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Groups acting on sets

1 Deﬁnition and examples

Most of the groups we encounter are related to some other structure. For example, groups arising in geometry or physics are often groups of a geometric object (such as Dn) or transformation groups of (such as SO3). The idea underlying this relationship is that of a action: Deﬁnition 1.1. Let G be a group and X a . Then an action of G on X is a function F : G × X → X, where we write F (g, x) = g · x, satisfying:

1. For all g1, g2 ∈ G and x ∈ X, g1 · (g2 · x) = (g1g2) · x. 2. For all x ∈ X, 1 · x = x.

When the action F is understood, we say that X is a G-set.

Note that a is not the same thing as a binary structure. In a binary structure, we combine two elements of X to get a third element of X (we combine two apples and get an apple). In a group action, we combine an element of G with an element of X to get an element of X (we combine an apple and an orange and get another orange).

Example 1.2. 1. The trivial action: g · x = x for all g ∈ G and x ∈ X.

2. Sn acts on {1, . . . , n} via σ · k = σ(k) (here we use the deﬁnition of multiplication in Sn as ). More generally, if X is any set, SX acts on X, by the same formula: given σ ∈ SX and x ∈ X, deﬁne σ · x = σ(x). To see that this is an action as we have deﬁned it, note that, given σ1, σ2 ∈ SX and x ∈ X,

σ1 · (σ2 · x) = σ1 · (σ2(x)) = σ1(σ2(x)) = (σ1 ◦ σ2)(x) = (σ1 ◦ σ2) · x,

since the group operation on SX is function composition. Clearly IdX ·x = IdX (x) = x for all x ∈ X. Thus SX acts on X. Note that

1 SX acts on many other objects associated to X, such as the P(X), the set of all of X, by the formula that, for all σ ∈ SX and A ⊆ X, σ · A = σ(A) = {σ(a): a ∈ A}.

Since #(σ·A) = #(A), if A is ﬁnite, SX also acts on the of P(X) consisting of all subsets of X with 2 elements, or with 3 elements, or with k elements for any ﬁxed k.

∗ n 3. The group R acts on the R by : ∗ n n given t ∈ R and v ∈ R , let t · v = tv ∈ R be scalar multiplication. That this is an action follows from familiar properties of scalar multi- ∗ n plication: t1(t2v) = (t1t2)v and 1v = v, for all t1, t2 ∈ R and v ∈ R . (Of course, these properties hold for t1, t2 ∈ R as well, but R is not a group under multiplication. Also, scalar multiplication has additional properties having to do with addition of scalars or vectors.)

n 4. GLn(R) acts on R by the usual rule A · v = Av is the multiplication of the matrix A on the vector v, and is the same thing as F (v), where n n F : R → R is the linear function corresponding to A. Similarly, On n and SOn act on R . In addition, On and SOn act on the (n−1)-sphere Sn−1 of radius 1 deﬁned by

n−1 n S = {v ∈ R : kvk = 1}.

1 2 2 Note that S = U(1) is the circle in R and S is the unit sphere 3 in R . An (n − 1)-sphere of radius r is deﬁned similarly. 2 5. Let Pn be a regular n-gon in R , n ≥ 3. For example, we could take Pn to be centered at the origin and to have vertices

pk = (cos(2πk/n), sin(2πk/n)), k = 0, 1, . . . , n − 1.

The dihedral group Dn acts Pn and on the set of vertices {p0,..., pn−1}, as well as on the set of edges {p0p1, p1p2,..., pn−1p0}. With the above notation, it is easy to see (as we have described in various home- work problems) that

Dn = {A ∈ O2 : A(Pn) = Pn}.

6. In a partial analogy with the previous example, let S be a regular solid 3 (or ) in R , known to Euclid, Plato, and before Plato to

2 the Pythagoreans. We will not give a precise deﬁnition. We view S 2 as centered at the origin. Here, unlike the case of R where there is a regular n-gon for every n ≥ 3, there are just 5 regular solids. A regular solid is an example of a , which has vertices, edges and faces. If we tabulate this information, we have the following list of the regular solids (where v is the number of vertices, e is the number of edges, and f is the number of faces):

name v e f n tetrahedron 4 6 4 3 8 12 6 4 octahedron 6 12 8 3 dodecahedron 20 30 12 5 icosahedron 12 30 20 3

Here n is the number of edges of a , which is a regular n-gon. This can be determined by the above data, since each meets exactly two faces, and thus 2e = nf. For example, the faces of a dodecahedron are regular pentagons. Note Euler’s formula, which in this case says that v − e + f = 2. To every regular solid S there is an associated dual solid S∨, where the number of vertices of S∨ is equal to the number of faces of S, and vice versa. Here the tetrahedron is its own dual, while the dual of the cube is the octahedron and the dual of the dodecahedron is the icosahedron. Given a regular solid S, we deﬁne its G(S) by

G(S) = {A ∈ SO3 : A(S) = S}.

Then G(S) acts on S, and on the sets of vertices, edges, or faces of S. It is not hard to show that G(S) = G(S∨). Note that (unlike the case of Dn where we allow elements of O2) we only consider elements of SO3. The group G(S) is always ﬁnite, and we shall say a little more about it later.

7. The remaining two examples are more directly connected with . If G is a group, then G acts on itself by left multiplication: g · x = gx. The axioms of a group action just become the fact that multiplication in G is associative (g1(g2x) = (g1g2)x) and the deﬁnition

3 of the identity (1x = x for all x ∈ G). More generally, if H ≤ G is a , not necessarily normal, then G acts on the set of left G/H via: g · (xH) = (gx)H. The argument that this is indeed an action is similar to the case of left multiplication.

−1 8. G acts on itself by conjugation ig: ig(x) = gxg . (We write it this way instead of as g · x to avoid confusion with the left multiplication action.) To see that this is an action, note that, for all g1, g2 ∈ G, −1 −1 −1 −1 −1 ig1 (ig2 (x)) = ig1 (g2xg2 ) = g1(g2xg2 )g1 = (g1g2)x(g2 g1 ) −1 = (g1g2)x(g1g2) = ig1g2 (x),

−1 −1 −1 where we have used the familiar fact that (g1g2) = g2 g1 . Since clearly −1 i1(x) = 1x1 = 1x1 = x, conjugation does give an action of G on itself. This action is the trivial action ⇐⇒ gxg−1 = x for all g, x ∈ G ⇐⇒ gx = xg for all g, x ∈ G ⇐⇒ G is abelian. One principle that we have seen implicitly in some of the above examples is the following: Proposition 1.3. If X is a G-set and f : G0 → G is a homomorphism, then X becomes a G0-set via g0 · x = f(g0) · x. In particular, if H ≤ G, then a G-set X is also an H-set via the inclusion homomorphism of H in G. 0 0 0 Proof. Given g1, g2 ∈ G , 0 0 0 0 0 0 0 0 g1 · (g2 · x) = g1 · (f(g2) · x) = f(g1) · (f(g2) · x) = ((f(g1)f(g2)) · x 0 0 0 0 = f(g1g2) · x = (g1g2) · x, using the fact that f is a homomorphism. Also, if 10 is the identity in G0, then f(10) = 1 is the identity in G, and thus, for all x ∈ X, 10 · x = f(10) · x = 1 · x = x. It follows that the formula g0 · x = f(g0) · x deﬁnes an action of G0 on X.

Example 1.4. If G is a group, then SG acts on G and on the set P(G) of all subsets of G. Thus, so does Aut G, the group of automorphisms of G (i.e. from G to G), which is a subgroup of G under composition. Note that Aut G also acts on the set of all of G, which is a subset of P(G), whereas SG does not act on this set (because a bijection from G to itself will not in general take a subgroup to a subgroup).

4 Using Proposition 1.3, we can give a partial generalization of Cayley’s theorem. Recall that, for the action of G on itself by left multiplication, we deﬁne a bijection `g : G → G by:

`g(x) = gx.

More generally, let G act on a set X, and deﬁne `g : X → X by the formula

`g(x) = g · x. Lemma 1.5. With notation as above,

(i) For all g1, g2 ∈ G, `g1 ◦ `g2 = `g1g2 .

(ii) `1 = IdX .

(iii) For all g ∈ G, `g is a bijection from X to X, i.e. `g ∈ SX for all g ∈ G, and the inverse of `g is `g−1 .

Proof. (i) We must check that, for all x ∈ X, `g1 ◦ `g2 (x) = `g1g2 (x). By deﬁnition,

`g1 ◦ `g2 (x) = `g1 (`g2 (x)) = `g1 (g2 · x)

= g1 · (g2 · x) = (g1g2) · x = `g1g2 (x).

(ii) Clearly, for all x ∈ X, `1(x) = 1 · x = x, and thus `1 = IdX . −1 (iii) It is enough to prove that (`g) = `g−1 , i.e. that `g ◦ `g−1 = `g−1 ◦ `g = IdX . Using (i) and (ii),

`g ◦ `g−1 = `gg−1 = `1 = IdX , and similarly `g−1 ◦ `g = IdX . Note in particular that, if y = g · x, then x = g−1 · y.

Corollary 1.6. If X is a G-set, then the function f : G → SX deﬁned by f(g) = `g is a homomorphism from G to SX .

Proof. By (iii) above, `g ∈ SX . The equation `g1 ◦ `g2 = `g1g2 says that f(g1g2) = f(g1) ◦ f(g2), in other words that f is a homomorphism. Remark 1.7. For the left multiplication action of G on itself, the homo- f : G → SG is easily seen to be injective; this is the content of Cayley’s theorem. In general, though, f need not be injective. For example, if G acts on X by the trivial action g · x = x for all g ∈ G, then `g = IdX for all g ∈ G, so that f is the trivial homomorphism.

5 We can also reverse the construction of the corollary: Given a homo- morphism f : G → SX , since SX acts on X, X becomes a G-set by Propo- sition 1.3. Finally, the two constructions just described (passing from an action of G on X to a homomorphism G → SX , and passing from a homo- morphism G → SX to an action of G on X) are inverse constructions. Thus, the concept of a G-set X is equivalent to the concept of a homomorphism G → SX .

Example 1.8. If G is a group, the Aut G is a subgroup of SG, and thus Aut G acts on G. We have the conjugation homomorphism f : G → Aut G −1 deﬁned by f(g) = ig, where as usual ig(x) = gxg . The composition G → Aut G → SG is the same thing as the action of G on itself by conjugation. Note that G acts on the set of all subgroups of G by conjugation, since Aut G does (Example 1.4).

Deﬁnition 1.9. If X is a G-set, then a G-subset Y of X is a subset Y ⊆ X such that, for all g ∈ G and y ∈ Y , g · y ∈ Y .A G-subset is itself a G-set.

Deﬁnition 1.10. If X1 and X2 are G-sets, an f from X1 to X2 of G-sets, or brieﬂy a G-isomorphism is a bijection f : X1 → X2 such that f(g · x) = g · f(x) for all g ∈ G and x ∈ X. In this case we say that X1 and X2 are are isomorphic as G-sets or G-isomorphic, and write this as ∼ X1 =G X2. Clearly IdX is an isomorphism of G-sets. If f : X1 → X2 is an isomorphism of G-sets, then so is f −1. Likewise the composition of two isomorphisms is again an isomorphism. Thus, as with the usual deﬁnition ∼ of isomorphism, the relation =G is reﬂexive, symmetric, and transitive.

2 Orbits and subgroups

Deﬁnition 2.1. If X is a G-set and x ∈ X, the of X (under G) is the set G · x = {g · x : g ∈ G}. Thus G · x ⊆ X. Clearly G · x is a G-subset of X and is the smallest G-subset of X containing x.

Example 2.2. In the case of the action of Sn on {1, . . . , n}, given σ ∈ Sn, we have previously deﬁned the orbits Oσ(i) of σ. The link with the current deﬁnition is as follows: the orbits of σ in the previous sense are the orbits of hσi acting on {1, . . . , n} as a subgroup of Sn. In other words, Oσ(i) = hσi·i. In fact, both sides are equal to

a {σ (i): a ∈ Z}.

6 We deﬁned the orbits Oσ(i) by an , and so it is natural to try to do the same thing for the orbit G · x.

Proposition 2.3. Let G act on a set X, and deﬁne x ∼G y ⇐⇒ there exists a g ∈ G such that g · x = y. Then ∼G is an equivalence relation, and the containing x is the orbit G · x. Thus, two orbits of G are either disjoint or identical.

Proof. Reﬂexive: x = 1 · x, hence x ∼G x. Symmetric: if x ∼G y, then by deﬁnition there exists a g ∈ G such that g · x = y. We have seen that, ion −1 this case, g · y = x. Thus y ∼G x. Transitive: Say that x ∼G y and that y ∼G z. Then there exists a g1 ∈ G such that g1 · x = y and there exists a g2 ∈ G such that g2 · y = z. Thus z = g2 · y = g2 · (g1 · x) = (g1g2) · x, and so x ∼G z. The remaining statements, that the equivalence class containing x is the orbit G · x and that two orbits of G are either disjoint or identical, are then clear by deﬁnition and from general properties of equivalence classes.

Deﬁnition 2.4. If X is a G-set and if G · x = X for one (or equivalently all) x ∈ X, we say that G acts transitively on X.

Example 2.5. (1) Sn acts transitively on {1, . . . , n}. This just says that, for all k ∈ {1, . . . , n}, there exists a σ ∈ Sn such that σ(1) = k, hence Sn · 1 = {1, . . . , n} and there is just one orbit. Likewise, it is easy to see that An acts transitively on Sn for n ≥ 3, but not for n = 2. But if σ ∈ Sn, then the subgroup hσi acts transitively on {1, . . . , n} ⇐⇒ there is just one orbit of σ and it has n elements ⇐⇒ σ is an n-cycle. n n (2) GLn(R) acts on R . There are two orbits: {0} and R − {0}. SOn n acts on R . For n ≥ 2, the orbits are {0} and the (n − 1)-spheres of radius r > 0 centered at the origin. In particular, for n ≥ 2, SOn acts transitively on Sn−1. Because this action is transitive, the geometry of Sn−1 is homogeneous, i.e. it looks the same at every point.

(3) The dihedral group Dn acts transitively on {p0,..., pn−1}, the set of vertices of Pn. (4) The group G acts on itself by left multiplication. This action is transitive, since for example the orbit G · 1 is clearly G. (5) The group G acts on itself by conjugation. The orbit of x ∈ G is the of x, the subset C(x) of G consisting of all elements conjugate to x. Thus by deﬁnition C(x) = {gxg−1 : g ∈ G}.

7 For example, C(1) = {1}, so that the conjugation action is never transitive as long as G is not the trivial group.

Deﬁnition 2.6. If X is a G-set and x ∈ X, the isotropy subgroup Gx is the set {g ∈ G : g · x = x}.

Lemma 2.7. Gx is a subgroup of G.

Proof. Closure: if g1, g2 ∈ Gx, then

(g1g2) · x = g1 · (g2 · x) = g1 · x = x.

Identity: as 1·x = x, 1 ∈ Gx for every x. Inverses: if g ∈ Gx, then g·x = x by −1 −1 deﬁntion. As we have seen, g · x = x, hence g ∈ Gx. Thus Gx ≤ G. Deﬁnition 2.8. If X is a G-set, then the ﬁxed set XG is the set {x ∈ X : g · x = x for all g ∈ G}. It is the largest G-subset of X for which the G G-action is trivial. Clearly x ∈ X ⇐⇒ Gx = G ⇐⇒ G · x = {x} ⇐⇒ the orbit G · x contains exactly one element. Proposition 2.9. (i) If X is a G-set, x ∈ X, and y = g · x ∈ G · x, then −1 Gy = gGxg . In other words, the isotropy groups of x and y are conjugate by g. (ii) If X is a G-set and x ∈ X, then there is an isomorphism of G-sets from G · x to G/Gx, where G acts on the set of left cosets of Gx in the usual way (by left multiplication of cosets). Proof. (i) Given h ∈ G, h · y = y ⇐⇒ h · (g · x) = g · x ⇐⇒ (hg) · x = g · x −1 −1 −1 ⇐⇒ g · ((hg) · x) = x ⇐⇒ (g hg) · x = x ⇐⇒ g hg ∈ Gx ⇐⇒ −1 −1 h ∈ gGxg . Thus Gy = gGxg .

(ii) It is simplest to deﬁne a function F : G/Gx → G · x and prove ﬁrst that it is a bijection and then that it is an isomorphism of G-sets. Given a gGx, deﬁne F (gGx) = g ·x (note that this is indeed an element of G·x). We must show that F is well-deﬁned, i.e. independent of the representative g of the coset gGx. Any other element of gGx is of the form gh with h ∈ Gx, and hence (gh) · x = g · (h · x) = g · x, since by the deﬁnition of Gx h · x = x. Hence F is well-deﬁned and it is surjective by the deﬁnition of G · x. Next, we claim that F is injective. Suppose that F (g1Gx) = F (g2Gx). Then by deﬁnition g1 · x = g2 · x, so that

−1 −1 −1 x = g1 · (g1 · x) = g1 · (g2 · x) = (g1 g2) · x.

8 −1 Thus g1 g2 ∈ Gx, and so g1Gx = g2Gx. Thus F is injective, hence a bijection. Finally, we must check that F is an isomorphism of G-sets. This follows since, for all g ∈ G and cosets hGx ∈ G/Gx, by the deﬁnition of the action of G on G/Gx,

F (g · (hGx)) = F ((gh)Gx) = (gh) · x = g · (h · x) = g · F (hGx).

Thus F is an isomorphism of G-sets.

Corollary 2.10. Suppose that G is ﬁnite. Let X be a G-set and let x ∈ X. Then #(G) = #(Gx) · #(G · x). Equivalently, (G : Gx) = #(G)/#(Gx) = #(G · x). Hence the order of an orbit of G in X divides the order of G. In particular, if G acts transitively on X, then #(G) = #(Gx) · #(X), or equivalently #(X) = (G : Gx).

Example 2.11. (1) Sn acts transitively on {1, . . . , n} and the isotropy sub- group of n is ∼ Hn = {σ ∈ Sn : σ(n) = n} = Sn−1.

Thus Sn/Hn is Sn-isomorphic to {1, . . . , n}. Note that #(Sn) = n!, #(Hn) = #(Sn−1) = (n − 1)!, and #({1, . . . , n}) = n = n!/(n − 1)! = #(Sn)/#(Hn). n−1 (2) SOn acts transitively on S and the isotropy subgroup of en is easily n−1 seen to be SOn−1. Thus SOn/SOn−1 is SOn-isomorphic to S . In topol- ogy, this is an important relationship between the (n − 1)-sphere and the group SOn.

(3) Dn acts transitively on the set of vertices {p0,..., pn−1} of Pn and the isotropy subgroup of p0, say, is the reﬂection in p0. This gives another argument that #(Dn) = 2n. (4) Let S be the dodecahedron and G(S) the group of of S. By experimenting with a model of S, it is plausible that G(S) acts transitively on the 12 faces of S and that the isotropy group of a face has order 5, corresponding to the possible of a pentagon. Thus we expect that #(G(S)) = 60. We would get a similar conclusion by looking at the action of G(S) on the 20 vertices, where the order of the isotropy group is 3, or on the 30 edges, where the order of the isotropy group is 2.

9 (5) If H is a subgroup of G, then G acts transitively on G/H since the orbit G · H = G/H. The isotropy subgroup of H is by deﬁnition {g ∈ G : gH = H} = H. The isotropy subgroup of xH is {g ∈ G : gxH = xH} = xHx−1, since gxH = xH ⇐⇒ x−1gx ∈ H ⇐⇒ g ∈ xHx−1. (6) If G acts on itself by conjugation, then the ﬁxed set GG is the center Z(G), the orbit of x ∈ G is, as we have seen, the conjugacy class C(x) = {gxg−1 : g ∈ G}, and the isotropy group of x is the centralizer of x, namely the subgroup −1 ZG(x) = {g ∈ G : gxg = x}.

3 The class equation and applications

Suppose X is a ﬁnite G-set with orbits O1 = G·x1,...,Ok = G·xk (for some choices of x1, . . . , xk ∈ X). Then, since every element of X is in exactly one G-orbit, k X #(X) = #(Oi). i=1 We rewrite this by grouping together the one-element orbits into XG, as

G X #(X) = #(X ) + #(Oi),

#(Oi)>1 where the second sum is over all of the orbits which have more than one element. Note that, if G is ﬁnite and if #(Oi) > 1, then #(Oi) is a nontrivial divisor of #(G), i.e. #(Oi) divides #(G) and #(Oi) 6= 1. In particular, if G is a ﬁnite group and we let G act on itself by conju- G gation, then the ﬁxed set G is just the center of G and the orbits G · xi are the conjugacy classes −1 C(xi) = {gxig : g ∈ G}.

The isotropy subgroup of xi ∈ G for the G-action of conjugation is the centralizer ZG(xi), and #(C(xi)) = (G : ZG(xi)), the index of the centralizer of xi in the group G. Thus we get the class equation: X X #(G) = #(Z(G)) + #(C(xi)) = #(Z(G)) + (G : ZG(xi)), i i

10 where the sum is over the distinct conjugacy classes C(xi) which have more than one element (i.e. for which xi ∈/ Z(G)).

Example 3.1. In Sn, the conjugacy classes are described by the “shapes” of a product of disjoint cycles. In other words, given σ ∈ Sn, there exist integers k1, . . . , kr ≥ 2 with k1 + ··· + kr ≤ n such that σ is a product of disjoint cycles of lengths k1, . . . , kr, and every two such products with the same k1, . . . , kr are conjugate. Here 1 is the empty product. Thus, for example in S3, there are three conjugacy classes:

C(1) = {1},C((1, 2)) = {(1, 2), (1, 3), (2, 3)},C((1, 2, 3)) = {(1, 2, 3), (1, 3, 2)}.

Note that the sum of all the elements is #(S3) = 6, and that the order of each conjugacy class divides 6. The situation in S4 is already much more complicated. Here, C(1) = {1} has just one element. The conjugacy class of 4 a transposition is the set of all transpositions, and there are 2 = 6 of these. The conjugacy class of a product of two disjoint transpositions is the set of all products of two disjoint transpositions, and we have seen that there are 1 4 2 2 = 3 of these. The conjugacy class of a 3-cycle is the set of all 3-cycles, and there are (4 · 3 · 2)/3 = 8 of these. The conjugacy class of a 4-cycle is the set of all 4-cycles, and there are (4 · 3 · 2)/4 = 6 of these. Thus, the order of each conjugacy class divides 24 = #(S4), and the total number is

1 + 6 + 3 + 8 + 6 = 24.

Returning to the case of a general action of a ﬁnite group G on a ﬁnite set X, we have:

Proposition 3.2. If #(G) = pr, where p is a prime, and if X is a ﬁnite G-set, then #(X) ≡ #(XG) (mod p).

Proof. In the formula #(X) = #(XG) + P #(O ) above, all of the #(Oi)>1 i r terms #(Oi) for #(Oi) > 1 are nontrivial divisors of p , hence are of the s form p for 1 ≤ s ≤ r. Thus, if #(Oi) > 1, then #(Oi) ≡ 0 (mod p) and hence #(X) ≡ #(XG) (mod p).

Corollary 3.3. Let p be a prime number. If #(G) = pr with r ≥ 1, then Z(G) 6= {1}. In particular, if #(G) > p, then G is not simple (and hence, by induction, solvable).

11 Proof. In this case, we let X = G with the conjugation action. Then the ﬁxed set XG is just the center of G, so that

#(Z(G)) ≡ #(G) = pr ≡ 0 (mod p).

Thus Z(G) is a subgroup of G with order divisible by p, and hence Z(G) 6= {1}. We have seen that Z(G)CG. Thus, either Z(G) is a proper (nontrivial) subgroup of G, or Z(G) = G. In this last case, G is abelian and every subgroup of G is normal, so that G is simple ⇐⇒ there are no proper nontrivial subgroups of G ⇐⇒ G is isomorphic to Z/pZ (by a homework problem).

Corollary 3.4. Let p be a prime number. If #(G) = p2, then G is abelian.

Proof. By the preceding corollary, the center Z(G) is a subgroup of G with #(Z(G)) = p or #(Z(G)) = p2. If #(Z(G)) = p2, then G = Z(G) and hence G is abelian. So let us assume that #(Z(G)) = p. We will show that in this case G is also abelian, a contradiction since then Z(G) = G. If #(Z(G)) = p, then Z(G) C G and G/Z(G) is a group of order p. But every group of order p is cyclic, so in particular G/Z(G) is a . By a homework problem, G is then abelian.

Remark 3.5. In fact, it is easy to show that, if #(G) = p2, then either ∼ 2 ∼ G = Z/p Z is cyclic or G = (Z/pZ) × (Z/pZ).

12