MAT 322: Algebra with Galois Theory
Professor Claus Sorensen
Problem Set 3
Rik Sengupta
October 9, 2009
1.
Show that
Q
/
Z
is isomorphic to
μ
∞
(
C
)
, the group of
z
∈
C
*
such that
z
n
= 1
for some positive integer
n
. Is it a finitely generated abelian group?
What is the torsion subgroup of
Q
/
Z
?
Solution.
We define a homomorphism from
Q
to
μ
∞
(
C
) by
ϕ
:
Q
→
μ
∞
(
C
), with the following mapping:
p
7→
e
2
πip
. This is obviously a ho
momorphism, because
ϕ
(
p
+
q
) =
e
2
πi
(
p
+
q
)
=
e
2
πip
e
2
πiq
=
ϕ
(
p
)
ϕ
(
q
). In
fact, the image space of the homomorphism has been chosen because we
can simply raise the image of the homomorphism to the denominator of
the rational number in question, and then we will get 1, as follows:
[
ϕ
(
p
q
)]
q
=
e
2
πi
p
q
q
=
e
2
πip
= 1
Thus the image space is welldefined, in the sense that all the images of
the map lie in
μ
∞
(
C
).
Note that the kernel of this homomorphism is the set of points
p
∈
Q
for
which
ϕ
(
p
) = 1, i.e.
e
2
πip
= 1. But this happens exactly when
p
∈
Z
. So,
the kernel of the homomorphism is simply
Z
.
Furthermore, the map is surjective, because given any element
z
∈
μ
∞
(
C
),
we know
z
n
= 1 for some
n
, and hence,
z
is one of the
n
th roots of unity.
So we can simply write
z
as
e
2
πik
for some rational
k
. Then,
k
is in
Q
, so
it is a preimage in the homomorphism, i.e. our map is surjective.
So then, from the First Isomorphism Theorem, we immediately get
Q
/
Z
is isomorphic to
μ
∞
(
C
).
It is not finitely generated. Suppose it is in fact a finitely generated group,
generated by
p
1
Z
, p
2
Z
, . . . p
r
Z
for finitely many rationals
p
i
. Then we can
simply take the denominators of all the
p
i
and find some natural number
that is coprime to them all. Then, take any rational
s
whose denominator
in its lowest form is the natural number we just found. Then,
s
Z
is not
1
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in the group generated by the finitely many elements, even though it is in
Q
/
Z
. Hence,
Q
/
Z
is not a finitely generated abelian group.
Every element in the group
Q
/
Z
has some finite order that is the denom
inator of the rational number. This means that since all the elements in
the group have finite order, all the elements are also present in the tor
sion subgroup. Furthermore, the torsion subgroup, being a subgroup, is
contained within the actual group. This is possible only when the torsion
subgroup equals the group itself.
Hence, (
Q
/
Z
)
tor
=
Q
/
Z
.
2.
Let
G
be a group, and let
G
der
be the subgroup generated by all commuta
tors
[
x, y
] =
xyx

1
y

1
, for
x, y
∈
G
. Show that
G
der
is a normal subgroup
of
G
, and that the corresponding quotient group
G
ab
is abelian. Describe
the group
D
ab
n
up to isomorphism, for all integers
n
≥
3
.
Solution.
We will show that for each
x
∈
G
der
,
gxg

1
is also in
G
der
whenever
g
∈
G
.
Since
G
der
is the subgroup generated by all commutators in
G
, for each
x
∈
G
der
we have
x
=
c
1
c
2
. . . c
m
– a word of commutators – so
c
i
= [
a
i
, b
i
]
for all
i
.
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 '09
 CLAUSSORENSEN
 Algebra, Subgroup, Dn, isomorphism theorem, Q/Z, torsion subgroup

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