aa r X i v : . [ m a t h . G R ] D ec Almost all permutations power to a prime length cycle
W. R. Unger ∗ School of Mathematics and Statistics,University of Sydney,Sydney, AustraliaDecember 3, 2019
Abstract
We show that almost all permutations have some power that is a cycle of primelength. The proof includes a theorem giving a strong upper bound on the proportionof elements of the symmetric group having no cycles with length in a given set.
In this note we prove that almost all permutations have some power that is a cycle ofprime length. Equivalently, that almost all permutations have a cycle of prime length, p say, where p does not divide the length of any other cycle of the permutation.This question arises from C. Jordan’s results of the 1870s, where he showed that if G isa primitive permutation group of degree n , and G contains a p -cycle, where p ≤ n − p is prime, then G contains the alternating group of degree n . Jordan’s result has beengeneralised several times, see for instance [4] for a generalisation to non-prime cycles.When computing with permutation groups such results are used as a test for a primitivepermutation group to contain the alternating group, and so merit special treatment. It isof interest to know that as the degree grows, it becomes easier to find such elements byrandom search in the giant permutation groups A n and S n .Thee next section gives terminology and notation. Following that we give a simple butquite strong upper bound on the proportion of permutations in S n having no cycle lengthsin some fixed set. The last section proves the result of the title, and some easy corollaries. We use S n to denote the symmetric group on the set { , . . . , n } , and A n to denote thealternating subgroup of S n . ∗ The author acknowledges the support of Australian Research Council grant DP160104626 P is some property of permutations we say that P holds for almost all permuta-tions when lim n →∞ p n = 1, where p n is the proportion of elements of S n having property P . When considering properties of a permutation σ , we will use n to denote the degree ofthe containing group S n .When we refer to the cycles of a permutation, we refer to the factors in the productof disjoint cycles decomposition of the permutation. We will say that a permutation is a cycle when the cycles of the permutation consist of one cycle of length greater than 1, andsome number (perhaps zero) of fixed points. When the non-trivial cycle has length ℓ , thepermutation is called an ℓ -cycle.A summation with index p indicates a sum over all primes p in the given range.We use the Bachmann-Landau symbols O () and o () (big-O and little-o) as defined in[3, § e . In this section we fix a degree n and a set C of possible cycle lengths, and consider theproportion of elements of S n having no cycles with length in C . We will prove an upperbound on this proportion.The parameter µ = P k ∈ C /k is important here. An upper bound of 1 /µ for theproportion was given in [1, Theorem VI], while [5, p39], gives an upper bound of exp( γ − µ )(1 + 1 /n ). The following result is a very small improvement on this last. The articlereferred to in [5] was inaccessible to the current author, which led to the proof below, andthe simplicity of the bound is striking. Theorem 1
Let n be a positive integer, let C be any subset of { , , . . . , n } , and put µ = P k ∈ C /k . Then the proportion of elements of S n having no cycle with length in C isless than exp( γ − µ ) , where γ is Euler’s constant. Proof:
Define, for k = 1 , , . . . n , a k = 0 for k ∈ C , with a k = 1 otherwise. Set A ( z ) = n X k =1 a k k z k , and F ( z ) = exp( A ( z )) . This F ( z ) is an entire function and we take the Taylor series for F ( z ) about 0 to be F ( z ) = ∞ X k =0 p k z k . We can identify p k , for all k ≥
0, as the proportion of elements of S k having all cycles withlength ≤ n , and having no cycles with length in C . This may be seen by [2, Proposition II.4],or by applying the methods of [6, Chapter 3]. With this notation we are trying to prove that p n < exp( γ − µ ). efine E ( n ), for n ≥
1, by n X k =1 k = log n + γ + E ( n ) . It is well known that 0 < E ( n ) < / n , and so exp( E ( n )) < /n . Now A (1) = n X k =1 k − µ = log n + γ − µ + E ( n ) . Using these relations we get F (1) < n exp( γ − µ ) (cid:18) n (cid:19) = ( n + 1) exp( γ − µ ) . Now consider the derivative of F ( z ), F ′ ( z ) = A ′ ( z ) F ( z ) . (1)Observe that A ′ ( z ) = P nk =1 a k z k − and equate the coefficients of z n − on either side of (1): np n = n − X k =0 a n − k p k ≤ n − X k =0 p k ≤ F (1) − p n . It follows that ( n + 1) p n ≤ F (1) < ( n + 1) exp( γ − µ ) , which proves the result. If µ > µ − γ ) > exp( µ − ≥ µ , so exp( γ − µ ) < /µ . This shows that thebound of Theorem 1 is always stronger than the bound of [1, Theorem VI].Consider the sequence of examples C n = { , , . . . n − } . The elements of S n with nocycles having length in C n are the n -cycles. As n → ∞ , the upper bound of Theorem 1is asymptotic to 1 /n , which is the exact proportion of n -cycles in S n . For these examplesthe upper bound looks very good.When C is a very small set, [5, Theorem B] implies that the proportion of permutationswith no cycle lengths in C will be close to e − µ . The extreme examples here are C = { ℓ } forsome fixed ℓ . It is known that the proportion of elements without an ℓ -cycle is asymptoticto e − /ℓ as n → ∞ , while our upper bound is e γ − /ℓ , so the e γ = 1 . . . . multiplier lookslarger than necessary. Indeed the upper bound given above is useless (that is >
1) for ℓ ≥ C where n/ | C | ≈ log n and e − µ ≈ n/ log n , a situation between the extreme examples above.3 The Main Theorem
We will now prove the title statement. We first show that almost all permutations have acycle of prime length where the prime is not too small. As we are looking at properties of S n for large n , we will assume that n > f ( n ) = (log n ) . A different f could be used. For theproof of the next lemma, f must not grow too quickly as n → ∞ (log f ( n ) = o (log n )),and, for the proof of the Main Theorem, f must not grow too slowly (log n = o ( f ( n ))). Lemma 4.1
For almost all permutations σ , there exists a prime p with p > f ( n ) , suchthat σ has a cycle of length p . Proof:
A theorem of Mertens [3, Theorem 427] tells us that there exists a constant M so that X p ≤ x p = log log x + M + o (1) (2)as x → ∞ . Now considering primes beteween f ( n ) and n we have X f ( n )
Theorem 2
For almost all permutations σ , there is some power of σ that is a cycle ofprime length. Proof:
We continue to use f ( n ) = (log n ) , and n >
10 so f ( n ) > T n ⊆ S n be the set of permutations σ such that there exists a prime p , with p > f ( n ),such that σ has a cycle of length p . Lemma 4.1 implieslim n →∞ | T n | n ! = 1 . (4)Fix a prime p > f ( n ), and consider U n,p ⊆ S n consisting of the permutations that have acycle of length p , and have another cycle of length divisible by p . Then U n,p ⊆ T n , and note that p > n/ U n,p is empty.We find an upper bound for | U n,p | by counting all structures consisting of a permutation from S n having a first distinguished cycle of length p , and a second distinguished cycle with lengthdivisible by p . This will be an upper bound for | U n,p | as every element of U n,p gives rise to uch a structure, some giving more than one structure by distinguishing different cycles. In thefollowing, the right hand side of (5) counts the structures mentioned, where the length of thesecond distinguished cycle is kp , and p > p > | U n,p | ≤ ( p − np ! ⌊ n/p ⌋− X k =1 n − pkp ! ( kp − n − p − kp )! (5)= n ! p ⌊ n/p ⌋− X k =1 k . | U n,p | n ! ≤ p ⌊ n/p ⌋− X k =1 k < p (cid:18) np (cid:19) < log np . (6)Now we bound the following sum over primes greater than x > X p>x p < ∞ X k = ⌈ x ⌉ k < x + Z ∞ x dtt = 1 x + 1 x < x . (7)It can be shown that P p>x /p is asymptotic to 1 / ( x log x ) as x → ∞ , so this bound is far frombest possible, but it is easy to derive and is enough for what follows.Let U n = S p>f ( n ) U n,p . We have U n ⊆ T n and, using (6) and (7), | U n | n ! ≤ X p>f ( n ) | U n,p | n ! < X p>f ( n ) log np < nf ( n ) = 2log n . (8)We deduce from (8) that lim n →∞ | U n | n ! = 0 . (9)Put V n = T n \ U n . Let σ ∈ T n , so σ has a cycle of prime length p with p > f ( n ). If no powerof σ is a p -cycle then σ has some other cycle with length divisible by p , so σ ∈ U n . It follows thatevery element of V n has the property that some power is a cycle of prime length. Furthermore,from (4) and (9), lim n →∞ | V n | n ! = lim n →∞ | T n | − | U n | n ! = 1 . Relating this to C. Jordan’s result, we have:
Corollary 3
Almost all permutations have a power that is a p -cycle with p prime and p ≤ n − . Proof:
The proportion of elements in S n having a cycle of length n , n −
1, or n − O (1 /n ).So almost all permutations have no cycle of length > n −
3, thus almost all permutations powerto a prime length cycle with cycle length ≤ n − The following shows that the results above also hold for almost all even permutations,so apply to algorithms recognising the alternating group as well as the symmetric group.5 orollary 4
Let q n be the proportion of elements of the alternating group A n that have apower that is a p -cycle with p prime and p ≤ n − . Then lim n →∞ q n = 1 . Proof:
Let r n be the proportion of elements of S n that have the given property. By Corollary 3,lim n →∞ r n = 1. As A n has index 2 in S n , we have 0 ≤ − q n ≤ − r n ), and the result followsby taking the limit n → ∞ . Finally, using the results above we may prove a sharper version of the main theorem.
Theorem 5
The proportion of elements in S n that do not power to a cycle of prime lengthis at most O (log log n/ log n ) . Proof:
We use notation from the proof of Theorem 2. It follows from Theorem 1 and equation (3)that | S n \ T n | n ! = O (cid:18) log log n log n (cid:19) . The proof of Theorem 2 gives | U n | n ! = O (cid:18) n (cid:19) = o (cid:18) log log n log n (cid:19) . Thus we find | S n \ V n | n ! = O (cid:18) log log n log n (cid:19) . Since all elements of V n power to a prime cycle this proves the result. References [1] P. Erd˝os and P. Tur´an. On some problems of a statistical group-theory II.
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