Lecture 5: Burnside s Lemma and the Pólya Enumeration Theorem

Transcription

1 Mth 6 Professor: Pric Brtlett Lecture 5: Burnsie s Lemm n the Póly Enumertion Theorem Weeks 8-9 UCSB 205 We finishe our Möbius function nlysis with question bout seshell necklces: Question. Over the weeken, you collecte stck of seshells from the seshore. Some of them re tn n some re blck; you hve tons of ech color. You wnt to rrnge them in necklce! You consier two necklces to be the sme if one cn be rotte so tht it is the other (you on t llow flips, though, becuse then your seshells woul be bckwrs.) For exmple, here re ll of the istinct necklces you cn mke with four shells: How mny ifferent necklces on n shells cn you mke, for ny n? We nswere this problem by performing Möbius inversion on the ivisor poset; however, t the time, mny of you note tht there seeme to be ifferent connection to group theory here, n in specific to the ie of groups n group ctions! Mny of you hve seen some of these ies before (the ie of group, if nothing else, is something you ve likely seen in Mth 8!) However, it my hve been while, so we review these ies here: Groups n Group Actions. Groups: Bsic Definitions Definition. A group is set G long with some opertion tht tkes in two elements n outputs nother element of our group, such tht we stisfy the following properties: Ientity: there is unique ientity element e G such tht for ny other g G, we hve e g = g e = g. In other wors, combining ny group element g with the ientity vi our group opertion oes not chnge g! You know mny objects like this: if we work with the rel numbers R n think of ition s our group opertion, then 0 is our ientity, s 0 + x = x for ny x. Similrly, if we consier the rel numbers gin but tke our opertion to be multipliction, then is our ientity, s x = x for ny x.

2 Inverses: for ny g G, there is unique g such tht g g = g g = e. In other wors, if we strt t ny group element g, we cn lwys fin something to combine with g using our group opertion to get bck to the ientity! Agin, you know severl objects like this: with R n ition, the inverse of ny number x is just its negtive x, while if we consier the set of nonzero rel numbers n multipliction, the inverse for ny x is just /x. Associtivity: for ny three, b, c G, (b c) = ( b) c. In other wors, the orer in which we group combintions together oesn t mtter, s long s the sequence tht we hve those objects groupe together in oes not chnge! I.e. we cn combine with b c, or first fin b n then combine tht with c. Agin, most of the nturl opertions you re fmilir with (ition, multipliction) re ssocitive: it is perhps more interesting to point out some things tht re nonssocitive. For exmple, exponentition is nonssocitive opertion: 2 (34 ) = , while (2 3 ) 4 = 8 4 = It bers noting tht this oes not sy tht b = b : tht is ifferent property, clle commuttivity, n is not property tht groups nee to hve (s we will show in the exmples!) Groups tht re commuttive re clle belin groups, fter the mthemticin Niels Henrik Abel. We list number of exmples of groups, s well s some nonexmples. Here, we on t give forml proofs tht ny of these objects re groups; inste, we list them rpi-fire to give you list of exmples to think bout in your he! (If you re intereste, you cn prove tht ny of these objects stisfy the clime properties, though!) Exmple. As note bove, the rel numbers with respect to ition, which we enote s R, +, is group: it hs the ientity 0, ny element x hs n inverse x, n it stisfies ssocitivity. Nonexmple. The rel numbers with respect to multipliction, which we enote s R,, is not group: the element 0 R hs no inverse, s there is nothing we cn multiply 0 by to get to! Exmple. The nonzero rel numbers with respect to multipliction, which we enote s R,, is group! The ientity in this group is, every element x hs n inverse /x such tht x (/x) =, n this group stisfies ssocitivity. Exmple. The integers with respect to ition, Z, + form group! Nonexmple. The integers with respect to multipliction, Z, o not form group: for exmple, there is no integer we cn multiply 2 by to get to. Nonexmple. The nturl numbers N re not group with respect to either ition or multipliction. For exmple: in ition, there is no element N tht we cn to to get to 0, n in multipliction there is no nturl number we cn multiply 2 by to get to. 2

3 Exmple. GL n (R), the collection of ll n n invertible rel-vlue mtrices, is group uner the opertion of mtrix multipliction. Notice tht this group is n exmple [ of] 0 nonbelin group, s there re mny mtrices for which AB BA: consier 0 0 [ ] [ ] [ ] [ ] [ ] = versus = Exmple. SL n (R), the collection of ll n n invertible rel-vlue mtrices with eterminnt, is lso group uner the opertion of mtrix multipliction; this is becuse the property of being eterminnt is preserve uner tking inverses n multipliction for mtrices. These exmples re ll gret, but they re not relly wht I m intereste in for this clss. Inste, I wnt to focus on exmples of finite groups! In the next section, we stuy exmples of these groups in more epth thn bove, n explin more bout why they form group..2 Finite Groups Definition. The object Z/nZ, +, is efine s follows: Your set is the numbers {0,, 2,... n }. Your ition opertion is the opertion ition mo n, efine s follows: we sy tht + b c mo n if the two integers + b n c iffer by multiple of n. For exmple, suppose tht n = 3. Then + 2 mo 3, n mo 3. Similrly, our multipliction opertion is the opertion multipliction mo n, written b c mo n, n hols whenever + b n c iffer by multiple of n. For exmple, if n = 7, then mo 7, mo 7, n mo 7. This is commuttive group! This is not the only wy in which moulr rithmetic cn mke group: Exmple. (Z/pZ) = {,... p } is commuttive group with respect to the opertion of multipliction mo p, if n only if p is prime. (If you hven t proven this before; o it t home, or emil me!) There re other finite groups beyon those me out of moulr rithmetic! In prticulr, there re severl notble exmples of noncommuttive groups: we escribe some of those here. Exmple. The symmetric group S n is the collection of ll of the permuttions on the set {,... n}, where our group opertion is composition. In cse you hven t seen this before: A permuttion of set is just bijective function on tht set. For exmple, one bijection on the set {, 2, 3} coul be the mp f tht sens to 2, 2 to, n 3 to 3. 3

4 One wy tht people often enote functions n bijections is vi rrow nottion: i.e. to escribe the mp f tht we gve bove, we coul write f : This, however, is not the most spce-frienly wy to write out permuttion. A much more conense wy to write own permuttion is using something clle cycle nottion. In prticulr: suppose tht we wnt to enote the permuttion tht sens 2, 2 3,... n n, n, n oes not chnge ny of the other elements (i.e. keeps them ll the sme.) In this cse, we woul enote this permuttion using cycle nottion s the permuttion ( n ). To illustrte this nottion, we escribe ll of the six possible permuttions on {, 2, 3} using both the rrow n the cycle nottions: i : 2 3 (2) : 2 3 (3) : 2 3 (23) : (23) : (32) : It s worth noting tht some permuttions nee to be represente with multiple cycles. For exmple, (43)(25) = , 4 5 becuse mps to 4, 4 mps to 3, n 3 mps to, giving us (43), n 2 mps to 5 mps to 2, giving us (25). Becuse the composition of ny two bijections is still bijection, we hve in prticulr tht the composition of ny two permuttions is nother permuttion: so our group opertion oes inee combine group elements into new group elements. Composing ny mp f with the ientity mp i(x) = x oes not chnge the mp f, so i(x) is n ientity element. Associtivity is oble to check: tke ny three bijections f : C D, g : B C, h : A B. We wnt to check tht (f g) h : A D is the sme mp s f (g h) : A D. 4

5 To o this, it suffices to show tht they sen the sme elements to the sme plces, s this is exctly wht it mens for two functions to be equl. Tke ny A, n notice tht ((f g) h)() = (f g)(h()) = f(g(h())) (f (g h))() = (f(g(h()))) = f(g(h())). So we stisfy ssocitivity! Finlly, to see tht we hve inverses, notice tht ny bijection f : X Y hs n inverse function f Y X efine by f (y) = the unique x such tht f(x) = y. Notice tht f f(x) = x for ny x: in other wors, their composition is the ientity! Therefore, ny bijection hs n inverse, n thus S n is group. While the bove work certinly shows tht S n is group, it my not o gret job of giving us feel for how to combine its elements. To o this, let s emonstrte smll yet useful property bout S n : Definition. A permuttion σ S n is clle trnsposition if we cn write σ = (b), for two istinct vlues, b {,... n}. Clim. We cn write ny σ S n s prouct of trnspositions. Proof. To illustrte wht our clim is, let s work it out for ll of the elements of S 3. We first note tht (2), (3) n (23) re ll trivilly covere by this proposition, s they re themselves trnspositions; s well, we cn trivilly write i s the prouct (2)(2), mongst other things, s i : 2 3 = (2)(2) : 2 3, becuse both mps sen to, 2 to 2, n 3 to 3 (just follow the rrows!) To work this out for (23) n (32) tkes not much more work. Simply notice tht to o the permuttion (23), we coul strt with the swp n 2, to get to mp to the right thing; from there, we currently hve 2 n 3 3, when we wnt 2 3 n 3. Swpping n 3 fixes these issues, n gives us the swp we wnt! Becuse function compositions re re right-to-left (becuse in f(g(x)), you pply g to x before pplying f!), we woul write this s (3)(2): n inee, we cn check tht 5

6 (23) = = (3)(2) =. Using similr logic, we cn see tht (32) cn be written s (2)(3): (32) : = (2)(3) = This hopefully gives us bit more of feel for wht we re oing here! Just to illustrte tricker cse, however, let s try longer permuttion: wht bout σ = (2345) S 5? Well: let s use similr logic to our erlier cses. We know tht σ sens 2; so we cn strt with (2) s our first trnsposition. Now s going to the right plce! If we consier 2, it s currently going to, when it shoul be going to 3. Therefore, if we follow up our first switch with (3), it will sen 2 to 3, while not chnging where mps to (s we in t touch its trget from the first step, 2!) This process continues: 3 is mpping to now, n it shoul mp to 4; so we shoul pply (4). Then, 4 is mpping to, n it shoul mp to 5; so we pply (5). Now we hve 5 mpping to, which is correct; in other wors, ll of our elements re mpping to 6

7 the right plce! This is esily verifie with quick igrm: (2345) = (5)(4)(3)(2) = In generl, suppose we hve ny cycle ( 2... n ). I clim tht we cn write this cycle s the prouct ( 2... n ) = ( n )... ( 4 )( 3 )( 2 ); we prove this by inuction. Our bse cse, when n = 2, is trivilly true: so we move to our inuctive step. Assume tht our cse hols for length-n cycles, n consier cycle ( 2... n n+ ) of length n +. So: consier the prouct ( n+ )( n )... ( 4 )( 3 )( 2 ). We wnt to show tht this permuttion is precisely the cycle ( 2... n n+ ). By inuction, we know our prouct of trnspositions is equl to ( n+ )( 2... n ). Consier where this permuttion sens elements: If we look t ny of the elements k {,... n }, ech k gets sent to k+ by the cycle ( 2... n ); becuse none of the elements 2,... n re in the trnsposition ( n+ ), it oes not interct further with ny of these elements. If we look t n, it is sent to by the cycle ( 2... n ); becuse is then sent to n+ by the trnsposition ( n+ ), in totl we hve tht n is sent to n+. Finlly, n+ is not touche by the cycle ( 2... n ), n is then sent to by the trnsposition ( n+ ). In totl, we hve tht ech k is sent to k+, with the exception of n+, which is sent to. In other wors, our prouct of trnspositions is precisely ( 2... n n+ ), s clime. Vi cycle nottion, we cn write ny permuttion s some prouct of cycles: pplying this result to ech cycle in turn lets us write ech permuttion s prouct of cycles, s clime. 7

8 It bers noting tht given permuttion cn be written s prouct of trnspositions in multiple wys: for exmple, (23) = (3)(2) = (2)(23) = (2)(2)(2)(23), for exmple. However, you cn prove tht the prity of the number of trnspositions neee to write ny permuttion is invrint: Theorem. Tke ny permuttion π S n. Suppose tht you cn write π s prouct of trnspositions in two ifferent wys; one tht uses s trnspositions, n nother tht uses t. Then either s, t re both o, or s, t re both even. We leve this theorem for you to prove if you hven t seen it before! Not ll finite groups re s lgebric s S n! Our lst exmple, for instnce, is beutifully geometric in nture: Exmple. Consier regulr n-gon. There re number of geometric trnsformtions, or similrities, tht we cn pply tht sen this n-gon to itself without stretching or tering the shpe: i.e. there re severl rottions n reflections tht when pplie to n-gon o not chnge the n-gon. For exmple, given squre, we cn rotte the plne by 0, 90, 80, or 270, or flip over one of the horizontl, verticl, top-left/bottom-right, or the top-right/bottom-left xes: (rotterbyr0 ) b b (rotterbyr90 ) b b c c (rotterbyr80 ) b c c c (rotterbyr270 ) b c b (fliproverrhorizontl) b c c c (fliproverrverticl) b b b c (fliproverrul/drrigonl) b b c c (fliproverrur/dlrigonl) b c b c b c c 8

9 Given two such trnsformtions f, g, we cn compose them to get new trnsformtion f g. Notice tht becuse these two trnsformtions ech iniviully sen the n-gon to itself, their composition lso sens the n-gon to itself! Therefore composition is wellefine opertion tht we cn use to combine two trnsformtions. b b c c b (rotte by 90 ) (flip over verticl) c = b c b c (flip over UR/DL igonl) Notice tht the trivil rottion by 0, when compose with ny other mp, oes not chnge tht mp: so, uner the opertion of composition, rottion by 0 is n ientity! Similrly, notice tht performing the sme flip twice in row returns us bck to the ientity, so every flip hs n inverse given by itself! (I.e. if f is flip, f f = i: i.e. f = f.) As well, if we rotte by k egrees, rotting by 360 k egrees results in totl rottion by 360: i.e. rottion by 0. So ll rottions hve inverses s well! Finlly, notice tht becuse function composition is ssocitive (s shown bove!), this opertion is ssocitive s well. Consequently, the collection of ll symmetries of regulr n-gon forms group uner the opertion of function composition! We cll this lst group D 2n, becuse it is the group of orer 2n. Algebrists will usully use this terminology; geometers, however, will often write the sme group s D n, s they cre bout they object whose symmetries we re stuying ( n-gon) more thn the number of symmetries themselves (2n). One prticulrly useful kin of group here is the ie of subgroup:.3 Subgroups n Cosets Definition. Tke ny group G,, n subset H G. We sy tht H is subgroup of G if the following three properties hol:. Closure: For ny two elements h, h 2 H, h h 2 is lso in H. 2. Ientity: The ientity i of G is in H. 3. Inverses: If h H, then h is lso in H. Notice tht ny subgroup of group is group in its own right! This is becuse the ssocitivity property is inherite from the lrger group, n we ve lrey checke ll of the other properties for being group. The orer of group is the number of elements in tht group! 9

10 Exmples. The set of even numbers is subgroup of the integers uner ition! This is becuse 0 is n even number (ientity), the sum of ny two even numbers is even (closure), n the itive inverse of ny even number is lso even (inverses.) Conversely, the set of ll o numbers is not subgroup of the integers uner ition: becuse 0 is not o, this subset oes not stisfy the ientity property! The set of ll powers of 2, {2 n n Z} is subgroup of the nonzero rel numbers uner multipliction! This is becuse = 2 0 (ientity,) 2 n 2 m = 2 n+m (closure), n 2 n 2 n = 2 0 = (inverses) ll hol for this subgroup. Conversely, the set of ll even integers is not subgroup of the nonzero rels uner multipliction, s there is no multiplictive inverse for 2 in this set! The set of ll rottions, long with the ientity is subgroup of the group D 2n of symmetries of n-gon: to see this, notice tht we hve the ientity by efinition, the inverse of rotting by θ is nother rottion (by 2π θ), n the composition of two rottions (by θ, ϕ) is just the rottion given by summing their ngles (θ + ϕ.) Given the ie of subgroup, nturl extension of this ie is the concept of cosets: Definition. Suppose tht G, is group, s G is some element of G, n H is subgroup of G. We efine the left coset of H corresponing to s s the set Hs = {sh h H}. We will often omit the left prt of this efinition n simply cll these objects cosets. Exmple. Consier the group G = Z, +. One subgroup of this group is the collection of ll multiples of 5: i.e. This subgroup hs severl left cosets: s = 0: this forms the coset which is just H itself. s = : this forms the coset s = 2: this forms the coset s = 3: this forms the coset H = {... 5, 0, 5, 0, 5, 0, 5...} 0 + H = {... 5, 0, 5, 0, 5, 0, 5...}, + H = {... 4, 9, 4,, 6,, 6...}. 2 + H = {... 3, 8, 3, 2, 7, 2, 7...}. 3 + H = {... 2, 7, 2, 3, 8, 3, 8...}. 0

11 s = 4: this forms the coset 4 + H = {..., 6,, 4, 9, 4, 9...}. Notice tht this collection of cosets bove is inee the collection of ll of the possible cosets of H within G: if we tke ny other element in Z, like sy 3, we ll get one of the five cosets bove: i.e. 3 + H = {... 2, 3, 8, 3, 8...} = 3 + H. In generl, H + x = H + y for ny x y mo 5. Exmple. Consier the group G = (Z/7Z),, i.e. the nonzero integers mo 7 with respect to the multipliction opertion. This hs the set H = {, 6} s subgroup (check this if you on t see why!) This group hs the following cosets: s =, which cretes the cosets H = H, s = 2, which cretes the coset s = 3, which cretes the coset s = 4, which cretes the coset 2 H = {2, 5}. 3 H = {3, 4}. 4 H = {4, 3}. Notice tht this coset is the sme s H 3. s = 5, which cretes the coset 5 H = {5, 2}. Notice tht this coset is the sme s H 2. s = 6, which cretes the coset Notice tht this coset is the sme s H. So, in totl we hve three istinct cosets! 6 H = {6, }.

12 Exmple. Consier the group S 3. This group hs the subgroup H = {i, (23), (32)} s subgroup. This subgroup hs two possible istinct cosets: i H = (23) H = (32) H re ll the sme coset, which is just H. (2) H = (3) H = (23) H = {(2), (3), (23)}. Cosets hve number of remrkbly useful properties: Theorem. Tke ny finite group G, n ny subgroup H.. For ny s G, the left coset sh is equl to H if n only if s H. 2. Two cosets sh, th re either completely ienticl or completely isjoint. 3. The vrious possible cosets of H prtition G into collection of isjoint subsets. (In prticulr, this proves tht the number of elements in H must ivie the number of elements in G.) 4. All cosets of H re the sme size: i.e. sh = H for ny s G. 5. If there re k istinct cosets of H, then k H = G. Proof. Tke ny group G, n ny subgroup H. We prove these properties in orer.. Tke ny s G. If s H, then becuse H is subgroup we hve tht s is in H, n therefore tht for ny h H we hve s h in H. Consequently, every member of H is in sh. As well, ny member of sh is prouct of two elements of H; therefore by closure, sh is subset of H. Therefore we hve proven tht Hs = H, s clime. Conversely, when s / H, we wnt to show tht sh H. But this is trivil: becuse H contins the ientity, we know tht s = s i sh, but s / H; so these re ifferent sets. 2. Tke ny two cosets sh, th. If they re not completely isjoint, then there is some element they shre in common: in other wors, there is some h, h 2 H such tht sh = th 2. But this mens tht t s = h h 2, s t = h h 2, n therefore tht t s, s t re elements in H. So, tke ny th 3 th. We cn write th 3 = t(t s)(t s) h 3 = s(t s) h 3 ; i.e. th 3 is n element of the form s something in H! So th sh. Similrly, for ny sh 3 sh, we cn write sh 3 = s(s t)(s t) h 3 = t(s t) h 3 ; i.e. sh 3 is of the form t something in H! So sh th s well, n we ve proven our clim. 3. Simply notice tht for ny element g G, gh contins g; this is becuse the ientity is in H, n therefore tht g i = g gh. But this mens tht every element is in some coset: if we combine this with (2), we hve tht every element is in exctly one coset. But this is wht it mens to be prtition! 2

13 4. On one hn, we know tht sh = {sh h H} mens tht we cnnot hve more elements in sh thn we hve in H, becuse we only get one element in Hs for ech element of H. But if we ever h sh = sh 2, we coul multiply by s on the left to get h = h 2 : consequently, we cn conclue tht we hve exctly s mny elements in sh s we o in H! 5. By point 3, the cosets of H prtition G: tht is, if we up the sizes of ll of our cosets, we will get the size of G. On the other hn, by point 4, ll of these cosets re the sme size; so the number of elements in ll of our cosets is just k H. So we ve proven our clim! Point 5 hs very fmous corollry, in the form of Lgrnge s theorem: Theorem. (Lgrnge.) Let G, be finite group, n H be ny subgroup of G. Then the orer of H ivies the orer of G..4 Group Actions Severl of the groups we efine in these notes S n, D 2n re interesting becuse they re groups efine by how they ct on some outsie object. For exmple, elements of S n re permuttions of the set {, 2,... n}; tht is, they re efine by how they shuffle the elements of set roun. Similrly, elements of D 2n re symmetries of n-gon; tht is, they re efine by how they move the vertices of some n-sie shpe roun! This motivtes us to introuce the ie of group ction: Definition. A group ction of group G, on set X is ny mp : G X X tht stisfies the following two properties:. Ientity: If e is the ientity in G, then for ny x X, we hve e x = x. 2. Comptible with the group opertion: For ny two elements g, h G n ny element x X, we hve g (h x) = (g h) x. The ie with this secon opertion is tht it sys tht it oesn t mtter whether we pply g, h one-by-one using to x, or if we combine them first in G with n then pply the result to x. In both cses we get the sme thing! We list severl useful exmples of group ctions: Exmple. Tke the group S n, n consier the following ction of this group on the set {, 2,... n}: for ny σ S n, k {,... n}, set σ k = σ(k). For exmple, if σ = (23) n k = 2, we woul sy σ(2) = 3. This is group ction! To see this, we just check our efinition:. Ientity: The ientity of S n is just the ientity mp i on {,... n}. Therefore, for ny k {,... n} we hve i k = i(k) = k: so we stisfy the ientity property! 3

14 2. Comptible with the group opertion: Tke ny two permuttions σ, ϕ S n, n ny k {,... n}. Notice tht (σ ϕ) (k) = σ(ϕ(k)), n σ (ϕ k) = σ (ϕ(k)) = σ(ϕ(k)) is just the sme thing; therefore we re comptible with our group s opertion! Exmple. Tke the group D 2n, n consier the following wy in which it cts on the set of vertices {v,... v n }, liste in orer, of n-gon: if f is symmetry of our n-gon, then we sy tht f v k is just f(v k ). In other wors, our ction sens v k to whtever vertex f mps v k to. Show tht this is group ction of D 2n on {v,... v n }. This is group ction! To see this, we just check our efinition:. Ientity: The ientity of D 2n is just the ientity mp i on {v,... v n }. Therefore, for ny k {v,... v n } we hve i k = i(k) = k: so we stisfy the ientity property! 2. Comptible with the group opertion: Tke ny two symmetries f, g S n, n ny v i {v,... v n }. Notice tht (f g) (v i ) = f(g(v i )), n f (g v i ) = f (g(v i )) = f(g(v i )) is just the sme thing; therefore we re comptible with our group s opertion! Exmple. For ny group G,, consier the following wy in which G coul ct on itself: for ny g G, h G, set g h = g h. This is group ction of G on G:. Ientity: Suppose tht the ientity of G is e. Then, for ny g G, we hve e g = e g = g; so we stisfy the ientity property! 2. Comptible with the group opertion: Tke ny two elements g, h G, n ny k G. Notice tht (g h) (k) = g h k, n g (h k) = g (h k) = g h k is just the sme thing; therefore we re comptible with our group s opertion! 2 Burnsie s Lemm n Necklce Problems 2. Orbits, Stbilizers, n Fixe Points It might seem like we ve introuce n wful lot of nottion for reltively simple necklce problem; but if you look t the mchinery we ve ssemble, it s ll ctully pretty nturl. In our necklce problem, we hve the following: A set given by ll of the 2 n necklces in two colors on n shells. A wy for the group of rottionl symmetries on n-gon to ct on this set; given ny necklce, we ct on it by rottion by rotting tht necklce! From here, we wnt to know wht re ll of the elements tht re istinct uner these rottions. To mke this more forml, consier the following efinitions: Definition. Suppose tht X is set n G is group tht cts on X by some opertion. For ny x X, mke the following efinitions: 4

15 The orbit of x, enote Orb(x), is the collection of ll elements in X tht cn be reche by strting t x n multiplying x by n pproprite element: tht is, Orb(x) = {g x g G}. The stbilizer of x, enote Stb(x), is the collection of ll elements in G tht on t move x: tht is, Stb(x) = {g G g x = x}. The fixe points of n element g, enote F ix(g) for ny g G, is the collection of ll of the elements of x unmove by G: tht is, Notice the following properties: F ix(g) = {x X g x = x}. Proposition. Tke ny group G cting on set X. Then the sets Orb(x) prtition the set X; tht is, every element of X is in some set Orb(x), n for ny x, y X, the sets Orb(x), Orb(y) re either completely isjoint or ienticl. Proof. The first property is esy to check. Let e enote the ientity of our group G; then, for ny x X, we hve e x = x by efinition. But this mens tht x Orb(x); so every x X is in one of these sets. For the secon property: tke ny two sets Orb(x), Orb(y). We clim tht these two sets re either isjoint or ienticl; to prove this, we cn simply look t the cse where they re not isjoint n prove tht they must be ienticl. Let z X be n element in both Orb(x), Orb(y); then there re group elements g, g 2 G with g x = g 2 y = z. By comptibility, we cn see tht x = e x = (g g ) x = g (g x) = g (g 2 y) = (g g 2 ) y, n therefore tht x Orb(y). But this in prticulr mens tht for ny h G, we hve h x = (h g g 2 ) y; i.e. ny element in Orb(x) is in Orb(y)! Similrly, we cn solve for y in terms of x to get tht for ny h G, (h g 2 g ) x = h y, n cn similrly conclue tht ny element in Orb(y) is in Orb(x). Proposition. Tke ny group G cting on set X. For ny x X, the set Stb(x) is subgroup of G. Proof. To check tht subset of group is subgroup, we just nee to check three things: 5

16 Ientity: We nee tht the ientity e of our group is in Stb(x). This is by efinition; for ny x X, e x = x, n therefore e Stb(x). Closure: For ny g, h Stb(x), we nee g h Stb(x). This is by comptibility; if g, h Stb(x), then g x = h x = x, n therefore (g h) x = g (h x) = g x = x. Inverses: For ny g Stb(x), we nee g Stb(x). This is lso by comptibility; if g Stb(x), then g x = x, n therefore g x = (g g) x = e x = x, so g is lso in Stb(x). In this sense, our necklce problem is specific instnce of the following problem: given group G cting on set X, how mny ifferent orbits re there? (I.e. how mny ifferent necklces re there, given tht n orbit of necklce is just ll of the things equivlent to tht necklce uner rottion?) We nswer this with the lst two bits of lgebr neee to solve our problem: the Orbit-Stbilizer Theorem n Burnsie s Lemm! 2.2 Theorems! Theorem. (Orbit-Stbilizer Theorem.) Let G be finite group tht cts on set X, n tke ny element x X. Then G = Orb(x) Stb(x). Proof. Let G/Stb(x) enote the collection of ll of the cosets of the subgroup Stb(x) in G. For ech coset K of Stb(x), pick n representtive element g K in tht coset; then we cn write K = g Stb(x) (this is becuse K, g Stb(x) re two cosets tht both contin g, n therefore must be equl by our results erlier.) This mkes it esier to work with our cosets! Consier the following function f : G/Stb(x) Orb(x): for ny coset g Stb(x) with representtive g, efine f(g Stb(x)) = g x. I clim tht f is bijection. To see this, we just check injectivity n surjectivity: Injectivity: Suppose tht g Stb(x), h Stb(x) re two cosets tht get mppe to the sme element; then g x = h x. But this mens tht (g h) x = (g g)x = e x = x; i.e. tht g h Stb(x). But this mens tht g g h = h gstb(x); tht is, tht g Stb(x), h Stb(x) re not isjoint! Therefore they re the sme. Consequently, we ve prove injectivity, s we ve shown tht the only wy for two cosets to mp to the sme element is if those two cosets were ienticl. Surjectivity: Tke ny h x Orb(x). Becuse the cosets of Stb(x) prtition G, there is some coset g Stb(x) with h g Stb(x). But this just mens tht h = g k for some k Stb(x); consequently, we hve tht h x = (g k) x = g (k x) = g x = f(g Stb(x)). So we mp to h x, n re therefore surjective! 6

17 Therefore, we hve tht the number of cosets of Stb(x) is the sme size s the number of elements in Orb(x). But by our work erlier, we know tht the number of cosets is just G / Stb(x) ; combining gives us s clime. G = Orb(x) Stb(x), Theorem. (Burnsie s 2 Lemm.) Let G be finite group cting on set X. Let k enote the number of istinct orbits of elements of X uner this ction, n let Ω,... Ω k enote these k ifferent orbits. Then k G = g G F ix(g). Proof. Consier the set S of ll pirs (g, x) where g fixes x: tht is, S = {(g, x) g x = x}. If we group pirs (g, x) by their first element, we cn see tht for fixe g G the number of pirs (g, x) with g x = x is just the number of elements x X tht re fixe by G: tht is, we hve S = g G {x g x = x} = g G F ix(g). On the other hn, if we group pirs (g, x) by their secon element, we cn see tht for fixe x X the number of pirs (g, x) where g x = x is just the number of elements g G tht fix x: tht is, S = x X Stb(x). If we split X prt into its k isjoint orbits Ω,... Ω k, we hve S = k i= x Ω i Stb(x). By the orbit-stbilizer result erlier, we know tht Stb(x) = G Ω i, for ny x in the orbit Ω i ; so we ctully hve S = k i= x Ω i G k Ω i = G i= Ω i. x Ω i 2 Often known s the Cuchy-Frobenius lemm, or the lemm tht is not Burnsie s, s Burnsie i not prove this result; he simply cite it n ttribute it to Frobenius in book he wrote on finite groups. History! 7

18 But, for ny Ω i, the sum x Ω i Ω i =, s we re just ing up the quntity Ω i totl of Ω i -mny times! So we ctully hve S = G k = k G. i= Combining this with our first result, we hve k G = g G F ix(g), s clime. 2.3 Applictions We strt by using this mchinery to solve our necklce problem: Question. Tke ny prime number n. Suppose we look t the set X of ll 2 n necklces on n shells, where ech shell is either blck or white; enote this collection s the set X = {(s,... s n ) s i {W, B}}. Let the group Z/nZ, + ct on this set by sying tht for ny Z/nZ, we hve (s,... s n ) = (s +, s 2+,... s n+ ), where the rithmetic in the subscripts bove is clculte mo n. (So, for exmple, (s, s 2, s 3 ) woul be (s 2, s 3, s ).) How mny necklces re istinct uner this ction; in other wors, how mny ifferent orbits exist for this group ction? Answer. By Burnsie s lemm, we hve k G = g G F ix(g), where k is the number of orbits (i.e. wht we wnt to fin,) n G = Z/nZ = n. So it suffices to unerstn the sets F ix(g), for ll g G. Notice tht necklces in X come in two ifferent kins: Some necklces specificlly, two necklces, the ll-white n ll-blck necklces hve ll of their shells the sme color. These necklces re fixe by every element of G, s they re the sme uner ny rottion. 8

19 Other necklces hve shells of ifferent colors. I clim tht these necklces re not fixe by ny element of G other thn the ientity element (i.e. 0) To see why, tke ny necklce (s,... s n ), n ny element 0 Z/nZ such tht (s,... s n ) = (s,... s n ). Then, for ny k, we hve (s,... s n ) = (s,... s n ) = (s,... s n ) = (s,... s n )... = k times {}}{... (s,... s n ) k times {}}{ = ( ) (s,... s n ) = (k ) (s,... s n ). But n is prime number, n 0; so, in the multiplictive group (Z/nZ),, hs n inverse element; cll it Z/nZ). Then, for ny b Z/Z, if we let k = b, we hve (s,... s n ) = (b ) (s,... s n ) = (b) (s,... s n ) = (s +b,... s n+b ). In other wors, we hve s = s +b for ny b; i.e. every shell in our necklce is the sme! This mens tht the F ix(g) sets come in two ifferent kins: g = 0. The ientity fixes every element in X by ssumption; so F ix(0) = 2 n, the set of ll of our necklces. g 0. Then, s shown bove, there re only two necklces tht g fixes; so F ix(g) = 2. Therefore, kn = g G F ix(g) = F ix(0) + g 0 Z/nZ n therefore tht the number of istinct orbits is Success! F ix(g) = 2 n + 2 n + 2(n ). n 9 g 0 Z/nZ 2 = 2 n + 2(n ),

20 We turn here to problem you will likely remember from our first clss: Question. Suppose tht you hve set of n ifferent postcrs, out of which you wnt to choose k. In how mny wys cn you o this, if you on t cre bout the orer in which you picke your crs? Answer. So, on one hn, we lrey know the nswer here: it s ( n k). To show this t the time, though, we h to o something firly o with equivlence clsses (or inee when we stuie this gin with generting functions, we h to fin some recurrence reltions!) Some people ske t the time if there ws connection here between our nswer n the ie of quotient groups, n t the time I si tht something close to tht hel; wht I ment t the time ws use Burnsie s lemm, n tht s wht we ll o here! Specificlly: let s unimgintively lbel our n postcrs {, 2,... n} Let X consist of ll of the wys to orer our set of n postcrs; there re n! elements in X, s we ve proven in mny wys in this clss. To select k postcrs, then, we cn simply choose ny orering from X n tke the first k postcrs in tht orering! So the elements of X correspon to wy to pick out k postcrs, where we cre bout the orering of wht we choose n wht we o not choose. How cn we get from here to our esire counting problem, where we on t cre bout the orering of wht we choose or the orering of wht we on t choose? We strt, s lwys, with few efinitions. Given two groups G, H, efine their irect prouct s follows: Definition. Given two groups G,, H, we efine their irect prouct G H, s follows: The set: G H = {(g, h) g G, h H}. The opertion: for ny (g, h ), (g 2, h 2 ) G H, set (g, h ) (g 2, h 2 ) = (g g 2, h h 2 ), where the opertions insie the prentheses come from the originl groups G, H. It s not hr to check tht this stisfies the properties of being group; o so if you re curious! For this problem, I specificlly wnt to consier the crtesin prouct G = S k S n k ; tht is, the group of ll permuttion (σ, π), where σ S k, π S n k. This group cts on our set X of postcr-orerings s follows: for ny postcr orering P = (p,... p n ) X, efine (σ, π) P s the orering given by using σ to permute the first k postcrs, n using π to permute the remining n k postcrs. In symbols, we write this s (σ, π) (p, p 2,... p k, p k+,... p n ) = (p σ(), p σ(2),... p σ(k), p k+π(), p k+π(2),... p k+π(n k) ). We like this group ction becuse it cptures exctly wht we ment by we on t cre bout the orering: given ny two orerings P, P 2 X, we consier P, P 2 to be the sme wys to pick out k postcrs if they hve the sme first k elements in some orer: in other wors, we consier P n P 2 to be the sme if there is some reorerings (σ, π) of P s crs tht yiels P 2! 20

21 So, our problem is now Burnsie s lemm problem: we hve set X n group G cting on X, n we wnt to count the number of elements tht re istinct uner ction by G; i.e. the number of istinct orbits of X uner this group ction. Therefore, it suffices to unerstn the fixe points of our group elements uner our group ction! This is pretty simple: The ientity, s lwys, fixes everything: so F ix(i) = n!, the number of elements in X. Any other group element, however, fixes nothing! This is not hr to see; tke ny pir of permuttions (σ, π) (i, i). One of σ, π must not be the ientity; therefore, when it cts on ny orere set of postcrs, it will shuffle those crs roun (n in prticulr sen tht orere set to something ifferent!) So F ix(g) = 0 for ny non-ientity element g. Applying Burnsie s lemm gives us the following: the number of istinct orbits of X uner ction by G, i.e. the number of wys to choose k crs out of n without cring bout the orer, is just F ix(g) = F ix(0) + F ix(g) G S k S n k g G g 0 Z/nZ = k!(n k)! n! + 0 = So we ve nswere our question. n! k!(n k)!. g 0 Z/nZ We close by stuying one lst exmple of Burnsie s lemm: counting chors! Question. Consier the stnr Western musicl scle of tones, where we consier two tones to be equivlent if they iffer by n octve: tht is, consier the tones rrnge in circle s follows: C, D, D, E, E, F, G, G, A, A, B, B B C D B D A E A E G G F 2

22 We cn ientify these tones with the group Z/2Z s follows: = B 0 = C = D 0 = B 2 = D 9 = A 3 = E 8 = A 4 = E 7 = G 6 = G 5 = F A tri in this sense is ny chor (i.e. subset of our tones) me out of three istinct tones; i.e. it is ny orere triple of three istinct elements of Z/2Z. In music, we often consier two triples to be equivlent if one cn be trnslte (musiclly, the wor here is trnspose) to become nother; i.e. we consier (C, E, G) = (0, 4, 7) n (F, A, C) = (5, 9, 0) = (0 + 5, 4 + 5, 7 + 5) mo 2 to be equivlent becuse if we re only cpble of iscerning reltive pitch, the spces between these notes re ll equl! Similrly, we often consier two triples to be equivlent if the tones of one cn be permute to become the notes of the other: i.e. we consier (C, E, G) = (0, 4, 7) n (G, C, E) = (7, 4, 0) to be equivlent becuse if ll three tones re soune simultneously these two chors re inistinguishble. (Musiclly, such chors re si to be inversions of ech other.) How mny istinct musicl tris re there, given tht we consier two tris to be equivlent if one cn be trnsforme into the other uner trnsltion n permuttion? Answer. If we phrse this s Burnsie s lemm-style problem, the nswer here is reltively esy to etermine! Let X be the set of ll unorere istinct triples of elements from Z/2Z. It is not hr to see tht X = ( ) 2 3, s we re just consiering ll of the wys to pick out three tones from set of 2 tones without cring bout the orering. This is lmost n nswer to our question: it counts ll of the tris up to inversions (i.e. permuttions.) So, to finish our problem, we just nee to el with trnsltion s well! 22

23 To o this, we simply consier the group ction of G = Z/2Z on our set of triples, where for ny k,, b, c Z/2Z, we hve k {, b, c} = { + k, b + k, c + k}, where our ition is one mo 2. This is group ction; moreover, we consier two tris in X equivlent if n only if one cn be trnslte into the other vi our group ction, so etermining the number of istinct orbits here will nswer our question bout the number of ifferent tris! As before, we consier the elements of our group cse-by-cse: As lwys, the ientity 0 Z/2Z fixes everything, so we hve F ix(0) = X = ( 2 3 ). If we consier trnsltion by 4, we hve tht {, b, c} is fixe point if n only if 4 {, b, c} = { + 4, b + 4, c + 4} = {, b, c}. In prticulr, this forces our set to be of the form {, + 4, + 8} fter some thought, n implies tht there re precisely four such triples {0, 4, 8}, {, 5, 9}{2, 6, 0}, {3, 7, }. So F ix(4) = 4. By completely ienticl resoning, F ix(8) = 4 s well! For ny other trnsltion, I clim tht there re no fixe points. In fct, we cn prove stronger clim: Lemm. Suppose tht we re working in -tone system (i.e. in Z/Z,) n stuying n-tuples of tones. Suppose tht A = {,... n } is n unorere n-tuple of tones, n k Z/Z is such tht k A = A. Then k n is multiple of. Proof. If k A = A, then if i A, we must hve tht i + k is in k A, n therefore in A itself. Repeting this logic tells us tht i + mk A for ny m, provie tht (s lwys) we o our rithmetic mo. Notice tht we get precisely /gc(k, )-mny istinct elements vi this opertion, s we get repet precisely when mk is multiple of for the first time. Tke ny A, n let A = { + mk m N}. If A is ll of A, stop; otherwise, there is some other 2 still in A not in A. Define A 2 = { 2 +mk m N}. Notice tht A, A 2 hve no elements in common, s + mk = 2 + m k + (m m )k = 2, n we woul hve then h 2 in A. Repet this process until we cn t go ny further. A,... A l, ech of size /gc(, k). This gives us bunch of sets Becuse we hve n elements in A, we must lso hve n elements in k A; therefore we hve l gc(, k) = n l k gc(, k) = kn. 23

24 k gc(,k) is lwys n integer, s it is k ivie by some of its fctors; so the LHS bove is two integers times. Therefore the RHS is multiple of, n we ve proven our clim! Therefore, in ny -tone system where we re picking out n-tuples of tones, the only vlue k in Z/Z where we cn possibly hve fixe points re when kn is multiple of. In prticulr, for our problem here, we re consiering n = 3-tuples of tones; so the only possible fixe points re when 3k is multiple of 2; i.e. when k = 0, 4, 8! So no other group elements hve fixe points in this problem; i.e. F ix(g) = 0 for g 0, 4, 8. Therefore, by Burnsie s lemm, the number of istinct orbits (i.e. number of istinct tris) is G F ix(g) = ( 2 ) 2 ( F ix(0) + F ix(4) + F ix(8) ) = g G = = 9. (Similr methos cn enumerte ll of the ifferent 4-tuples of tones, or in generl ll of the n-tuples for ny n; try it if you re intereste!) 24