GROUP ACTIONS ON SETS WITH APPLICATIONS TO FINITE GROUPS


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1 GROUP ACTIONS ON SETS WITH APPLICATIONS TO FINITE GROUPS NOTES OF LECTURES GIVEN AT THE UNIVERSITY OF MYSORE ON 29 JULY, 01 AUG, 02 AUG, 2012 K. N. RAGHAVAN Abstract. The notion of the action of a group on a set is a fundamental one, perhaps even more so than that of a group itself: groups derive their interest from their actions. We first define this notion and give some examples. The structure of an action can be understood by means of orbits and stabilisers. We introduce these and make some observations about them. While in themselves so elementary as to be selfevident, these observations nevertheless lead to interesting consequences when combined effectively in the right context. After discussing Lagrange s theorem in the language of group actions, we come to the class equation, which is the main technical result in these notes. We apply the class equation to derive some standard basic results in the theory of finite groups: that the centre of a pgroup is nontrivial, theorems of Cauchy and Sylow on the existence of subgroups of prime power orders, and those of Sylow about the number and conjugacy of subgroups named after him. Up to date version of these notes is available from the author s home page at Corrections and other suggestions for improvement are solicited and may please be ed to 1
2 Group actions on sets Notation. Throughout these notes, G denotes a group and X, Y denote sets. We use symbols g, g 1, g,... to denote elements of G; similarly x, x 1, x,..., to denote elements of X, and y, y 1, y,... to denote elements of Y. The basic definitions. An action of G on X is a map G X X denoted (g, x) gx such that 1x = x and g(hx) = (gh)x for all x in X and g, h in G. Given an action of G on X, we call X a Gset. A Gmap between Gsets X and Y is a map f : X Y of sets that respects the Gaction, meaning that, f(gx) = gf(x) for all x in X and g in G. To give an action of G on X is equivalent to giving a group homomorphism from G to the group of bijections of X. Examples. Here are some examples: Trivial action: the prescription gx = x for all g and x defines a Gaction on X. The defining action: Most groups come with a natural action that helps define them. For example, the symmetric group S X is defined as the set of bijections from X to itself, the multiplication being composition. By the very definition we have a map S X X X, namely (f, x) f(x), that satisfies the axioms for an action. The group GL 2 (R) of invertible 2 2 matrices with real entries acts on the vector space of column matrices of size 2 1 with real entries: the action map is just the usual matrix multiplication. More generally, the group GL n (R) of invertible n n matrices with real entries acts by usual matrix multiplication on the space of column matrices of size n 1 with real entries. The previous example is in fact properly thought of as a defining action. If V is a vector space of finite dimension n over the real numbers, then GL(V ) is defined as the group of vector space isomorphisms from V to itself (multiplication is composition). So there is a defining action of GL(V ) on V. By the choice of a basis for V, we may identify V with real column matrices of size n 1 and GL(V ) with invertible real matrices of size n n. The action of GL(V ) on V is then identified as the action by matrix multiplication of GL n (R) on real n 1 column matrices. Left regular action: G acts on itself: putting X = G, it is readily checked that the map G G G defining the multiplication operation of G satisfies the action axioms. Right regular action: Again put X = G. Define ρ : G G G by ρ(g, x) := xg 1. We have ρ(1, x) = x1 1 = x and ρ(g, ρ(h, x)) = (ρ(h, x))g 1 = (xh 1 )g 1 = x(h 1 g 1 ) = x(hg) 1 = ρ(hg, x), so ρ defines an action. Conjugation action: Yet again put X = G. This time define G G G by (g, x) g x := gxg 1. It is easily checked that 1 x = x and g ( h x) = (gh) x. Action on left cosets: Let H be a subgroup of G. The set G/H of left cosets is naturally a Gset: G G/H G/H is given by (g, xh) gxh. New Gsets from old. From a given Gset X, we may cook up other related Gsets. In fact, any set obtained from X by a natural set operation is also naturally a Gset. 1 Here are some examples: the Cartesian square X X with action being defined by g(x, x ) := (gx, gx ); more generally, X n := X X(n times) with Gaction g(x 1,..., x n ) := (gx 1,..., gx n ) becomes a Gset. the power set 2 X of X: for Y a subset of X, we let gy := {gy y Y }. The set of all functions (say, complex valued) on X: if f is such a function and g an element of G, define gf by (gf)(x) := f(g 1 x) for x in X. 1 Although the words natural and naturally in this sentence have precise mathematical meanings, the first time reader may want not to dwell too much upon them, preferring instead to take them in the sense of everyday language. 2
3 if Y is a Gset too, the set X Y of maps from Y to X is naturally also a Gset: given f : Y X and g G, we define gf by (gf)(y) := g(f(g 1 y)). Restricting the action. The following remarks may seem innocuous but can be used to good effect: if H is a subgroup of G, any Gset may be considered to be also a Hset by just restricting the action to H. a subset Y of a Gset X is Ginvariant if gy Y whenever y Y. A Ginvariant subset is naturally itself a Gset. Here are some examples of the use of the second item above: Consider the collection of all subsets of a given cardinality of a Gset X. Being a G invariant subset of the power set of X, this itself is a Gset. Consider the action of GL(V ) on a vector space V discussed above. The subspace of linear functionals is a GL(V )invariant in the space of all complex valued functions on V, and so is a GL(V )set. Let V be a vector space of finite dimension n (over some field). For an integer r, 0 r n, consider the collection denoted G(r, V ) of all linear subspaces of dimension r of V. For example, G(0, V ) is a singleton whose only element is {0}; and G(1, V ) is the collection of all lines through the origin in V. Now, G(r, V ) is a GL(V )invariant subset of the power set of V, and so in its own right a GL(V )set. Ginvariant subset Notation. As already mentioned, G denotes a group throughout. In what follows, X denotes a Gset. Neither G nor X is assumed to be finite for now: finiteness assumptions will be explicitly mentioned wherever imposed. Orbits and their structure, stabilizers. For x an element of X, the orbit of x or the orbit through x is the subset Gx := {gx g G} of X. We have: x is in the orbit through x; if y is in the orbit through x, then x is in the orbit through y; if y is in the orbit through x and z in the orbit through y, then z is in the orbit through x. Thus, the relation on X defined by y x if y is in the orbit through x is an equivalence relation. The equivalence classes are called orbits. Being equivalence classes, the orbits partition X into a union of pairwise disjoint subsets: (1) X = orbits Orbits are Ginvariant subsets of X (in the sense defined above of Ginvariance), and are therefore themselves Gsets: in fact, a subset of X is Ginvariant precisely when it is a union of orbits. Equation (1) therefore has the following implication, at least philosophically if not also practically: to understand the structure of any Gset, it is enough if we understand the structures of all possible Gorbits, for after all each Gset is pieced together from its orbits. For x in X, define its stabilizer G x by G x := {g G gx = x} It is clearly a subgroup of G. Consider the space G/G x of left cosets of G x with its natural Gaction (g, g G x ) gg G x. The Gset structure of the orbit Gx of x can be understood in terms of that of G/G x. In fact, as is readily verified, we have a Gset isomorphism: (2) G/G x Gx as Gsets via gg x gx The upshot is: 3 The orbits form a partition of the Gset X. stands for disjoint union. stabilizer G x Structure of an orbit
4 To understand the Gset structure on an arbitrary Gorbit, it is enough to understand the Gset structure of the left coset spaces G/H for all subgroups H of G. Transitive actions and homogeneous spaces. The whole of X is a single Gorbit if and only if, given any two elements x and x of X, there exists an element g of G such that gx = x. In this case, the Gaction is said to be transitive, and X is called a homogeneous space for G. In view of (2), every homogeneous space for G is of the form G/H for some subgroup H. Lagrange s theorem Lagrange s theorem. Let H be a subgroup of G. Consider G as an Hset with the left regular action: (h, g) hg. The following facts are readily verified: the orbits for this action are precisely the right cosets of H; each right coset has cardinality the same as that of H: if Hx is a right coset, h hx defines a bijection between H and Hx. By equation (1), G is a disjoint union of right cosets of H, so 2 (3) G = H H\G where H\G denotes the set of right cosets of H Considering instead of the left regular action the right one of H on G given by (h, g) gh 1, we conclude similarly that (4) G = H G/H By (2), any Gorbit is of the form G/H for some subgroup H, so in view of (4): (5) the order of any Gorbit divides the order of the group G From (3) and (4) we also conclude: (6) the order of any subgroup H divides the order of the group; the number of left cosets of H equals the number of its right cosets. The class equation. Let us rewrite in a slightly modified form the right side of equation (1). Divide the orbits into two classes: singleton orbits and nonsingleton orbits. Obviously, an element x forms an orbit by itself if and only if it is fixed by every element of G: gx = x for all fixed point g in G. Let us denote by X G the fixed point set of X, that is the collection of all fixed points: set X G X G := {x X gx = x g G} With this notation, equation (1) can be written as: (7) X = X G nonsingleton orbits By taking cardinalities of both sides of the equation, 3 we get Centre of G := {g G gh = hg h G} The class equation (8) X = X G + orbit (where the sum is over nonsingleton orbits) Let us now apply the above considerations to the situation when X = G acted upon by conjugation. The orbits are the conjugacy classes, and the fixed point set X G is the centre of G. The last equation in this situation becomes (9) G = centre(g) + conjugacy class (sum is over nonsingleton classes) 2 Here and in the sequel, we will be considering relations between cardinalities of various sets. While cardinalities make sense even without finiteness assumptions, the reader may just want to assume the sets whose cardinalities are in consideration to be finite. 3 As already mentioned, this operation makes sense even when X is not finite, but the reader may just want to think of the special case when X is finite. 4
5 Application to the structure of finite groups. Assume the group G to be finite. Suppose that for a prime p the following hypothesis is satisfied: 4 (10) every proper subgroup of G has index divisible by p Then, by (2), every nonsingleton Gorbit has cardinality divisible by p. This applies in particular to every nonsingleton conjugacy class in G. Reading equations (8) and (9) modulo p, we get: (11) X X G mod p centre(g) 0 mod p Since centre(g) is positive the identity element of the group always belongs to the centre we conclude that the centre is nontrivial. When G is a pgroup (that is, G is a power of a prime p), the hypothesis (10) is satisfied (by (6)). From the previous paragraph we conclude: (12) the centre of a pgroup is nontrivial (that is, has cardinality more than 1) Groups of order p 2 are in fact abelian see item (2) in the appended tutorial sheet. There do exist groups of order p 3 that are not abelian (irrespective of p) see item (8) in the tutorial sheet. On the existence or lack thereof of subgroups of given orders: theorems of Cauchy and Sylow. We continue to assume that G is finite. By Lagrange (6), the order of any subgroup H is a factor of the order of G. It is natural to ask whether, given a factor of G, there exists a subgroup of that order. This is false in general, as easy examples show: see item (9) in the tutorial sheet. However, we have the following: (13) (14) If G is abelian and d divides G, a subgroup of G of order d. If q = p e (e 0) is a prime power dividing G, a subgroup of G of order q. In the proofs of these statements, we use the following observation: If N is a normal subgroup of G, then the preimages in G of elements of G/N under the natural quotient map G G/N are the (left/right) cosets of N; so they form a partition of G into subsets each of cardinality N. In particular, the preimage in G of a subgroup K of G/N is a subgroup of cardinality N K. Let us first prove (13) in the case when d is a prime p. Proceed by induction on G : there being no prime dividing 1, we assume that G > 1. It is enough to find an element z in G whose order f is divisible by p, for then z f/p has order p. Choose 1 x in G. If p divides the order of x, we are done. If not, consider G/X where X is the subgroup generated by x. By induction, there exists y in G/X of order p. Any preimage z in G of y has order a multiple of p, and (13) is proved in the case when d is a prime. For the proof of (13) in the general case, proceed by induction on d. The case d = 1 being obvious, assume d 2, let p be a prime dividing d, and choose subgroup N of G of order p. By induction, there exists subgroup of order d/p of G/N. Its preimage in G is a subgroup of order d, and we are done. We now prove (14). Proceed by induction on G. The case G = 1 being obvious, assume G > 1. We are done by induction if there is a proper subgroup with index coprime to p. So we may assume that hypothesis (10) holds for G. By (11), we conclude that p divides the cardinality of the centre of G. Applying (13), choose subgroup N of order p of the centre of G. By induction, there exists a subgroup of G/N of order q/p. Its preimage in G is a subgroup of order q, and we are done. The case e = 1 of (14) is Cauchy s theorem; the case when q is the highest power of p dividing G is Sylow s theorem. 4 At the end of the day see (14) below it turns out that the only groups that satisfy this hypothesis are those of order a power of p. The hypothesis is thus merely a provisional one that enables the proofs. 5
6 Sylow psubgroups: their conjugacy and number. Let G be finite and p a prime. Write G = p e m with m coprime to p. Thus p e is the highest power of p dividing G, the case e = 0 not being excluded from consideration. A subgroup of order p e of G is called a Sylow psubgroup in honour of Sylow who proved not only their existence (14) but also: (15) (16) Any two Sylow psubgroups are conjugate. The number of Sylow psubgroups divides m and is congruent to 1 mod p. We will prove a stronger result than (15), to state which, let P be an arbitrary Sylow psubgroup and H an arbitrary psubgroup. 5 Then: (17) H is contained in some conjugate of P. In proving (17) and (16), we make use of the following observation: (18) If H normalizes P, then H P. We first prove (18). Consider the subgroup P H of G. Being a quotient of P H, it is a pgroup. On the other hand it contains P. Since p e is the highest power of p that divides G, we conclude that P H = P, in other words that H P, and (18) is proved. To prove (17), consider the conjugation action of G on its set of subgroups. Since the cardinality of a subgroup doesn t change under the Gaction, the Sylow psubgroups form a Ginvariant subset. The conjugates of P form a Gorbit, which let us denote X. The stabiliser of P being its normalizer N(P ), we have X G/N(P ) as Gsets by (2). Since P N(P ), it follows that G/N(P ) divides m: (19) X divides m In particular, X is nonzero modulo p. Now consider X as an Hset and apply (11) to conclude that X H is nonzero, which means that there exists a conjugate of P, say P, that is normalized by H. By (18), we have H P, and (17) is proved (and so in particular is (15)). To prove (16), let us look back at the above proof of (17) in the light of (15). We see that X must be the set of all Sylow psubgroups. So the first half of (16) follows from (19). To proof the second half, consider X as a P set and apply (11). By (18), the only Sylowp subgroup that P normalizes is itself, so X P = {P }, and we are done. 5 A psubgroup is a subgroup which is a pgroup, in other words, a subgroup whose order is a power of p. 6
7 Tutorial sheet: Group actions on sets (1) Let G be a group, X a Gset, and x, y be two elements of X. If x and y belong to the same orbit of G, then their stabilzers G x and G y are conjugate: in fact, writing y = gx for g in G, we have G y = G gx = g G x. On the other hand, if the stabilzers G x and G y are conjugate, say g G x = G y, then the orbits Gx and Gy are isomorphic as Gsets: in fact, hgx hy gives an isomorphism. (2) Let H be a subgroup of a group G. We may consider the group H H acting on G as follows: (h, k) g := hgk 1. Let H\G/H denote the set of orbits for this action. (These orbits are called the double cosets: each is of the form HgH for some g G.) Let G/H denote, as usual, the set of left cosets of H. There is a natural action of G on G/H: g.xh := gxh. Consider the induced action of G on the Cartesian product G/H G/H. The Gorbits of G/H G/H are in bijection with H\G/H: (xh, yh) Hxy 1 H. (3) An action of G on X is said to be faithful if gx = x for all x implies g = 1; equivalently if the map from G to the group of bijections of X is an injection. It is said to be simply transitive if it is both transitive and faithful. The action of G on itself by left multiplication is faithful. If G acts simply transitively on X, then X G as Gsets noncanonically (where G is a Gset by the left regular action). (4) Let G be a finite group and X a finite Gset. Prove the following result of Burnside: # of Gorbits in X = 1 X g G where X g := {x X gx = x}. 6 Solution: Imagine a matrix whose columns are indexed by elements of G and whose rows by elements of X, the entry in column g and row x being 1 or 0 precisely as whether or not g fixes x (we say g fixes x if gx = x). The desired equality comes from two different ways of computing the sum of all the entries of the matrix: first along rows and then summing the rowsums; or first along columns and then summing the columnsums. The sum of the elements of in row x is G x where G x := {g G gx = x} is the stabilizer of x. As x varies over a Gorbit, G x is fixed: the stablizers are conjugate along an orbit, by item (1) above. Moreover, G x = G /# orbit (see equation (2)). So the sum of the rowsums as x runs over a Gorbit is G, independent of the orbit. Since the orbits partition X, the sum of all rowsums is the sum over all orbits of the sum over each orbit of the rowsums. Thus, the sum of all entries of the matrix is G # of Gorbits in X. On the other hand, the sum of the elements in column g is clearly X g. So the sum of the columnsums is g G Xg. g G Notation: For g G and H a subgroup of G, g H := ghg 1, the conjugate of H by g. The first assertion in item (3) is called Cayley s theorem. (5) (from Artin s Algebra p. 196, problem 8) Considering colourings of the vertices of the regular octogon by two colours, say red and blue. Consider two colurings to be equivalent if one is obtained by the other by either a rotation by an integer multiple of the angle π/4 or by reflection in an axis of symmetry (a line passing through two opposite vertices, or the midpoints of two opposite edges). Use Burnside s result above to determine the number of nonequivalent colourings. Solution: Let G be the dihedral group of symmetries of the regular octogon. The natural action of G on the vertices of the octogon also induces an action on colourings: we can identify colourings with the set X of 4element subsets of the set of vertices for example, by letting those elements of such a subset be coloured red and those in the complement blue. The number of nonequivalent colourings is thus the number of orbits for the action of G on X. 6 Alternative solution (using basic facts about linear representations of finite groups): consider the linear representation of G on the complex vector space CX with X as basis. The projection to the trivial isotypical component is given by (1/ G ) g G g, whose trace clearly is (1/ G ) g G Xg. The dimension of the trivial isotypical component on the other hand is the number of Gorbits in X: the elements x orbit x form a basis for this component as we vary over orbits. The trace of a projection equals the dimension of the image, so Burnside s equality follows. 7
8 To reduce the problem to an application of Burnside s result, we need to compute the cardinality of X g for all g in G: If g is the identity, then X g = X, so X g = X = ( 8 4) = 70. If g is rotation by an angle π/4, either clockwise or anticlockwise, X g is empty since g acts transitively on the vertices. If g is rotation by an angle π/2, either clockwise or anticlockwise, then X g consists of 2 elements: the four vertices that are, so to speak, in the cardinal directions form an orbit under g, so the set of these four vertices as also that of the other four are both fixed by g. If g is rotation by an angle 3π/4, either clockwise or anticlockwise, X g is again empty since g acts transitively on the vertices. If g is the rotation by angle π, then each vertex is sent to its antipode by g. There are 4 pairs of antipodal vertices, and a 4element vset of vertices is fixed by g if and only if it consists of two of these pairs. Thus there are ( 4 2) = 6 elements of X fixed by g. Now let g be the reflection in a line joining the midpoints of two opposite edges. Each vertex being sent to its reflection in the line, and there being 4 pairs of vertices that are reflections of each other in the line, an element of X is fixed under g precisely when it contains two of these 4 pairs. Thus there are ( 4 2) = 6 elements of X fixed by g. Let g be the reflection in a line through two antipodal vertices. Each vertex is sent to its reflection in the line; there are 3 pairs of vertices that are reflections of each other in the line; and the two vertices lying on the line are fixed. For an element of X to be fixed, it must therefore consist of two of those pairs reflective pairs or one of those pairs and the two vertices on the line. Thus there are ( 4 2) = 6 fixed points in X fixed by g. Thus we have g G X g = ( ) + ( ) = 128 By Burnside, the number of orbits is thus 128/16 = 8. (6) Let p be a prime and G a pgroup. Determine all nonnegative integers n such that there is a set of cardinality n on which G acts transitively. Solution: Let G = p e. Let X be a transitive Gset. Then X G/H for some subgroup H of G, by (2). So X has cardinality p f where f is an integer such that 0 f e. Conversely, given such an integer f, there exists a subgroup H of G or cardinality p e f, by (14). The action of G on G/H is transitive and G/H = p f. (7) Let N be a central subgroup of a group G (this means that N is a subgroup contained in the centre). Then N is clearly normal. Suppose that G/N is cyclic. Show that G is abelian. Combine (12) with this argument to conclude that groups of order p 2 are abelian (where p is a prime). (8) Let p be a prime, q = p e a power of p with e 1, and F q the finite field with q elements. Let GL n (F q ) denote the group of invertible n n matrices with entries in F q. GL n (F q ) has order (q n 1)(q n q)(q n q 2 ) (q n q n 2 )(q n q n 1 ). In particular, the cardinality of a Sylowp subgroup of GL n (F q ) is q n(n 1)/2. Consider the subgroup U n (F q ) consisting of those elements of GL n (F q ) that are upper triangular and all of whose diagonal entries are equal to 1. The cardinality of U n (F q ) is clearly q n(n 1)/2, so that it is a Sylowp subgroup of GL n (F q ). U 3 (F p ) has order p 3 and is not abelian. (9) The alternating group A 4 of even permutations of 4 letters has no subgroup of order 6. (10) Let G be a finite group, p a prime, and P a Sylow psubgroup of G. It follows from (15) that P is normal if and only if it is the only Sylow psubgroup. In particular, P is the unique Sylow psubgroup of its normalizer N(P ) in G. Show that the normalizer of N(P ) in G equals N(P ). (11) With notation as in item 8, let B n (F q ) be the subgroup of GL n (F q ) consisting of upper triangular matrices. Show that B n (F q ) is the normalizer in GL n (F q ) of U n (F q ). Deduce from exercise (10) that B n (F q ) is its own normalizer. 8
9 (12) Find the orbits and stabilizers for the action of GL n (R) on n 1 real column matrices. (13) Let V be the real vector space of all real n 1 column matrices. Let B denote the collection of those subsets of cardinality n of V that form a basis for V. The induced action of GL n (R) on the power set of V leaves B invariant. Detemine the orbits and stablizers for the action of GL n (R) on B. (14) Let S n be the symmetric group on n letters. Elements of S n are called permutations. Every permutation is a product of disjoint cycles. The cardinalities of the cycles in such a decomposition (counted with multiplicity and including the singleton cycles) determines a partition of n, called the cycle type of the permutation. Two permutations are conjugate if and only if they have the same cycle type, and every partition arises as a cycle type. Thus conjugacy classes of permutations are in bijection with partitions. For λ = 1 m 1 2 m 2... a partition, the centralizer of an element with cycle type λ has cardinality 1 m 1 m 1! 2 m 2 m 2!. (15) Let X be a finite set, say [n] := {1, 2,..., n}. Then the group G = S X = S n of bijections of X acts on X (naturally of course). Consider the induced action of G in turn on each of the following. Determine in each case the orbits, stabilisers, and their cardinalities. (a) the power set of X. (b) Cartesian square X X. (c) complex valued functions on X. (d) higher Cartesian powers X 3, X 4, etc. (e) functions from X to X. Warning: Answers to (d), (e) are a bit involved. Solution: (a) There are n + 1 orbits: the null set by itself is an orbit; the whole set by itself is an orbit; more generally, for each nonnegative integer k with k n, all subsets of cardinality k form an orbit, which let us denote by ( ) ( [n] k. The cardinality of the orbit [n] ) ( k is n ) k. The partition ( 2 X = [n] ) 0 k n k of 2 X into disjoint orbits may be thought of as a realization of the familiar binomial identity 2 n = ( n 0 k n k). For Y a subset of X, the stabilizer is S Y S X\Y, the cardinality of which is of course Y! (n Y )!. (b) The diagonal in X X is the subset consisting of elements of the form (x, x). There are two orbits: and its complement (except, in the degenerate case when X is a singleton, the complement is empty). The respective cardinalities are n and n 2 n. For a point (x, x) in the diagonal, the stabilizer is S X\{x} (the cardinality of which is (n 1)!). For a point (x, y) in the complement of the diagonal, the stabilizer is S X\{x,y} (the cardinality of which is (n 2)!). (c) Given a function f on X and a complex number α, let us denote by m(f, α) the number of elements of X that have value α under f. Two functions f and g belong to the same orbit under S X if and only if m(f, α) = m(g, α) for every complex number α. For a given function f, the list of nonzero values of m(f, α) as α varies over C form a partition of n. If m 1,..., m r be this list, the number of elements in the orbit is ( ) n m 1 m 2... m r := n! m. 1!m 2! m r! An element (x 1, x 2,..., x n ) of C n defines a partition of {1, 2,..., n} into subsets: i and j belong to the same subset if and only if x i = x j. If Y 1,..., Y r are these subsets, the stabilizer of the point is the subgroup S Y1 S Yr (whose cardinality is Y 1! Y r!). (d) For X 3, we consider various partitions of the set {1, 2, 3} into subsets. There are 5 of these: {{1}, {2}, {3}} {{1, 2}, {3}}; {{1, 3}, {2}}; {{2, 3}, {1}} {{1, 2, 3}} the first of type , the three partitions in the second item of type 2 + 1, and the one in the last item of type 3. 9
10 We may use the partitions of {1, 2, 3}, all of which are listed in the previous paragraph, to index orbits of X 3 under S X : there is one orbit corresponding to each partition. 7 The orbit corresponding to {{1, 3}, {2}} for example is the subset of X 3 consisting of elements (x 1, x 2, x 3 ) such that x 1 = x 3 x 2 ; that corresponding to {{1}, {2}, {3}} is the subset of elements (x 1, x 2, x 3 ) of that form x 1 x 2, x 2 x 3, x 1 x 2. The cardinalities of the orbits are: n(n 1)(n 2) for the orbit of type , n(n 1) each for the three orbits of type 2 + 1, and n for that of type 3. The stabilizer of the element (n, n 1, n 2) in X 3 is S {1,2,...,n 3} = S n 3 (whose cardinality is (n 3)!), of each of the elements (n, n, n 1), (n, n 1, n), and (n, n 1, n 1) is S {1,2,...,n 2} = S n 2 (whose cardinality is (n 2)!), and of the element (n, n, n) is S {1,2,...,n 1} = S n 1 (whose cardinality is (n 1)!). For other elements, the stabilizers will be the appropriate conjugates of those listed here see the first item in this exercise set. We may unify these assertions by the following single assertion: for a point (x 1, x 2, x 3 ) in X 3 (with possibly there being equalities among x 1, x 2, and x 3 ), its stablizer is S X\Y where Y is the subset {x 1, x 2, x 3 } of X. Observe that the three orbits of type are isomorphic to each other as Gsets (by the first item in this exercise set). (d) (Continued) For X 4, we consider various partitions of the set {1, 2, 3, 4} into subsets. There are 15 of these: 1 of type : {{1}, {2}, {3}, {4}} 6 of type : {{1, 2}, {3}, {4}}; {{1, 3}, {2}, {4}}; {{1, 4}, {2}, {3}}; {{2, 3}, {1}, {4}}; {{2, 4}, {1}, {3}}; {{3, 4}, {1}, {2}} 3 of type 2 + 2: {{1, 2}, {3, 4}}; {{1, 3}, {2, 4}}; {{1, 4}, {2, 3}} 4 of type 3 + 1: {{1, 2, 3}, {4}}; {{1, 2, 4}, {3}}; {{1, 3, 4}, {2}}; {{2, 3, 4}, {1}}. 1 of type 4: {{1, 2, 3, 4}} Assuming n 4, the orbits of X 4 are in bijection with the partitions listed above. 8 The orbit corresponding to {{1, 3}, {2, 4}} for example is the subset consisting of elements (x 1, x 2, x 3, x 4 ) such that x 1 = x 3 x 2 = x 4. The cardinality of an orbit depends upon the number of parts in the corresponding partition. It is n(n 1)(n 2)(n 4) for the partition of type ; n(n 1)(n 2) for partitions of type ; n(n 1) for partitions of type and 3 + 1; n for the partition of type 4. The decomposition of X 4 into disjoint orbits therefore implies the following identity: n(n 1)(n 2)(n 3) + 6n(n 1)(n 2) + 3n(n 1) + 4n(n 1) + n = n 4 7 If n 2, this assertion and others below in this item are to be taken with a pinch of salt: for example, since we cannot find three distinct elements in X in that case, there would be no orbit corresponding to the parition {{1}, {2}, {3}}. However, the formulas for the cardinalities of the orbits would still hold good, so we just need to interpret the statements appropriately for them to be correct: to be precise, there is one orbit corresponding to each partition of {1, 2, 3} into subsets at most n in number. 8 Generally, there will as many orbits of X k as there are partitions of the set {1, 2,..., k} into nonempty subsets at most n in number. The number of partitions of the integer k into at most n parts is the coefficient of t k in the power series expansion of the following 1 (1 t)(1 t 2 ) (1 t n ) Such a partition λ can be written as: λ 1 = λ 2 =... = λ a > λ a+1 =... = λ b > λ b+1 =... = λ c > λ c+1 =... >... >... = λ p > λ p+1 =... = λ q where the above is to be understood as follows: there are q parts in λ arranged in nonincreasing order, the first a of which are equal, the next b a are equal, the next c b are equal,..., the last q p are equal. The number of partitions of type λ of {1, 2,..., k} into nonempty subsets is ( ) k λ 1 λ 2... λ q 1 a!(b a)!(c b)! (q p)! ( ) k where := λ 1 λ 2... λ q k! λ 1!λ 2! λ q! 10
11 The stabilizer of an element (x 1, x 2, x 3, x 4 ) of X 4 (equalities among x 1, x 2, x 3, x 4 notwithstanding), is S X\Y where Y is the subset {x 1, x 2, x 3, x 4 }. Observe that not only is an orbit of a certain type isomorphic to another of the same type but also to any other of type with the same number of parts: so for example, orbits of type and are isomorphic. (d) (Continued) The orbits of S X on X k are naturally indexed by partitions of the set {1, 2,..., k} into nonempty subsets at most n in number. If there are p subsets in the partition, the corresponding orbit has cardinality n(n 1) (n p + 1). The stabiliser of a point (x 1,..., x k ) is S X\Y, where Y is the subset {x 1,..., x k }. The cardinality of S X\Y is of course X \ Y!. 11
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