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Homework 3
September 27, 2010
Problem 1
(a) Inverses exist, since matrices
A
with det(
A
)
Ó
= 0 are invertible. If
A
is invertible, then
det(
A
) = det(
A

1
)

1
, so it is closed under inverses. The identity is the identity matrix.
Generally, for matrices
A,B
in the group, det(
AB
) = det(
A
) det(
B
) =
±
1, so the group
is closed under multiplication, satisfying the last requirement.
(b) Again, inverses exist because the determinant is nonzero. Again, since det(
A
) = 1 for
some matrix
A
in the group, we have that det(
A

1
)

1
= det(
A
) = 1, so 1
/
det(
A

1
) = 1
so det(
A

1
) = 1, showing that the inverses are indeed in the group. The identity is again
the identity matrix, and the argument for closure is the same as before.
Problem 2
The Klein 4 group
Z
2
×
Z
2
is not isomorhic to
Z
4
, since all nonidentity elements in
Z
2
×
Z
2
have order 2, while e.g.
3
∈
Z
4
has order 4.
Problem 3
We wish to show that every element of the Heisenberg group has inﬁnite order. Consider some
matrix in the group
A
=
1
a b
0 1
c
0 0 1
,a,b,c
∈
R
,
then we can write out powers of
A
:
A
2
=
1
a b
0 1
c
0 0 1
1
a b
0 1
c
0 0 1
=
1 2
a
2
b
+
ac
0
1
2
c
0
0
1
A
3
=
1 2
a
2
b
+
ac
0
1
2
c
0
0
1
1
a b
0 1
c
0 0 1
=
1 3
a
3
b
+ 3
ac
0
1
3
c
0
0
1
A
4
=
1 3
a
3
b
+ 3
ac
0
1
3
c
0
0
1
1
a b
0 1
c
0 0 1
=
1 4
a
4
b
+ 6
ac
0
1
4
c
0
0
1
.
In general, we ﬁnd that
A
n
=
1
na nb
+ [2(
n

1)]
ac
0
1
nc
0
0
1
.
For this to be the identity matrix, we would require, for
n
∈
N
, that
na
= 0
nc
= 0
nb
+ [2(
n

1)]
ac
= 0
.
1/5
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 Summer '11
 gelfand
 Chemistry

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