Group Theory: Math30038, Sheet 4

GCS

1. Suppose that G acts on a set Ω. If α ∈ Ω, we let

G_{α} = {g ∈ G | αg = α}.

Now suppose that β, γ ∈ Ω are such that βh = γ for some h ∈ G. Show that G_{γ} = h^{−1}G_{β}h. Solution: Suppose that x ∈ G_{β}, then γh^{−1}xh = βh · h^{−1}xh = βxh = βh = γ so h^{−1}G_{β}h ⊆ G_{γ}. Now β = γh^{−1} so a similar argument shows that hG_{γ}h^{−1} ⊆ G_{β}. Premultiplying by h^{−1} and postmultiplying by h it follows that G_{γ} ⊆ h^{−1}G_{β}h. We have an inclusion and its reverse, so G_{γ} = h^{−1}G_{β}h.

2. Let P be a group of order p^{n} where p is a prime number. Suppose that P acts on a finite set Q of size q where p does not divide q. Show that this action of P has a fixed point (i.e. there is α ∈ Q such that αg = α∀g ∈ P). Solution: If ξ ∈ Q, then the stabilizer (isotropy group) of ξ is denoted P_{ξ}, and the orbit of ξ has size |P : P_{ξ}| which is a power of p. Now count Q by adding up the sizes of the orbits of P acting on Q to discover that at least one orbit must have size 1, else q would be divisible by p.

3. In how many essentially different ways can one colour the edges of a regular octahedron using c colours (where each edge is monochromatic, and two colourings are deemed the same if one can moved to the other by a rigid motion – and reflections are not allowed). Solution: A regular octahedron has 8 identical equilateral faces, and its group of rotational symmetries G has order 24. There are 4 axes of symmetry through the centres of faces and the centre of the opposite face, giving rise to 8 rotations A of order 3. There are 3 axes of symmetry through a vertex and the opposite vertex, giving rise to 6 rotations B of order 4 and 3 rotations C of order 2. There are 6 axes of symmetry through the centre of an edge and the centre of the opposite edge, giving rise to 6 rotations D of order 2, and there is the identity map E. We have described 8 + 6 + 3 + 6 + 1 = 24 all of which are different, and so have a list of all the elements of G. Now let Ω be the set of coloured octahedra, a set of size c^{12}. We seek to count the orbits of G acting on Ω by using the counting principle not due to Burnside.

Element type Number of this type |Fix|
A  8  c^{4}
B  6  c^{2}
C  3  c^{4}
D  6  c^{4}
E  1  c^{12}

The number of essentially different edge colourings of the octahedron is therefore

\frac{c^{12} + 17c^{4} + 6c^{2}}{24} .

For example when c = 2 this is 183.

4. Let G be a group with subgroups H and K, each of finite index in G. Prove that H ∩ K has finite index in G. Solution It follows from Sheet 3, Question 4, that |K : H ∩ K| ≤ |G : H| < ∞. Now |G : H ∩ K| = |G : K| · |K : K ∩ H| < ∞.

5. Let G be a group and suppose that H ≤ G and |G : H| < ∞. By considering the groups g^{−1}Hg as g ranges over G (or otherwise), prove that G has a normal subgroup N with |G : N| < ∞ and N ≤ H ≤ G. Solution: The solution to the previous question shows that the intersection of two subgroups of finite index in G is of finite index in G. A straightforward induction yields that the intersection of finitely many subgroups of finite index in G is of finite index in G. Now, it is a routine matter to check (please do it) that each set g^{−1}Hg is a subgroup of G (where g is an arbitrary element of G). Let Ĥ = ∩_{g∈G}g^{−1}Hg ≤ G. If x ∈ G and y ∈ Ĥ, then x^{−1}yx ∈ (gx)^{−1}Hgx for every g ∈ G so x^{−1}yx ∈ g^{−1}Hg for every g ∈ G. Thus x^{−1}Ĥx ≤ Ĥ. Similarly xĤx^{−1} ≤ Ĥ and therefore Ĥ ≤ x^{−1}Ĥx for every x ∈ G so Ĥ is a normal subgroup of G.

If T is a right transversal for H in G, then it is easy to verify that g^{−1}T is a right transversal for g^{−1}Hg in G (please do it). Therefore each group g^{−1}Hg is of finite index in G. We will be finished if we can show that there are only finitely many groups g^{−1}Hg as g ranges over G. Suppose that a, b ∈ G and that a^{−1}Ha ≠ b^{−1}Hb so (ab^{−1})^{−1}Hab^{−1} ≠ H it follows that ab^{−1} ∉ H and so a and b are in different right cosets of H in G. There are only finitely many right cosets of H in G so there are only finitely many groups g^{−1}Hg as g ranges over G.

6. Let G be a group and suppose that x, y ∈ G. Prove that o(xy) = o(yx). Solution: y(xy)^{n} = (yx)^{n}y so (xy)^{n} = 1 if and only if (yx)^{n} = 1. Thus the orders of xy and yx co-incide.