Derangements and p -elements in permutation groups Peter J. Cameron p.j.cameron@qmul.ac.uk Groups and their Applications Manchester, 14 February 2007
In the beginning . . . A derangement is a permutation with no fixed points.
In the beginning . . . A derangement is a permutation with no fixed points. 1. The proportion of derangements in the symmetric group S n is approximately 1/e. More precisely, the number of derangements in S n is the nearest integer to n !/e.
In the beginning . . . A derangement is a permutation with no fixed points. 1. The proportion of derangements in the symmetric group S n is approximately 1/e. More precisely, the number of derangements in S n is the nearest integer to n !/e. 2. (Jordan) A transitive permutation group of degree n > 1 contains a derangement. In fact (Cameron and Cohen) the proportion of derangements in a transitive group G is at least 1/ n . Equality holds if and only if G is sharply 2-transitive, and hence is the affine group { x �→ ax + b : a , b ∈ F , a � = 0 } over a nearfield F .
In the beginning . . . A derangement is a permutation with no fixed points. 1. The proportion of derangements in the symmetric group S n is approximately 1/e. More precisely, the number of derangements in S n is the nearest integer to n !/e. 2. (Jordan) A transitive permutation group of degree n > 1 contains a derangement. In fact (Cameron and Cohen) the proportion of derangements in a transitive group G is at least 1/ n . Equality holds if and only if G is sharply 2-transitive, and hence is the affine group { x �→ ax + b : a , b ∈ F , a � = 0 } over a nearfield F . The finite nearfields were determined by Zassenhaus. They all have prime power order.
Why do we care? The presence of derangements in a permutation group has important implications in number theory and topology. See Serre’s beautiful paper “On a theorem of Jordan”, in Bull. Amer. Math. Soc. 40 (2003), 429–440.
Why do we care? The presence of derangements in a permutation group has important implications in number theory and topology. See Serre’s beautiful paper “On a theorem of Jordan”, in Bull. Amer. Math. Soc. 40 (2003), 429–440. ◮ Let f be an integer polynomial of degree n > 1, irreducible over Q . Then f has no roots mod p for infinitely many primes p (indeed, for at least a proportion 1/ n of all primes).
Why do we care? The presence of derangements in a permutation group has important implications in number theory and topology. See Serre’s beautiful paper “On a theorem of Jordan”, in Bull. Amer. Math. Soc. 40 (2003), 429–440. ◮ Let f be an integer polynomial of degree n > 1, irreducible over Q . Then f has no roots mod p for infinitely many primes p (indeed, for at least a proportion 1/ n of all primes). ◮ Let π : T → S be a covering map of degree n ≥ 2, and suppose that T is arcwise connected but not empty. Then there is a continuous closed curve in S which cannot be lifted to T .
Why do we care? The presence of derangements in a permutation group has important implications in number theory and topology. See Serre’s beautiful paper “On a theorem of Jordan”, in Bull. Amer. Math. Soc. 40 (2003), 429–440. ◮ Let f be an integer polynomial of degree n > 1, irreducible over Q . Then f has no roots mod p for infinitely many primes p (indeed, for at least a proportion 1/ n of all primes). ◮ Let π : T → S be a covering map of degree n ≥ 2, and suppose that T is arcwise connected but not empty. Then there is a continuous closed curve in S which cannot be lifted to T . ◮ The Fein–Kantor–Schacher theorem (see later) is equivalent to the statement that the relative Brauer group of any finite extension of global fields is infinite. (The proof uses the classification of finite simple groups.)
So find one then . . . A subgroup of S n can be generated by at most n − 1 elements, and such a generating set can be found efficiently (with polynomial delay) (Jerrum). So such a subgroup can be described by O ( n 2 log n ) bits.
So find one then . . . A subgroup of S n can be generated by at most n − 1 elements, and such a generating set can be found efficiently (with polynomial delay) (Jerrum). So such a subgroup can be described by O ( n 2 log n ) bits. Problem: Given a subgroup of S n , does it contain a derangement?
So find one then . . . A subgroup of S n can be generated by at most n − 1 elements, and such a generating set can be found efficiently (with polynomial delay) (Jerrum). So such a subgroup can be described by O ( n 2 log n ) bits. Problem: Given a subgroup of S n , does it contain a derangement? This problem is NP-complete, even for elementary abelian 2-groups. There is a simple reduction from the known NP-complete problem 3-SAT. Indeed, the argument shows that counting the derangements in a subgroup of S n is #P-complete.
and in a transitive group . . . Given generators for a subgroup G of S n , we can check quickly whether H is transitive. If it is (and n > 1), then we know that G contains a derangement.
and in a transitive group . . . Given generators for a subgroup G of S n , we can check quickly whether H is transitive. If it is (and n > 1), then we know that G contains a derangement. Problem: Suppose that G is transitive. Find a derangement in G .
and in a transitive group . . . Given generators for a subgroup G of S n , we can check quickly whether H is transitive. If it is (and n > 1), then we know that G contains a derangement. Problem: Suppose that G is transitive. Find a derangement in G . There is an efficient randomised algorithm for this problem. Since at least a fraction 1/ n of the elements of G are derangements, we can do this by random search: in n trials we will have a better-than-even chance of finding one, and in n 2 trials we will fail with exponentially small probability.
and in a transitive group . . . Given generators for a subgroup G of S n , we can check quickly whether H is transitive. If it is (and n > 1), then we know that G contains a derangement. Problem: Suppose that G is transitive. Find a derangement in G . There is an efficient randomised algorithm for this problem. Since at least a fraction 1/ n of the elements of G are derangements, we can do this by random search: in n trials we will have a better-than-even chance of finding one, and in n 2 trials we will fail with exponentially small probability. Problem: Can it be done deterministically?
and in a transitive group . . . Given generators for a subgroup G of S n , we can check quickly whether H is transitive. If it is (and n > 1), then we know that G contains a derangement. Problem: Suppose that G is transitive. Find a derangement in G . There is an efficient randomised algorithm for this problem. Since at least a fraction 1/ n of the elements of G are derangements, we can do this by random search: in n trials we will have a better-than-even chance of finding one, and in n 2 trials we will fail with exponentially small probability. Problem: Can it be done deterministically? The answer is likely to be “yes” – this is theoretically interesting but the randomised algorithm will almost certainly be more efficient!
Groups with many derangements Although the lower bound | G | / n for the number of derangements in a transitive group G is attained (by sharply 2-transitive groups), there are many groups with a higher proportion of derangements. For example, if G is regular, than all but one of its elements are derangements!
Groups with many derangements Although the lower bound | G | / n for the number of derangements in a transitive group G is attained (by sharply 2-transitive groups), there are many groups with a higher proportion of derangements. For example, if G is regular, than all but one of its elements are derangements! The argument of Cameron and Cohen gives a lower bound of about ( r − 1 ) / n for the proportion of derangements in a transitive group G , where r is the permutation rank (the number of orbits of G on ordered pairs).
Groups with many derangements Although the lower bound | G | / n for the number of derangements in a transitive group G is attained (by sharply 2-transitive groups), there are many groups with a higher proportion of derangements. For example, if G is regular, than all but one of its elements are derangements! The argument of Cameron and Cohen gives a lower bound of about ( r − 1 ) / n for the proportion of derangements in a transitive group G , where r is the permutation rank (the number of orbits of G on ordered pairs). Can anything be said about families of (say, primitive) groups in which the proportion of derangements is bounded away from zero?
An example Example: There is a constant α k > 0 so that the proportion of derangements in S n acting on k -sets tends to α k as k → ∞ . (For example, α 1 = e − 1 = 0.3679 . . . , while α 2 = 2e − 3/2 = 0.4463 . . . . There is a formula for α k as a sum over subsets of the partitions of k . But most of the terms cancel, so I suspect there is a much simpler formula!
An example Example: There is a constant α k > 0 so that the proportion of derangements in S n acting on k -sets tends to α k as k → ∞ . (For example, α 1 = e − 1 = 0.3679 . . . , while α 2 = 2e − 3/2 = 0.4463 . . . . There is a formula for α k as a sum over subsets of the partitions of k . But most of the terms cancel, so I suspect there is a much simpler formula! Problem: Is it true that α k → 1 monotonically as k → ∞ ?
Recommend
More recommend
