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Mathematics 1214: Introduction to Group Theory Solutions to Exercise Sheet 11

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Mathematics 1214: Introduction to Group Theory Solutions to Exercise Sheet 11 Mathematics 1214: Introduction to Group Theory Solutions to exercise sheet 11 1. Let n ≥ 3 and let Dn be the dihedral group of order 2n. Writing ρ = ρ2π/n, show that hρi is a −1 normal subgroup of Dn. [Hint: if r is reflection in the x-axis, then r ◦ρ = ρ ◦r, and the reflections j in Dn are rj = ρ ◦ r for 0 ≤ j < n.] What is the order of the quotient group Dn/hρi? Solution Let θ : Dn → {−1, 1} be given by defining θ(α) = 1 if α is a proper motion, and θ(α) = −1 if α is an improper motion. By homework exercise sheet 5, question 5(d), θ is then a homomorphism into the group {−1, 1} with multiplication. Moreover, the proper motions in Dn are 2 n−1 precisely e,ρ,ρ ,...,ρ (since these are orientation-preserving elements of Dn, and the remaining elements are all reflections, hence orientation-reversing), so ker θ = {α ∈ Dn : θ(α) = 1} = hρi, hence hρi is a normal subgroup of Dn by Theorem 47. Alternatively, we could avoid having to dream up a suitable homomorphism, and instead use the definition of “normal subgroup” directly. Let N = hρi. We have to check that if n ∈ N and g ∈ Dn then gng−1 ∈ N. If g ∈ N then gng−1 ∈ N, since N is a subgroup. j −1 j −1 If g 6∈ N then g is one of the reflections in Dn, say g = rj = ρ ◦ r. Then g = (ρ ◦ r) = r−1◦(ρj)−1 = r◦ρ−j. Moreover, n ∈ N = hρi, so n = ρk for some k ∈ Z. So gng−1 = ρj ◦r◦ρk ◦r◦ρ−j. Since r ◦ ρ = ρ−1 ◦ r, we have r ◦ ρ2 = r ◦ ρ ◦ ρ = ρ−1 ◦ r ◦ ρ = ρ−1 ◦ ρ−1 ◦ r = ρ−2 ◦ r, and similarly, we have r ◦ ρk = ρ−k ◦ r. Hence gng−1 = ρj ◦ ρ−k ◦ r ◦ r ◦ ρ−j = ρj ◦ ρ−k ◦ ρ−j = ρj−k−j = ρ−k, which is in N = hρi. So gng−1 ∈ N. Hence N is normal in Dn. |Dn| 2n Finally, we have |Dn| = 2n and |hρi| = o(ρ)= n, so |Dn/hρi| = |hρi| = n = 2. 2. Give an example of a normal subgroup of GL(2, R), and give an example of a subgroup of GL(2, R) which is not normal. Solution Trivial examples of normal subgroups are GL(2, R) itself, and {eGL(2,R)} = {I}. A less trivial example is SL(2, R), which is the kernel of the determinant and so a normal subgroup (since the determinant is a homomorphism). 1 x R R Consider H = { 0 1 : x ∈ }. This is a subgroup of GL(2, ) [you should check this] but taking 1 1 1 0 R −1 1 0 −1 0 1 X = 0 1 ∈ H and A = 1 1 ∈ GL(2, ) gives A = −1 1 and AXA = −1 2 6∈ H. So H is not normal in GL(2, R). 3. Let G be a group and let N be a normal subgroup of G. (a) Prove that the mapping η : G → G/N defined by η(a) = Na for a ∈ G is a homomorphism. What are the kernel and image of η? (b) If H is another group and θ : G → H is a homomorphism with kernel N, and ϕ: G/N → H is defined by ϕ(Na) = θ(a), then ϕ is a well-defined homomorphism (see the proof of the Fundamental Homomorphism Theorem). Prove that θ = ϕ ◦ η. Solution (a) For all a, b ∈ G we have η(ab)=(Na)(Nb) = N(ab) = η(ab), so η is a homo- morphism. The image of η is η(G) = {η(a): a ∈ G} = {Na: a ∈ G} = G/N and the kernel is ker η = {a ∈ G: Na = eG/N = N} = {a ∈ G: a ∈ N} = N since Na = N ⇐⇒ a ∈ N (if you like, by Theorem 35 applied with H = N and b = e). (b) For any a ∈ G we have (ϕ ◦ η)(a)= ϕ(η(a)) = ϕ(Na)= θ(a) by the definitions of the mappings η and ϕ. Since ϕ ◦ η has the same domain and codomain as θ and we’ve just seen that they take the same values at each point of their domains, this shows that ϕ ◦ η = θ. 4. Let θ : R → C×, θ(x)= eix. Prove that θ is a homomorphism with kernel h2πi, and that R/h2πi ≈ {z ∈ C× : |z| = 1}. Solution For any x, y ∈ R we have θ(x+y)= ei(x+y) = eixeiy = θ(x)·θ(y). So θ is a homomorphism from (R, +) to (C×, ·). Its kernel is ix ker θ = {x ∈ R: θ(x)= eC× = 1} = {x ∈ R: e = 1} = {n · 2π : n ∈ Z} = h2πi. By the fundamental homomorphism theorem (actually, this is Corollary 50), we have R/h2πi ≈ θ(R)= {eix : x ∈ R} = {z ∈ C× : |z| = 1}. 5. Let G be a group with identity element e. (a) Show that {e} ⊳ G and G/{e} ≈ G. (b) Show that G ⊳ G and G/G ≈ {e}. Solution (a) Let θ : G → G be given by θ(a) = a for all a ∈ G (so that θ is just the identity mapping on G). Then θ is a surjective homomorphism with kernel {e}, so {e} ⊳ G by Theorem 47, and G/{e} ≈ G by the fundamental homomorphism theorem. (b) Let θ : G → {e} be given by θ(a)= e for all a ∈ G. Then θ is a surjective homomorphism with kernel G, so G ⊳ G by Theorem 47 and G/G ≈ {e} by the fundamental homomorphism theorem. 6. Let G be a group. (a) Show that if N is a normal subgroup of G, then G/N is an abelian group if and only if aba−1b−1 ∈ N for all a, b ∈ G. (b) Give an example of a non-abelian group G and a normal subgroup N of G such that G/N is a finite abelian group. [You can always do this easily with N = G. So if you want a little more of a challenge, try to do it with N =6 G.] (c) Give another example of a non-abelian group G and a normal subgroup N of G such that G/N is an infinite abelian group. Solution (a) If a, b ∈ G then we have (Na)(Nb)=(Nb)(Na) ⇐⇒ N(ab)= N(ba) ⇐⇒ (ab)(ba)−1 ∈ N ⇐⇒ aba−1b−1 ∈ N where we have used Theorem 35 in the middle. So G/N is abelian ⇐⇒ (Na)(Nb)=(Nb)(Na) for all a, b ∈ G ⇐⇒ aba−1b−1 ∈ N for all a, b ∈ G. 2 (b) For a trivial example, take N = G = S3; then G is non-abelian and G/N = G/G ≈ {e} and {e} is finite and abelian, so G is finite and abelian. For a less trivial example, take G = S3 × Z2 and let N = S3 × {[0]}. Then G is non-abelian since S3 is non-abelian. Moreover, N is a normal subgroup of G and G/N ≈ Z2. [You should justify the three statements in the last sentence by considering the map π : S3 × Z2 → Z2, π((α, [k])) = [k] for (α, [k]) ∈ S3 × Z2]. Since Z2 is a finite and abelian, this shows that G/N is finite and abelian. (c) Let G = S3 × Z and let N = S3 × {[0]}. Then (much as in (b)) G is non-abelian, N ⊳ G and G/N ≈ Z is an infinite abelian group. 7. If G is a cyclic group and N is a subgroup of G, explain why N is a normal subgroup of G and prove that G/N is a cyclic group. [Hint: for the second part, if G = hai, prove that G/N = hNai.] Solution If G is cyclic then G is abelian. Any subgroup of an abelian group is normal. So any subgroup N of G is normal. Since G is cyclic, we have G = hai = {ak : k ∈ Z} for some a ∈ G. So G/N = {Nx: x ∈ G} = {N(ak): k ∈ Z} = {(Na)k : k ∈ Z} = hNai. (if you like, you can justify the third equality by applying Exercise 3(a) with Theorem 40(c).) Hence G/N is cyclic. 8. Let G be a group and let N ⊳ G. (a) If G is a finite group, prove that G/N is a finite group. (b) If G is an infinite group and N is a finite normal subgroup, prove that G/N is an infinite group. (c) Show (by giving examples) that if G is an infinite group and N is an infinite normal subgroup, then G/N may be finite, or it may be infinite. Solution (a) G/N is a partition of G, so if G is finite then so is G/N. Alternatively, if n = |G| then we may write G = {a1,...,an} and so G/N = {Na: a ∈ G} = {Na1,...,Nan} is a set of size at most n. [Note that G/N will usually have size much smaller than n, since for many values of i, j we will have Nai = Naj]. |G| Alternatively, simply recall that if G is finite then |G/N| = |N| . Since |N| ≥ 1, we have |G/N| ≤ |G|, so G/N is a finite group. (b) Again, G/N is a partition of G. Since G is an infinite set and each coset Na in G/N has the same size as N, which is finite, there must be infinitely many sets in this partition. So G/N is infinite. (c) If G is any infinite group, then G/G is finite (in fact, it has order 1).
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