Group Theory: Math30038, Sheet 9 Solutions

GCS

1. Suppose that G is a simple group of order 60.

(a) Show that G has a subgroup A of order 12. Solution Since G is simple there must be more than one Sylow 2-subgroup of G. Let P and Q be distinct subgroups of G of size 4. If P ∩ Q = T has order 2, then T has index 2 in both P and Q and so is normalized by each of them. Let H = N_{G}(T) which has order more than 4, divisible by 4 and dividing 60. Now H ≠ G else T ⊴G. Also |G : H| ≠ 3 else a Poincaré argument would have 60 dividing 3! = 6. Thus |N_{G}(T)| = 12. Let A = N_{G}(T).

(b) Show that A has exactly 5 different conjugates.

Solution |N_{G}(A)| divisible by 12 and divides 60. However, A is not normal in G and so N_{G}(A) = A. Let G act on the conjugates of A via conjugation. The size of the orbit is |G : N_{G}(A)| = 5.

(c) Show that there is an injective homomorphism from G to S_{5}. Solution Let the image of this map be H, a subgroup of index 2 in S_{5}. Let Ω be the set of conjugates of A, a set of size 5. The conjugation action of G on Ω induces a group homomorphism θ : G → Sym(Ω). This homomorphism has non-trivial image, and trivial kernel by simplicity of G. This gives a monomorphism from G to Sym(Ω). Labelling the elements fo Ω using the first five natural numbers we obtain a monomorphism from G into S_{5}. The image X is a copy of G and has index 2 in S_{5}.

(d) Show that both A_{5} and H both contain every element of S_{5} of the form g^{2} and therefore every 5-cycle and every 3-cycle. Solution Suppose that Z is any subgroup of S_{5} of index 2. Then Z is normal in S_{5} and the quoent group S_{5}/Z is cyclic of order 2. Thus if g ∈ G, then (gZ)2 = Z so g^{2} ∈ Z. Thus all squares are in Z. Now all 3-cycles and 5-cycles are squares because (a, b, c) = (a, c, b)2 and (a, b, c, d, e) = (a, d.b, e, c)^{2}.

(e) Show that H = A_{5}.

Solution There are 20 three cycles and 24 5-cycles in S_{5} so A_{5} ∩ H is a subgroup of each of A_{5} and H and has size at least 44. Therefore A_{5} ∩ H has order 60 by Lagrange’s theorem. Thus H = A_{5}.

(f) Deduce that any simple group of order 60 must be isomorphic to A_{5}. Solution That is what we have shown.

Incidentally, here is another way to show that G is isomorphic to A_{5}, and this method cuts out (d), (e) and (f): Now let Y = X ∩ A_{5}. Now from an early sheet, |A_{5} : Y | ≤ |S_{5} : X| = 2 so either |A_{5} : X| is 2 or 1. It cannot eb 2, else then Y would be normal in A_{5} which would violate the simplicity of A_{5}. Therefore the index is 1 so A_{5} ≤ Y. Now each of A_{5} and Y has order 60 so Y = A_{5}.

2. Show that the following presentations all describe the trivial group.

(a) ⟨x | x^{2}, x^{3}⟩, Solution Let x denote the image of x in the presented group G. Thus x = x^{3}(x^{2})^{−1} = 1. Since G is cyclic and generated by x, then G is trivial.

(b) ⟨x, y | xy = yx, xyx = yxy, x^{5}, y^{6}⟩ Solution As before we regard x and y as elements of the presented group G. Now G is abelian since xy = yx. Therefore y = x and G is cyclic. Now 1 = x^{5} = y^{6} = x^{6} so x = 1 = y and G is trivial.

(c) ⟨x, y | x^{2}, y^{4}, xyx = y^{2}, xyxyxy⟩. Solution Working in the established way in G we discover that (xyx)^{2} = y^{4} = 1. Thus xyxxyx = 1 so xyyx = 1 so yy = xx = 1. Thus xyx = y^{2} = 1 so y = xx = 1. Now xyxyxy = x^{3} = 1 = x^{2} so x = 1 in G and G is the trivial group.

3. We discuss the group G = ⟨x, y | x^{2} = y^{2}, (xy)^{2} = x^{2}, x^{4}⟩ in stages:

(a) Let T = ⟨x^{2}⟩ ≤ G so |T | ≤ 2. Show that T ⊴ G. Solution T = {1, x^{2}} = {1, y^{2}}. Thus in G the element or elements of T commute with both x and y and therefore with all elements of G. Thus T is central in G and is therefore normal in G.

(b) Let H = G/T. Show that xT and yT commute so that H is abelian. Solution Now yxT is the inverse of xyT and so is xyT so xyT = yxT. Thus xT and yT commute so G/T is abelian.

(c) Show that |H| ≤ 4.

Solution It is easy to verify that the group generated by xT and yT is {T, xT, yT, xyT } and so H has size at most 4.

(d) Deduce that |G| ≤ 8.

Solution |G| = |G : T | · |T | ≤ 8.

(e) Find a group of order 8 containing elements a, b such that a^{2} = b2, (ab)^{2} = a^{2} and a^{4} = 1. Apply von Dyck’s theorem to deduce that G has a homomorphic image of order 8, and by part (d) deduce that this homomorphism is an isomorphism. Solution The quaterion group will do the trick, in the usual notation i and j satisfy the relations (for x and y respectively). By Von Dyck’s theorem there is an epimorphism from G to a group of order 8, so G has size at least 8. Thus G has order 8, and the homomorphism from G onto the quaternion is an isomorphism.

4. Discuss the (language) group L presented by the small roman alphabet with relations consisting of all homonyms. Thus we have relations such as too = two, practice = practise, sea = see and so on. The group you get in my ideolect is C_{2} (South London with a transparent veneer of civilization), whereas in South African English you get the trivial group. On the other hand, I understand that in German the spelling of a word is determined by its pronunciation, so L is a free group. To get you started in L for English, since two = too in L, then t^{−1}twoo^{−1} = t^{−1}tooo^{−1} and thus w = o in L. Also too = to so o = 1 and therefore w = 1. Here 1 denotes the identity element of L for English. Solution Well, this is best done for fun. I reckon South African Englsih can kill off the group with felt = veldt.