8 Group Actions Actions on Sets Action: Let G be a multiplicative group and let Ω be a set. An action of G on Ω is a group homomorphism G ! Sym(Ω). So, each element g 2 G is associated with a permutation of Ω, and for convenience, we let g(x) denote the image of an element x 2 Ω under this permutation. The fact that we have a group homomorphism from G to Sym(Ω) is equivalent to g(h(x)) = (gh)(x) for g; h 2 G and x 2 Ω. Note: above we have defined a left action. For a right action we would denote the image of x under g by x · g and then x · (gh) = (x · g) · h. We will use left actions exclusively. Examples: 1. The group Sn has a natural action on [n] since each element of Sn is a permutation. More generally Sym(Ω) acts on Ω. 2. The group GL(n; F) acts on Fn by matrix multiplication, that is, if A 2 GL(n; F) and ~x 2 Fn then A(~x) = A~x. 3. For any group G, we have that G acts on itself by the rule that g(x) = gx for all x 2 G and g 2 G. 4. For any group G and subgroup H ≤ G we define G=H = fgH : g 2 Gg, that is, the set of all left H-cosets. Now, G has a natural action on G=H by the rule that g 2 G applied to g0H is gg0H. Faithful: We say that the action of G on Ω is faithful if the kernel of the homomorphism from G to Sym(Ω) is trivial. Equivalently, the action is faithful if any two distinct elements g; h 2 G give distinct permutations of Ω (otherwise gh−1 is in the kernel). Note: if we have a faithful group action, then we have represented G as a subgroup of Sym(Ω). If our action is unfaithful, and H is the kernel of our group homomorphism, then H/G and G=H has a faithful action on Ω. 2 Orbit: Let x 2 Ω. The orbit of x is the set Ωx = fy 2 Ω: g(x) = y for some g 2 Gg: We let Ω=G denote the set of all orbits. Stabilizer: The stabilizer of x 2 Ω is the set Gx = fg 2 G : g(x) = xg: Proposition 8.1 Let x; y 2 Ω let h 2 G and assume that h(x) = y. Then: (i) fg 2 G : g(x) = yg = hGx −1 (ii) Gy = hGxh (iii) jΩxj · jGxj = jGj. Proof: For (i), note that if g 2 Gx then hg(x) = h(x) = y (which proves "⊇") and conversely, −1 −1 −1 if g(x) = y then h g(x) = h (y) = x so h g 2 Gx which implies g 2 hGx (thus proving −1 "⊆"). Similarly for (ii), note that if g 2 Gx then hgh (y) = hg(x) = h(x) = y (proving −1 −1 −1 "⊇ ") and conversely if g 2 Gy then h gh(x) = h g(y) = h (y) = x which implies −1 g 2 hGxh (proving "⊆"). Part (iii) is an immediate consequence of (i) since each element of Ω which is the image of x under a group element is an image under exactly jGxj group elements. Transitive: The action of G on Ω is transitive if there is a single orbit. Theorem 8.2 Let G act transitively on Ω. Then there exists H ≤ G so that the action of G on Ω is isomorphic to the action of G on G=H. Proof: Choose a point x0 2 Ω and set H = Gx0 . Now, apply Proposition 8.1 to choose for every xi 2 Ω a group element gi 2 G so that giH is the subset of G which maps x0 to xi. We now show that this correspondence between Ω and G=H yields an isomorphism. For this, we must prove that if xi; xj 2 Ω and h(xi) = xj then hgiH = gjH. But this is immediate, if h(xi) = xj then hgi(x0) = h(xi) = xj so hgi 2 gjH but then hgiH = gjH. 3 Polya Counting Motivation: How can we count the number of essentially distinct ways of colouring the faces of an Octahedron using fred; yellow; blueg, where two colourings are considered equivalent if there is a rotational symmetry of the Octahedron which takes one to the other? Actions on Colourings: Let A; B be finite sets, and let the group G act on A. We regard B as a set of colours, so we think of a function f : A ! B as a colouring of A. Now, the group G inherits an action on the set BA (the colourings of A) by the rule that σ 2 G applied A −1 A to f 2 B is given by σ(f) = f ◦ σ . To check this, let σ1; σ2 2 G let f 2 B and note that −1 −1 −1 (σ2σ1)(f) = f ◦ (σ2σ1) = f ◦ σ1 ◦ σ2 = σ2(σ1(f)) (note here that the −1 is necessary to have a group action). In the above problem, A is the set of faces of the Octahedron, G is the rotational symmetry group acting on A, and B is the set of colours fred; yellow; blueg. Now, two colourings f; f 0 2 BA are equivalent if there exists σ 2 G so that f 0 = f ◦ σ or in other words f 0 = σ−1(f). So, the number of essentially different colourings is precisely jBA=Gj (i.e. the number of orbits of the action of G on BA). Fixed Points: For every g 2 G we let F ix(g) = fx 2 Ω: g(x) = xg. Theorem 8.3 (Burnside's Lemma) If G acts on the finite set Ω then 1 X jΩ=Gj = jF ix(g)j jGj g2G Proof: We have 1 X 1 X jF ix(g)j = jG j jGj jGj x g2G x2Ω X 1 = jΩxj x2Ω = jΩ=Gj Theorem 8.4 (Polya) Let A; B be finite sets, let G act on A, and let ck denote the number of group elements σ 2 G which have exactly k cycles in their action on A. Then 1 1 X jBA=Gj = c jBjk: jGj k k=1 4 Proof: Considering the action of G on BA, we observe that a colouring f 2 BA is a fixed point of σ 2 G if and only if f is constant on each cycle of σ. It follows that the number of colourings fixed by σ is precisely jBjk where k is the number of cycles of σ. The theorem follows immediately from this and Burnside's Lemma. Problem Solution: The 24 rotational symmetries of the Octahedron consist of: 1. One identity (8 cycles) 2. Six rotations by π about an axis through antipodal edges (4 cycles). π 3. Six rotations by ± 2 about an axis through antipodal vertices (2 cycles). 4. Three rotations by π about an axis through antipodal vertices (4 cycles). 2π 5. Eight rotations by ± 3 about an axis through antipodal faces (4 cycles). Using the notation from Polya's theorem this gives c2 = 6, c4 = 17 and c8 = 1 so the number of essentially distinct colourings is 1 1 (6 · 32 + 17 · 34 + 1 · 38) = (54 + 1377 + 6561) = 333 24 24 The Number Six Motivating Problem: In what ways can the group Sn act faithfully on a set Ω of size n? There are many obvious actions of this type: just label the points of Ω with 1::n and let the permutation π 2 Sn act accordingly. Could there ever be another such action? Conjugation: If g; h 2 G then we call ghg−1 the conjugate of h by g. Define a relation on G by declaring two elements to be equivalent if one is a conjugate of the other. It is immediate that this is an equivalence relation and we call the equivalence classes conjugacy classes. Observation 8.5 A conjugacy class in Sn consists of all permutations with the same cycle structure (i.e. the same number of cycles of each length). Proof by Example: Given σ = (123)(4567)(8)(9X) and τ = (abc)(defg)(h)(ij), the function −1 π given by the rule π(1) = a; π(2) = b; : : : ; π(X) = j satisfies σ = π τπ. 5 Group Automorphism: An automorphism of a group G is a group isomorphism φ : G ! −1 −1 G. Let g 2 G and let φg : G ! G be given by φg(x) = gxg . Then φg(xy) = g(xy)g = −1 −1 (gxg )(gyg ) = φg(x)φg(y) so φg is an automorphism. We say that any automorphism of this type is inner and any other automorphism is outer. Motivating Problem, version 2: We return to the original problem, but now set Ω = [n]. A faithful action of Sn on [n] is, by definition, an injective group homomorphism from Sn to Sym([n]) = Sn. So this is precisely a group automorphism of Sn. Now, the "obvious" actions are given by relabeling: if g 2 Sn then we may have Sn act on itself by the rule that −1 x 2 Sn gives the permutation gxg . However, this is precisely an inner automorphism of Sn. So, our motivating problem is equivalent to the question: Does there ever exist an outer automorphism of Sn? Lemma 8.6 If φ is an automorphism of Sn and φ maps the conjugacy class of transpositions to itself, then φ is inner.