HW 6A
Melody Mott
March 20, 2019
1. Theorem (3.4#12) LetGbe the following set of matrices over R:
1 0
0 1
;
1 0
0 1
;
1 0
0 1
;
1 0
0 1
.
Show that Gis isomorphic to Z
2 x
Z
2.
Proof. Adirect product ofG
1 and
G
2, denoted
G
1 x
G
2 is the set of all
ordered pairs ( x
1; x
2) such that
x
1 2
G
1and
x
2 2
G
2.
If we can show that G is isomorphic and Z
2 x
Z
2 is isomorphic, then it follows
that the mapping of Gto Z
2 x
Z
2 is also isomorphic.
First, we will show that Gis isomorphic.
One way we can show that Gis isomorphic is to take the inverse of each
matrices in G.
A
1
=
a b
c d
1
= 1 ad
bc
d b
c a
1 0
0 1
1
= 1 (1)(0)
(0)(0)
1 0
0 1
=
1 0
0 1
;
1 0
0 1
1
= 1 (
1)(1) (0)(0)
1 0
0 1
=
1 0
0 1
;
1 0
0 1
1
= 1 (1)(
1) (0)(0)
1 0
0 1
=
1 0
0 1
;
1 0
0 1
1
= 1 (
1)( 1) (0)(0)
1 0
0 1
=
1 0
0 1
So we see that Gis isomorphic.
Now, we construct an addition table for Z
2 x
Z
2
1
+ (0,0) (1,0) (0,1) (1,1)
(0,0) (0,0) (1,0) (0,1) (1,1)
(1,0) (1,0) (0,0) (1,1) (0,1)
(0,1) (0,1) (1,1) (0,0) (1,0)
(1,1) (1,1) (0,1) (1,0) (0,0)
We see in this table that the order of mapping from row to column do not e ect
the outcome. This is shown by the diagonals, which all have the same ordered
pairs.
Since Gand Z
2 x
Z
2 are individually isomorphic, the mapping of
Gto Z
2 x
Z
2
is also isomorphic. 2.
Theorem (3.2.22) Compute the center of GL
2(
R ), denoted by Z(G ) is given
by Z(G ) = fx 2 G :xg =gx for all g2 Gg.
Proof. For all elements g2 G, we want to show that xg=gx when x2 G.
Let a; b; c; d 2R.
Let M =
a b
c d
and I =
1 0
0 1
.
So M I =
b a
d c
and I M=
c d
a b
.
M I =bc ad and I M=cb da
So M I =I M because multiplication is commutative in GL
2(
R ).
Repeating this process for each element of GL
2(
R ), we
nd that M g
n=
g
n M
for all g
n 2
GL
2R
.
Therefore M is the center of GL
2(
R ), where M=
a 0
0 a
such that a6
= 0. 3.
Theorem (3.4.1) Show that the multiplicative group Zx
10 is isomorphic to the
additive group Z
4. De
ne
:Z ! Zx
10 by
([ n]
4 ) = [
a]n
10 and
nd a generator
[ a ]
10 of
Zx
10 .
Proof. We will show that the multiplicative group Zx
10 is isomorphic to the
additive group Z
4.
First, we will
nd a generator [ a]
10 .
Let (n ) = an
(mod 10), where n2 Z
Z x
10 = 1
;3 ;7 ;9( mod 10)
Let a= 7 and n2 Z
Then 7 1
= 7, 7 2
= 9, 7 3
= 3, and 7 4
= 1.
So (n ) = 7 n
(mod 10), where n2 Z.
Therefore, Z
4 !
Zx
10 . 2
4.
Theorem (3.3.1) FindH KinZx
16 , if
H=< [3] >and K=< [5] >.
Proof. We will compute H K. //H K =an
(mod 16). 3