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Orthogonal and Special Linear Groups, Group Homomorphisms, and Cosets, Exams of History of Africa

Solutions to homework exercises from a university-level mathematics course. Topics covered include defining the orthogonal and special linear groups, group homomorphisms, and cosets. Proofs are provided for each exercise.

Typology: Exams

2010/2011

Uploaded on 08/25/2011

milwaukeeschoolofengineering7640
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Mathematics 113: Homework # 2
Curtis Tekell
Dr. Poirier
September 22, 2010
Exercise 1.
(a) Define the orthogonal group O(n) = {MGL(n, R) : det(M) = ±1}. Then
G=hO(n),×i is a group.
(b) Define the special linear group SL(n, R) = {MGL(n, R) : det(M)=1}.
Then G=hSL(n, R),×i is a group.
Proof.
(a) We need only show that O(n)GL(n,R). Let A, B O(n). Then as
det(B) = ±16= 0, B1exists, and clearly AB1GL(n, R). Additionally,
det(AB1) = det(A)det(B1) = det(A)
det(B)=±1
±1=±1,
thus AB1O(n). It follows O(n)GL(n, R).
(b) We need only show that SL(n, R)GL(n, R). Let A, B SL(n, R). Then
as det(B)=16= 0, B1exists, and clearly AB1GL(n, R). Additionally,
det(AB1) = det(A)det(B1) = det(A)
det(B)=1
1= 1,
thus AB1SL(n, R). It follows SL(n, R)GL(n, R).
Exercise 2. Is Z2Z2
=Z4?
Proof. No. Z2Z2has 3 non-identity elements of order 2: (1,1), (0,1), and
(1,0), whereas Z4has two non-identity elements of order 4: 1 and 3, and one of
order 2: 2. It follows they are not isomorphic.
1
pf3

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Mathematics 113: Homework # 2

Curtis Tekell

Dr. Poirier

September 22, 2010

Exercise 1.

(a) Define the orthogonal group O(n) = {M ∈ GL(n, R) : det(M ) = ± 1 }. Then G = 〈O(n), ×〉 is a group.

(b) Define the special linear group SL(n, R) = {M ∈ GL(n, R) : det(M ) = 1}. Then G = 〈SL(n, R), ×〉 is a group.

Proof.

(a) We need only show that O(n) ≤ GL(n, R). Let A, B ∈ O(n). Then as det(B) = ± 1 6 = 0, B−^1 exists, and clearly AB−^1 ∈ GL(n, R). Additionally,

det(AB−^1 ) = det(A)det(B−^1 ) =

det(A) det(B)

thus AB−^1 ∈ O(n). It follows O(n) ≤ GL(n, R).

(b) We need only show that SL(n, R) ≤ GL(n, R). Let A, B ∈ SL(n, R). Then as det(B) = 1 6 = 0, B−^1 exists, and clearly AB−^1 ∈ GL(n, R). Additionally,

det(AB−^1 ) = det(A)det(B−^1 ) =

det(A) det(B)

thus AB−^1 ∈ SL(n, R). It follows SL(n, R) ≤ GL(n, R).

Exercise 2. Is Z 2 ⊕ Z 2 ∼= Z 4?

Proof. No. Z 2 ⊕ Z 2 has 3 non-identity elements of order 2: (1, 1), (0, 1), and (1, 0), whereas Z 4 has two non-identity elements of order 4: 1 and 3, and one of order 2: 2. It follows they are not isomorphic.

Exercise 3. Every non-identity element in the Heisenburg Group H is of infi- nite order.

Proof. Let A ∈ H. It is easy to see that for n ∈ Z+,

An^ =

1 a b 0 1 c 0 0 1

n

1 na ∗ 0 1 nc 0 0 1

where ∗ is, at this point in our proof, complicated and irrelevent. Hence if A is of finite order n, we must have a = c = 0, and we can see that

An^ =

1 0 b 0 1 0 0 0 1

n

1 0 nb 0 1 0 0 0 1

and we must have b = 0 as well, or A = I. It follows the only element of finite order in H is I, and thus all non-identity elements are of infinite order.

Exercise 4. Let f : G → H be a group homomorphism. Let K = Ker(f )

(a) K ≤ G.

(b) Im(f ) ≤ H.

(c) K E G.

Proof.

(a) Let a, b ∈ K, then f (ab−^1 ) = f (a)f (b)−^1 = (1)(1)−^1 = 1, and ab−^1 ∈ K, thus K ≤ G.

(b) Let a, b ∈ Im(f ), then there are some x, y ∈ G such that f (x) = a and f (y) = b, hence f (xy−^1 ) = f (x)f (y)−^1 = ab−^1 , and ab−^1 ∈ Im(f ), thus Im(f ) ≤ H.

(c) Let k ∈ K, and g ∈ G. Then f (gkg−^1 ) = f (g)f (k)f (g)−^1 = f (g)(1)f (g)−^1 = f (g)f (g)−^1 = 1, and gKg−^1 ⊆ K, thus K E G.

Exercise 5. Is f : Z 6 → Z 8 , defined by a 7 → a, a group homomorphism?

Proof. No. In fact, f is not even well defined:

2 = f (2) = f (8) = 8 = 0

is a contradiction.