GROUP ACTIONS EMMANUEL KOWALSKI

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1 GROUP ACTIONS EMMANUEL KOWALSKI Definition 1. Let G be a group and T a set. An action of G on T is a map a: G T T, that we denote a(g, t) = g t, such that (1) For all t T, we have e G t = t. (2) For all t T, all g 1, g 2 in G, we have g 1 (g 2 t) = (g 1 g 2 ) t. Remark 2. Let a be an action of G on T. For g G, let α g be the map from T to T such that α g (t) = g t. Then for any g 1 and g 2 in G, we get by the second condition α g1 α g2 = α g1 g 2. By the first, we get α g α g 1 = α g 1 α g = Id T, so α g is a bijection, with inverse α g 1. Consequently we have a map α: G Sym(T ), where Sym(T ) is the group of bijections of T (automorphisms in the category of sets). From condition (1) again, the map α is a group homomorphism. Conversely, let α: G Sym(T ) be a group homomorphism. Define a(g, t) = α(g)(t) T. Then a is a group action of G on T. To conclude: it is equivalent to give an action of G on T, or to give a group homomorphism G Sym(T ). Note that for a given set T, there may well exist more than one action of G on T. We want to make actions of a fixed group G on sets into a category. The objects are group actions, and if G acts on T 1 and T 2, a morphism T 1 T 2 of actions of G is a map of sets f : T 1 T 2 such that f(g t 1 ) = g 2 f(t 1 ), where the first action is in T 1, and the second in T 2. We also say that f is a G-morphism. If G acts on T, then the identity Id T is a G-morphism from T to T. Moreover if G acts on T 1, T 2 and T 3 and f 1 : T 1 T 2 and f 2 : T 2 T 3 are G-morphisms, then f 2 f 1 : T 1 T 3 is also one. One can then easily check that there is a category (G sets) with objects the actions of G on varying sets, morphisms the G-morphisms, and the usual composition of functions as composition of morphisms. Note that if f : T 1 T 2 is a bijective G-morphism, then from f(g t 1 ) = g f(t 1 ), we deduce that g t 1 = f 1 (g f(t 1 )), and if we apply this to t 2 = f(t 1 ), we get g f 1 (t 2 ) = f 1 (g t 2 ). In other words, f 1 is a G-morphism. Hence a G-morphism f is an isomorphism if and only if f is bijective, and its inverse in the category of G-sets is the reciprocal bijection. Date: November 17, 2017, 16:56. 1

2 Example 3. (1) For any set T, we can make G act on T by g t = t for all t. Then α is the morphism g Id T. It is called the trivial action of G on T. (2) If G acts on T, then any subgroup H of G also acts on T. Many interesting actions are obtained in this manner. (3) For H G any subgroup, we claim that G acts on G/H by g (xh) = gxh. To check this, we must first check that this formula gives a well-defined map G G/H G/H. This means that we must check that gxh doesn t depend on the choice of the element x in the left coset xh. But if we replace x by xh, for some h H, then we replace gxh by gxhh = gxh since H is a subgroup. One we know that this map is well-defined, it is elementary that it is a group action of G on G/H. Indeed, we have by definition and e G xh = (e G x)h = xh, g 1 (g 2 xh) = g 1 (g 2 xh) = g 1 (g 2 x)h = (g 1 g 2 )xh = g 1 g 2 xh for any x G and g 1, g 2 in G, which establishes the conditions for a group action. In particular, taking H = {e G }, the group G acts on itself by left multiplication x y = xy. (4) For any integer n 1, the symmetric group S n acts on {1,..., n} by σ i = σ(i). (5) For any field K and any integer n 1, the group GL n (K) acts on K n by g x = gx (matrix times vector product). (6) The group G G acts on G by multiplication on both sides: (x 1, x 2 ) y = x 1 yx 1 2. The group G acts on G by conjugation, namely g x = gxg 1. In that case, the corresponding homomorphism G Sym(G) is the group morphism sending an element of G to the associated inner automorphism. (7) Let G act on T. Then we can define other associated actions. For instance, let n 1 be an integer and T 1 = T n. Then the group G acts on T n by g (t 1,..., t n ) = (g t 1,..., g t n ). Let T 2 be the set of subsets of T. Then G acts on T 2 by g X = {g x x X} = α g (X) T. (8) For instance, G acts by conjugation on subsets of G. In fact, it also acts by conjugation on the set of subgroups of G (this set is a subset of T 2 for the action of G by itself on conjugation), because a conjugate xhx 1 of a subgroup of G is also a subgroup of G. (On the other hand, if we consider the action of G on itself by left multiplication, then we do not obtain an action on the set of subgroups, because gh is not necessarily a subgroup of G if H is one). Now we introduce various items of terminology: If G acts on T and S T is such that g s S for all s S, then G also acts on S by the same formula as it does on T, but with the second variable restricted to S. One says that S is a G-invariant subset of T (for the given action). In particular, if S = {t} has a single element and is invariant, then we say that t is a fixed point of the action. This means that g t = t for all g G. 2

3 Let G acts on T. If the homomorphism α: G Sym(T ) is injective, then we say that the action is faithful. This means that e G G is the only element of G such that g t = t for all t T. Assume that G acts on T. Let t T. The map γ t : G T such that γ t (g) = g t is called the orbit map associated to t. Its image γ t (G) = {g t g G} is called the G-orbit of t in T. (Note that γ t : G T is a G-morphism when G is viewed as a set on which G acts by multiplication on the left. Indeed, for g G and x G, we compute γ t (g G x) = γ t (gx) = gx t = g (x t) = g f(x), where we denote g G x the action of G on itself to clarify the computation. This is the definition of a G-morphism in that case). If H is a subgroup of G, the H-orbit of t in T is the orbit for the restricted action of H, namely the subset γ t (H) of T. Lemma 4. Let G act on T. The relation on T defined by t 1 t 2 if and only if there exists g G such that t 2 = g t 1 is an equivalence relation. Its equivalence classes are the orbits of G in T. Proof. (1) We check the condition for an equivalence relation: We have t t since t = e G t. If t 1 t 2, with t 2 = g t 1, then t 1 = g 1 t 2, so t 2 t 1. If t 1 t 2 and t 2 t 3 with t 2 = g t 1 and t 3 = h t 2, then t 3 = hg t 1 so t 1 t 3. (2) Let t 1 T be given. The equivalence class of t 1 is {t 2 T t 1 t 2 } = {t 2 T t 2 = g t 1 for some g G} which is the G-orbit of t 1 in T. Conversely, if S T is the orbit of some element t 1 T, then S is the equivalence class of t 1 (since any g t 1 S is equivalent to t 1 ). Definition 5. Let G act on T. (1) The set of G-orbits of T is denoted G\T (or sometimes T/G). (2) If there is a unique G-orbit in T, then we say that the action of G on T is transitive. Example 6. (1) The action of S n on {1,..., n} is transitive (for instance, i = σ i 1 where σ i (1) = i, σ i (i) = 1 and σ i (j) = j for all j / {1, i}). (2) Let K be a field. The action of GL n (K) on K n has two orbits, namely {0} and K n {0} (if a vector x is non-zero, then there is a basis where it is the first basis vector, and the change of basis matrix from the canonical basis maps the first canonical basis vector e 1 to x, which shows that the orbit of e 1 is equal to K n {0}). (3) Let G act on T. Any orbit S T is G-invariant (since g (h t) = (hg) t S), and the action of G on S is transitive. (4) Let H be a subgroup of G. Then G acts transitively on G/H: for any xh G/H, we have xh = x H, so all cosets are in the orbit of the coset H. In fact, the last example gives all possible examples of transitive G-actions, up to isomorphism (in the category of sets with G-actions). 3

4 Theorem 7 (Orbit-stabilizer Theorem). Let G be a group and T a non-empty set with a G-action. Let t 0 T be a fixed element of T and T 0 T the G-orbit of t 0. Let H = {g G g t 0 = t 0 }, called the stabilizer of t 0 in G. (1) The set H is a subgroup of G. (2) There is a well-defined G-morphism f from G/H to T such that f(xh) = x t 0 for all x G. (3) The map f induces a G-isomorphism of G/H with T 0. In particular, if G is a finite group, then T 0 is finite, its cardinality divides Card(G), and we have Card(T 0 ) = Card(G) Card(H). Proof. (1) follows from the conditions defining an action. (2) Suppose that h H. Then (xh) t = x (h t) = x t by definition of H, so f(xh) = x t is well-defined. (3) First of all, by definition of the action on G/H, we have f(y xh) = f(yxh) = yx t = y (x t) = y f(xh), which means that f is a G-morphism. By construction, the set of values of f is the orbit of x, hence is equal to T since the action is assumed to be transitive. So f is surjective. Finally, let x 1 H and x 2 H be two left cosets of H in G. We have f(x 1 H) = f(x 2 H) if and only if x 1 t = x 2 t, which means that x 1 2 x 1 H, so that x 1 x 2 H, or equivalently x 1 H = x 2 H. So f is also injective, and consequently it is a G-isomorphism. In general, we denote by Stab G (t 0 ) the stabilizer of an element t 0 in a G-set T. Example 8. (1) Let G act on itself by conjugation. Then the stabilizer of x 0 G is the centralizer C G (x 0 ), namely Stab G (x 0 ) = {g G gx 0 g 1 = x 0 } = {g G gx 0 = x 0 g}. The orbit of x 0 is its conjugacy class Cl(x 0 ). Hence we obtain the relation Card(Cl(x 0 )) = Card(G) Card(C G (x 0 )) if the group is finite. (2) Let G act on the set of subgroups of G by conjugation. The stabilizer of a subgroup H 0 G is called the normalizer of H 0 in G, and is denoted N G (H 0 ). The subgroup H 0 is normal in N G (H 0 ), and in fact N G (H 0 ) is the largest subgroup of G with this property. In particular, H 0 is normal in G if and only if N G (H 0 ) = G. (3) As an application, consider a finite group G with cardinality equal to p n for some prime number p and some integer n 1. Consider the action of G on itself by conjugation. Since the orbits form a partition of G, we have Card(G) = T 4 Card(T ),

5 where T runs over the orbits of G, namely over the conjugacy classes in G. If T = Cl(x 0 ) is the conjugacy class of some x 0 G, we have Card(T ) = Card(G) Card(C G (x 0 )). For each x 0, the cardinality of C G (x 0 ) is a divisor of p n, hence has the form p m(x 0) for some integer m(x 0 ) such that 0 m(x 0 ) n. Two things may happen: either C G (x 0 ) = G, or C G (x 0 ) is a proper subgroup of G. In that second case, m(x 0 ) < n and Card(T ) is divisible by p. In the sum over T, we conclude that all terms are divisible by p, except those where T contains a single element. These correspond to elements in the center Z(G) of G (since C G (x 0 ) = G). So the formula shows that there exists k Z such that Card(G) = Card(Z(G)) + kp. Since Card(G) is also divisible by p, we deduce that Card(Z(G)) is divisible by p. In particular Card(Z(G)) p. We conclude that G has a non-trivial center. (4) We use the orbit-stabilizer theorem to compute the cardinality of GL n (F p ) by induction on n 1. For n = 1, we have Card(F p ) = p 1. Now suppose n 2. Then let GL n (F p ) act on the set T = F n p {0} with p n 1 elements. We know that this action is transitive. Consider the first canonical basis vector e 1. Its stabilizer is the subgroup H of all g GL n (F p ) such that g(e 1 ) = e 1, which means the subgroup of matrices ( ) 1 a 0 h where 0 is a column vector of size n 1, where a is a row vector of size n 1, and h is an element of GL n 1 (F p ). So we have Card(GL n (F p )) = Card(H) Card(T ) = (p n 1)p n 1 Card(GL n 1 (F p )). For instance, with n = 2, we obtain Card(GL n (F p )) = (p 2 1)p(p 1) = p(p 1) 2 (p + 1). In general, one can easily deduce by induction on n that Card(GL n (F p )) = p n(n 1)/2 n j=1 (p j 1). (5) Let K be a field and n 1. A flag in K n is a family (V 0,..., V n ) of subspaces of K n such that dim(v i ) = i and V i V i+1 for 0 i n 1. (In particular, V 0 = {0} and V n = K n ). Let T be the set of flags in K n. Then the group GL n (K) acts on T by g (V 0,..., V n ) = (gv 0,..., gv n ), which is again a flag. For instance, let (e 1,..., e n ) be the canonical basis of K n. Then F 0 = ({0}, Ke 1, Ke 1 Ke 2,..., Ke 1 Ke n ) is a flag. The action of G on T is transitive. Indeed, given a flag F = (V 0,..., V n ), there is for 1 i n a vector f i V i that is not in V i 1 ; then the family (f 1,..., f n ) is a basis of K n. 5

6 (Indeed, one checks by induction on i that (f 1,..., f i ) generates V i ). Consider the matrix g such that g(e i ) = f i. Then g GL n (K), and g F 0 = F, so any flag is in the orbit of F 0. Now we compute the stabilizer of F 0 in G = GL n (K). This is the set of g GL n (K) such that g(e i ) Ke 1 Ke i for 1 i n. This means that g is upper-triangular, so a 1,1 { 0 a 2,2 } Stab G (F 0 ) = 0 0 a 3, a n,n Consider now again K = F p for p prime. compute the cardinality of the set of flags: Card(T ) = Card(GL n(f p )) Card(Stab G (F 0 )) = pn(n 1)/2 n j=1 (pj 1) (p 1) n p n(n 1)/2 = We can use the orbit-stabilizer theorem to n j=1 p j 1 p 1. Corollary 9. Let G be a finite group acting on a finite set T. Let F be the set of fixed points of the action. We have Card(T ) = Card(F ) + x [G : Stab G (x)] where x runs over representatives of the non-trivial orbits of G in T. Proof. The orbits of the action of G on T form a partition of T, so the cardinality of T is the sum of the sizes of the orbits. We separate the orbits with a single element, which are exactly the fixed points, from the others, and use for the non-trivial orbits the fact that the size of an orbit T 0 is the index of the stabilizer of some element x in T 0. Here are further illustrations of the use of group actions. Theorem 10 (Sylow). Let G be a finite group of cardinality n and p a prime number. Write n = p k m where k 0 and p does not divide m. (1) There exists a subgroup of cardinality p k in G. (2) If H is a subgroup of G of cardinality a power of p, then there exists a subgroup of G of cardinality p k that contains H. (3) All subgroups of G of cardinality p k are conjugate. Lemma 11. For n = p k m with m coprime to p, the integer ( ) n is not divisible by p. p k Proof. Let S = Z/p k Z {1,..., m}. This is a finite set with n elements. Let T be the set of subsets of S of cardinality p k. This is a finite set with ( ) n p elements. The group Z/p k Z k acts on S by x (y, j) = (x + y, j), and then it acts on T by x X = {x z z X}. 6

7 We claim that the set F T of fixed points of this action is the set with elements Z/p k Z {j} for 1 j m. Indeed, we see first that each such set is indeed fixed by the action. Conversely, let X T be a fixed point of the action. Let (y, j) X be any element. Then for all x Z/p k Z, we have x X = X, hence in particular x (y, j) = (x + y, j) X. Therefore X contains the set Z/p k Z {j}, and we must have X = Z/p k Z {j} because each has the same number of elements. So the cardinality of F is m, which is coprime to p. On the other hand, writing the formula Card(T ) = Card(F ) + R Card(R) where R runs over the orbits of G in T that are not fixed points, we deduce that so Card(T ) is coprime to p. Card(T ) Card(F ) mod p, Proof of the theorem. (1) Let T be set of subsets of G of cardinality p k. This is a finite set with ( ) n p elements. Consider the action of G on T by g X = gx for X T. Since the k cardinality of T is not divisible by p, according to the lemma, there exists some element X T whose orbit T X has cardinality coprime to p. Let P be the stabilizer of X in G. Then Card(G)/ Card(P ) = [G : P ] = Card(T X ) is coprime to p, which means that p k must divide the cardinality of P. But, on the other hand, the elements g of P satisfy g X = X, so that fixing some x X, we have P Xx 1, which has cardinality p k, so Card(P ) p k. Hence we conclude that Card(P ) = p k, and we have found the desired subgroup. (2) Let P be a subgroup of G of cardinality p k, which exists by (1). Consider the action of H on G/P. Since H has order a power of p, all non-trivial orbits have cardinality divisible by p, so the number of fixed points is congruent to Card(G/P ) modulo p. But Card(G/P ) = m is coprime to p, so there is at least one fixed point xp. The condition hxp = xp for all h H translates into x 1 hx P for all h H, so H xp x 1. Since xp x 1 has cardinality p k, we have the desired result. (3) Let again P be a fixed subgroup of order p k. By (2), a subgroup H of cardinality p k is contained in a conjugate xp x 1 of P. Since H and xp x 1 have the same finite number of elements, they are equal. If G is a finite group and p a prime number, then a subgroup as in (1) is called a p-sylow subgroup of G. Of course, these are only interesting when p divides the cardinality of G. Example 12. Let p be a prime number and let n 1 be an integer. Consider the group G = GL n (F p ). Let U G be the subgroup of upper-triangular matrices with diagonal coefficients equal to 1. Since the coefficients a i,j of g U can be arbitrary, we have Card(U) = p n(n 1)/2. This subgroup is a p-sylow subgroup of G, since (by Example??, (4)) the size of G is p n(n 1)/2 n j=1 (p j 1), where the product over j is coprime to p. (Note that this p is the same prime as the cardinality of the finite field! The other Sylow subgroups of G can be mode difficult to describe!) 7

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