Asymptotics of generating the symmetric and alternating groups John D. Dixon School of Mathematics and Statistics Carleton University, Ottawa, Ontario K2G 0E2 Canada jdixon@math.carleton.ca October 20, 2005 Abstract The probability that a random pair of elements from the alternating group A n generates all of A n is shown to have an asymptotic expansion of the form /n /n 2 4/n 3 23/n 4 7/n 5.... This same asymptotic expansion is valid for the probability that a random pair of elements from the symmetric group S n generates either A n or S n. Similar results hold for the case of r generators (r > 2). MSC 2000: Primary 20B30; Secondary 20P05 05A6 20E05 Submitted: Jul 8, 2005; Accepted: Oct 8, 2005; Published: Oct 3, 2005 Introduction In [5] I proved that the probability that a random pair of elements from the symmetric group S n will generate either S n or A n is at least 2/(log log n) 2 for large enough n. This estimate was improved by Bovey and Williamson [3] to exp( p log n). Finally Babai [] showed that the probability has the asymptotic form /n + O(/n 2 ). Unlike the earlier estimates, the proof of Babais result uses the classication of nite simple groups. Babais result depends on two elementary results from [5], namely: the probability t n that a pair of elements in S n generates a transitive group is /n + O(/n 2 ); and the probability that a pair of elements generates a transitive, imprimitive group of S n is n2 n/4. Using the classication, he shows that the probability that a pair of elements generates a primitive subgroup of S n di erent from A n or S n is < n pn /n! for all su ciently large n. Thus the probability that a pair of elements of S n generates a transitive group but does not generate either S n or A n is O(n2 n/4 + n pn /n!) = O(n k ) for all k. the electronic journal of combinatorics (2005), #R00
The object of the present paper is to show that there is an asymptotic series of the form t n» + P c k /n k so that t n = + c /n + c 2 /n 2 +... + c m /n m + O(/n m+ ) for m =, 2,.... By what we have just said, the same asymptotic series is valid for the probability that a pair of elements of S n generates either A n or S n. We shall also show that this asymptotic series is valid for the probability that a pair of elements in A n generates A n. More precisely, we shall prove the following. Theorem The probability t n that a random pair of elements from S n generates a transitive group has an asymptotic series of the form described above. The rst few terms are t n» n n 4 2 n 23 3 n 7 4 n 542.... 5 n 6 The same asymptotic series is valid for the probability that the subgroup generated by a random pair of elements from S n is either A n or S n. Theorem 2 If a n is the number of pairs (x, y) 2 A n A n which generate a transitive subgroup of A n and s n is the number of pairs (x, y) 2 S n S n which generate a transitive subgroup of S n, then s n 4a n = ( ) n 3 (n )! for all n. Thus for n 2 the probability 4a n /(n!) 2 that a random pair of elements from A n generates a transitive subgroup is equal to t n 3/(n n!). Hence the probability that a random pair of elements from A n generates A n has the same asymptotic expansion as given above for t n. Remark 3 The sequence ft n g also appears in other contexts. Peter Cameron has pointed out to me that a theorem of M. Hall shows that the number N(n, 2) of subgroups of index n in a free group of rank 2 is equal to n!nt n (see () below and [6]). On the other hand, a result of Comtet [4] (quoted in [8, page 48] and [7, Example 7.4]) implies that n!nt n = c n+ for all n where c n is the number of indecomposable permutations in S n (in this context x 2 S n is called indecomposable if there is no positive integer m < n such that x maps f, 2,..., mg into itself). We shall discuss a generalisation to more than two generators at the end of this paper. 2 Lattice of Young subgroups In the present section we shall prove Theorem 2. Consider the set P of all (set) partitions of f, 2,..., ng. If = f,... k g is a partition with k parts then, as usual, we dene the Young subgroup Y ( ) as the subgroup of S n consisting of all elements which map each of the parts i into itself. The set of Young subgroups of S n is a lattice, and we dene an ordering on P by writing 0 when Y ( ) Y ( 0 ). Under this ordering the greatest the electronic journal of combinatorics (2005), #R00 2
element of P is := ffg, f2g,..., fngg and the least element is 0 := ff, 2,..., ngg. Consider the Möbius function µ on P (see, for example, [8, Section 3.7]), and write µ( ) in place of µ( 0, ). By denition, µ( 0 ) = and P 0 µ( 0 ) = 0 for all > 0. Example 3.0.4 of [8] shows that µ( ) = ( ) k+ (k )! whenever has k parts. Now let f A ( ) (respectively f S ( )) be the number of pairs (x, y) of elements from A n (respectively, S n ) such that the parts of are the orbits of the group hx, yi generated by x and y. Similarly let g A ( ) and g S ( ), respectively, be the number of pairs for which the parts of are invariant under hx, yi; that is, for which x, y 2 Y ( ). Every Young subgroup Y ( ) except for the trivial group Y ( ) contains an odd permutation and so we have g A ( ) = 4 jy ( )j2 = 4 g S( ) for 6= and g A ( ) = g S ( ) =. We also have g A ( ) = X 0 f A ( 0 ) and g S ( ) = X 0 f S ( 0 ). Since µ( ) = ( ) n+ (n )!, the Möbius inversion formula [8, Propositon 3.7.] now shows that s n = f S ( 0 ) = X µ( )g S ( ) = 4 X µ( )g A ( ) 3µ( ) = 4f A ( 0 ) 3µ( ) = 4a n + ( ) n 3(n )! as claimed. 3 Asymptotic expansion It remains to prove Theorem and obtain an asymptotic expansion for t n = s n /(n!) 2. It is possible that this can be done with a careful analysis of the series f S ( 0 ) = P µ( )g S( ) since the size of the terms decreases rapidly: the largest are those when has the shapes [, n ], [2, n 2], [ 2, n 2],...; but the argument seems to require considerable care. We therefore approach the problem from a di erent direction using a generating function for t n which was derived in [5]. Consider the formal power series E(X) := n!x n and T (X) := n=0 n!t n X n. n= Then Section 2 of [5] shows that E(X) = exp T (X) and so T (X) = log E(X). () We shall apply a theorem of Bender [2, Theorem 2] (quoted in [7, Theorem 7.3]): the electronic journal of combinatorics (2005), #R00 3
Theorem 4 (E.A. Bender) Consider formal power series A(X) := P n= a nx n and F (X, Y ) where F (X, Y ) is analytic in some neighbourhood of (0, 0). Dene B(X) := F (X, A(X)) = P n=0 b nx n, say. Let D(X) := F Y (X, A(X)) = P n=0 d nx n, say, where F Y (X, Y ) is the partial derivative of F with respect to Y. Now, suppose that all a n 6= 0 and that for some integer r we have: (i) a n /a n! 0 as n! ; and (ii) P n r k=r ja ka n k j = O(a n r ) as n!. Then Xr b n = d k a n k + O(a n r ). k=0 Using the identity () we take A(X) = E(X), F (X, Y ) = log( + Y ), D(X) = /E(X) and B(X) = T (X) in Benders theorem. Then condition (i) is clearly satised and (ii) holds for every integer r since for n > 2r Xn r k!(n k)! 2r!(n r)! + (n 2r )(r + )!(n r )! < f2r! + (r + )!g (n r)!. k=r Thus we get and hence Xr n!t n = d k (n k)! + O((n r)!) k=0 t n = + Xr k= d k [n] k + O(n r ) where [n] k = n(n )...(n k + ). The Stirling numbers S(m, k) of the second kind satisfy the identity S(m, k)x m X k = ( X)( 2X)...( kx) m=k where the series converges for jxj < (see [8, page 34]). Thus for n k > 0 we have [n] k = n k ( /n)( 2/n)...( (k )/n) = X m=k S(m, k ) n. m+ This shows that t n has an asymptotic expansion of the form + P k= c kn k where c k = P k i= S(k, i)d i+ since S(m, 0) = 0 for m = 0. To compute the numerical values of the coe cients we can use a computer algebra system such as Maple to obtain and then D(X) = /E(X) = X X 2 3X 3 3X 4 7 5 46 6 3447X 7... t n» [n] [n] 2 3 [n] 3 3 [n] 4 7 [n] 5 46 [n] 6...» n n 2 4 n 3 23 n 4 7 n 5 542 n 6.... the electronic journal of combinatorics (2005), #R00 4
4 Generalization to more than two generators In view of the theorem of M. Hall mentioned in Remark 3 there is some interest in extending the analysis for t n to the case of r generators where r 2. Let t n (r) be the probability that r elements of S n generate a transitive group (so t n = t n (2)). A simple argument similar to that in Section 2 of [5] shows that the generating function T r (X) := P n= (n!)r t n (r)x n satises the equation T r (X) = log E r (X) where E r (X) := P n=0 (n!)r X n. Now, following the same path as we did in the previous section, an application of Benders theorem leads to t n (r)» + k= d k (r) ([n] k ) r where the coe cients d k (r) are given by /E r (X) = P k=0 d k(r)x k. For example, we nd that /E 3 (X) = X 3X 2 29X 3 499X 4 30 5... so References t n (3)» [n] 2 3 [n] 2 29 2 [n] 2 499 3 [n] 2 4» n 2 3 n 4 6 n 5.... 30 [n] 2... 5 [] L. Babai, The probability of generating the symmetric group, J. Combin. Theory (Ser. A) 52 (989) 4853. [2] E.A. Bender, An asymptotic expansion for some coe cients of some formal power series, J. London Math. Soc. 9 (975) 45458. [3] J. Bovey and A. Williamson, The probability of generating the symmetric group, Bull. London Math. Soc. 0 (978) 996. [4] L. Comtet, Advanced Combinatorics, Reidel, 974. [5] J.D. Dixon, The probability of generating the symmetric group, Math. Z. 0 (969) 99205. [6] M. Hall, Jr., Subgroups of nite index in free groups, Canad. J. Math. (949) 87 90. the electronic journal of combinatorics (2005), #R00 5
[7] A.M. Odlyzko, Asymptotic enumeration methods, in Handbook of Combinatorics (Vol. II) (eds.: R.L. Graham, M. Grötschel and L. Lovász), M.I.T. Press and North- Holland, 995 (pp. 063229). [8] R.P. Stanley, Enumerative Combinatorics (Vol. ), Wadsworth & Brooks/Cole, 986 (reprinted Cambridge Univ. Press, 997). the electronic journal of combinatorics (2005), #R00 6