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

134 Problems of Chapter 1
134. (see 133) Hence
135. Let and be two arbitrary elements of the group G. Thus
136. If and then (by the
definition of cosets)
137. See 57. 1) If and belong to ker then
and Therefore also belongs to ker
2) (see 133) so belongs to ker 3) If then
(see 134)
also belongs to it.
Hence if belongs to ker
138. Let be an arbitrary element of the kernel ker and an
arbitrary element of the group G. Thus and
(see 134)
Hence the element
also belongs to ker
the group G.
and therefore ker is a normal subgroup of
139. Suppose that the two elements and belong to the same
coset ker Thus there exist in ker two elements and such that
and Thus
Vice versa let In this case we
have (see 134) Hence
where is an element of the kernel ker It follows that
and both elements and belong to the coset ker
140. Let be an arbitrary element of F. Since is a surjective
mapping, there exists an element of the group G such that
Let A be the coset containing Thus by definition
141. Let and let be two elements, representant of
the classes A and B. Thus It follows (see
139) that A = B.
142. Let A and B be two arbitrary cosets and and their representant
elements. Thus element belongs to the coset AB. Applying
the definition of the mapping we obtain
is a homomorphism) The mapping being
bijective (see 141), is an isomorphism.