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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.
