THE ORDER OF ELEMENTS IN SYLOW p-subgroups OF THE SYMMETRIC GROUP

THE ORDER OF ELEMENTS IN SYLOW p-subgroups OF THE SYMMETRIC GROUP JAN-CHRISTOPH SCHLAGE-PUCHTA Abstract. Define a random variable ξ n by choosing a conjugacy class C of the Sylow p-subgroup of S p n by random, and let ξ n be the logarithm of the order of an element in C. We show that ξ n has bounded variance and mean order log n + O, which differs greatly from the average order of elements chosen with equal probability. In a sequence of papers [2] [8], P. Erdős and P. Turán developed a statistical theory of the symmetric group S n on n letters. Here a statistical theory describes properties of almost all elements of a sequence of groups, or statements on densities of certain elements or other associated structures, e.g. characters. P. Turán posed the problem of developing a statistical theory for subgroups of S n, in particular for Sylow subgroups. This was done by P. P. Pálfy and M. Szalay [9], [0], []. In particular, in the second of these papers they proved the following: Theorem. Define the random variable ζ n as follows: Choose an element g of P n, the Sylow p- subgroup of S p n, by random, and define ζ n = og, where og is the order of g. Then ζ n has bounded variance, and there are positive constants c, c 2, such that c n < M n < c 2 n, where M n denotes the mean value of ζ n. Recently M. Abért and B. Virág [] showed, that the mean value is in fact asymptotically equal to c p n, where c p is given as solution of the equation cp /c p log cp + log c p = log /p. In this note the analogous question is studied, however, we choose a conjugacy class instead of an element. Define h n k to be the number of conjugacy classes of P n consisting of elements of order p k. Our approach is based on the following Theorem, the second half of which was proven by P. P. Pálfy and M. Szalay [9]. Theorem 2. For n, k we have the recurrence relation h n+ k = p hn k p h n k + h n k + p h n k. In particular, for, the total number of conjugacy classes, we have = [p γ+ p Pn γ] δ p, where 0 < γ < is a constant depending on p, and δ p = if p = 2, and 0 otherwise. Using this recurrence relation we will prove the following theorem. Define ξ n to be oc, where C is a conjugacy class chosen among the conjugacy classes of P n at random. This research was supported through a European Community Marie Curie Fellowship. Disclaimer: The author is solely responsible for information communicated and the European Community is not responsible for any view or result expressed. Keywords: Symmetric group, Sylow subgroups, statistical group theory MSC-Index 20P05

2 JAN-CHRISTOPH SCHLAGE-PUCHTA Theorem 3. The random variable ξ n has mean value log n + O and has bounded variance. Comparing Theorem and Theorem 3, one sees that there are few conjugacy classes of order c p n, containing almost all elements, whereas almost all conjugacy classes are of order log n but contain a neglectable proportion of all elements. It is interesting to note that P. Erdős and P. Turán discovered the converse phenomenon in the symmetric group S n : Almost all elements of S n have order e /2+o log2 n, whereas almost all conjugacy classes have order e c+o n, where c is some positive constant. To prove Theorem 2, note that P n+ = P n C p, where C p is the cyclic group of order p. To count the conjugacy classes in P n+, we distinguish three cases: Conjugacy classes, which are contained in the base group P n P n, and which are embedded in the diagonal of this product, conjugacy classes which are non-diagonally contained in the base group, and conjugacy classes, which are not contained in the base group. Clearly, these cases are disjoint and cover all possibilities. The conjugacy classes of the first kind correspond to conjugacy classes of P n, and the action of C p leaves these conjugacy classes invariant, hence, there are h n k conjugacy classes of elements of order p k of the first kind. There are h n k p h n k conjugacy classes in P n P n which do not contain diagonal elements, and the action of C p does not leave any such class invariant, hence, there are p h nk p h n k conjugacy classes of the second kind. To count conjugacy classes of the third kind, write an element of P n+ as f, σ, where f : {,..., p} P n, and σ C p. Conjugation cannot change σ, hence, the total number of conjugacy classes of the third kind equals p times the number of conjugacy classes of elements having σ = 2. For an element x P n and an integer i p, define the function g by gi = x, gj = for j i. Then we find f, σ g,id = h, σ, where fj, j i, i + hj = xfi, j = i. fi + x, j = i + In particular, every element of the third kind is conjugated to an element f, σ with fi = for all i. Moreover, choosing gi = x for all i {,..., p}, we find that conjugacy classes of the third kind correspond to conjugacy classes of P n. Finally, the order of f, σ in P n+ equals p times the order of p i= fi in P n, hence, there are h n k conjugacy classes of elements of order k in P n+ of the third kind. Adding up the contribution of these three classes yields the recurrence relation. The formula for follows from this by induction. A different proof of this formula counting characters was given by P. P. Pálfy and M. Szalay [9]. We begin the proof of Theorem 3 with two simple remarks: First it is obvious from the definition that for fixed n, h n k is increasing with k. Further note that h n+ > h n + p, thus, for n, h n tends to. We will write α n k := h nk and β n k := h p nk hn k. h n k h n k We first collect some properties of α and β. Proposition 4. Define α and β as above, and assume that k n. We have β n k = + p h n k p + pp h n k 2h n k p + p h n k p + pp h n k h n k p 2 2 We have log β n k h n k p. 3 We have α n k = β k k pn k β k+ kβ k+ k + pn k β n k β n n.

THE ORDER OF ELEMENTS IN SYLOW p-subgroups OF THE SYMMETRIC GROUP 3 4 We have β n n = p n h nn. Furthermore, for p = 2, we have hn 2 n 2 β n n = + O. 5 As k tends to, the asymptotic log α n k p n k p k h k k holds. Before we prove this, we indicate how Theorem 3 follows from Proposition 4. Obviously, for fixed n, α n k is increasing with k, α n = h 0, and α nn nn + =, hence, for n there is a unique integer k 0 n with the property α n k 0 /2 < α n k o +. By 4.5 we get p k0 n k0 log 2 log α n k 0 p h k0 k 0 Taking logarithms again and using the value for h k0 k 0 given by Theorem, we get log log 2 + o n k 0 + k 0 logp γ + p γ pk0 p. From this inequality it is easy to see that k 0 log n + O. In the same way, starting from the inequality α n k 0 + > /2, we get k 0 log n + O, thus k 0 = log n + O. We want to show that k 0 is close to the mean value of ξ n, and that the variance of ξ n is bounded. Both statements follow at once, if we can show that the mean value of ξ n k 0 2 is bounded. We estimate this value as follows. Eξ n k 0 2 = h n k h n k k k 0 2 k=0 = α n k + α n kk k 0 2 k=0 k 0 k=0 α n k + k k 0 2 + k=k 0+ α n kk k 0 2. To estimate these sums, we note first that α n k 2 > α n k for k greater than some absolute constant, for by 4.5 we have log α n k log α n k p h k k = + o p h k k, and since the right hand side tends to, it will eventually become 2. Using this fact together with the definition of k 0, we obtain the estimates α n k 0 d < 2 2d, α n k 0 + d > 2d /2 = log 2 2 d + O2 2d. Hence, both sums in the estimate for Eξ n k 0 2 can be estimated by converging sums of the form d 2 2 d, and we obtain Eξ n k 0 2, which proves Theorem 3. Therefore, it remains to prove Proposition 4.. In the definition of β n k, replace the values of h in the first fraction with the recurrence relation to obtain β n k = h n k p p + p h n k + pp h n k 2 hn k h n k p + p h n k + pp h n k h n k = + p h n k p + pp h n k 2h n k p + p h n k p + pp h n k h n k p. 2. We use the 2. To give an upper bound, we estimate the denominator by. In the numerator we replace h n k 2 by h n k, which increases the fraction, too. Thus we obtain the upper bound

4 JAN-CHRISTOPH SCHLAGE-PUCHTA + p 2 h n k p. In the same way we obtain a lower bound, and both bounds together yield a bound for log β n k. 3. This statement follows from the definition of α by a simple computation: p h n k hn k = β n k h n k h n k = β n kβ n k p β k k pn k α n k = h nk h n k h n k h n k + hnn = β n kβ n k p β k k pn k βn k + β n k + p β k+ k + pn k β n n β n n p β n n. Now rearranging terms according to the exponent proves our claim. 4. To compute β n n, it suffices to compute h n n. We have h n n = p hn n p h n n + h n n + p h n n 2 = p h n n h n n 2. From this we deduce by induction that h n n = p n h h 0. But h h 0 = p, since P is cyclic of order p, and we get Now assume that p = 2. Then we have β n n = h nn = p n. h n n 2 = 2 hn n 2 2 h n n 2 + h n n 2 + h n n 3 = 2 hn n 2 hn n + h n n + h n n 3 = h n n h n n h n n 2 + h n n 3 = 2h n n + h n n 3, which implies h n n 2 h n n ; using this estimate to bound h n n h n n 3, we obtain From this we obtain Now we can compute β n n : β n n = h nn 2 h n n h n n 2 = h n n + Oh n 2 n 2. hn n h n n 2 2 = h nn h n n + Oh n 2 n 2 hn n h n n = h n n hn 2 n 2 2 + O + h n n + O = + 2h nn h n n 2 hn 2 n 2 + O h n n hn 2 n 2 = + O, where in the last line we used the recurrence relation in the form = 2 h n n 2 + Oh n n. 2 h n n 2

THE ORDER OF ELEMENTS IN SYLOW p-subgroups OF THE SYMMETRIC GROUP 5 5. By 4.3 we can express α in terms of β, and then we will use 4.2 and 4.4 to estimate the resulting expression. We have log α n k = p n k log β k k + p n k log β k+ k + log β k+ k + + + log β n k +... + log β n n 3 n k p k = p + O n p n k p 2k k k k 2 κ + O p n κ log β κ ν. h k k κ=k+ The first error term is of lesser order than the main term, provided that k. To bound the second error term, we first consider the sum over the range k + 2 κ n. Using 4.2, we find κ κ p n κ log β κ ν p n κ h κ ν p κ=k+2 ν=k κ=k+2 κ=k+2 ν=k p n κ κ k + h κ k p n k h k+ k. As can be seen from 4.2 and 4.3, we have h n n 2 as n, hence, the last quantity is p of the same order of magnitude as n k h k+ k+, which is negligible compared to the main term provided that k. To bound the term coming from κ = k +, we distinguish the cases p = 2 and p 3. In the latter case, 4.2 implies that log β k+ k + log β k+ k + h k k p h k k 2. As in the proof of 4.4 we find that h k k h k k = p k = oh k k, hence, for k, h k k 2 = oh k k, and the contribution of this term is negligible. If on the other hand p = 2, we use 4.4 to deduce log β k+ k + log β k+ k + h k+ k + + O hk k. h k+ k + Since h k+ k + h k k 2 h k k 4, we obtain log β k+ k + log β k+ k + h k k 2 + h k k 3/2 = oh kk. Hence, the main term in 3 dominates the error terms as k, and 4.5 is proven. ν=k References [] M Abért, B. Virág, Dimension and randomness in groups acting on rooted trees, to appear. [2] P. Erdős, P. Turán, On some problems of a statistical group theory. I. Z. Wahrscheinlichkeitstheorie und Verw. Gebiete 4 965, 75 86. [3] P. Erdős, P. Turán, On some problems of a statistical group theory. II. Acta Math. Acad. Sci. Hungar. 8 967, 5 63. [4] P. Erdős, P. Turán, On some problems of a statistical group theory. III. Acta Math. Acad. Sci. Hungar. 8 967, 309 320. [5] P. Erdős, P. Turán, On some problems of a statistical group theory. IV. Acta Math. Acad. Sci. Hungar 9 968, 43 435. [6] P. Erdős, P. Turán, On some problems of a statistical group theory. V. Period. Math. Hungar. 97, 5 3. [7] P. Erdős, P. Turán, On some problems of a statistical group theory. VI. J. Indian Math. Soc. 34 97, 75 92. [8] P. Erdős, P. Turán, On some problems of a statistical group theory. VII. Period. Math. Hungar. 2 972, 49 63. [9] P. P. Pálfy, M. Szalay, The distribution of the character degrees of the symmetric p-groups, Acta Math. Hungar. 4 983, 37 50. [0] P. P. Pálfy, M. Szalay, On a problem of P. Turán concerning Sylow subgroups, in: Studies in pure mathematics to the memory of P. Turán, 53 542, Birkhäuser, Basel, 983. [] P. P. Pálfy, M. Szalay, Further probabilistic results on the symmetric p-groups, Acta Math. Hugar. 53 989, 73 95.

6 JAN-CHRISTOPH SCHLAGE-PUCHTA J.-C. Schlage-Puchta Mathematisches Institut Eckerstr. 7904 Freiburg Germany