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MATH 74 HOMEWORK 9
Due Wednesday April 24th at 3:10pm. Throughout, let
G
be a group. Many of the following problems are
from (or adapted from) Herestein’s
Topics in Algebra
.
1 Let
Z
(
G
) be the center of
G
and suppose that
G/Z
(
G
) is cyclic. Prove that
G
is abelian.
Solution.
Since
Z
(
G
) is a subgroup,
G
can be written as the disjoint union of the cosets of
Z
(
G
) in
G
. Since the quotient is cyclic (say with generator
g
) these cosets are precisely
Z
(
G
)
g
k
for
k
and
integer. Then any elements
a,b
can be written as
a
=
z
1
g
n
,
b
=
z
2
g
m
for
z
1
,z
2
∈
Z
(
G
) and
m,n
integers. Note that
z
1
and
z
2
commute with all elements of
G
by virtue of being in the center of
G
.
So we compute
ab
=
z
1
g
n
z
2
g
m
=
z
1
z
2
g
m
+
n
=
z
2
z
1
g
m
g
n
=
z
2
g
m
z
1
g
n
=
ba,
so
G
is abelian.
2 Prove that any group
G
of order 15 is cyclic.
Solution.
By Cauchy’s theorem, there are elements
x
and
y
of orders 3 and 5 respectively. If
x
and
y
commute with each other, we obtain a homomorphism,
Z
/
3
Z
×
Z
/
5
Z
→
G,
(
i,j
)
7→
x
i
y
j
with trivial kernel. Then
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 Spring '09
 DANBERWICK
 Algebra, Division

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