MAT301-Fall2023-Term-Test-2-sol
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University of Toronto *
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Course
348
Subject
Mathematics
Date
Nov 24, 2024
Type
Pages
5
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(1) For each of the questions a)-d) decide if
H
is not a subgroup, a
subgroup which is not normal or a normal subgroup in
G
.
(a)
G
= (
C
*
,
·
)
, H
=
{
z
∈
C
*
| |
z
| ∈
Q
}
.
Solution
Let us check the subgroup test for
H
.
Let
z
1
, z
2
∈
H
.
This
means that
z
1
6
= 0
, z
2
6
= 0 and
|
z
1
|
=
q
1
,
|
z
2
|
=
q
2
are rational.
Then
z
1
z
2
6
= 0 and
|
z
-
1
z
2
|
=
|
z
1
| · |
z
2
|
=
q
1
q
2
∈
Q
. Hence
H
is
closed under the operation. also
|
1
z
|
=
1
|
z
|
is rational if
z
6
= 0 and
|
z
|
is rational. Hence
H
is closed under inverses. Thus
H
⊂
G
is a subgroup. Since
G
is abelian
H
is a normal subgroup.
Answer:
H
C
G
is a normal subgroup
(b)
G
=
GL
(3
,
R
)
, H
=
{
A
∈
G
|
det
A >
0
}
.
Solution
Let us check the subgroup test for
H
.
Let
A, B
∈
H
.
then
det
A >
0
,
det
B >
0. Therefore det(
AB
) = det
A
·
det
B >
0
also.
Thus
H
is closed under matrix multipolication.
Next
det(
A
-
1
) =
1
det
A
>
0 is det
A >
0.
Thus
H
is closed under
inverses. This shows that
H
⊂
G
is a subgroup. Let us check
the normal subgroup test. Let
A
∈
H, B
∈
G
then
det(
BAB
-
1
) = det
B
·
det
A
·
det(
B
-
1
) =
detB
·
det
A
·
1
det
B
= det
A >
0
This shows that
BHB
-
1
⊂
H
and therefore
H
C
G
.
Answer:
H
C
G
is a normal subgroup
(c)
G
=
E
(2)
, H
=
{
g
∈
G
| |
g
|
<
∞}
.
Solution
Let
l
1
, l
2
be lines through 0 such that the angle between
l
1
and
l
2
is
α
such that
α/π
is irrational. Let
F
1
, F
2
be reflections in
lines
l
1
, l
2
respectively. Then both
F
1
and
F
-
2 belong to
E
(2)
and have order 2. But
F
1
F
2
=
R
2
α
has infinite order. Thus
h
is not closed under the operation. hence it is not a subgroup of
G
.
Answer:
H
is not a subgroup of
G
.
(d)
G
=
S
4
⊕
S
4
, H
=
{
(
g, g
)
|
g
∈
S
4
}
.
Solution
1
2
Let us check the subgroup test for
H
.
Since
G
is finite it’s
enough to check that
H
is closed under the operation.
let
x
= (
g
1
, g
1
)
, b
= (
g
2
, g
2
)
∈
H
then
xy
= (
g
1
g
2
, g
1
g
2
)
∈
H
as
well. This shows that
H
⊂
G
is a subgroup.
Let us check the normal subgroup test for
H
.
Let
x
= ((12)
, ε
)
∈
G
and let
h
= ((13)
,
(13))
∈
H
.
Then
xhx
-
1
= (12)(13)(12)
, ε
·
(13)
·
ε
) = ((23)
,
(12))
/
∈
H
. We see
that
H
is not closed under conjugation by elements of
G
and
hence it is not normal in
G
.
Answer:
H
is a subgroup of
G
which is not normal.
(e) What is the number of normal subgroups of order 2 in
D
10
?
Solution
Subgroups of order 2 are generated by elements of order 2.
There are 11 elements of order 2 in
D
1
0, namely
R
180
and 10
reflections. the subgroup
h
R
180
i
=
{
R
0
, R
180
}
=
Z
(
D
10
) is nor-
mal in
D
10
.
OTOH none of the subgroups
h
F
i
where
F
is a reflection are
normal. Indeed Let
α
=
2
π
10
=
π
5
. Then
R
α
∈
D
10
. let
F
=
F
l
∈
D
10
.
By a formula from problem 1 on assignment 2
R
α
FR
-
1
α
=
R
α
FR
-
α
=
F
R
α
(
l
)
6
=
F
l
since
l
∈
R
α
(
l
). Thus
h
F
i
=
{
R
0
, F
}
does not satisfy the normal subgroup test and hence it is not
normal in
D
10
.
Therefore the is only one normal subgroup of order 2 in
D
10
,
namely
h
R
180
i
=
{
R
0
, R
180
}
.
Answer:
There is exactly 1 normal subgroup of order 2 in
D
10
.
(2) (10 pts) Find a subgroup of
Z
75
⊕
Z
8
isomorphic to
Z
12
⊕
Z
5
.
Solution
Since
gcd
(12
,
5) = 1 we have that
Z
12
⊕
Z
5
∼
=
Z
12
·
5
=
Z
60
.
Let
g
= (
¯
5
,
¯
2)
∈
Z
75
⊕
Z
8
. then
|
g
|
=
lcm
(
|
¯
5
|
,
|
¯
2
|
) =
lcm
(
75
5
,
8
2
) =
lcm
(15
,
4) = 60.
Hence
H
=
h
g
i
is cyclic of order 60 and therefore is isomorphic
Z
60
and hence to
Z
12
⊕
Z
5
.
Answer:
H
=
h
(
¯
5
,
¯
2)
i
.
(3) (10 pts) Let
G
be a finite abelian group such that
G
contains at least
three elements of order 3. Prove that
|
G
|
is divisible by 9.
Solution
3
Let
a
∈
G
be an element of order 3.
Let
H
=
h
a
i
.
Then
H
=
{
e, a, a
2
}
. Since there are at least 3 elements in
G
of order 3 there
is
b
∈
G
of order 3 such that
b /
∈
H
. Let
K
=
h
b
i
.
Then
H
6
=
K
. By a result from class two subgroups of prime order
are either equal or intersect at identity. Since
H
6
=
K
it follows that
H
∩
K
=
{
e
}
.
Let
L
=
HK
⊂
G
.
Then
L
is a subgroup of
G
.
This was proved in class and in the book in example 7 on page 64.
here is the proof.
we just need to check the finite subgroup test.
let
g
1
=
h
1
k
1
, g
2
=
h
2
k
2
∈
HK
, where
h
i
∈
H, k
i
∈
K
.
Then
g
1
g
2
=
h
1
h
2
k
1
k
2
= (
h
1
h
2
)(
k
1
k
2
)
∈
HK
. We used that
G
is abelian.
Hence
HK
⊂
G
is a subgroup.
Also by the general formula
|
HK
|
=
|
H
|·|
K
|
|
H
∩
K
|
=
3
·
3
1
= 9. This shows that
G
contains a subgroup of
order 9. Therefore
|
G
|
is divisible by 9 by Lagrange’s theorem.
(4) Let
G
be a group and let
H
C
G
. Prove that [
H, H
]
C
G
.
Solution
Since [
H, H
] is a subgroup of
H
and
H
is a subgroup of
G
it fol-
lows that [
H, H
] is a subgroup of
G
.
Thus we only need to check
normality.
Let us verify that the normal subgroup test holds for [
H, H
].
Let
x
∈
[
H, H
]
, g
∈
G
.
We need to show that
gxg
-
1
∈
[
H, H
].
let
ϕ
g
:
G
→
G
be conjugation by
g
.
We know that this is an
automorphism of
G
.
Since
x
∈
[
H, H
] it can be written as
x
=
[
h
1
, u
1
]
·
. . .
·
[
h
n
, u
n
] for some
h
i
, u
i
∈
H, i
= 1
, . . . , n
.
Since
ϕ
g
preserves operation we have
ϕ
g
(
x
) =
ϕ
g
([
h
1
, u
1
])
·
. . .
·
ϕ
g
([
h
n
, u
n
])
Since [
H, H
] is a subgroup and is closed under the operation it is
enough to show that
ϕ
g
([
h
i
, u
i
])
∈
[
H, H
] for every
i
.
It was shown in class that
ϕ
g
([
h
i
, u
i
]) = [
ϕ
g
(
h
i
)
, ϕ
g
(
u
i
)] = [
gh
i
g
-
1
, gu
i
g
-
1
]
Since
H
C
G
is normal we have that
gh
i
g
-
1
=
h
0
i
∈
H
and
gu
i
g
-
1
=
u
0
i
∈
H
. Hence
ϕ
g
([
h
i
, u
i
]) = [
h
0
i
, u
0
i
]
∈
[
H, H
]
as claimed. Therefore
gxg
-
1
∈
[
H, H
] for any
g
∈
G, x
∈
[
H, H
]
which verifies the normal subgroup test. Hence [
H, H
]
C
G
.
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4
(5) (10 pts) Let
G
=
S
9
, Let
σ
=
1
2
3
4
5
6
7
8
9
2
5
7
6
1
4
9
8
3
, τ
=
1
2
3
4
5
6
7
8
9
8
5
1
7
2
9
4
3
6
. Does there exist an inner automor-
phism
ϕ
of
G
such that
ϕ
(
σ
) =
τ
? If yes, find such
ϕ
. If no, prove
that no such
ϕ
exists.
Solution 1
Let us write both
σ
and
τ
as products of disjoint cycles:
σ
= (1 2 5)(3 7 9)(4 6)
,
τ
= (1 8 3)(2 5)(4 7)(6 9)
Let
ϕ
=
ϕ
α
be conjugation by some
α
∈
S
9
.
Recall the general formula for conjugation of of cycles in
S
n
,
α
(
i
1
, i
2
, . . . , i
k
) = (
α
(
i
1
)
, α
(
i
2
)
, . . . , α
(
i
k
))
Using this formula and the fact that
ϕ
preserves the operation we
get
ϕ
(
σ
) =
ϕ
((1 2 5))
ϕ
((3 7 9))
ϕ
((4 6)) = (
α
(1)
, α
(2)
, α
(5))(
α
(3)
, α
(7)
, α
(9))(
α
(4)
, α
(6))
Since
α
:
{
1
, . . . ,
9
} → {
1
, . . . ,
9
}
is a bijection we see that all thee
cycles in decomposition of
ϕ
(
σ
) are disjoint. I.e.
ϕ
(
σ
) is a product
of two disjoint 3 cycles and a a disjoint 2 -cycle. On the other hand
τ
is a product of one 3-cycle and three 2-cycles (all disjoint). Since
decomposition into disjoint cycles is unique up to reordering we see
that it’s impossible to have
ϕ
(
σ
) =
τ
.
Answer:
There is no inner automorphism
ϕ
of
S
9
such that
ϕ
(
σ
) =
τ
.
Solution 2
Note that
τ
(8) = 8. Let
ϕ
=
ϕ
α
be an inner automorphism of
S
9
.
Then
ϕ
α
(
σ
) =
ασα
-
1
. let’s evaluate it on
α
(8) we have
ασα
-
1
(
α
(8)) =
ασ
(
α
-
1
(
α
(8)) =
ασ
(8) =
α
(
σ
(8)) =
α
(8).
Thus we see that
ϕ
α
(
σ
) maps
α
(8) to itself. However, we are given
that
τ
(
i
)
6
=
i
for any
i
= 1
, . . .
9.
Therefore we can not have that
ϕ
α
(
σ
) =
τ
.
Answer:
There is no inner automorphism
ϕ
of
S
9
such that
ϕ
(
σ
) =
τ
.
(6) (10 pts) Let
G
=
Z
4
⊕
Z
4
and let
H
=
h
(
¯
3
,
¯
2)
i
. Is
G/H
cyclic?
Solution
5
First, we compute that
|
(
¯
3
,
¯
2)
|
=
lcm
(4
,
2) = 4 and hence
H
has
order 4.
more explicitly we see that 2(
¯
3
,
¯
2) = (
¯
6
,
¯
4) = (
¯
2
,
¯
0) and
3(
¯
3
,
¯
2) = (
¯
9
,
¯
6) = (
¯
1
,
¯
2), 4(
¯
3
,
¯
2) = (
¯
12
,
¯
8) = (
¯
0
,
¯
0).
Hence
H
=
{
(
¯
0
,
¯
0)
,
(
¯
3
,
¯
2)
,
(
¯
2
,
¯
0)
,
(
¯
1
,
¯
2)
}
,
Therefore
|
G/H
|
=
|
G
|
/
|
H
|
= 16
/
4 = 4. To see if
G/H
cyclic we
have to check if it contains an element of order 4.
Recall that by a result from class in general the order of
a
+
H
in
G/H
is the smallest
n >
0 such that
na
∈
H
.
let
a
= (
¯
0
,
¯
1) Let us compute the order of
a
+
H
in
G/H
.
We
have 2
a
= (
¯
0
,
¯
2)
/
∈
H,
3
a
= 3(
¯
0
,
¯
1) = (
¯
0
,
¯
3)
/
∈
H
and 4
a
= (
¯
0
,
¯
0)
∈
H
.
Hence
|
a
+
H
|
= 4 and therefore
G/H
is cyclic.
Answer:
Yes,
G/H
is cyclic.