Group Theory: Math30038, Sheet 7

Solutions – GCS

1. Suppose that n ≥ 3. Let g ∈ G = S_{n} be an n-cycle. Recall that all n-cycles are conjugate in S_{n}.

(a) Prove that C_{G}(g) = ⟨g⟩.

Solution Since the conjugacy class of g has size (n−1)! it follows that C_{G}(g) has order n and so is ⟨g⟩.

(b) Deduce that Z(G) = 1.

Solution Let g_{1} = (1, 2, 3, …, n) and g_{2} = (2, 1, 3, … n) (so g_{1} and g_{2} only differ in their effects on 1 and 2). Now Z(G) ≤ C_{G}(g_{1}) ∩ C_{G}(g_{2}) = ⟨g_{1}⟩ ∩ ⟨g_{2}⟩ = 1.

(c) Show that Aut(G) has a subgroup which is isomorphic to G. Solution Let τ : G ⟶ Aut(G) be the homomorphism which sends each group element x to the automorphism τ_{x} which is “conjugation by x”. Now Ker τ = Z(G) = 1 so τ is a monomorphism and Im τ is a subgroup of Aut(G) which is isomorphic to G as required.

2. Suppose that G is a group and that Aut(G) is the trivial group.

(a) Show that G must be abelian. Solution If x ∈ G − Z(G), then τ_{x} (conjugation by x) will be a non-trivial automorphism of G. Thus G − Z(G) = ∅ so G = Z(G) is abelian.

(b) By considering the inversion map, deduce that g^{2} = 1 for all g ∈ G. Solution The inversion map is a non-trivial automorphism unless g = g^{−1} for every g ∈ G, so g^{2} = 1 for all g ∈ G.

(c) Write G additively, so that G becomes a vector space over the field ℤ_{2} of “integers modulo 2”. You may assume that every vector space has a basis. Deduce that G must either be cyclic of order 2 or the trivial group. Solution Pick a basis B of the vector space G. If dim G > 1, then we can choose two distinct elements u, v of B. Now the map which swaps u and v and fixes all other elements of B extends uniquely to a non-trivial linear isomorphism from G to G. This linear isomorphism will also be a non-trivial group automorphism of G. We deduce that G has dimension 0 or 1, and so is either the trivial group or the cyclic group of order 2. Finally we observe that these two groups do indeed have trivial groups of automorphisms.

3. Suppose that G is a group and that Aut(G) is finite. Prove that |G : Z(G)| must be finite. Solution G/Z(G) is isomorphic to a subgroup of Aut G, and so is finite. Thus |G : Z(G)| is finite.

4. (Challenge) Suppose that G is a group and that Aut(G) is cyclic. Prove that G must be abelian. Solution Since a subgroup of a cyclic group is cyclic, it follows that G/Z(G) is a cyclic group. Say G/Z(G) is generated by tZ(G). Thus every element of G lies in a coset t^{k}Z(G) for some integer k, and therefore G = ⟨{t} ∪ Z(G)⟩. Thus G is abelian.

5. Prove that for a finite group, the notions of isomorphism, monomorphism and epimorphism all co-incide. Solution This was poorly worded. I was thinking of homomorphisms from a finite group G to itself (so called endomorphism of G). Since G is finite, the notions of injection, surjection and bijection all co-incide.

6. In each case give an example of a group G and a map ζ : G → G which satisfies the specified property.

(a) An epimorphism (i.e. a surjective homomorphism) but not an isomorphism (suggestion for G: fix a prime number p and consider those complex numbers z such that z^{p^{n}} = 1 for some natural number n which may vary as z varies). Solution The map whic sends each element to its p-th power is a homomorphism, and is surjective but not injective.

(b) A monomorphism (i.e. an injective homomorphism) but not an isomorphism (suggestion for G: ℤ under addition). Solution The map which doubles every integer is an injective homomorphism, but is clearly not surjective.

7. Suppose that G is a finitely generated group (i.e. there is a finite subset X of G such that ⟨X⟩ = G) and that every conjugacy class of G is finite. Prove that |G : Z(G)| is finite. Solution Since every conjugacy class is finite, each centralizer of an element is of finite index in G. Now an element is in the centre of G if and only if it commutes with every element of a generating set. Thus

Z(G) = ∩_{x∈X}C_{G}(x).

Now, we have proved that the intersection of two (and therefore finitely many) subgroups of finite index is of finite index. Since X is finite it follows that |G : Z(G)| < ∞.

8. Let V be a vector space of dimension n over a field F which has q elements. Let GL(V) denote the group of linear isomorphisms from V to V (a group under composition of maps). Determine |GL(V)|. Solution Pick and fix a basis for use both in domain and codomain. The invertible linear maps on V are in bijective correspondence with the invertible n by n matrices. we count such matrices row by row. The first row (viewed as an element of F^{n}) can be any non-zero vector, of which there are q^{n} −1. The second row must not be a linear multiple of the first, yielding q^{n} − q choices. The first two rows are now linearly independent, and their span must have size q^{2}. The third row can be any vector not in the span of the first two rows, giving us q^{n} − q^{2} possibilities.

Continuing in this fashion, we deduce that

|GL(V)| = (q^{n} − 1)(q^{n} − q)(q^{n} − q^{2}) ⋯ (q^{n} − q^{n−1}).

9. (a) Suppose that G is a group acting on a set Ω. Define a map η : G ⟶ Sym(Ω) by (g)η : ω ↦ ω · g for all ω ∈ Ω and for all g ∈ G. Prove that η is a homomorphism of groups. Solution Suppose that u, v ∈ G, then (uv)η : ω ↦ ω · (uv) = (ω · u) · v. On the other hand (u)η ∘ (v)η : ω ↦ (ω · u) · v.

Now for each g ∈ G we have (g)η ∘ (g^{−1})η = (g^{−1})η ∘ (g)η = (1)η which is the identity map. Therefore each (g)η has a twosided inverse and so is an element of Sym(Ω). We have already established the multiplicativity property which ensures that η is a group homomorphism.

(b) Suppose that η : G ⟶ Sym(Ω) is a group homomorphism. Show that one may define a group action of G on the set Ω by defining ω · g to be (ω)((g)η) for all ω ∈ Ω and for all g ∈ G. Show that this is indeed a group action. Solution Now suppose that η : G → Sym(Ω) is a group homomorphism. We define an action of G on Ω by ω · g = (ω)(g)η. We check that this is an action: ω · 1 = (ω)(1)η = (ω)Id_{Ω} = ω for every ω ∈ Ω. Also if u, v ∈ G, then ω · (uv) = (ω)(uv)η = (ω)((u)η(v)η) = ((ω)(u)(η))(v)(η) = (ω · u) · v. Thus we have an action.

(c) Show that the procedures outlined in parts (a) and (b) are mutually inverse (i.e. if you perform one, then the other, then you recover the situation in which you started). Solution Suppose that we have an action denoted by a dot which gives rise to the group homomorphism η : G → Sym(Ω) outlined in part (a). Now we use part (b) to define an action of G on Ω denoted ⋆. The definition of ⋆ is that if ω ∈ Ω and g ∈ G, then ω ⋆ g = (ω)((g)η) = ω · g. Thus ⋆ is the same action as the original one. Conversely suppose that you have a homomorphism η : G → Sym(Ω). This gives rise to an action as outlined in part (b). According to part (a) this gives rise to a homomorphism ξ : G → Sym(Ω) defined by (g)ξ : ω ↦ ω · g = (ω)(g)η. Thus the maps (g)ξ and (g)η have the same domains and codomains, and agree at all arguments. Thus (g)ξ = (g)η for every g, and so ξ = η.