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

Solutions 125
i.e., there exists an element in N such that Hence the
element belongs to N and therefore N is a normal subgroup of the
group G.
Let now N be a normal subgroup of the group G. We will prove that
for every element of the group G. Let be an arbitrary
element of Thus where is an element of N. Thus
and it follows that (and therefore the entire belongs to
Now let be an arbitrary element of Thus
where is an element of N. Hence and (and therefore the
entire belongs to So and coincide.
102. Hint. In this case both the left and the right partitions contain
two classes: one is the given subgroup, the other contains all the remaining
elements. See also Theorem 2 (§1.10).
103. Let be the given normal subgroups of the group
G and N their intersection. If is an arbitrary element of N then
belongs to all the Hence if is an arbitrary element of the group G
then belongs to all the and therefore to N. This means that
N is a normal subgroup of G.
104. Let be an arbitrary element of the group G. Since
belongs to the centre. If belongs to the centre then Multiplying
both members of this equality by on the left one obtains
Thus also belongs to the centre. If and belong to the centre then
and Therefore and thus
also belongs to the centre. By virtue of the result of Problem 57 the
centre is a subgroup.
Let be any element of the centre and an arbitrary element of the
group G. Thus the element too, belongs to the
centre. The centre is therefore a normal subgroup.
105. Let be two arbitrary elements of and and
be two arbitrary elements of and respectively. Thus the element
belongs to whereas the element belongs to Consequently
the element
belongs to It follows that is a normal
subgroup of
106. Since belongs to the class also (by hypothesis) belongs
to the class This means that there exists in N an element such
that Similarly we obtain that there exists in N an element