positive real numbers under multiplication.
53. Let be an arbitrary element of group G. Thus
Hence is a surjective mapping of the group G onto the
group G. If then and hence is a bijective
mapping of the group G into itself.
54. for every element
therefore Since
for every element therefore Hence we
have a group of transformations (see §1.2).
55. Consider a mapping such that Since
and if The mapping is therefore
bijective. Moreover,
(see solution 54)
It follows that is an isomorphism.
56. a) Let and be the unit elements respectively in the group G
and in the subgroup H. In the subgroup H one has the identity
By the definition of subgroup this identity also holds in the group
G. Moreover, in the group G we also have the identity from
which we obtain that in G and (see 24)
b) Let be any element of the subgroup H, and let and be
its inverse elements respectively in the group G and in the subgroup H.
In the subgroup H we thus have By the
definition of subgroup this identity also holds in the group G. Moreover,
in the group G we have from which we obtain that in G
and
57. The necessity follows from the result of Problem 56 and from the
definition of subgroup.
Sufficiency. From property 1 the binary operation of the group G
is also the binary operation of H. The element which belongs to
H by property 2, is the unit element in H because
for every element of the group G and, in particular, for all elements
of H. If is an arbitrary element of H then the element which
belongs to H by property 3, is the inverse element of in H because
The associativity is obviously satisfied: H is
thus a subgroup of the group G.