Download 103B - Homework 1 Solutions - Roman Kitsela Exercise 1. Q6 Proof

103B - Homework 1 Solutions - Roman Kitsela
Exercise 1. Q6
Proof. We need to determine whether {π n : n ∈ Z} is a group under addition.
When I see this questions I immediately notice that the set {π n : n ∈ Z} is
defined multiplicatively. Meaning that you generate new elements in the set
by multiplying existing elements by π. So my suspicion is that this cannot
cannot be a group under addition.
I need to check check whether the additive group axioms holds:
1. closure
2. associativity
3. identity
4. inverses
The easiest of these to check is identity, so we need to check whether 0 is in
the set {π n : n ∈ Z}? i.e. can find some integer k such that 0 = π k ?
Clearly this is impossible and so {π n : n ∈ Z} is not a group under addition.
Exercise 2. Q8
Proof. We need to determine whether the n × n real matrices with determinant 2 form a subgroup of GL(n, R).
The group operation is matrix multiplication (inherited from GL(n, R)) and
the main thing that can go wrong in subgroup problems is the closure property fails (if I have two elements both in the subgroup will their ”product”
be in the subgroup?)
So my first thought is: ”If I multiply two matrices with a determinant of 2
will their product have determinant 2?” That doesn’t sound right to me, so
my suspicions are that this is not a subgroup.
To prove this we need to quote a result from linear algebra:
Given two n × n matrices:
det(AB) = det(A)det(B)
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103B - Homework 1 Solutions - Roman Kitsela
Now if we take two elements A, B in the set {M : M ∈ GL(n, R) and det(M ) =
2}:
det(AB) = det(A)det(B) = 2 × 2 = 4
So AB is not in {M : M ∈ GL(n, R) and det(M ) = 2}
Hence {M : M ∈ GL(n, R) and det(M ) = 2} is not a subgroup of GL(n, R)
(The closure property fails)
Exercise 3. Q23
Proof. We want to describe the cyclic group of GL(2, R) generated by:
1 1
M=
0 1
Calculating we find that:
1 2
2
M =
0 1
1 3
3
M =
0 1
1 4
4
M =
0 1
At this point it should be clear what the pattern is in the entries of M k :
1 k
k
for k positive integer.
Claim: M =
0 1
I will prove this claim by induction.
1 1
Base case: When k = 1, M 1 =
so result is obvious.
0 1
1 k
k
Inductive step: I will assume that M =
is true, and I want to show
0 1
1 k+1
that M k+1 =
0
1
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103B - Homework 1 Solutions - Roman Kitsela
We can do this by a simple calculation:
M
k+1
1 k
=M ·M =M ·
0 1
1 k+1
=
0
1
k
This handles all positive powers of M , it is not hard to see that the result
holds for negative values of k, i.e.
1 −1
−1
M =
0 1
1 −2
−2
M =
0 1
1 k
k
holds for negative
and so on. You should try and prove that M =
0 1
k.
1 1
Combining the results for positive and negative k: the cyclic subgroup
0 1
1 k
in GL(2, R) is the collection of matrices
:k∈Z
0 1
Exercise 4. 26
Proof. G1 is cyclic, generated by 1 or −1.
G2 is not cyclic.
G3 is not cyclic.
G4 is cyclic, generated by 6 or −6.
G5 is cyclic, generated by 6 or 6−1 .
G6 is not cyclic.
Exercise 5. Q27
Proof. Want to calculate the order (size) of the subgroup of Z4 (integers
mod 4) generated by 3.
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103B - Homework 1 Solutions - Roman Kitsela
The group action inherited from Z4 is (modular) addition. So we calculate:
3+3=6=2
2+3=5=1
1+3=4=0
So the subgroup generated by 3 and denoted by h3i = {3, 2, 1, 0} = Z4 , so
has order 4.
Exercise 6. Q36(c)
Proof. h1i and h5i
These are the numbers that are coprime to 5, think about why this is the
case.
Exercise 7. Q18
Proof. Want to calculate the order of h30i in Z42 .
This is given by:
42
42
=
=7
GCD(42, 30)
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Exercise 8. Q19
Proof. hii = {i, i2 , i3 , i4 } = {i, −1, −i, 1}
So order of hii in (C∗ , ·) is 4.
Exercise 9. Q4
Proof. This is a homomorphism, just perform the straightforward check
φ(ab) = φ(a)φ(b) for all a, b ∈ Z6 .
NOTE:
When we write φ(ab) = φ(a)φ(b) we are using multiplicative notation to
denote the group action. What we really mean is:
φ(a ∗1 b) = φ(a) ∗2 φ(b)
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103B - Homework 1 Solutions - Roman Kitsela
where ∗1 is the group action of Z6 (i.e. addition modulo 6) and ∗2 is the
group action of Z2 (i.e. addition modulo 2). So the check you need to
perform is:
Given a and b in Z6 is it true that φ(a +1 b) = φ(a) +2 φ(b)? Where +1 is
addition modulo 6 and +2 is addition modulo 2.
Look at Q5 to see how this may fail to be a homomorphism!
Exercise 10. Q6
Proof. This is a homomorphism, we can check quickly:
Let a and b be two real numbers:
φ(a + b) = 2a+b = 2a 2b = φ(a)φ(b)
Exercise 11. Q10
Proof. This one looks weird but carefully applying the criterion is not too
bad:
Let f and g be elements of F , so they are continuous real functions. We are
told that F is the additive group of real continuous functions so the group
action is addition.
Z 4
Z 4
Z 4
Z 4
φ(f +g) =
(f +g)(x)dx =
(f (x)+g(x))dx =
f (x)dx+
g(x)dx
0
0
0
0
So φ(f + g) = φ(f ) + φ(g)
Exercise 12. Q18
Proof. φ : Z −→ Z10 is determined completely by where it sends 1, in this
case φ sends 1 to 6. This means:
φ(18) = φ(1 + 1 + · · · + 1) = 18φ(1) = 18 × 6 = 108 = 8 (in Z10 )
By definition:
Ker(φ) = {x ∈ Z : φ(x) = 0}
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103B - Homework 1 Solutions - Roman Kitsela
We can calculate:
φ(1) = 6
φ(2) = 12 = 2
φ(3) = 8
φ(4) = 14 = 4
φ(5) = 10 = 0
So it is not hard to see that φ(5k) = 0 for all integers k, i.e.
Ker(φ) = {5k : k ∈ Z} = 5Z
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