HW6 - Answers Included PDF

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Homework 6 There will be a quiz on this homework on Friday, March 16.

1. Let A and B be normal subgroups of a group H such that AB = H. Prove that there is an isomorphism ∼ = H/(A ∩ B) − → (H/A) × (H/B ). Solution: Define a function f : H → (H/A) × (H/B ) by h 7→ (hA, hB ). f is obviously a homomorphism. We have ker(f ) = A∩B, since f (h) = 1(H/A)×(H/B ) = (A, B) if and only if h ∈ A and h ∈ B. Thus, by the First Isomorphism Theorem, it suffices to show f is surjective. Let (h1 A, h2 B) ∈ (H/A) × (H/B ). Write h1−1 = a1 b1 and h2 = a2 b2 ; this is possible since H = AB. Then −1 −1 f (a2 b−1 1 ) = (a2 b1 A, a2 b1 B).

Notice that a2 b1−1B = a2 B = a2 b2 B = h2 B. Also, a2 b1−1A = b1−1b1 a2 b1−1 A. Since A is normal, −1 −1 −1 −1 −1 −1 b1 a2 b−1 1 ∈ A, so b1 b1 a2 b1 A = b1 A = b1 a1 A = h1 A. Thus, f (a2 b1 ) = (h1 A, h2 B), and so f is surjective. 2. Let G be a group. Recall that Z(G) denotes the center of G. Prove that G/Z(G) is cyclic if and only if G is abelian. Solution: (⇐) Since G is abelian, Z(G) = G, so G/Z(G) is the trivial group. In particular, G/Z (G) is cyclic. (⇒) Let gZ (G) be a generator of G/Z (G). Then every element of G may be written as g k z for some k ∈ Z and z ∈ Z(G). Let x, y ∈ G, and choose k1 , k2 ∈ Z, z1 , z2 ∈ Z(G) such that x = g k1 z1 and y = g k2 z2 . Then xy = g k1 z1 g k2 z2 . Since g k1 , z1 , g k2 , and z2 all commute with one another, we have g k1 z1 g k2 z2 = g k2 z2 g k1 z1 = yx. 3. If G is a group, H is an abelian, normal subgroup of G, and K is any subgroup of G, prove that H ∩ K E HK. Note: HK is indeed a subgroup of G, since H E G. Solution: First of all, since H E G, KH is a subgroup of G. This means HK = KH, and so HK is a subgroup of G as well. Let hk ∈ HK and x ∈ H ∩K. We have hkx(hk)−1 = hkxk −1 h−1 . Notice that kxk −1 ∈ K . Also, kxk −1 ∈ H, since H is normal and x ∈ H. Thus, kxk −1 ∈ H ∩ K. Finally, since H is abelian, we have hkxk −1 h−1 = kxk −1 hh−1 = kxk −1 ∈ H ∩ K .

4. Let G = hgi be a cyclic group of order n < ∞, and let d be a positive divisor of n. Prove that the quotient G/hg d i is cyclic of order d. Note: cyclic groups are abelian, and subgroups of abelian groups are always normal, so the quotient G/hg d i makes sense. Solution: g hg d i is a generator of G/hg d i, so G/hg d i is cyclic. Since |g d | = Proposition 18), Lagrange’s Theorem implies |G/hg d i| = [G : hg d i] =

n = d. n/d

n (n,d)

=

n d

(by...