STATISTICAL GROUP THEORY

Transcription

1 STATISTICAL GROUP THEORY ELAN BECHOR Abstract. Ths paper examnes two major results concernng the symmetrc group, S n. The frst result, Landau s theorem, gves an asymptotc formula for the maxmum order of an element n S n, and the second, Dxon s Theorem, settled an open conjecture concernng the probablty that two randomly selected elements of S n wll generate A n or S n. 1. Introducton The feld of statstcal group theory was born n the 1960s wth the seres of papers by Erdős and Turán that gave a statstcal characterzaton of the symmetrc group (see, for example, [5]). It s an easy exercse n nducton to show that, for example, f and element of S n s chosen unformly at random, the length of the cycle contanng 1 s dstrbuted unformly. On the other hand, the two authors show that the order of elements n S n are asymptotcally normally dstrbuted. Ths paper wll gve two examples of proofs on the symmetrc group: one classcal, the other statstcal n the style of Erdős and Turán. Landau s theorem, proved n 1902, states that the maxmum order of an element n S n s asymptotc to e n log n. Dxon s theorem states that as n goes to nfnty, f two elements are selected from S n unformly at random, the probablty that these two elements generate A n or S n approaches 1. Both proofs are nterestng n that they use both combnatoral and number-theoretcal technues. 2. A Proof of Landau s Theorem The order of an element x n a group s defned as the smallest non-negatve nteger n such that x n = 1. It s not dffcult to prove that for an element x n S n wth cycle lengths a 1, a 2,..., a k, the order of x s lcm (a 1, a 2,..., a k ). Landau s Theorem, proved n 1902, states that the maxmum order of an element n S n s asymptotc to e n log n. The proof n ths paper follows that of Mller [6]. Defnton 2.1. The Landau functon G(n) s defned as the maxmum order of an element n S n. Euvalently, G(n) = max{lcm (a 1, a 2,..., a k ) : a = n} Defnton 2.2. If m s any natural number greater than 1 and has prme factorzaton e1 1 e e k k, defne S(m) = k e The followng lemma wll allow us to relate the sum of the prme factors of a gven number to the Landau functon. Lemma 2.3. If lcm (a 1, a 2,..., a k ) = m, S(m) k a Date: 29 July

2 2 ELAN BECHOR Proof. Gven m, take a 1, a 2,..., a k wth lcm (a 1, a 2,..., a k ) = m such that a s mnmal. Each a 1 s greater than 1 snce f a = 1 for any, then a can be removed to get a smaller sum, contradctng mnmalty. Moreover, each a s a prme power. For f not, we may decompose a nto ntegers b and c such that a = bc and gcd(a, b) = 1 where nether a nor b are 1. Assumng wthout loss of generalty c > b > 1, we have that b + c b + c(b 1) = bc + (b c) < bc and hence we can construct a new mnmal set, a 1, a 2,..., a 1, b, c, a +1,..., a k wth least common multple m (snce b and c were relatvely prme). Fnally, the p s must be dstnct, snce only the p wth the largest exponent contrbutes to the least common multple, so any smaller power may be removed to get a smaller sum, contradctng mnmalty. After these reductons, then, we have the lcm(p e1 1, pe2 2,..., pe k k ) = m, so the set of a wth mnmal sum s exactly the prme factorzaton of m. Ths means that S(m) = k a for the mnmal case and hence that n general S(m) k a. Hence we have the followng two corollares, whch reduce calculatng G(n) to a purely number theoretc ueston. Corollary 2.4. S n contans an element of order m f and only f S(m) n Proof. If S(m) n, then f m has prme decomposton e1 1,..., e k k, there s an element of S n contanng one cycle each of lengths e for = 1,..., k wth the remanng ponts fxed. On the other hand, f there s a partton of n wth least common multple of m, then by the prevous lemma, S(m) n. Ths gves us the followng: Corollary 2.5. G(n) = max m S(m) n Defnton 2.6. Defne P to be the unue prme such that p<p p n whle p P p > n and let F (n) = p. p<p F (n) represents the order of an obvous canddate for G(n),.e. the one wth dsjont cycles wth lengths correspondng to the frst π(p 1) prmes. The goal s to prove that log F (n) log G(n) and then to show that log F (n) n log n. Lemma 2.7. Let 1 <... < s be the prmes dvdng G(n). Then s log < 2 + log F (n) + log P Proof. We note that s j S(G(n)) n < j=1 p P Defne 1,..., t to be the prmes dvdng G(n) that do not exceed P, and p 1,..., p r to be the odd prmes not dvdng G(n) that do not exceed P. Then by subtractng t from above, we have s r 2 + =t+1 By defnton, j > P whenever j > t, and for all = 1,..., r, p < P. We note that log x a x s a decreasng functon, so that f a b, log a log b b and a b log b log a. p p.

3 STATISTICAL GROUP THEORY 3 log P Thus log P and p P log P log p. Hence we have s log 2 log P r P + log p. =t We then add t log to both sdes to get the concluson. Lemma 2.8. Let be prme. If e > 1 and e G(n), then e 2P and 2P. Proof. Denote by Q the smallest prme not dvdng G(n). We actually show that e 2Q. Suppose e > 2Q, and let N be the smallest nteger such that Q N >. Snce by defnton Q N 1 <, we have < Q N < Q. Now let m = QN G(n). Thus m > G(n). Because does not dvde Q, m has the same prme factorzaton of G(n) except that t ncludes a factor of Q N and the exponent of s (e 1) nstead of e. Thus we have the followng euaton for S(m): S(m) = S(G(n)) + (Q N e + e 1 ). Note that t s suffcent to prove that (Q N e + e 1 ) < 0, snce ths would mply S(m) S(G(n)) n, whle smultaneously m > G(n). Ths would contradct 2.5. If < Q, then N = 1 and e + e 1 e /2 < (2Q)/2 < Q, a contradcton. On the other hand, f > Q, snce e > 1, then Q N e + e 1 < Q ( 1) Q Q = 0, also a contradcton. Q cannot eual, of course, snce dvdes G(n) and Q does not. Remark 2.9. The lemma s nterestng n that t shows the prme factors of G(n) are bunched up; once one s skpped, the number of prmes larger than Q dvdng G(n) s bounded by π(2q) π(q). Theorem log F (n) log G(n). Proof. Splt up the factorzaton of G(n) nto prmes wth exponent 1 and prmes whch appear wth exponent greater than 1. The latter set contans at most 2P prmes, for otherwse we would have at least one prme factor greater than 2P rased to a power of two or greater, contradctng Lemma 2.8. Thus log F (n) log G(n) 2 + log F (n) + log P + 2P log 2P. It thus suffces to prove that log F (n) > cp for some constant c. For ths t s suffcent to show that A(x) = p x log p x. Abel s summaton formula says that f a(x) s the ndcator functon of the prmes, x x π(x) A(x) = a(j) log j = π(x) log(x) π(1) log(1) x. j=1 By the prme number theorem, then, we have that [ A(x) = x ] [ π(x) x x x x + π(x) x ] 1 x x x x log x + 1. x log x 2 The theorem s proved wth smple bounds on the last term above. Theorem p n. p< n log n 2 2

4 4 ELAN BECHOR Proof. We wll need a statement euvalent to the proof of the Prme Number Theorem (whose euvalence we wll not prove but can be found n [6]). In partcular, f we defne A (x) = p xp, then we have A(x) x2 2 log x. The proof s mmedate when one substtutes n log n for x. Corollary G(n) e n log n. Proof. By Theorem 2.11, we have that log F (n) n log n. Hence F (n) e n log n. By Theorem 2.10, G(n) e n log n. 3. Dxon s Theorem We wll now prove Dxon s theorem, whch settled affrmatvely a 70-year-old conjecture that as n tends to nfnty, the probablty that randomly selected pars of elements of S n generate ether A n or S n tends to 1. The theorem employs some algebrac machnery, combnatoral analyss, and some of the results of Erdős and Turán. We begn wth some standard defntons about group actons that wll help classfy the subgroups of S n that canddate pars wll generate. Defnton 3.1. The orbt of an element a A under the acton of a group G s the set {a g : g G}. Defnton 3.2. A subset of S {1, 2,..., n} s called a block f for any element g G, S g S =. Defnton 3.3. A subgroup G S n s called transtve f t has one orbt. Defnton 3.4. A subgroup G S n s called prmtve f t s transtve and has no non-trval blocks, where sngletons and the whole set are consdered trval. Defnton 3.5. t n and p n are defned as the proporton of the (n!) 2 pars of elements n S n that generate a transtve group and prmtve group, respectvely. The dea of the proof s to show that p n s nearly 1, whle at the same tme usng prmtvty as part of a suffcent condton for generatng A n and S n. n Lemma 3.6. We have the recursve formula n = t Proof. Choose any partton of {1, 2,..., n}, say, A 1, A 2,..., A k. Then the number of pars of elements (x, y) that have exactly A 1, A 2,..., A k as fxed blocks n the subgroup x, y s n (k!) 2 t k where k = A. We next observe that the number of parttons of {1, 2,..., n} that have j parts of sze s n n! v j1,j 2,...,j n = (!) j j!.

5 STATISTICAL GROUP THEORY 5 Ths, of course, comes from the multnomal coeffcent wth k ndces of, further dvded by n j! because subsets of the same sze are not dstngushed. Thus by summng over all n-tuples (j 1, j 2,..., j n ) such that n j = n, we arrve at the dentty: Substtutng for v j1,j 2,...,j n we get (n!) 2 = v j1,j 2,...,j n n n! = n (!t ) j. j! ((!) 2 t ) j Next, we multply each sde by X n and formally sum over all nonnegatve ntegers n, gvng the followng formal denttes: [ ] ( n n!x n (!t ) k = X n (!t X ) j ) = = exp!t X. k! j! n=0 n=0 =0 j=0 The second eualty can be justfed as follows: the seres on the rght wll contrbute 1 to the coeffcent of X n f and only f t s the product of coeffcents whose power terms where the ns are the k s. We recall that (e f(x) ) = f (x)e f(x), and hence we have: n!nx n 1 =!t X 1 k!x k. n=0 Thus by takng the formal product of the two seres on the rght and euatng the (n 1)st coeffcents, we get n n = t as desred. k=0 Lemma 3.7. We have the relaton t n = 1 1 n + O(n 2 ) Proof. Defne r n = n(1 t n ) and c n = n 1 ( n ) 1. Then n n r n = n(1 t n ) = (1 t ) = c n Snce ( ) ( n = n ) n, we may wrte cn = n 1 ( n 1 n 1 ) = n ( n ) ( 3 n ) ( c, c n 3 and n 3 n ( n =0 r ) 1. Snce also ) = O(n 3 ), we have the followng upper bound: n 1 c n n = O(n 1 ) 3 Snce r 0, r n c n by defnton. Hence t n = 1 r n n 1 c n n whch goes to 1 as n goes to nfnty. Defnton 3.8. An mprmtve group s a transtve group whch s not prmtve. The proporton of pars of elements n S n whch generate an mprmtve group s denoted by n.

6 6 ELAN BECHOR Proposton 3.9. There are at most (m!) 2 t m (d!) 2m pars (x, y) S n S n that generate an mprmtve group wth m blocks. Proof. As n the frst part of the proof, there are exactly (m!) 2 t m pars of elements whch generate a group that has m blocks of eual sze of d. Each element x and y acts transtvely on these blocks, so the acton of x or y on a sngle block determnes the acton of x or y on the whole set. Ths gves at most (d!) 2m possble choces (wth each factor d! comng from the nternal varaton wthn each block). Lemma The proporton of pars of (x, y) that generate an mprmtve group n S n s at most n2 n 4. n! Proof. There are (d!) m m! ways to partton {1, 2,..., n} nto m blocks of sze d. Each mprmtve group has some block structure, even f t s not unue, so summng over each par m, d such that md = n, we have at most n!(m!)2 t m(d!) 2m (d!) m m! pars of generators, and hence we have: n md=n (m!) 2 t m (d!) 2m n!(d!) m m! We need the dentty m ( ) 1 d m = d We also note that (d d)!d! (d)! (d d)!d! (d)! = d 1 j=1 md=n d j d j. Snce each d j d j 1, ths product s bounded above by m (d 1) = (m!) (d 1). m!(d!) m n! = m!(d!)m. n! Snce we reure m 2 and d 2, we have m! 2 m 2 and (d 1) d 2. Hence m!(d!) m n! (2 m 2 ) d 2 = 2 n 4. So we arrve at: n 2 n 4 n2 n 4 = O(n 2 ). md=n We have shown that almost all pars of elements generate a prmtve group. Ths s very useful when combned wth Jordan s Theorem, whch gves a suffcent condton for a subgroup to generate A n or S n. Theorem (Jordan, 1893). A prmtve subgroup of S n s eual to ether A n or S n whenever t contans at least one permutaton whch s a -cycle for some prme n 3. Proof. Throughout the proof we wll make use of a few lemmas whose proofs are elementary and can be found n [7]. To begn, we must gve a multdmensonal analogue to transtvty: Defnton A subgroup G S n s called k-fold transtve f for every par of ordered k-tuples S and B, there exsts g G such that S g = B.

7 STATISTICAL GROUP THEORY 7 Suppose that some element g G s a -cycle; wthout loss of generalty g = ( ). Throughout the proof we use the followng notaton: Ω = {1, 2,..., n}, Γ = {1, 2,..., }, and = { + 1,..., n}, where of course 3 by hypothess. For any set A, we also denote the pontwse stablser of A by G A. Observe that g, havng order, s a Sylow subgroup of G, whch has order dvdng!. It s not hard to see that G s prmtve on Γ. Thus we can apply the followng lemma from [7]: Lemma If G s prmtve on {1, 2,..., n} and G s prmtve on Γ, where 1 < Γ = m < n, then G s (n m + 1)-fold transtve and prmtve on Ω. We next use the followng lemma, whch can also be found n [7]: Lemma Let G be k-fold transtve on Ω, and a subgroup U G Γ be a subgroup such that for any V such that U s conjugate to V n G, U s conjugate to V n G Γ. Then the normalzer N = N (U) s k-fold transtve on the set of ponts left fxed by U. Snce g s a Sylow subgroup, t satsfes the hypotheses of the lemma and hence the normalzer N of g s k-fold transtve on G. If G S denotes the acton of G on the subset S, then ths amounts to N = S, snce k-fold transtvty on a set of cardnalty k generates the entre S. The followng stronger fact s also true: Lemma If α Γ, N α = S. Proof. By the work above, N = S, so for any k-tuple τ S, there exsts σ N such that σ = τ on. But f we take the composton g c σ for some approprate exponent c, then g c σ(α) = α, where stll g c σ = τ on, snce g c leaves these ponts fxed. The next fact about N Γ α wll gve us a complete characterzaton of the commutator subgroup of N α : Lemma N Γ α s abelan. Proof. Nα Γ = {σ S n : σ(α) = α, σ = d, σ g σ 1 = g } In partcular, the frst and thrd condton gve exactly one σ S n for each g g such that σ g σ 1 = g. Ths s because for each g c g, we reure (σ(1)σ(c)σ(2c + 1)... ασ(α + c)... ) σ g σ 1. Snce the elements n the cycle must be eually spaced, together these unuely determne some element n S n. Nα Γ s somorphc to some subgroup of the automorphsms of g ; n other words, N Γ α = T Aut( g ). Snce Aut( g ) s abelan, the result follows. We just need one more well-known fact [3], namely that Lemma Any prmtve group G contanng a 3-cycle s the alternatng group or the symmetrc group. Proof. Denote by Λ the largest subset of Ω such that Alt(Λ) G, where Alt(Λ) denotes the subgroup of A Ω whch leaves ponts outsde of Λ fxed. Such a set s non-empty snce G contans a 3-cycle. For the sake of contradcton, assume Λ Ω. Λ s not a block, so for some g G, Λ Λ g or Λ. Suppose ths ntersecton

8 8 ELAN BECHOR conssts of at most one pont for each g. Call ths pont a and let the orgnal 3-cycle be (abc). We note that snce g Alt(Λ)g 1 = Alt(Λ g ) G, some other 3-cycle (ade) s n G as well. It s easly checked that (abc) (ade) (abc) 1 (ade) 1 = (abd). Now suppose that the ntersecton Λ Λ g conssts of two ponts or more ponts, ncludng a and b, for some g G, choose d Λ g Λ. We know there s such a pont snce G s prmtve. Snce Alt(Λ g ) G, G contans the 3-cycle (abd). Now suppose Alt(Φ) = Λ {d}. We wsh to show Alt(Φ) G, whch wll gve the desred contradcton. To do ths, note that we only need to show that for all elements σ Alt(Φ) such that σ(d) d, we have σ G. Of course, f wthout loss of generalty a = σ(d) d but σ Alt(Φ), then σ(d) Λ. But then we have some τ Alt( ) such that τ(σ(d)) = b, so π = τσ Alt( ). However, then we have both (abd) πσ and (abd) π n G snce the former fxes d. Then clearly σ G. Note that the commutator of a drect product follows a rather nce rule: [N α, N α ] = [ Nα, Nα ] [ N Γ α, Nα Γ ] [ = S, S ] = A Thus G contans a 3-cycle, snce a subgroup of a subgroup of G contans a 3-cycle, and we apply Lemma Ths completes the proof of Jordan s theorem. Defnton For a prme satsfyng (log n) 2 n 3, defne T n = S n σ contans a -cycle and ts other cycle lengths are relatvely prme to }. {σ Corollary Suppose there s some z T n wth order h, where h s dvsble by. Snce ord(z h ) = h =, then any prmtve group G contanng z generates gcd( h,h) A n or S n by Jordan s theorem. 4 We shall prove that the proporton of T n n S n s at least 1 3 log log n to prove Dxon s theorem. Three more lemmas wll fnsh the proof. The frst s an asymptotc estmate for p 1 p that I wrote. Two come from Erdős and Turán s semnal paper from 1967 [5]. Lemma For prme p n, p 1 p log log n. Proof. We wrte p 1 p = 1 x n a(x) 1 x, where a(n) s the ndcator functon of the prmes. Now, n =0 a() = π(n), so by Abel s summaton formula, 1 p = π(x) 1 x x + π(x) x 2. The estmate π(x) x log x p 2 of the Prme Number Theorem proves the lemma. Lemma Let 1 a 1 a 2 a n n. Then the proporton d n of elements n S n wthout any cycles of length a 1, a 2,..., a n s at most ( n 1 a ) 1. Lemma The proporton of elements n S n wth order relatvely prme to s bounded below by e 1 log n for suffcently large n.

9 STATISTICAL GROUP THEORY 9 Proof. It s clear that f we take the product over all postve ntegers such that p does not dvde v and for any z < 1, ( ) z v n 1 v n! = ( ) z v exp v v v n=0 the nth coeffcent gves the number of ways to sum to n usng no multple of p. But n takng ths product, we are merely excludng multples of p, so v exp( zv v ) = exp( v=0 z v v v=0 z pv pv ). But usng the facts that log(1 z) = v=0 zv v and log(1 zp ) p smplfy ths to (1 z p ) 1 p 1 z p 1 =0 z Usng the representaton (1 z p ) p 1 p get the exact formula, where m = = ( (1 z) p 1 ) p 1 1 p = ( =0 z )(1 z p ) p 1 p. = v=0 zpv pv, we = 1 + m=1 zmp m k=1 (1 1 kp ), we actually m 1 d n = n. It s not dffcult to show that, snce (log n) 2, m 1 exp( log n log ) exp( log n ) exp( 1 log n ) We now have all the tools to prove Dxon s celebrated result. Theorem (Dxon, 1969.) The proporton of pars of elements n S n whch generate ether A n or S n s at least 1 2 (log log n) 2. Proof. Let u n be the proporton of elements n S n whch le n T n. Combnng the prevous two lemmas, we see that for prme, (log n) 2 n 3, and n suffcently large, u n (1 ( 1 ) 1 )e 1 log n By the estmate on the sum of prme recprocals, for suffcently large n 1 log log(n 3) log log(log n)2 > 4 log log n. 5 In the prevous part of the proof, we showed that the proporton of elements whch generate a prmtve group, p n 1 2 n for large enough n. Thus the proporton of pars of elements whch generate A n or S n s at least p n (1 u n ) 2 (1 2 n ) 16 9(log log n) (log log n) 2.

10 10 ELAN BECHOR Remark Snce publshng hs paper n 1969, Dxon has sharpened hs result wth respect to the rate at whch the probablty approaches one. In fact, he fnds an asymptotc formula [4]: 1 1 n 1 n 2 4 n 3 23 n 4... Ths result s an mprovement on Baba s 1989 theorem [1] whch also gave a lnear error term. Both of these papers use the classfcaton of fnte smple groups. References [1] Baba, László. The probablty of generatng a symmetrc group. Journal of Combnatoral Theory [2] Dxon, John. The probablty of generatng the symmetrc group. Math. Z [3] Dxon, John. Permutaton Groups. Sprnger [4] Dxon, John. Asymptotcs of generatng the symmetrc and alternatng groups. jdxon/generatng-an.pdf [5] Erdős, Paul. On some problems of a statstcal group-theory II. Acta Mathematca [6] Mller, Wllam. The maxmum order of an element of a fnte symmetrc group. The Amercan Mathematcal Monthly [7] Welandt, Helmut. Fnte Permutaton Groups. Academc Press