68
Solution.
Suppose, on the contrary, that
G
/
Z
(
G
)
is cyclic, say, with gen
erator
aZ
(
G
)
. Since
G
is not abelian,
a
/
∈
Z
(
G
)
. On the other hand, we
now show that
a
commutes with every element in
g
∈
G
. By hypothesis,
there is an integer
i
and an element
z
∈
Z
(
G
)
with
g
=
a
i
z
. Therefore,
ag
=
aa
i
z
=
a
i
az
=
a
i
za
=
ga
,
and this contradicts
a
/
∈
Z
(
G
)
.
2.99
Let
G
be a
f
nite group, let
p
be a prime, and let
H
be a normal subgroup of
G
. Prove that if both

H

and

G
/
H

are powers of
p
, then

G

is a power
of
p
.
Solution.
If

H
=
p
h
and

G
/
H
p
m
, then

G
=
G
/
H

H
p
h
+
m
.
2.100
Call a group
G
fnitely generated
if there is a
f
nite subset
X
⊆
G
with
G
=h
X
i
. Prove that every subgroup
S
of a
f
nitely generated
abelian
group
G
is itself
f
nitely generated.
Solution.
Use induction on
n
≥
1, where
X
={
a
1
,...,
a
n
}
. The inductive
step should consider the quotient group
G
/
h
a
n
+
1
i
.
2.101
(i)
Let
π
:
G
→
H
be a surjective homomorphism with ker
π
=
T
.
Let
H
X
i
, and, for each
x
∈
X
, choose an element
g
x
∈
G
with
π(
g
x
)
=
x
. Prove that
G
is generated by
T
∪{
g
x
:
x
∈
X
}
.
Solution.
Absent.
(ii)
Let
G
be a group and let
T
C
G
. If both
T
and
G
/
T
are
f
nitely
generated, prove that
G
is
f
nitely generated.
Solution.
Absent.
2.102
Let
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '11
 KeithCornell
 Normal subgroup, Ker, ker γ

Click to edit the document details