Fall2023-term-test2-practice-problems-sol
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Course
301
Subject
Mathematics
Date
Nov 24, 2024
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MAT301 Solutions to Practice Problems for Term Test 2
(1) Prove that
C
*
/
R
*
has infinite order.
Solution 1
Note that since
C
*
is abelian the subgroup
R
*
C
C
*
is normal and
hence the factor group
C
*
/
R
*
is well defined.
Next note that cosets given
z
1
, z
2
∈
C
*
the cosets
z
1
R
and
z
2
R
are equal iff
z
1
z
2
∈
R
.
Therefore to show that
C
*
/
R
*
is infinite it’s
enough to produce
z
n
6
= 0
, n >
0 such that
z
n
z
m
/
∈
R
for any
n
6
=
m
.
Let
z
n
=
n
+
i, n
= 1
,
2
,
. . . .
. Then for
n
6
=
m
we have
z
n
z
m
=
n
+
i
m
+
i
=
(
n
+
i
)(
m
-
i
)
(
m
+
i
)(
m
-
i
)
=
mn
+ 1 +
i
(
m
-
n
)
m
2
+ 1
/
∈
R
Therefore there are infinitely many distinct left cosets of
R
*
in
C
*
and hence
C
*
/
R
*
is infinite.
Solution 2
Let
z
= cos
α
+
i
sin
α
where
α
=
√
2
·
2
π
.
Then
z
n
= cos(
n
√
2
·
2
π
) +
i
sin(
n
√
2
·
2
π
).
Note that since
√
2 is irrational we can never have that
n
√
2
·
2
π
=
πk
for any integer
k, n
with
n
6
= 0.
Since sin
x
= 0 only for
x
=
πk
it follows that sin(
n
√
2
·
2
π
)
6
= 0
for any
n
6
= 0 and therefore
z
n
6∈
R
for any
n
6
= 0 and hence
z
R
*
has
infinite order in
C
*
/
R
*
.
(2) Let
G
= (
R
*
,
·
).
Find all subgroups of
G
of index 2.
Solution
Let
H
⊂
G
be a subgroup of index 2.
Since
G
is abelian
H
C
G
is normal.
Therefore
G/H
∼
=
Z
2
and every element
gH
in
G/H
satisfies (
gH
)
2
=
eH
which is equivalent to
g
2
∈
H
. Therefore for
any
x
∈
R
*
we have that
x
2
∈
H
. Since for any
x >
0 we have that
x
= (
√
x
)
2
it follows that all positive real numbers are in
H
.
We claim that
H
= (0
,
∞
).
Indeed, if
H
contains a negative
number
x
0
then any other negative number
x
can be written as
x
=
x
0
·
t
where
t
=
x
x
0
>
0.
since both
x
0
, t
∈
H
it follows that
x
∈
H
also.
but this means that
H
=
G
which contradicts the
assumption that
H
has index 2. Therefore
H
contains no negative
numbers and hence
H
= (0
,
∞
).
Lastly, observe that (0
,
∞
) is indeed a subgroup of
G
of index 2.
1
2
Answer:
The only subgroup of
G
of index 2 is
H
= (0
,
∞
).
(3) Let
n
≥
3 and
H
C
S
n
be a normal subgroup such that (12)
∈
H
.
Prove that
H
=
S
n
.
Hint:
Use that every element of
S
n
can be written as a product
of transpositions.
Solution
Let
σ
∈
S
n
be arbitrary. Recall that by the formula for conjugation
on
S
n
proved in class we have that
σ
(12)
σ
-
1
= (
σ
(
i
)
, σ
(
j
)).
Since
H
C
S
n
we have that
σ
(12)
σ
-
1
= (
σ
(
i
)
, σ
(
j
))
∈
H
. Since
σ
can be chose to be arbitrary for any transposition (
ij
)
∈
S
n
we can
find a
σ
such that
σ
(1) =
i, σ
(2) =
j
. Therefore (
ij
)
∈
H
for any
i
6
=
j,
1
≤
i
6
=
j
≤
n
. Since every element of
S
n
can be written as
a product of transposition and
H
is closed under multiplication it
follows that
H
=
S
n
.
(4) Let
G
= (
Z
,
+). Let
H
=
h
5
i
and
K
=
h
7
i
.
(a) Prove that
G
=
HK
Solution
We need to show that any element of
¯
Z
can be written as
h
+
k
where
h
= 5
m, k
= 7
n
for some integer
m, n
. In other words we
need to show that any integer can be written as 5
m
+ 7
n
. This
is true since
gcd
(5
,
7) = 1. Explicitly, 1 = 5
·
3
-
7
·
2. Therefore
any
l
∈
Z
can be written as
l
= 5
·
(3
l
)
-
7
·
(2
l
).
(b) Is
G
the internal direct product of
H
and
K
?
Solution
No, because
H
∩
K
6
=
{
0
}
. In particular 35 = 7
·
5
∈
H
∩
K
.
Answer:
No.
(5) Let
G
be a non abelian group of order 125 such that
|
Z
(
G
)
|
>
1.
Prove that
G/Z
(
G
)
∼
=
Z
5
⊕
Z
5
.
Solution
We have that
|
G
|
= 125 = 5
3
. By Lagrange’s theorem
|
Z
(
G
)
|
divides
|
G
|
= 125 so it can only be equal to 1, 5, 25 or 125. Since we are given
that
|
Z
(
G
)
|
>
1 we know that
|
Z
(
G
)
| 6
= 1. Since
G
is not abelian
we know that
|
Z
(
G
)
| 6
= 125. So either
|
Z
(
G
)
|
= 5 or
|
Z
(
G
)
|
= 25.
Suppose
|
Z
(
G
)
|
= 25.
Then
|
G/Z
(
G
)
|
= 125
/
25 = 5 which is
prime. Hence
G/Z
(
G
) is cyclic. But then by Theorem 9.3 (this the-
orem was also proved in class)
G
is abelian. This is a contradiction
and therefore
|
Z
(
G
)
|
= 5.
3
Hence
|
G/Z
(
G
)
|
= 125
/
5 = 25 = 5
2
. By classification of groups
of order
p
2
for prime
p
(Theorem 9.7 which was also proved in calss)
it follows that
G/Z
(
G
)
∼
=
Z
5
⊕
Z
5
or
G/Z
(
G
)
∼
=
Z
25
.
Suppose
G/Z
(
G
)
∼
=
Z
25
.
Since
Z
25
is cyclic, again using Theorem 9.3 we
conclude that
G
is abelian. This is a contradiction and therefore
G/Z
(
G
)
∼
=
Z
5
⊕
Z
5
.
(6) Let
H, K
⊂
G
be subgroups. Suppose
H
C
G, K
C
G
. Prove that
HK
is a normal subgroup of
G
.
Solution
First let us show that
HK
is a subgroup of
G
.
Let us verify the two-step subgroup test.
Let
h
1
k
1
, h
2
∈
HK
where
h
1
, h
2
∈
H, k
1
, k
2
∈
K
.
Then (
h
1
k
1
)(
h
2
k
2
) =
h
1
(
k
1
h
2
)
k
2
=
h
1
(
h
0
2
k
1
)
k
2
= (
h
1
h
0
2
)(
k
1
k
2
)
∈
HK
where we used that
k
1
H
=
Hk
1
since
H
C
G
. This shows that
HK
is closed under the operation.
Next let
hk
∈
HK
where
h
∈
H, k
∈
K
. Then (
hk
)
-
1
=
k
-
1
h
-
1
=
h
0
k
-
1
∈
HK
where
h
0
∈
H
because
k
-
1
H
=
Hk
-
1
.
This shows that
HK
⊂
G
is a subgroup.
Let us verify that it is normal.
Let
g
∈
G, h
∈
H, k
∈
K
. Then
g
(
hk
)
g
-
1
=
ghg
-
1
gkg
-
1
=
h
0
k
0
where
h
0
=
ghg
-
1
∈
H
since
H
C
G
and
k
0
=
gkg
-
1
∈
K
since
K
C
G
.
Hence
g
(
hk
)
g
-
1
∈
HK
and therefore
g
(
HK
)
g
-
1
∈
HK
.
Since
g
∈
G
was arbitrary, by the normal subgroup test this implies
that
HK
C
G
.
(7) Let
G
be a group of order 30. Suppose
Z
(
G
) =
{
e
}
.
Prove that
G
can not be represented as the internal direct product
of two nontrivial subgroups.
Hint:
First show that if
G
=
H
×
K
and both
|
H
|
>
1 and
|
K
|
>
1
then
|
H
|
or
|
K
|
is prime.
Solution
Suppose
G
=
H
×
K
where
|
H
|
>
1
,
|
K
|
>
1.
Then 30 =
|
G
|
=
|
H
| · |
K
|
. The only ways to factor 30 as a product of two integers
bigger than 1 are 30 = 2
·
15 = 3
·
10 = 5
·
6.
Note that in each
of the cases one of the factors is prime.
Therefore
|
H
|
or
|
K
|
is
prime. WLOG
|
H
|
=
p
is prime where
p
= 2
,
3 or 5. Since groups
of prime order are cyclic it follows that
H
is abelian.
But then
any element
h
∈
H
commutes with any element of
H
and it also
commutes with any element of
K
because
G
=
H
×
K
by a theorem
about internal direct products from class.
Therefore
H
⊂
Z
(
G
).
Indeed if
g
∈
G
then
g
=
h
0
k
0
where
h
0
∈
H, k
0
∈
K
and hence
gh
= (
h
0
k
0
)
h
=
h
0
k
0
h
=
h
0
hk
0
=
hh
0
k
0
=
hg
.
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4
This is a contradiction with the assumption that
Z
(
G
) =
{
e
}
.
Therefore
G
can not be represented as the internal direct product of
two nontrivial subgroups.
(8) Suppose
G
1
⊕
G
2
⊕
. . .
⊕
G
n
is cyclic.
Prove that each
G
i
is cyclic.
Solution 1
Let
G
=
G
1
⊕
G
2
⊕
. . .
⊕
G
n
.
Then each
G
i
is isomorphic top
a subgroup of
G
via
ϕ
i
:
G
i
→
G
the canonical map onto the i’th
factor;
ϕ
i
(
x
) = (
e
1
, . . . , e
i
-
1
, x, e
i
+1
, . . . , e
n
).
here
e
k
∈
G
k
is the unit.
Since a subgroup of a cyclic group is cyclic it follows that each
G
i
is
cyclic.
Solution 2
Let
a
= (
a
1
, . . . , a
n
) be a generator for
G
then for any
g
=
(
g
1
, . . . , g
n
)
∈
G
there is a
k
∈
Z
such that
a
k
=
g
which means
that
g
i
=
a
k
i
for all
i
= 1
, . . . n
.
Since
g
i
∈
G
i
can be arbitrary it
follows that any element of
G
i
is a power of
a
i
and hence
G
i
=
h
a
i
i
is cyclic.
(9) Let
ϕ
:
Z
30
→
Z
6
⊕
Z
5
be an isomorphism such that
ϕ
(
¯
3) = (
¯
3
,
¯
2).
Find all the possibilities for
ϕ
(
¯
1). Justify your answer.
Solution
Let
ϕ
(
¯
1) = (
¯
k,
¯
l
). Then
ϕ
(
¯
3) = (3
¯
k,
3
¯
l
) = (
¯
3
,
¯
2). Hence 3
k
= 3
mod 6
,
3
k
-
3 = 6
n, k
-
1 = 2
n, k
= 1
mod 2. Also 3
l
= 2
mod 5.
Multiplying by 2 we get 6
l
= 4
mod 5
, l
= 4
mod 5.
This means that
ϕ
(
¯
1) must be one of the following (
¯
1
,
¯
4)
,
(
¯
3
,
¯
4)
,
(
¯
5
,
¯
4).
By the general formula
|
(
g
1
, g
2
)
|
=
lcm
(
|
g
1
|
,
|
g
2
|
).
Applying it we
get
|
(
¯
1
,
¯
4)
|
=
lcm
(6
,
5) = 30 =
|
Z
6
⊕
Z
5
|
. Therefore
|h
(
¯
1
,
¯
4)
i|
= 30 =
|
Z
6
⊕
Z
5
|
and hence
h
(
¯
1
,
¯
4)
i
=
Z
6
⊕
Z
5
.
Which means that (
¯
1
,
¯
4) is a generator of the cyclic group
Z
6
⊕
Z
5
.
It was proved in lectures that given two cyclic groups of the same
order
G
1
=
h
a
i
, G
2
=
h
b
i
with
|
a
|
=
|
b
|
=
n
the map
G
1
→
G
2
given
by
ϕ
(
a
k
) =
b
k
is an isomorphism. Note that it sends
a
to
b
.
Therefore there exists an isomorphism
ϕ
:
Z
30
→
Z
6
⊕
Z
5
such
that
ϕ
(
¯
1) = (
¯
1
,
¯
4).
Similarly, we compute
|
(
¯
1
,
¯
5)
|
=
lcm
(6
,
5) = 30.
By the same
argument as above there is an isomorphism
ϕ
:
Z
30
→
Z
6
⊕
Z
5
such
that
ϕ
(
¯
1) = (
¯
1
,
¯
5).
5
Lastly, we compute
|
(
¯
3
,
¯
4)
|
=
lcm
(2
,
5) = 10
6
= 30
.
Since isomor-
phisms preserve orders of elements there is no isomorphism
ϕ
:
Z
30
→
Z
6
⊕
Z
5
such that
ϕ
(
¯
1) = (
¯
3
,
¯
4).
Therefore the only possibilities for
ϕ
(
¯
1) are (
¯
1
,
¯
4) and (
¯
5
,
¯
4)
Answer:
The possibilities for
ϕ
(
¯
1) are (
¯
1
,
¯
4) and (
¯
5
,
¯
4).
(10) Determine the order of (
Z
⊕
Z
)
/
h
(2
,
2)
i
. Is this group cyclic?
Solution
Let
G
=
Z
⊕
Z
, H
=
h
(2
,
2)
i
. Note that
H
=
{
(2
l,
2
l
)
|
l
∈
Z
}
. Let
a
= (1
,
0)
∈
G
. We claim that
|
aH
|
=
∞
in
G/H
. Indeed
|
aH
|
=
k
if
k
is the smallest positive number such that
a
k
∈
H
. However, for
k >
0 we have that
a
k
= (
k,
0)
6
= (2
l,
2
l
) for any
k >
0
, l
∈
Z
. Hence
a
k
/
∈
H
for any
k >
0 and therefore
|
G/H
|
=
∞
.
We claim that
G/H
is not cyclic. Indeed if it’s cyclic then it must
be infinite cyclic since
|
G/H
|
=
∞
. But every infinite cyclic group is
isomorphic to
Z
. In particular it has no nontrivial elements of finite
order since
Z
doesn’t. On the other hand let
b
= (1
,
1) Then
b /
∈
H
but 2
b
∈
H
which means that
|
bH
|
= 2 in
G/Z
. Therefore
G/H
Z
and hence
G/H
is not cyclic.
Answer:
|
(
Z
⊕
Z
)
/
h
(2
,
2)
i|
=
∞
. (
Z
⊕
Z
)
/
h
(2
,
2)
i
is not cyclic.
(11) Prove that (1 3 5) belongs to [
A
5
, A
5
].
Solution
Let
a, b, c, d, e
be distinct numbers. Let
σ
= (
a b c
)
, τ
= (
c d e
). Let
us compute [
σ, τ
] We have
σ
-
1
= (
a c b
) and
τ
-
1
= (
c e d
).
Hence [
σ, τ
] =
στσ
-
1
τ
-
1
= (
a b c
)(
c d e
)(
a c b
)(
c e d
) = (
a d c
). Set-
ting
a
= 1
, d
= 3
, c
= 5
, b
= 2
, e
= 4 we get that [(1 2 5)
,
(5 3 4)] =
(1 3 5).
It was shown in class that a
k
-cycle with odd
k
is even. Therefore
both (1 2 5) and (5 3 4) are in
A
5
and hence (1 3 5) = [(1 2 5)
,
(5 3 4)]
is in [
A
5
, A
5
].
(12) Let
n >
2.
Prove that
|
U
(
n
)
|
is even.
Hint:
Use Lagrange’s theorem.
Solution
Note that
gcd
(
n
-
1
, n
) = 1 since if
d
divides both
n
and
n
-
1
then it also divides
n
-
(
n
-
1) = 1.
Therefore
n
-
1
∈
U
(
n
).
6
Next (
n
-
1) =
-
1
mod
n
and hence (
n
-
1)
2
= (
-
1)
2
= 1
mod
n
.
This can also be seen more directly since (
n
-
1)
2
=
n
2
-
2
n
+ 1 = 1
mod
n
.
Since
n >
2 we have that
n
-
1
>
1 and hence
n
-
1
6
=
¯
1.
This means that
n
-
1 has order 2 in
U
(
n
).
By a corollary to Lagrange’s theorem we have that 2 =
|
n
-
1
|
divides
|
U
(
n
)
|
.
.
(13) Let
m, n >
1 be relatively prime. Let
ϕ
:
Z
mn
→
Z
m
⊕
Z
n
be given
by
ϕ
(
k
mod
mn
) = (
k
mod
m, k
mod
n
)
Prove that
ϕ
is an isomorphism.
Solution
Let us first check that
ϕ
is well defined and preserves the operation.
if
k
1
=
k
2
mod
mn
then
mn
divides
k
1
-
k
2
.
Hence both
m, n
divide
k
1
-
k
2
and therefore
k
1
=
k
2
mod
m
and
k
1
=
k
2
mod
n
.
This shows that
ϕ
is well defined.
Next,
ϕ
(
¯
k
1
+
¯
k
2
) = ((
k
1
+
k
2
)
mod
m,
(
k
1
+
k
2
)
mod
n
)) =
(
k
1
mod
m
+
k
2
mod
m, k
1
mod
n
+
k
2
mod
n
) = (
k
1
mod
m, k
1
mod
n
)+(
k
2
mod
m, k
2
mod
n
) =
ϕ
(
¯
k
1
)+
ϕ
(
¯
k
2
). This shows that
ϕ
preserves operation.
Let us show that
ϕ
is 1
-
1.
Suppose
ϕ
(
¯
k
1
) =
ϕ
(
¯
k
2
).
This means that
k
1
=
k
2
mod
m
and
k
1
=
k
2
mod
m
. Therefore both
m
and
n
divide
k
1
-
k
2
.
Since
gcd
(
m, n
) = 1 this implies that
mn
divides
k
1
-
k
2
as well,
i.e.
¯
k
1
=
¯
k
2
in
Z
mn
.
This proves that
ϕ
is 1-1.
Since
|
Z
mn
|
=
mn
=
|
Z
m
⊕
Z
n
|
this means that
ϕ
is an injec-
tive map between two finite sets with the same number of elements.
Hence
ϕ
must be onto.
This verifies that
ϕ
is a bijection.
(14) Let
G
be a group such that
G/Z
(
G
) is abelian. Prove that for any
a, b, c
∈
G
it holds that [[
a, b
]
, c
] =
e
.
Solution
Let
H
=
Z
(
G
). Since
G/H
is abelian we have that [
aH, bH
] =
H
.
Recall that (
aH
)
1
=
a
-
1
H
and (
bH
)
-
1
=
b
-
1
H
.
Therefore
H
= [
aH, bH
] = (
aH
)(
bH
)(
aH
)
-
1
(
bH
)
-
1
= (
aH
)(
bH
)(
a
-
1
H
)(
b
-
1
H
) =
aba
-
1
b
-
1
H
= [
a, b
]
H
.
Therefore
z
= [
a, b
]
∈
Z
(
G
).
This means
that
z
commutes with any element of
G
and hence [
z, c
] =
e
i.e.
[[
a, b
]
, c
] =
e
.
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7
(15) Let
G
=
U
(20)
, H
=
h
¯
9
i
, K
=
h
11
i
be subgroups of
G
.
Are
G/H
and
G/K
isomorphic?
Solution
Recall that in general
|
aH
|
is the smallest positive
n
such that
(
aH
)
n
=
H
which is equivalent to
a
n
H
=
H
or
a
n
∈
H
.
Let us compute orders of various elements of
G/H
and
G/K
.
First, we have that
U
(20) =
{
¯
1
,
¯
3
,
¯
7
,
¯
9
,
11
,
13
,
17
,
19
} ⊂
Z
20
.
We have that
¯
3
2
=
¯
9
/
∈
K,
3
3
= 27 = 7
mod 20
/
∈
K,
3
4
= 81 = 1
mod 20. Hence
|
¯
3
K
|
= 4.
On the other hand
¯
3
2
=
¯
9
∈
H,
¯
7
2
= 49
mod 20 = 9
mod 20
∈
H,
¯
9
2
=
¯
1
∈
H,
11
2
=
-
9
2
=
¯
1
∈
H,
13
2
=
-
7
2
=
¯
9
∈
H,
17
2
=
-
3
2
=
¯
9
∈
H,
19
2
=
-
1
2
=
¯
1
∈
H
. This shows that every element of
G/H
has order at most 2.
Therefore
G/H
G/K
.
Answer:
G/H
G/K
.