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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 ∈ G is associated with a permutation of Ω, and for convenience, we let g(x) denote the image of an element x ∈ 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 ∈ G and x ∈ 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:
- The group Sn has a natural action on [n] since each element of Sn is a permutation. More generally Sym(Ω) acts on Ω.
- The group GL(n, F) acts on Fn^ by matrix multiplication, that is, if A ∈ GL(n, F) and ~x ∈ Fn^ then A(~x) = A~x.
- For any group G, we have that G acts on itself by the rule that g(x) = gx for all x ∈ G and g ∈ G.
- For any group G and subgroup H ≤ G we define G/H = {gH : g ∈ G}, that is, the set of all left H-cosets. Now, G has a natural action on G/H by the rule that g ∈ G applied to g′H is gg′H.
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 ∈ 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.
Orbit: Let x ∈ Ω. The orbit of x is the set
Ωx = {y ∈ Ω : g(x) = y for some g ∈ G}.
We let Ω/G denote the set of all orbits.
Stabilizer: The stabilizer of x ∈ Ω is the set
Gx = {g ∈ G : g(x) = x}.
Proposition 8.1 Let x, y ∈ Ω let h ∈ G and assume that h(x) = y. Then: (i) {g ∈ G : g(x) = y} = hGx
(ii) Gy = hGxh−^1
(iii) |Ωx| · |Gx| = |G|.
Proof: For (i), note that if g ∈ Gx then hg(x) = h(x) = y (which proves ”⊇”) and conversely, if g(x) = y then h−^1 g(x) = h−^1 (y) = x so h−^1 g ∈ Gx which implies g ∈ hGx (thus proving ”⊆”). Similarly for (ii), note that if g ∈ Gx then hgh−^1 (y) = hg(x) = h(x) = y (proving ”⊇ ”) and conversely if g ∈ Gy then h−^1 gh(x) = h−^1 g(y) = h−^1 (y) = x which implies g ∈ hGxh−^1 (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 |Gx| 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 x 0 ∈ Ω and set H = Gx 0. Now, apply Proposition 8.1 to choose for every xi ∈ Ω a group element gi ∈ G so that giH is the subset of G which maps x 0 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 ∈ Ω and h(xi) = xj then hgiH = gj H. But this is immediate, if h(xi) = xj then hgi(x 0 ) = h(xi) = xj so hgi ∈ gj H but then hgiH = gj H.
Proof: Considering the action of G on BA, we observe that a colouring f ∈ BA^ is a fixed point of σ ∈ 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 |B|k^ 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:
- One identity (8 cycles)
- Six rotations by π about an axis through antipodal edges (4 cycles).
- Six rotations by ±π 2 about an axis through antipodal vertices (2 cycles).
- Three rotations by π about an axis through antipodal vertices (4 cycles).
- Eight rotations by ±^23 π about an axis through antipodal faces (4 cycles).
Using the notation from Polya’s theorem this gives c 2 = 6, c 4 = 17 and c 8 = 1 so the number of essentially distinct colourings is
1 24 (6^ ·^3
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 ∈ Sn act accordingly. Could there ever be another such action?
Conjugation: If g, h ∈ 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)(def g)(h)(ij), the function π given by the rule π(1) = a, π(2) = b,... , π(X) = j satisfies σ = π−^1 τ π.
Group Automorphism: An automorphism of a group G is a group isomorphism φ : G → G. Let g ∈ G and let φg : G → G be given by φg(x) = gxg−^1. Then φg(xy) = g(xy)g−^1 = (gxg−^1 )(gyg−^1 ) = 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 ∈ Sn then we may have Sn act on itself by the rule that x ∈ Sn gives the permutation gxg−^1. 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.
Proof: Note that two distinct transpositions commute if and only if they transpose dis- joint pairs. Let (a 1 a 2 ) be the image of (12) under φ and consider the image of (23) under φ. Since (12) and (23) do not commute, we may assume (without loss) that the image of (23) is (a 2 a 3 ). Next, consider the image of (34). This transposition must commute with (ab) but not (bc) so without loss it is (cd). Continuing in this manner we find that (12), (23), (34),... , (n − 1 , n) map respectively to (a 1 a 2 ), (a 2 a 3 ), (a 3 a 4 ),... (anan− 1 ). Since the transpositions (12), (23),... , (n − 1 n) generate Sn (check!) it follows that φ is given by the rule φ(i) = ai.
Theorem 8.7 If n 6 = 6 then there is no outer automorphism of Sn.
Proof: Consider all conjugacy classes of involutions (elements of order 2). Any group au- tomorphism must map conjugacy classes to conjugacy classes (check!) and must preserve the order of each element (check!) so it follows that every automorphism of Sn sends each conjugacy class of involutions to another conjugacy class of involutions. Let Kn have vertex set [n]. Every involution has the form (ab)(cd)..(st)(u)(v)..(z) so we may identify it with the matching (set of pairwise nonadjecent edges) in Kn consisting of the edges ab, cd,... , st. This gives a natural bijection between the conjugacy class of
Figure 1: A 3-prism
Since there are exactly 15 synthemes and every pentad contains 5 synthemes, we deduce that the number of pentads is exactly six. Based on this construction, we now have an action of Sn on the six pentads. If σ is a transposition, then σ does not fix any pentad (check!) and σ^2 is the identity, so σ acts on the pentads by a permutation with cycle structure (··)(··)(··). Labelling the pentads 1..6 yields an outer automorphism of S 6.
Actions on Structures
Suppose we have a combinatorial structure defined on a ground set Ω. Then we say that a group G acts on this structure if G acts on Ω preserving all properties of the structure.
Examples:
- If Γ is a graph with vertex set V , we say that a group G acts on Γ if G acts on V preserving adjacency (i.e., if u, v ∈ V and g ∈ G then u, v are adjacent if and only if g(u), g(v) are adjacent.)
- if P = (X, ≤) is a poset we say that G acts on P if G acts on X preserving ≤ (i.e., if x, y ∈ X and g ∈ G then x ≤ y if and only if g(x) ≤ g(y).
- If I = (V, B, ∼) is an incidence structure then we say that G acts on I if there is an action of G on V and an action of G on B which preserves ∼ (i.e., if x ∈ V ,
∈ B and g ∈ G then x ∼ if and only if g(x) ∼ g(`)).
Automorphisms: An automorphism of a structure on a ground set Ω is a permutation of Ω which preserves all properties of the structure. So, for instance, an automorphism of a graph Γ with vertex set V is a permutation π of V with the property that u ∼ v if and only
if π(u) ∼ π(v) for every u, v ∈ V. We let Aut(Γ) denote the set of automorphisms of G (and we apply similar notation for other structures). Note that Aut(Γ) is a group.
PGL: The elements of P G(n, F) are subspaces of Fn+1^ and there is a natural action of GL(n + 1, F) on these subspaces: if V is a subspace of Fn+1^ and A ∈ GL(n + 1, F) then AV is another subspace of Fn+1. It is immediate that this action preserves subspace inclusion, so GL(n + 1, F) acts on P G(n, F). Setting Z = {sI : s ∈ F \ { 0 }} we find (check!) that Z is the kernel of this group action (i.e. Z is precisely the set of elements which give the trivial permutation of P G(n, F)). We define the projective general linear group P GL(n + 1 , F) = GL(n + 1, F)/Z and note that P GL(n + 1, F) acts faithfully on P G(n, F). As with homogeneous coordinates for vectors, we will write elements in P GL(n + 1, F) as invertible (n + 1)-dimensional matrices over F with the understanding that two such matrices are equivalent if they are scalar multiples.
PSL: The group SL(n+1, F) has a natural action on P G(n, F) (as a subgroup of GL(n+1, F)) and setting Z′^ = {sIn+1 : s ∈ F \ { 0 } and det(sIn+1) = 1} we find that Z′^ is the kernel of this action. We define the projective special linear group P SL(n + 1, F) = SL(n + 1, F)/Z′ and note that P SL(n + 1, F) acts faithfully on P G(n, F). Note that |Z′| = |{s ∈ F \ { 0 } : sn+1^ = 1}| which depends on the order of the group (in particular, for F = Fq we have |Z′| = (n + 1, q − 1)).
Theorem 8.9 The group P SL(n, Fq) is simple whenever n, q ≥ 2 except for P SL(2, F 2 ) ∼= S 3 and P SL(2, F 3 ) ∼= A 4.
Projective Line: We call P G(1, F) the projective line over F. Using homogeneous coordi- nates, each point in P G(1, F) may be denoted by a pair 〈x, y〉 with x, y not both zero. Since these coordinates are invariant under scalar multiples, we may think of 〈x, y〉 as a slope xy (indeed this is the familiar notion of slope for graphs of lines in R^2 ) with the usual convention that x 0 = ∞. Since slopes form a more convenient labelling of the points, we shall identify P G(1, F) with the set of slopes: F ∪ {∞}.
M¨obius Transformations: Consider the action of P GL(2, q) on P G(1, q). Using homoge- neous coordinates (here as column vectors) we have [ a b c d
] [
x y
]
[
ax + by cx + dy
]
Proof: Let (x, y, z) be a triple of distinct points in P G(1, F). First we shall show that there is a sequence of M¨obius transformations which map (x, y, z) to (∞, 0 , 1). If x 6 = ∞ then the map s → (^) s−^1 x maps x to ∞ and brings our triple to (∞, y′, z′). Now, the transform s → s − y′^ fixes ∞ and sends y′^ to 0 bringing our triple to (∞, 0 , z′′). Since z′′^6 = 0 the transform s → (^) z^1 ′′ s now fixes ∞ and 0 and sends z′′^ to 1 bringing our triple to (∞, 0 , 1) as desired. Composing a function which sends (x, y, z) to (∞, 0 , 1) with the inverse of a function which sends (x′, y′, z′) to (∞, 0 , 1) (in the right order) gives a function which sends (x, y, z) to (x′, y′, z′) thus showing that this action is 3-transitive. To show that this action is sharply 3-transitive, we need only check that G(∞, 0 ,1) is trivial (why?). However, the transformations which fix ∞ are precisely those of the form s → as + b (with a 6 = 0), those that fix (∞, 0) are precisely those of the form s → as and thus, only the identity can fix (∞, 0 , 1).
The Autmorphism Group: For a group G we let Aut(G) denote the set of all group automorphisms of G. Note that if φ, ψ ∈ Aut(G) then φ ◦ ψ ∈ Aut(G), so Aut(G) forms a group under composition.
Examples:
- Consider the (additive) group Zn and let k be relatively prime to n. Then the function φk : Zn → Zn given by φk(a) = ka is an automorphism. In this case Aut(Zn) = Z∗ n.
- Consider the (additive) group G = Z 2 × Z 2. This group has exactly three nonidentity elements, say α, β, γ, each is its own inverse, and the sum of any two is the third. It follows that any bijection from G to itself which fixes the identity is an automorphism, so Aut(G) ∼= S 3.
- If n 6 = 6 then every automorphism of Sn is of the form φg(x) = gxg−^1 for some g ∈ Sn. Then φg ◦φh(x) = g(hxh−^1 )g−^1 = (gh)x(gh)−^1 = φgh(x). It follows that Aut(Sn) ∼= Sn.
Centralizers & Normalizers: Let G be a group and let S ⊆ G. We let
Z(S) = {g ∈ G : gs = sg for every s ∈ S} N (S) = {g ∈ G : gS = Sg}
It is immediate from the definitions that Z(S) ≤ N (S) ≤ G.
Lemma 8.12 If H ≤ G then Z(H) / N (H) and N (H)/Z(H) is isomorphic to a subgroup of Aut(H).
Proof: Consider the action of N (H) on H by the rule that g ∈ N (H) applied to x ∈ H is given by g(x) = gxg−^1. The Kernel of this action is Z(H), so it follows that N (H)/Z(H) is embedded in Aut(H) (i.e. isomorphic to a subgroup of ).
Lemma 8.13 If G acts sharply 4-transitively on X then |X| ≤ 11.
Proof: We may assume (without loss) that X = [n] and then associate G with its image in Sn and let ι denote the identity permutation. We begin the proof with two easy claims.
Claim 1: If δ, ∈ G commute then δ fixes the set {x ∈ X : (x) = x}
This is immediate, if (x) = x and δ(x) = y then we have y = δ(x) = δ(x) = (y) so y is also a fixed point of .
Claim 2: If δ, ∈ G have order two, they are conjugate.
Now, since only the identity fixes four points, we may assume that δ has cycle structure (12)(34)... and that has cycle structure (ab)(cd).. .. Now choose an element φ ∈ G so that φ(1) = a, φ(2) = b, φ(3) = c and φ(4) = d. Now we have that δφ−^1 φ fixes 1, 2 , 3 , 4 so it is the identity. Thus δ = φ−^1 φ and these elements are conjugate as desired.
Now, let α, β, γ be the unique elements of G with α = (1)(2)(34).. ., β = (12)(3)(4).. ., and γ = (12)(34).. .. Then α^2 , β^2 , γ^2 all fix the first four elements, so they are the identity. Furthermore H = {ι, α, β, γ} is a subgroup of G which is isomorphic to Z 2 × Z 2. If α has a fixed point other than 1,2, then we shall denote it by ∞. Note that since α, β, γ commute it follows from the first claim that if ∞ exists, then it is also fixed by β and γ. Now, γ must fix two other points 5, 6 6 = ∞ (since it is conjugate to α - and therefore has the same cycle structure). It then follows that α = (1)(2)(34)(56).. ., β = (12)(3)(4)(56).. ., γ = (12)(34)(5)(6).. .. Next we shall prove that Z(H) = H. To see this, suppose that δ ∈ Z(H). Now by the first claim δ must fix the sets { 1 , 2 , ∞}, { 3 , 4 , ∞}, { 5 , 6 , ∞} so it must be that δ fixes ∞ (if it exists) and either transposes or fixes each of { 1 , 2 }, { 3 , 4 }, { 5 , 6 }. If it fixes two of these sets, then δ = ι, and otherwise it has the same behavior as one of α, β, γ on 4 elements. Thus
