Math 4124
Wednesday, April 20
Eighth Homework Solutions
1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order
2704.
First we write 2704 as a product of prime powers, namely 2
4
·
13
2
. To find the number
of abelian groups of order 16, we find the number of partitions of 4. The partitions
are (4), (3,1), (2,2), (2,1,1), (1,1,1,1) (so for example (3,1) corresponds to the group
Z
/
8
Z
×
Z
/
2
Z
). Thus there are 5 abelian groups of order 16. Also there are 2 abelian
groups of order 9.
Therefore the total number of abelian groups of order 2704 is
5
·
2
=
10.
2. Exercises 5.2.2(e) and 5.2.3(e). List the elementary divisors and invariant factors for
the abelian groups of order 44100.
44100
=
2
2
·
3
2
·
5
2
·
7
2
. Thus the number of abelian groups of order 44100 is 2
4
=
16.
In the table below
n
will indicate
Z
/
n
Z
; thus using this notation there are two abelian
groups of order
p
2
, namely
p
2
and
p
,
p
.
Elementary Divisors
Invariant Factors
2
2
,
3
2
,
5
2
,
7
2
44100
2
2
,
3
2
,
5
2
,
7
,
7
6300,7
2
2
,
3
2
,
5
,
5
,
7
2
8820,5
2
2
,
3
,
3
,
5
2
,
7
2
14700,3
2
,
2
,
3
2
,
5
2
,
7
2
22050,2
2
2
,
3
2
,
5
,
5
,
7
,
7
1260,35
2
2
,
3
,
3
,
5
,
5
,
7
2
2940,15
2
,
2
,
3
,
3
,
5
2
,
7
2
7350,6
2
2
,
3
,
3
,
5
2
,
7
,
7
2100,21
2
,
2
,
3
2
,
5
,
5
,
7
2
4410,10
2
,
2
,
3
2
,
5
2
,
7
,
7
3150,14
2
2
,
3
,
3
,
5
,
5
,
7
,
7
420,105
2
,
2
,
3
2
,
5
,
5
,
7
,
7
630,70
2
,
2
,
3
,
3
,
5
2
,
7
,
7
1050,42
2
,
2
,
3
,
3
,
5
,
5
,
7
2
1470,30
2
,
2
,
3
,
3
,
5
,
5
,
7
,
7
210,210
3. Exercise 7.2.7. Let
R
be a commutative ring with 1. Prove that the center of the ring
M
n
(
R
)
is the set of scalar matrices.
First we show that the scalar matrices are in the center of M
n
(
R
)
, so let
A
be a scalar
matrix, say
A
=
rI
where
r
∈
R
. Then the
i j
entry of
A
is
r
δ
i j
, where
δ
is the Kronecker
δ
. If
B
∈
M
n
(
R
)
has entries
b
i j
, then the
i j
entry of
AB
is
∑
k
r
δ
ik
b
k j
=
rb
i j
, whereas
the
i j
entry of
BA
is
∑
k
b
ik
r
δ
k j
=
b
i j
r
. Since
R
is commutative,
rb
i j
=
b
i j
r
for all
i
,
j
,
hence
A
commutes with
B
. It follows that
A
is in the center of M
n
(
R
)
.