Some Algorithmic questions in Finite Group Theory

Transcription

1 Some Algorithmic questions in Finite Group Theory V Arvind Institute of Mathematical Sciences, Chennai India arvind@imsc.res.in June 30, 2015

2 Plan of Talk Permutation groups: background and some basic algorithms.

3 Plan of Talk Permutation groups: background and some basic algorithms. Fixed point-free elements of a permutation group. Existence and computation.

4 Plan of Talk Permutation groups: background and some basic algorithms. Fixed point-free elements of a permutation group. Existence and computation. Computing small bases for permutation groups.

5 Plan of Talk Permutation groups: background and some basic algorithms. Fixed point-free elements of a permutation group. Existence and computation. Computing small bases for permutation groups. Random subproducts in groups.

6 What is in this talk for me? The group theory content is easy/elementary and is from a first course in algebra.

7 What is in this talk for me? The group theory content is easy/elementary and is from a first course in algebra. The algorithms are straightforward based on simple techniques taught in a first algorithms course.

8 What is in this talk for me? The group theory content is easy/elementary and is from a first course in algebra. The algorithms are straightforward based on simple techniques taught in a first algorithms course. What s new is probably the cocktail of group theory + algorithms.

9 What is in this talk for me? The group theory content is easy/elementary and is from a first course in algebra. The algorithms are straightforward based on simple techniques taught in a first algorithms course. What s new is probably the cocktail of group theory + algorithms. Teachers: Experimenting with similar cocktails in other courses like linear algebra can be interesting.

10 Permutation Groups: Definitions Let S n denote the group of all permutations on n elements, say {1, 2,..., n}. It is a group under permutation composition and has n! many elements.

11 Permutation Groups: Definitions Let S n denote the group of all permutations on n elements, say {1, 2,..., n}. It is a group under permutation composition and has n! many elements. A subgroup G of S n, denoted G S n, is a permutation group.

12 Permutation Groups: Definitions Let S n denote the group of all permutations on n elements, say {1, 2,..., n}. It is a group under permutation composition and has n! many elements. A subgroup G of S n, denoted G S n, is a permutation group. We can describe a permutation group G by listing down all its elements. A more compact description is to give a generating set for it.

13 Permutation Groups: Definitions Let S n denote the group of all permutations on n elements, say {1, 2,..., n}. It is a group under permutation composition and has n! many elements. A subgroup G of S n, denoted G S n, is a permutation group. We can describe a permutation group G by listing down all its elements. A more compact description is to give a generating set for it. The permutation group S, generated by a subset S S n of permutations, is the smallest subgroup of S n containing S.

14 Permutation Groups: Definitions Contd. Every finite group G has a generating set of size log 2 G. Because g 1 < g 1, g 2 <... < g 1, g 2,..., g k = G.

15 Permutation Groups: Definitions Contd. Every finite group G has a generating set of size log 2 G. Because g 1 < g 1, g 2 <... < g 1, g 2,..., g k = G. Each new generating element at least doubles the group size by Lagrange s theorem. Thus, k log 2 G.

16 Permutation Groups: Definitions Contd. Every finite group G has a generating set of size log 2 G. Because g 1 < g 1, g 2 <... < g 1, g 2,..., g k = G. Each new generating element at least doubles the group size by Lagrange s theorem. Thus, k log 2 G. So, giving a generating set for G is a succinct representation as it as algorithmic input.

17 Permutation Groups: Definitions Contd. For a permutation π S n, the image of a point i [n] is denoted by i π.

18 Permutation Groups: Definitions Contd. For a permutation π S n, the image of a point i [n] is denoted by i π. For a permutation π S n, a point i [n] is a fixed point if i π = i. Let fix(π) denote the number of points fixed by π.

19 Permutation Groups: Definitions Contd. For a permutation π S n, the image of a point i [n] is denoted by i π. For a permutation π S n, a point i [n] is a fixed point if i π = i. Let fix(π) denote the number of points fixed by π. A permutation group G S n partitions the domain [n] into orbits: i and j are in the same orbit precisely when i g = j for some g G.

20 Permutation Groups: Definitions Contd. For a permutation π S n, the image of a point i [n] is denoted by i π. For a permutation π S n, a point i [n] is a fixed point if i π = i. Let fix(π) denote the number of points fixed by π. A permutation group G S n partitions the domain [n] into orbits: i and j are in the same orbit precisely when i g = j for some g G. The group G is called transitive if there is exactly one orbit.

21 Permutation Group Algorithmics Each permutation π in S n can be represented as an n-tuple (1 π, 2 π,..., n π ) (or an array of n integers).

22 Permutation Group Algorithmics Each permutation π in S n can be represented as an n-tuple (1 π, 2 π,..., n π ) (or an array of n integers). Given π S n and a point i [n] we can compute i π in one step by looking up the i th entry of the array representing π. We can consider this a unit cost operation.

23 Permutation Group Algorithmics Each permutation π in S n can be represented as an n-tuple (1 π, 2 π,..., n π ) (or an array of n integers). Given π S n and a point i [n] we can compute i π in one step by looking up the i th entry of the array representing π. We can consider this a unit cost operation. Given two permutations π, ψ S n we can compute their product πψ by computing (i π ) ψ for each i. This operation costs n.

24 Permutation Group Algorithmics Each permutation π in S n can be represented as an n-tuple (1 π, 2 π,..., n π ) (or an array of n integers). Given π S n and a point i [n] we can compute i π in one step by looking up the i th entry of the array representing π. We can consider this a unit cost operation. Given two permutations π, ψ S n we can compute their product πψ by computing (i π ) ψ for each i. This operation costs n. What is an efficient algorithm on permutation groups?

25 Permutation Group Algorithmics Contd. Elements and subgroups of S n require encoding size n and n 2 log n respectively.

26 Permutation Group Algorithmics Contd. Elements and subgroups of S n require encoding size n and n 2 log n respectively. Roughly speaking, for algorithm dealing with permutation groups in S n :

27 Permutation Group Algorithmics Contd. Elements and subgroups of S n require encoding size n and n 2 log n respectively. Roughly speaking, for algorithm dealing with permutation groups in S n : Polynomial in n many operations = Efficient.

28 Permutation Group Algorithmics Contd. Elements and subgroups of S n require encoding size n and n 2 log n respectively. Roughly speaking, for algorithm dealing with permutation groups in S n : Polynomial in n many operations = Efficient. Exponential in n operations = Inefficient.

29 Computing Orbits Efficiently Given G S n by a generating set S, we can compute the orbit of any point i in (n S ) O(1) time.

30 Computing Orbits Efficiently Given G S n by a generating set S, we can compute the orbit of any point i in (n S ) O(1) time. Input: S = {g 1, g 2,..., g k } generators for G; O := {i}; while O changes do O := O {i g j i S, 1 j k}; endwhile

31 Computing Orbits Efficiently Given G S n by a generating set S, we can compute the orbit of any point i in (n S ) O(1) time. Input: S = {g 1, g 2,..., g k } generators for G; O := {i}; while O changes do O := O {i g j i S, 1 j k}; endwhile The loop runs for at most n steps. In the loop the number of operations is bounded by O(nk). Thus, O(n 2 k) operations in all.

32 The Membership Testing Problem Given as input π S n and a subgroup G = S S n test if π is in G. Express π in terms of the generators.

33 The Membership Testing Problem Given as input π S n and a subgroup G = S S n test if π is in G. Express π in terms of the generators. Writing π as a product of generators may be exponentially long!

34 The Membership Testing Problem Given as input π S n and a subgroup G = S S n test if π is in G. Express π in terms of the generators. Writing π as a product of generators may be exponentially long! Example Consider the cyclic group G = g, where g is a permutation of order 2 O( n log n) (by choosing g to be a product of cycles of prime length for different primes).

35 The Membership Testing Problem Given as input π S n and a subgroup G = S S n test if π is in G. Express π in terms of the generators. Writing π as a product of generators may be exponentially long! Example Consider the cyclic group G = g, where g is a permutation of order 2 O( n log n) (by choosing g to be a product of cycles of prime length for different primes). We need to have a more compact way of expressing π G in terms of its generators.

36 Membership Testing Contd. Elements of G are g b where b is t = O( n log n) bits. We compute g b by repeated squaring and multiplying the appropriate powers g 2i. Let b = t 1 i=0 b i2 i.

37 Membership Testing Contd. Elements of G are g b where b is t = O( n log n) bits. We compute g b by repeated squaring and multiplying the appropriate powers g 2i. Let b = t 1 i=0 b i2 i. The following straight-line program computes g b :

38 Membership Testing Contd. Elements of G are g b where b is t = O( n log n) bits. We compute g b by repeated squaring and multiplying the appropriate powers g 2i. Let b = t 1 i=0 b i2 i. The following straight-line program computes g b : x 0 := g; for i := 1 to t 1 do x i := x 2 i 1 ; x t := 1; for i := 1 to t 1 do x t+i := x t+i 1 x b i i ;

39 Srtaight-Line Programs Let π G = g 1, g 2,..., g k S n. A straight-line program for π consists of the following:

40 Srtaight-Line Programs Let π G = g 1, g 2,..., g k S n. A straight-line program for π consists of the following: The first k lines of the program has x i := g i, 1 i k as instructions, where g 1, g 2,..., g k are the generators of G.

41 Srtaight-Line Programs Let π G = g 1, g 2,..., g k S n. A straight-line program for π consists of the following: The first k lines of the program has x i := g i, 1 i k as instructions, where g 1, g 2,..., g k are the generators of G. Each subsequent line is an instruction of the form: Where j < i and k < i. x i := x j x k

42 Srtaight-Line Programs Let π G = g 1, g 2,..., g k S n. A straight-line program for π consists of the following: The first k lines of the program has x i := g i, 1 i k as instructions, where g 1, g 2,..., g k are the generators of G. Each subsequent line is an instruction of the form: Where j < i and k < i. x i := x j x k Nice Fact If π g 1, g 2,..., g k S n then it has a straight-line program of length polynomial in n and k, and the membership testing algorithm will find in poly(n) time.

43 Strong Generating Sets For G S n let G [i] denote its subgroup that pointwise stabilizes {1, 2,..., i}.

44 Strong Generating Sets For G S n let G [i] denote its subgroup that pointwise stabilizes {1, 2,..., i}. Consider the tower of stabilizers subgroups in G: {id} = G [n 1] < G [n 2] <... < G [1] < G [0] = G.

45 Strong Generating Sets For G S n let G [i] denote its subgroup that pointwise stabilizes {1, 2,..., i}. Consider the tower of stabilizers subgroups in G: {id} = G [n 1] < G [n 2] <... < G [1] < G [0] = G. Consider the right coset representative sets T i for G [i] in G [i 1], 1 i n 1. Their union forms a strong generating set for G.

46 Strong Generating Sets For G S n let G [i] denote its subgroup that pointwise stabilizes {1, 2,..., i}. Consider the tower of stabilizers subgroups in G: {id} = G [n 1] < G [n 2] <... < G [1] < G [0] = G. Consider the right coset representative sets T i for G [i] in G [i 1], 1 i n 1. Their union forms a strong generating set for G. Strong because every π G can be expressed uniquely as a short product π = π n 1 π n 2... π 1, where π i T i.

47 Back to Membership Testing Given a strong generating set for G S n, membership testing is easy and efficient. Let π S n : π G π G [1] π 1 for π 1 T 1. We can find π 1 T 1 easily and the problem reduces to checking if ππ 1 1 is in G [1].

48 Back to Membership Testing Given a strong generating set for G S n, membership testing is easy and efficient. Let π S n : π G π G [1] π 1 for π 1 T 1. We can find π 1 T 1 easily and the problem reduces to checking if ππ 1 1 is in G [1]. How do we find a strong generating set for G?

49 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2?

50 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2? Schreier s Lemma Let G = A and H G of finite index with R as the set of right coset representatives. Then H is generated by the set B = {r 1 ar2 1 H a A, r 1, r 2 R}.

51 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2? Schreier s Lemma Let G = A and H G of finite index with R as the set of right coset representatives. Then H is generated by the set B = {r 1 ar2 1 H a A, r 1, r 2 R}. Proof We know G = HR.

52 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2? Schreier s Lemma Let G = A and H G of finite index with R as the set of right coset representatives. Then H is generated by the set B = {r 1 ar2 1 H a A, r 1, r 2 R}. Proof We know G = HR. And we have RA BR

53 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2? Schreier s Lemma Let G = A and H G of finite index with R as the set of right coset representatives. Then H is generated by the set B = {r 1 ar2 1 H a A, r 1, r 2 R}. Proof We know G = HR. And we have RA BR Which implies RAA BRA BBR.

54 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2? Schreier s Lemma Let G = A and H G of finite index with R as the set of right coset representatives. Then H is generated by the set B = {r 1 ar2 1 H a A, r 1, r 2 R}. Proof We know G = HR. And we have RA BR Which implies RAA BRA BBR. Repeating, we get R A B R.

55 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2? Schreier s Lemma Let G = A and H G of finite index with R as the set of right coset representatives. Then H is generated by the set B = {r 1 ar2 1 H a A, r 1, r 2 R}. Proof We know G = HR. And we have RA BR Which implies RAA BRA BBR. Repeating, we get R A B R. Hence G = B R.

56 Finding a Strong Generating Set Given G = S, finding T 1 is easy. We can compute the orbit O of 1, and for each j O keep track of a π 1 G such that 1 π 1 = j. How do we find T 2? Schreier s Lemma Let G = A and H G of finite index with R as the set of right coset representatives. Then H is generated by the set B = {r 1 ar2 1 H a A, r 1, r 2 R}. Proof We know G = HR. And we have RA BR Which implies RAA BRA BBR. Repeating, we get R A B R. Hence G = B R. Since B H it follows that B = H.

57 Finding a Strong Generating Set Difficulty Applying Schreier s lemma, the number of generators for G [1] can be n A. For G [2] it can grow to n 2 A and so on...

58 Finding a Strong Generating Set Difficulty Applying Schreier s lemma, the number of generators for G [1] can be n A. For G [2] it can grow to n 2 A and so on... Solution: a reduce step Given G = g 1, g 2,..., g k, we can efficiently find a generating set of size O(n 2 ).

59 Finding a Strong Generating Set Difficulty Applying Schreier s lemma, the number of generators for G [1] can be n A. For G [2] it can grow to n 2 A and so on... Solution: a reduce step Given G = g 1, g 2,..., g k, we can efficiently find a generating set of size O(n 2 ). for i = 1 to n do while there are generators x, y fixing 1, 2,..., i 1 such that i x = i y do replace the pair x, y with the pair x, yx 1. end-while end-for

60 How many generators are needed? Exercise Given G = S S n as input we can efficiently compute a generating set of size at most n 1. Hint:

61 How many generators are needed? Exercise Given G = S S n as input we can efficiently compute a generating set of size at most n 1. Hint: For each g S, let i g [n] be the smallest point moved by g.

62 How many generators are needed? Exercise Given G = S S n as input we can efficiently compute a generating set of size at most n 1. Hint: For each g S, let i g [n] be the smallest point moved by g. Consider the graph X S on vertex set {1, 2,..., n} and edge set {(i g, i g g ) g S}.

63 How many generators are needed? Exercise Given G = S S n as input we can efficiently compute a generating set of size at most n 1. Hint: For each g S, let i g [n] be the smallest point moved by g. Consider the graph X S on vertex set {1, 2,..., n} and edge set {(i g, i g g ) g S}. As long as X S has cycles, we can apply a modified reduce step to shrink the size of S.

64 How many generators are needed? Exercise Given G = S S n as input we can efficiently compute a generating set of size at most n 1. Hint: For each g S, let i g [n] be the smallest point moved by g. Consider the graph X S on vertex set {1, 2,..., n} and edge set {(i g, i g g ) g S}. As long as X S has cycles, we can apply a modified reduce step to shrink the size of S. McIver-Neumann Every subgroup of S n has a generating set of size at most n/2. Proof uses CFSG. No efficient algorithm is known for it.

65 How many generators are needed? Exercise Given G = S S n as input we can efficiently compute a generating set of size at most n 1. Hint: For each g S, let i g [n] be the smallest point moved by g. Consider the graph X S on vertex set {1, 2,..., n} and edge set {(i g, i g g ) g S}. As long as X S has cycles, we can apply a modified reduce step to shrink the size of S. McIver-Neumann Every subgroup of S n has a generating set of size at most n/2. Proof uses CFSG. No efficient algorithm is known for it.

66 Finding a Strong Generating Set Theorem (Schreier-Sims) Let G < S n be input by some generating set. In polynomial time we can compute a strong generating set T i with the following properties: 1 Every element π G can be expressed uniquely as a product π = π 1 π 2... π n 1 with π i T i,

67 Finding a Strong Generating Set Theorem (Schreier-Sims) Let G < S n be input by some generating set. In polynomial time we can compute a strong generating set T i with the following properties: 1 Every element π G can be expressed uniquely as a product π = π 1 π 2... π n 1 with π i T i, 2 Membership in G of a given permutation can be tested in polynomial time.

68 Finding a Strong Generating Set Theorem (Schreier-Sims) Let G < S n be input by some generating set. In polynomial time we can compute a strong generating set T i with the following properties: 1 Every element π G can be expressed uniquely as a product π = π 1 π 2... π n 1 with π i T i, 2 Membership in G of a given permutation can be tested in polynomial time. 3 G can be computed in polynomial time.

69 Finding a Strong Generating Set Theorem (Schreier-Sims) Let G < S n be input by some generating set. In polynomial time we can compute a strong generating set T i with the following properties: 1 Every element π G can be expressed uniquely as a product π = π 1 π 2... π n 1 with π i T i, 2 Membership in G of a given permutation can be tested in polynomial time. 3 G can be computed in polynomial time.

70 Running Time Analysis Suppose G = S is the input group.

71 Running Time Analysis Suppose G = S is the input group. Initial reduce operation if S > n 2. For i = 1, 2,..., n 1 we compute i g for S many g. After that at most S n many replacements of x, y by x, yx 1.

72 Running Time Analysis Suppose G = S is the input group. Initial reduce operation if S > n 2. For i = 1, 2,..., n 1 we compute i g for S many g. After that at most S n many replacements of x, y by x, yx 1. In Schreier s lemma: computing orbit of i takes n 2 S n 4 operations, which also gives transversal R for G [i] in G [i 1].

73 Running Time Analysis Suppose G = S is the input group. Initial reduce operation if S > n 2. For i = 1, 2,..., n 1 we compute i g for S many g. After that at most S n many replacements of x, y by x, yx 1. In Schreier s lemma: computing orbit of i takes n 2 S n 4 operations, which also gives transversal R for G [i] in G [i 1]. R S = n 3 operations for computing generating set of G [i]. Applying reduce operation costs S n n 4 operations.

74 Running Time Analysis Suppose G = S is the input group. Initial reduce operation if S > n 2. For i = 1, 2,..., n 1 we compute i g for S many g. After that at most S n many replacements of x, y by x, yx 1. In Schreier s lemma: computing orbit of i takes n 2 S n 4 operations, which also gives transversal R for G [i] in G [i 1]. R S = n 3 operations for computing generating set of G [i]. Applying reduce operation costs S n n 4 operations. Overall costs is n 4 n plus O( S n) operations. Each operation costs O(n). Thus, O(n 6 + S n 2 ) is the time.

75 Orbit Counting Lemma i [n] is a fixpoint of g G if i g = i.

76 Orbit Counting Lemma i [n] is a fixpoint of g G if i g = i. fix(g) = number of fixpoints of g.

77 Orbit Counting Lemma i [n] is a fixpoint of g G if i g = i. fix(g) = number of fixpoints of g. orb(g) = number of orbits of G.

78 Orbit Counting Lemma i [n] is a fixpoint of g G if i g = i. fix(g) = number of fixpoints of g. orb(g) = number of orbits of G. Lemma (Orbit Counting Lemma) Let G S n and orb(g) denote the number of orbits of G. Then orb(g) = 1 fix(g) = E g G [fix(g)]. G g G

79 Orbit Counting Lemma i [n] is a fixpoint of g G if i g = i. fix(g) = number of fixpoints of g. orb(g) = number of orbits of G. Lemma (Orbit Counting Lemma) Let G S n and orb(g) denote the number of orbits of G. Then orb(g) = 1 fix(g) = E g G [fix(g)]. G g G Proof Define 0-1 matrix: M g,i = 1 iff i g = i. Equate row-wise and column-wise sums using O(i) = G / G i.

80 Fixpoint free elements in G Theorem (Jordan s Theorem) If G S n is transitive then the group G has fixpoint free elements.

81 Fixpoint free elements in G Theorem (Jordan s Theorem) If G S n is transitive then the group G has fixpoint free elements. Proof G has a single orbit and fix(g) = n for the identity. Since the expectation is 1 there are g such that fix(g) = 0.

82 Fixpoint free elements in G Theorem (Jordan s Theorem) If G S n is transitive then the group G has fixpoint free elements. Proof G has a single orbit and fix(g) = n for the identity. Since the expectation is 1 there are g such that fix(g) = 0. Problem: Given G = g 1, g 2,..., g k transitive, can we find a fixpoint free element in polynomial time?

83 Theorem (CC92) Cameron-Cohen s Theorem If G S n is transitive then the group G has at least G /n many fixpoint elements. Proof

84 Theorem (CC92) Cameron-Cohen s Theorem If G S n is transitive then the group G has at least G /n many fixpoint elements. Proof Let A denote the set of fixpoint free elements in G.

85 Theorem (CC92) Cameron-Cohen s Theorem If G S n is transitive then the group G has at least G /n many fixpoint elements. Proof Let A denote the set of fixpoint free elements in G. G = g G fix(g) = g G 1 fix(g) + g G\G 1 fix(g)

86 Theorem (CC92) Cameron-Cohen s Theorem If G S n is transitive then the group G has at least G /n many fixpoint elements. Proof Let A denote the set of fixpoint free elements in G. Which yields G = g G fix(g) = g G 1 fix(g) + G G /n + G 1 + G \ (A G 1 ) g G\G 1 fix(g) A G /n.

87 A Randomized Algorithm Let G = S be a permutation group given as input by generating set S.

88 A Randomized Algorithm Let G = S be a permutation group given as input by generating set S. We first compute a strong generating set T for G.

89 A Randomized Algorithm Let G = S be a permutation group given as input by generating set S. We first compute a strong generating set T for G. Using T we can sample π G uniformly at random, and check if π is fixpoint free.

90 A Randomized Algorithm Let G = S be a permutation group given as input by generating set S. We first compute a strong generating set T for G. Using T we can sample π G uniformly at random, and check if π is fixpoint free. Analysis Probability that we do not find a fixpoint free element in, say, n 2 trials is bounded by (1 1/n) n2 e n.

91 Deterministic Algorithm Let move(g) = n fix(g). Orbit counting lemma restated: E g G [move(g)] = 1 move(g) = n orb(g). G g G

92 Deterministic Algorithm Let move(g) = n fix(g). Orbit counting lemma restated: E g G [move(g)] = 1 move(g) = n orb(g). G g G For G transitive we have E g G [move(g)] = n 1.

93 Deterministic Algorithm Let move(g) = n fix(g). Orbit counting lemma restated: E g G [move(g)] = 1 move(g) = n orb(g). G g G For G transitive we have E g G [move(g)] = n 1. Since G 1 has at least two orbits, Let G = π R G 1 π. E g G1 [move(g)] n 2.

94 Deterministic Algorithm Let move(g) = n fix(g). Orbit counting lemma restated: E g G [move(g)] = 1 move(g) = n orb(g). G g G For G transitive we have E g G [move(g)] = n 1. Since G 1 has at least two orbits, Let G = π R G 1 π. E g G1 [move(g)] n 2. E g G [move(g)] = E π R E g G1 π[move(g)].

95 Deterministic Algorithm Let move(g) = n fix(g). Orbit counting lemma restated: E g G [move(g)] = 1 move(g) = n orb(g). G g G For G transitive we have E g G [move(g)] = n 1. Since G 1 has at least two orbits, Let G = π R G 1 π. E g G1 [move(g)] n 2. E g G [move(g)] = E π R E g G1 π[move(g)]. Thus, for some coset G 1 π of G 1 in G we must have E g G1 π[move(g)] > n 1.

96 Deterministic Algorithm Contd. For each coset representative π R we explain how to efficiently compute E g G1 π[move(g)]. Revisit the proof of the orbit counting lemma: Recall M gπ,i = 1 iff i gπ = i.

97 Deterministic Algorithm Contd. For each coset representative π R we explain how to efficiently compute E g G1 π[move(g)]. Revisit the proof of the orbit counting lemma: Recall M gπ,i = 1 iff i gπ = i. Number of 1 s in the i th column is {g G 1 i gπ = i}.

98 Deterministic Algorithm Contd. For each coset representative π R we explain how to efficiently compute E g G1 π[move(g)]. Revisit the proof of the orbit counting lemma: Recall M gπ,i = 1 iff i gπ = i. Number of 1 s in the i th column is {g G 1 i gπ = i}. This number is zero if i and i π 1 G 1,i otherwise. are in different G 1 -orbits, and is Thus, using the Schreir-Sims algorithm we can compute all column sums efficiently, and hence also the above expectation in polynomial time.

99 Method of Conditional Probabilities G G 1 π 1... G 1 π i G 1 π k G 1,2 τ 1... G 1,2 τ l

100 Method of Conditional Probabilities G G 1 π 1... G 1 π i G 1 π k G 1,2 τ 1... G 1,2 τ l At the d th level of the tree compute the coset G 1,...,d σ that maximizes E g G1,...,d σ[move(g)].

101 Remarks and related questions Method of conditional probabilities Erdös-Selfridge, Spencer, Raghavan.

102 Remarks and related questions Method of conditional probabilities Erdös-Selfridge, Spencer, Raghavan. Fixpoint free element checking in general permutation groups G = S S n is NP-complete.

103 Remarks and related questions Method of conditional probabilities Erdös-Selfridge, Spencer, Raghavan. Fixpoint free element checking in general permutation groups G = S S n is NP-complete. Fein-Kantor-Schacher: Every transitive permutation group G S n for n 2 has a fixpoint free element of prime power order. Is there an efficient algorithm for finding one?

104 Remarks and related questions Method of conditional probabilities Erdös-Selfridge, Spencer, Raghavan. Fixpoint free element checking in general permutation groups G = S S n is NP-complete. Fein-Kantor-Schacher: Every transitive permutation group G S n for n 2 has a fixpoint free element of prime power order. Is there an efficient algorithm for finding one? Isaacs-Kantor-Spaltenstein: For G S n and prime p dividing G, there are at least G /n many elements whose order is divisible by p. Deterministic algorithm?

105 Remarks and related questions Method of conditional probabilities Erdös-Selfridge, Spencer, Raghavan. Fixpoint free element checking in general permutation groups G = S S n is NP-complete. Fein-Kantor-Schacher: Every transitive permutation group G S n for n 2 has a fixpoint free element of prime power order. Is there an efficient algorithm for finding one? Isaacs-Kantor-Spaltenstein: For G S n and prime p dividing G, there are at least G /n many elements whose order is divisible by p. Deterministic algorithm?

106 Bases for Permutation Groups Let G S n be a permutation group. A subset of points B [n] is called a base for G if the subgroup G B of G that fixes every point of G is the identity.

107 Bases for Permutation Groups Let G S n be a permutation group. A subset of points B [n] is called a base for G if the subgroup G B of G that fixes every point of G is the identity. This generalizes bases for vector spaces and has proven computationally useful. There is a library of nearly linear-time algorithms for small base groups due to Akos Seress and others.

108 Bases for Permutation Groups Let G S n be a permutation group. A subset of points B [n] is called a base for G if the subgroup G B of G that fixes every point of G is the identity. This generalizes bases for vector spaces and has proven computationally useful. There is a library of nearly linear-time algorithms for small base groups due to Akos Seress and others. Finding minimum bases of permutation groups is NP-hard [Blaha 1992] even for cyclic groups and groups with bounded orbit size.

109 Bases for Permutation Groups Let G S n be a permutation group. A subset of points B [n] is called a base for G if the subgroup G B of G that fixes every point of G is the identity. This generalizes bases for vector spaces and has proven computationally useful. There is a library of nearly linear-time algorithms for small base groups due to Akos Seress and others. Finding minimum bases of permutation groups is NP-hard [Blaha 1992] even for cyclic groups and groups with bounded orbit size.

110 Approximating minimum bases If G = S S n has minimum base size b then G n b.

111 Approximating minimum bases If G = S S n has minimum base size b then G n b. Proof If {1, 2,..., b} is a minimum size base then G [i 1] / G [i] n and G [b] = 1.

112 Approximating minimum bases If G = S S n has minimum base size b then G n b. Proof If {1, 2,..., b} is a minimum size base then G [i 1] / G [i] n and G [b] = 1. An irredundant base B = {α 1, α 2,..., α d } is such that no α i is fixed by the pointwise stabilizer of all earlier points. Hence G 2 d.

113 Approximating minimum bases If G = S S n has minimum base size b then G n b. Proof If {1, 2,..., b} is a minimum size base then G [i 1] / G [i] n and G [b] = 1. An irredundant base B = {α 1, α 2,..., α d } is such that no α i is fixed by the pointwise stabilizer of all earlier points. Hence G 2 d. Thus, any irredudant base is of size d b log n.

114 A Better Approximation Algorithm Having picked {α 1, α 2,..., α i }, pick α i+1 from an orbit of largest size in the pointwise stabilizer of {α 1, α 2,..., α i }.

115 A Better Approximation Algorithm Having picked {α 1, α 2,..., α i }, pick α i+1 from an orbit of largest size in the pointwise stabilizer of {α 1, α 2,..., α i }. Claim This yields a base of size (log log n + O(1))b.

116 A Better Approximation Algorithm Having picked {α 1, α 2,..., α i }, pick α i+1 from an orbit of largest size in the pointwise stabilizer of {α 1, α 2,..., α i }. Claim This yields a base of size (log log n + O(1))b. If H G then H has an orbit of size at least H 1/b. Fixing a point in it makes H α H 1/b.

117 A Better Approximation Algorithm Having picked {α 1, α 2,..., α i }, pick α i+1 from an orbit of largest size in the pointwise stabilizer of {α 1, α 2,..., α i }. Claim This yields a base of size (log log n + O(1))b. If H G then H has an orbit of size at least H 1/b. Fixing a point in it makes H α H 1/b. Thus, by picking b log log n points G shrinks as G (1 1/b)b log log n e b. Pick O(b) more points irredundantly.

118 Other Finite Groups Let G GL n (F q ), where q is a prime power and F q is the finite field of size q.

119 Other Finite Groups Let G GL n (F q ), where q is a prime power and F q is the finite field of size q. Membership testing is likely to be hard.

120 Other Finite Groups Let G GL n (F q ), where q is a prime power and F q is the finite field of size q. Membership testing is likely to be hard. Given a, b F q, checking if a x = b(modp) is considered a computationally hard problem. Finding x is the so-called discrete log problem.

121 Other Finite Groups Let G GL n (F q ), where q is a prime power and F q is the finite field of size q. Membership testing is likely to be hard. Given a, b F q, checking if a x = b(modp) is considered a computationally hard problem. Finding x is the so-called discrete log problem. We cannot embed F q factors. in S n for small n if q 1 has large prime

122 References/further reading Peter J Cameron Permutation Groups, LMS Student Texts 45, Cambridge Univ Press. Eugene M Luks Permutation groups and polynomial-time computation, in Groups and Computation, DIMACS series in Discrete Mathematics and Theoretical Computer Science 11 (1993), Available online. Laszlo Babai Local expansion of vertex-transitive graphs and random generation in finite groups.

123 Other Finite Groups Contd. Suppose G = g 1, g 2,..., g k where we assume no structure about G. What can we compute efficiently? Randomness helps. Random subproducts of g 1, g 2,..., g k are elements of the form g ɛ 1 1 g ɛ g ɛ k k, ɛ i R {0, 1}.

124 Input G = g 1, g 2,..., g k. Testing Commutativity

125 Testing Commutativity Input G = g 1, g 2,..., g k. Check if g i g j = g j g i for all pairs i, j. This is an O(k 2 ) test, and the best possible deterministic test.

126 Testing Commutativity Input G = g 1, g 2,..., g k. Check if g i g j = g j g i for all pairs i, j. This is an O(k 2 ) test, and the best possible deterministic test. A randomized test Let x = g ɛ 1 1 g ɛ g ɛ k k and y = g µ 1 1 g µ g µ k k subproducts. Accept iff xy = yx. be two independent

127 Testing Commutativity Contd. Claim If H < G is a proper subgroup then Prob[g ɛ 1 1 g ɛ g ɛ k k H] 1/2.

128 Testing Commutativity Contd. Claim If H < G is a proper subgroup then Prob[g ɛ 1 1 g ɛ g ɛ k k H] 1/2. Prob[xy = yx] Prob[x Z(G)] + Prob[y C(x) x Z(G)] p + (1 p)/2 = (1 + p)/2 3/4

129 Testing Commutativity Contd. Claim If H < G is a proper subgroup then Prob[g ɛ 1 1 g ɛ g ɛ k k H] 1/2. Prob[xy = yx] Prob[x Z(G)] + Prob[y C(x) x Z(G)] p + (1 p)/2 = (1 + p)/2 3/4 Curious Fact If G is nonabelian then Prob x,y G [xy = yx] 5/8.

130 Erdös-Rényi Sequences Random subproducts come from an Erdös-Rényi paper titled Probabilistic methods in group theory.

131 Erdös-Rényi Sequences Random subproducts come from an Erdös-Rényi paper titled Probabilistic methods in group theory. A random subproduct g e 1 1 g e g e k k x G is ε-uniform in G if for all (1 ε)/ G Prob[g e 1 1 g e g e k k = x (1 + ε)/ G. If k 2 log G + 2 log(1/ε) + log(1/δ) and g 1, g 2,..., g k are randomly picked, then g e 1 1 g e g e k k is ε-uniform in G with probability 1 δ.

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133 G = g 1, g 2,..., g k. Straight-line programs for G

134 Straight-line programs for G G = g 1, g 2,..., g k. Define the cube C = {g e g e k k inverses of the elements in C. e i {0, 1}} and C 1 be the

135 Straight-line programs for G G = g 1, g 2,..., g k. Define the cube C = {g e g e k k inverses of the elements in C. e i {0, 1}} and C 1 be the If G = C 1 C, we have short st-line programs for all of G.

136 Straight-line programs for G G = g 1, g 2,..., g k. Define the cube C = {g e g e k k inverses of the elements in C. e i {0, 1}} and C 1 be the If G = C 1 C, we have short st-line programs for all of G. Otherwise, there is a generator g j such that C 1 Cg j C 1 C.

137 Straight-line programs for G G = g 1, g 2,..., g k. Define the cube C = {g e g e k k inverses of the elements in C. e i {0, 1}} and C 1 be the If G = C 1 C, we have short st-line programs for all of G. Otherwise, there is a generator g j such that C 1 Cg j C 1 C. Include an element g k+1 C 1 Cg j \ C 1 C to extend the sequence.

138 Straight-line programs for G G = g 1, g 2,..., g k. Define the cube C = {g e g e k k inverses of the elements in C. e i {0, 1}} and C 1 be the If G = C 1 C, we have short st-line programs for all of G. Otherwise, there is a generator g j such that C 1 Cg j C 1 C. Include an element g k+1 C 1 Cg j \ C 1 C to extend the sequence. As Cg k+1 C = we have doubled the size of the cube.

139 Short straight-line programs In l log G steps we obtain elements g k+1, g k+2,..., g k+l G such that for the cube C l = {g e g e k+l k+l e i {0, 1}}. we have G = C 1 l C l.

140 Short straight-line programs In l log G steps we obtain elements g k+1, g k+2,..., g k+l G such that for the cube C l = {g e g e k+l k+l e i {0, 1}}. we have G = C 1 l C l. Each g G has a straight-line program of length log G (2(k + i) + 1) + 2(k + log G ) = O((k + log G ) log G ) i=0 in terms of the original generators.

141 THANKS!