Báo cáo toán hoc:" Hurwitz Equivalence in Tuples of Dihedral Groups, Dicyclic Groups, and Semidihedral Groups" pptx

Hurwitz Equivalence in Tuples of Dihedral Groups, Dicyclic Groups, and Semidihedral Groups Charmaine Sia Department of Mathematics Massachusetts Institute of Technology Cambridge, MA 02139-4307, USA sia@mit.edu Submitted: Dec 27, 2008; Accepted: Jul 27, 2009; Published: Aug 7, 2009 Mathematics S ubject Classification: 20F36, 20F05 Abstract Let D 2N be the dihedral group of order 2N , Dic 4M the dicyclic group of or- der 4M, SD 2 m the semidihedral group of order 2 m , and M 2 m the group of order 2 m with presentation M 2 m = α, β | α 2 m−1 = β 2 = 1, βαβ −1 = α 2 m−2 +1 . We classify the orbits in D n 2N , Dic n 4M , SD n 2 m , and M n 2 m under the Hurwitz action. 1 Introduction Let B n denote the braid group on n strands, which is given by the presentation B n = σ 1 , . . . , σ n−1 | σ i σ j = σ j σ i , |i − j|  2; σ i σ i+1 σ i = σ i+1 σ i σ i+1 , 1  i  n − 2. For an arbitrary group G and n  2, there is an action of B n on G n , called the Hurwitz action, which is defined by σ i (a 1 , . . . , a n ) = (a 1 , . . . , a i−1 , a i+1 , a −1 i+1 a i a i+1 , a i+2 , . . . , a n ) for every 1  i  n − 1 and (a 1 , . . . , a n ) ∈ G n . Note that σ −1 i (a 1 , . . . , a n ) = (a 1 , . . . , a i−1 , a i a i+1 a −1 i , a i , a i+2 , . . . , a n ). Hence, if we write a = (a 1 , . . . , a n ) ∈ G n and define π(a) = a 1 · · · a n ∈ G, then π(a) is an invariant of the Hurwitz action on G n . An a ction by σ i or σ −1 i on G n is called a Hurwitz move. Two tuples (a 1 , . . . , a n ), (b 1 , . . . , b n ) ∈ G n are said to be (Hurwitz) equivalent, denoted (a 1 , . . . , a n ) ∼ (b 1 , . . . , b n ), if they lie in the same B n -orbit. The problem of classifying the orbits in G n under the Hurwitz action arose from the study of braid monodromy factorization (see, e.g., Kulikov and Teicher [5]). Clearly, this the electronic journal of combinatorics 16 (2009), #R95 1 problem is trivial for any abelian group G: two n-tuples a, b ∈ G n are equiva lent if and only if one is a permutation of the other. However, there are few results on the classification of B n -orbits in G n for nonabelian groups G. Ben-Itzhak and Teicher [1] determined all B n -orbits in S n m represented by (t 1 , . . . , t n ), where S m is the symmetric group of order m!, each t i is a transposition, and t 1 · · · t n = 1. Recently, Hou [3] determined completely the B n -orbits in Q n 2 m and D n 2p m , where Q 2 m is the generalized quaternion group of order 2 m and D 2p m is the dihedral group of o r der 2p m for some prime p. Clearly, if a 1 , . . . , a n ∈ G generate a finite subgroup, then the B n -orb! it of (a 1 , . . . , a n ) in G n is finite. Humphries [4] and Michel [6] proved a partial converse when G is the general linear group GL(R n ): if s 1 , . . . , s n ∈ GL(R n ) are reflections such that the B n -orbit o f (s 1 , . . . , s n ) is finite, then the group generated by s 1 , . . . , s n is finite. In this paper, we determine completely the B n -orbits in G n for four families of groups G: the dihedral g r oup D 2N of order 2N, the dicyclic gro up Dic 4M of order 4M, the semidi- hedral gr oup SD 2 m of order 2 m , and the group M 2 m = α, β | α 2 m−1 = β 2 = 1, βαβ −1 = α 2 m−2 +1  of order 2 m . Our method is to find a number of invariants of the Hurwitz action and show that these invariants completely determine the Hurwitz equivalence classes. The invariants and the strategies used to find a canonical representative equivalent to each tuple are essentially the same as those in [3]. The novel element of the present paper is the idea that when performing a series of Hurwitz moves to normalize a tuple in D n 2N with respect to a prime factor o f N, we can preserve certain congruence properties with respect to other factors of N that were obtained in earlier moves. This paper is organized as follows. In Section 2, we develop some preliminary results regarding the Hurwitz action on D n 2N . In Section 3, we classify the orbits in D n 2N under the Hurwitz action. In Section 4 , we classify the Hurwitz equivalence classes in Dic n 4M , SD n 2 m , and M n 2 m . 2 The Hurwitz Action on D n 2N In this section, we develop some preliminary results regarding the Hurwitz action o n D n 2N . With the exception of Lemma 2.1(iv), the results presented in this section are similar to those in [3, Section 2]. We use the following generators and relations for the dihedral group D 2N of order 2N: D 2N = α, β | α N = β 2 = 1, βαβ −1 = α −1 . Each element of D 2N can be uniquely written in the form α i β j , where 0  i < N and 0  j  1. Conjugating one element of D 2N by another g ives (α k β l ) −1 (α i β j )(α k β l ) = α (−1) l (i−2kj) β j , (2.1) (α i β j )(α k β l )(α i β j ) −1 = α (−1) j k+2il β l . (2.2) Therefore, a Hurwitz move in D n 2N yields one of the f ollowing two equivalences: (· · · , α i β j , α k β l , · · · ) ∼ (· · · , α k β l , α (−1) l (i−2kj) β j , · · · ), (· · · , α i β j , α k β l , · · · ) ∼ (· · · , α (−1) j k+2il β l , α i β j , · · · ). the electronic journal of combinatorics 16 (2009), #R95 2 To direct the reader’s attention to the Hurwitz moves that we consider, we shall occa- sionally omit common terms from two equivalent n-tuples a, b ∈ G n if there is a sequence of moves transforming a to b that does not involve any of those terms. For example, setting (j, l) = (0, 0), (0, 1), (1, 0), and (1, 1) respectively in the above equivalences and omitting common terms, we obtain (α i , α k ) ∼ (α k , α i ), (2.3)  (α i , α k β) ∼ (α k β, α −i ), (α i , α k β) ∼ (α k+2i β, α i ), (2.4)  (α i β, α k ) ∼ (α k , α i−2k β), (α i β, α k ) ∼ (α −k , α i β), (2.5)  (α i β, α k β) ∼ (α k β, α −i+2k β) = (α i+(k−i) β, α k+(k− i) β), (α i β, α k β) ∼ (α −k+2i β, α i β) = (α i−(k−i) β, α k−(k− i) β). (2.6) The following lemma sets forth some key equivalences that can be obtained through a sequence of Hurwitz moves. Lemma 2.1 (see Hou [3, Lemma 2.1]). (i) (α i , α j β) ∼ (α −i , α j+2i β) for all i, j ∈ Z. (ii) (α i β, α j β) ∼ (α i+h(j−i) β, α j+h(j−i) β) for all h, i, j ∈ Z. (iii) Let p 1 , . . . , p t be distinct prime divisors of N (not necessarily all the prime divisors of N) such that p k r r N for r = 1, . . . , t, and let 0  ν r  k r − 1 for r = 1, . . . , t. Let e, f ∈ Z such that e ≡ f (mod p r ) for r = 1, . . . , t. Then for all g ∈ Z such that g ≡ 0 (mod N/  t r=1 p k r r ) and τ ∈ Z, we have (α τ+e Q t r =1 p ν r r β, α τ+f Q t r =1 p ν r r β) ∼ (α τ+(e+g) Q t r =1 p ν r r β, α τ+(f +g) Q t r =1 p ν r r β). (iv) Let p 1 , . . . , p t be distinct prime divisors of N (not necessarily all the prime divisors of N) such that p k r r N for r = 1, . . . , t, and let 0  ν r  k r − 1 for r = 1, . . . , t. Then for all e ≡ f (mod p r ), there exists g ∈ Z such that (a) (α τ+e Q t r =1 p ν r r β, α τ+f Q t r =1 p ν r r β) ∼ (α τ+(e+g) Q t r =1 p ν r r β, α τ+(f +g) Q t r =1 p ν r r β), (b) p k r −ν r r | f + g, and (c) if p r ′ is another prime divisor of N, then f + g ≡ f (mod p k r ′ −ν r ′ r ′ ). In particular, p k r | (f + g)  t r=1 p ν r r , and if p r ′ is another prime divisor of N such that p k r ′ r ′ | f  t r=1 p ν r r , then p k r ′ r ′ | (f + g)  t r=1 p ν r r . Proof. (i) We have (α i , α j β) ∼ (α j β, α −i ) (using the first equivalence in (2.4)) ∼ (α −i , α j+2i β) (using the first equivalence in (2.5)). the electronic journal of combinatorics 16 (2009), #R95 3 (ii) This follows from (2.6). (iii) Setting i = τ + e  t r=1 p ν r r and j = τ + f  t r=1 p ν r r in (ii), we see that it suffices to find h ∈ Z satisfying h(f − e)  t r=1 p ν r r ≡ g  t r=1 p ν r r (mod N). This can be achieved by using the Chinese Remainder Theorem to choose h such t hat h ≡ g(f − e) −1 (mod p k r −ν r r ) for r = 1, . . . , t, h ≡ 0 (mod N/  t r=1 p k r r ). (iv) Setting i = τ + e  t r=1 p ν r r and j = τ + f  t r=1 p ν r r in (ii), we see that it suffices to find g, h ∈ Z satisfying the fo llowing system of congruences: h(f − e) t  i=1 p ν r r ≡ g t  i=1 p ν r r (mod N), g ≡ −f (mod p k r −ν r r ), g ≡ 0 (mod p k r ′ −ν r ′ r ′ ) for all other primes p r ′ dividing N. This can be achieved by using the Chinese Remainder Theorem to choose h such that h ≡ −f(f − e) −1 (mod p k r −ν r r ), h ≡ 0 (mod p k r ′ −ν r ′ r ′ ) for all other primes p r ′ dividing N. It is easy to see that corresponding to any choice of h, there is a unique value of g modulo N/  t i=1 p ν r r that satisfies the conditions in (iv). This proves the lemma. 3 B n -orbits in Tuples of Dihedral Groups In this section, we classify the orbits in D n 2N under the Hurwitz action. The main idea behind our proof is as follows. First, we partition D n 2N into subsets, each of which is invariant under the Hurwitz action. We then find a number of invariants of the Hurwitz action and show that these invariants completely determine the equivalence classes within each subset. For a = (α i 1 β j 1 , . . . , α i n β j n ) ∈ D n 2N , where 0  i k < N and 0  j k  1, let Λ(a) = the multiset {min{i k , N − i k } : j k = 0} and Γ(a) = {i k : j k = 1}. For example, if a = (α 12 , α 11 β, α 4 , α 3 ) ∈ D 4 30 , then Λ(a) = {3, 4, 3} and Γ(a) = {11}. It is easy to see that Λ(a) is invariant under each of the Hurwitz moves in (2.3)–(2.6), hence it is an invariant of t he Hurwitz action. the electronic journal of combinatorics 16 (2009), #R95 4 We fix a notational convention here. If N is odd, we write its prime factorization as N = p k 1 1 · · · p k m m ; if N is even, we write its prime factorization as N = 2 k 0 p k 1 1 · · · p k m m (i.e., we set p 0 = 2). Let v p r (i) denote the p r -adic order of a numb er i. We partition D n 2N into subsets as follows. Let A = {a ∈ D n 2N : Γ(a) = ∅}. For each odd prime divisor p r of N, for each 0  ν r  k r and 0  τ r < p ν r r , let B p r ν r ,τ r = {a ∈ D n 2N : min({v p r (i) : i ∈ Λ(a)} ∪ {k r }) = ν r , ∅ = Γ(a) ⊂ τ r + p ν r r Z} , and for each 0  ν r  k r − 1 and 0  τ r < p ν r r , let C p r ν r ,τ r = {a ∈ D n 2N : min({v p r (i) : i ∈ Λ(a)} ∪ {k r })  ν r + 1, ∅ = Γ(a) ⊂ τ r + p ν r r Z, ∃j, j ′ ∈ Γ(a) such that v p r (j − j ′ ) = ν r } . Then, for a ny odd prime divisor p r of N, we have D n 2N = A ⊔      0ν r k r 0τ r <p ν r r B p r ν r ,τ r     ⊔      0ν r k r −1 0τ r <p ν r r C p r ν r ,τ r     . (3.1) It is easy to check that each of A, B p r ν r ,τ r , and C p r ν r ,τ r is invariant under the Hurwitz moves in (2.3)–(2.6). Thus A, B p r ν r ,τ r , and C p r ν r ,τ r are invariant under the Hurwitz action. For a ∈ C p r ν r ,τ r , collect the components of a of the form α i β from left to right a nd let the result be (α i 1 β, . . . , α i t β), where 0  i k < N. Let e s ∈ Z p r , 1  s  t, be defined by i s ≡ τ r + p ν r r e s (mod p ν r +1 r ). Define σ p r (a) = t  s=1 (−1) s−1 e s . For example, let N = 135 = 3 3 · 5, p r = 3, ν r = 2, τ r = 3, n = 4, and let a = (α 7+3 2 ·13 β, α 3 2 ·6 , α 7+3 2 ·2 β, α 7+3 2 ·11 β) ∈ C 3 2,7 . Then σ 3 (a) = 13 − 2 + 11 = 1 ∈ Z 3 . It is easy to see from (2.3)–(2.6) that σ(a) is also an invariant under the Hurwitz equivalence. This allows us to further partition C p r ν r ,τ r into two invariant subsets C p r ν r ,τ r ,0 = {a ∈ C p r ν r ,τ r : σ p r (a) = 0} and C p r ν r ,τ r ,1 = {a ∈ C p r ν r ,τ r : σ p r (a) = 0}. Thus, the partition (3.1) can be further refined into D n 2N = A ⊔      0ν r k r 0τ<p ν r r B p r ν r ,τ r     ⊔      0ν r k r −1 0τ<p ν r r C p r ν r ,τ r ,0     ⊔      0ν r k r −1 0τ<p ν r r C p r ν r ,τ r ,1     (3.2) the electronic journal of combinatorics 16 (2009), #R95 5 for odd primes p r dividing N. If N is even, we require some additional definitions. For each 0  ν 0  k 0 and 0  τ 0 < 2 ν 0 , let B 2 ν 0 ,τ 0 = {a ∈ D n 2N : min({v 2 (i) : i ∈ Λ(a)} ∪ {k 0 − 1}) = ν 0 , ∅ = Γ(a) ⊂ τ 0 + 2 ν 0 Z} , and for each 0  ν 0  k 0 − 1 and 0  τ 0 < 2 ν 0 , let C 2 ν 0 ,τ 0 = {a ∈ D n 2N : min({v 2 (i) : i ∈ Λ(a)} ∪ {k 0 − 1})  ν 0 + 1, ∅ = Γ(a) ⊂ τ 0 + 2 ν 0 Z ∃j, j ′ ∈ Γ(a) such that v 2 (j − j ′ ) = ν 0 } . Then A, B 2 ν 0 ,τ 0 , and C 2 ν 0 ,τ 0 are all invariant under the Hurwitz equiva lence and D n 2N = A ⊔     0ν 0 k 0 0τ 0 <2 ν 0 B 2 ν 0 ,τ 0    ⊔     0ν 0 k 0 −1 0τ 0 <2 ν 0 C 2 ν 0 ,τ 0    . (3.3) For a = (α i 1 β j 1 , . . . , α i n β j n ) ∈ C 2 ν 0 ,τ 0 , where 0  i k  N and 0  j k  1, let u(a) = #{k : j k = 1 and i k ≡ τ 0 + 2 ν 0 (mod 2 ν 0 +1 )}. It is easy to check that u(a) is also invariant under the Hurwitz action. Having set up this fra mework, we are now ready to define our desired partition P of D n 2N . Let Q be the common refinement of the partitions (3.2 ) as p r varies over all the odd prime factors of N. If N is odd, then we take P = Q, so that any block of the partition P is either A or has the form X p 1 ν 1 ,τ 1 ∩ X p 2 ν 2 ,τ 2 ∩ · · · ∩ X p m ν m ,τ m , where each X p r ν r ,τ r stands for one of B p r ν r ,τ r , C p r ν r ,τ r ,0 , or C p r ν r ,τ r ,1 . If N is even, we take P to be the common refinement of Q and (3.3). Let R ⊔ S 0 ⊔ S 1 ⊔ T ⊔ U be a partition of the set of prime divisors of N, with the restriction that 2 ∈ R ∪ S 0 ∪ S 1 , and either T = U = ∅, (T, U) = ({2} , ∅), or (T, U) = (∅, {2}). Fo r convenience, we will denote the block   p r ∈R B p r ν r ,τ r  ∩   p r ∈S 0 C p r ν r ,τ r ,0  ∩   p r ∈S 1 C p r ν r ,τ r ,1  ∩   p r ∈T B p r ν r ,τ r  ∩   p r ∈U C p r ν r ,τ r  by Π(R, S 0 , S 1 , T, U)(ν r )(τ r ), where (ν r ) and (τ r ) represent vectors that record the num- bers ν r and τ r for each prime p r . For example, if p 0 = 2, p 1 = 3, p 2 = 5, and p 3 = 7, then Π({5, 7}, {3}, ∅, ∅, {2} )( 1, 2, 1, 1)(1, 8, 4, 0) = C 2 1,1 ∩ C 3 2,8,0 ∩ B 5 1,4 ∩ B 7 1,0 . By our remarks above, each block of P is invariant under the Hurwitz action, hence it suffices to find a set of representatives of the B n -orbits in A and in each of the blocks Π(R, S 0 , S 1 , T, U)(ν r )(τ r ). This is a chieved in Theorem 3.1 below. the electronic journal of combinatorics 16 (2009), #R95 6 Theorem 3.1. (i) The B n -orbits in A are represented by (α i 1 , . . . , α i n ), where 0  i 1  · · ·  i n < N. (ii) For each odd prime divisor p r of N, let 0  ν r  k r and 0  τ i < p ν i i ; if N is even, further let 1  ν 0  k 0 and 0  τ 0 < 2 ν 0 . The B n -orbits in Π(R, S 0 , S 1 , T, U)(ν r )(τ r ) are represented by (α i 1 , . . . , α i s , α τ+E β, α τ+F β, w    α τ+G β, . . . , α τ+G β, α τ β, . . . , α τ β    n−s−1 ), (3.4) where (a) 0  s < n and 0  i 1  · · ·  i s  N/ 2, (b) τ is the unique integer such that 0  τ <  p r |N p ν r r and τ ≡ τ r (mod p ν r r ) for each prime p r dividing N, (c) for each p r ∈ R, we have min{v p r (i 1 ), . . . , v p r (i s ), k r } = ν r , n − s − 1  0, p ν r r | E, p k r r | F , and p k r r | G, (d) for each p r ∈ S 0 , we have min{v p r (i 1 ), . . . , v p r (i s ), k r }  ν r + 1, n − s − 1  2, E ≡ p ν r r (mod p ν r +1 r ), F ≡ p ν r r (mod p k r r ), and p k r r | G , (e) for each p r ∈ S 1 , we have min{v p r (i 1 ), . . . , v p r (i s ), k r }  ν r + 1, n − s − 1  1, p ν r r E, p k r r | F , and p k r r | G, (f) if 2 ∈ T , then min{v 2 (i 1 ), . . . , v 2 (i s ), k 0 − 1} = ν 0 − 1, n − s − 1  0, 2 ν 0 | E, 2 k 0 | F , and 2 k 0 | G , (g) if 2 ∈ U, then min{v 2 (i 1 ), . . . , v 2 (i s ), k 0 − 1}  ν 0 , 2 ν 0 E, G ≡ 2 ν 0 (mod 2 k 0 ), and either (1) 2 k 0 | F and w = 0 (so u(a) = 1) or (2) F ≡ 2 ν 0 (mod 2 k 0 ) and n − s − 1  w + 2 (so u(a)  2). There are certain degenerate cases where terms of the form α τ+F or α τ+G do not ap- pear in (3.4); this occurs exactly when conditions (c)–(g) force F ≡ G ≡ 0 (mod N). The reason for our final comment is that a term of the form α τ+F arises only when S 0 ∪ U is nonempty, while terms of the f orm α τ+G arise only when U is nonempty. Let ϕ : D 2N → D 2N /α N/p k i i  ∼ = D 2p k i i be the canonical projection. We remark that under the map ϑ : D n 2N → D n 2p k i i , ϑ(a) = (ϕ(a 1 ), . . . , ϕ(a n )), the images of the representa- tives in (3.4) agree with the representatives in [3, Theorems 3.1 and 4.2] up to the ordering of α i 1 , . . . , α i s . Thus Theorem 3.1 can be viewed as a generalization of the results in [3]. Before pro ceeding with the proof of Theorem 3.1, we give two examples to familiarize the reader with the content of parts (ii)(b)–(g). Suppose N = 225 = 3 2 ·5 2 , p 1 = 3, p 2 = 5, the electronic journal of combinatorics 16 (2009), #R95 7 n = 2, and consider the block Π({3, 5}, ∅, ∅, ∅, ∅ ) (1, 1)(2, 3). Since S 0 = S 1 = T = U = ∅, only t he conditions in parts (a)–(c) apply; furthermore, there are no terms of the form α τ+F or α τ+G . From (ii)(b), we have 0  τ < 15, τ ≡ 2 (mod 3 ) , and τ ≡ 3 (mod 5), so τ = 8. From (ii)(c), min{v 3 (i 1 ), . . . , v 3 (i s ), 2} = 1 and min{v 5 (i 1 ), . . . , v 5 (i s ), 2} = 1, so we must have s = 1 and v 3 (i s ) = v 5 (i s ) = 1; a lso, 3 | E and 5 | E, so 15 | E. Finally, from (ii)(a), 0  i 1  2 25/2. Thus, by ( 3.4), the equivalence classes in this block are represented by (α 15i , α 8+15e β), where gcd(15, i) = 1, 1  i  15/2, and e ∈ Z. Now, suppose instead that N = 36 = 2 2 · 3 2 , p 0 = 2, p 1 = 3, n = 2, and consider the block Π({3}, ∅, ∅, ∅, {2} ) (1, 2)(0, 7). From (ii)(b), we have 0  τ < 18, τ ≡ 0 (mod 2), and τ ≡ 7 (mod 9), so τ = 16. From (ii)(g) , we have 2E. Now, (ii)(g)(2) would require that n  3, so we only need to consider (ii)(g)(1); this condition implies that there are no terms of the form α τ+F or α τ+G . Moreover, since 2 ∈ U, both terms must be of the form α i β. Finally, from (ii)(c), we have 3 2 | E, so E ≡ 18 (mod 36) . Thus, the (unique) equivalence class in this block is represented by (α 34 β, α 16 β). Proof of Theorem 3.1. (i) This is clear. (ii) First, we observe t hat different tuples in (3.4) have different combinations of invari- ants Λ(a), π(a), σ p r (a), and u(a) (whenever these invariants are defined for a). Thus, different tuples in (3.4) are inequivalent. Next, we show that every a ∈ Π(R, S 0 , S 1 , T, U)(ν r )(τ r ) is equivalent to one of the tuples in (3.4). Since we can use a sequence of Hurwitz moves to shift all the terms of the for m α i to the front, we may as well assume that a has the form a = (α i ′ 1 , . . . , α i ′ s , α i ′ s+1 β, . . . , α i ′ n β). The general idea behind our proof is to write a in the form a = (α i ′ 1 , . . . , α i ′ s , α τ+e 1 Q p r |N p ν r r β, . . . , α τ+e t Q p r |N p ν r r β) and consider the effects of Hurwitz moves on the numbers e 1 , . . . , e t modulo p k r −ν r r for each prime p r dividing N. To avoid cluttering up expressions, we shall use the notation  p ν r r to mean  p r |N p ν r r in the sequel; if a different product is intended, it will be specified in the subscript of the product symbol. Note that the existence and uniqueness of τ is a direct consequence of the Chinese Remainder Theorem. Because the case p r = 2 must be handled differently f rom the case of odd p r , we shall first prove the t heorem for odd values of N, and then show how the proof can be modified to work for even values of N. Observe that it suffices to prove that we can obtain the conditions in parts (c)–(g), since we can then use (2.3) and Lemma 2.1(i) repeatedly to ensure that part (a) is also satisfied. the electronic journal of combinatorics 16 (2009), #R95 8 First suppose that N is odd, so that we only need to prove that we can obtain the conditions in parts (c)–(e). We proceed by induction on t, the number of terms of the form α i β in a. The case t = 1 is trivial. Suppose t = 2. Write a in the form a = (α i ′ 1 , . . . , α i ′ s , α τ+e 1 Q p ν r r β, α τ+e 2 Q p ν r r β). Note that by the definition of C p r ν r ,τ r ,0 , we cannot have a ∈ C p r ν r ,τ r ,0 for any prime divisor p r of N (because t = 2). Hence, we must have e 1 ≡ e 2 (mod p r ) for every prime p r ∈ S 0 ∪S 1 . Suppose that p r ∈ R. By the definition of B p r ν r ,τ r , either Λ(a) = ∅ and at least one of v p r (i ′ 1 ), . . . , v p r (i ′ s ), say v p r (i ′ k ), is equal to ν r , or Λ(a) = ∅. First suppo se that we are in the fo r mer case. Applying (2.3) a nd (2.4) multiple times, we can shift the term α i ′ k to the right until the la st three terms of a are (α i ′ k , α τ+e 1 Q p ν r r β, α τ+e 2 Q p ν r r β). If e 1 ≡ e 2 (mod p r ), then applying Lemma 2.1(i) to the first two terms yields (α −i ′ k , α τ+e ′ 1 Q p ν r r β, α τ+e 2 Q p ν r r β), where e ′ 1 ≡ e 2 (mod p r ). Thus we may assume that e 1 ≡ e 2 (mod p r ) for all prime divisors p r of N. Now, by Lemma 2.1(iv), we have (α τ+e 1 Q p ν r r β, α τ+e 2 Q p ν r r β) ∼ (α τ+f 1 Q p ν r r β, α τ+f 2 p k r −ν r r Q p ν r r β) (3.5) for some f 2 such that if p r ′ is another prime divisor of N such that p k r ′ −ν r ′ r ′ | e 2 , then p k r ′ −ν r ′ r ′ | f 2 also. If Λ(a) = ∅ instead, then ν r = k r by definition of B p r ν r ,τ r and we obtain (3.5) without any additional work. Repeating this argument for each prime p r dividing N, we have (α τ+e 1 Q p ν r r β, α τ+e 2 Q p ν r r β) ∼ (α τ+E β, α τ β). This completes the case t = 2. Now assume t > 2. Again, we write a in the for m a = (α i ′ 1 , . . . , α i ′ s , α τ+e 1 Q p ν r r β, . . . , α τ+e t Q p ν r r β). First consider p r ∈ R. As before, we wish to apply a sequence of Hurwitz moves to obtain an n-tuple a ′ = (α j 1 , . . . , α j s , α τ+f 1 Q p ν r r β, . . . , α τ+f t−1 Q p ν r r β, α τ+f t Q p ν r r β) ∼ a such that if p r ′ is another prime divisor of N such that p k r ′ −ν r ′ r ′ | e t , then p k r ′ −ν r ′ r ′ | f t also. Using a similar argument as above, we may assume that e t−1 ≡ e t (mod p r ) for every p r ∈ R, and hence by Lemma 2.1(iv), we have (α τ+e t−1 Q p ν r r β, α τ+e t Q p ν r r β) ∼ (α τ+f t−1 Q p ν r r β, α τ+f t p k r −ν r r Q p ν r r β) the electronic journal of combinatorics 16 (2009), #R95 9 for some f t such that if p r ′ is another prime divisor of N such that p k r ′ −ν r ′ r ′ | e t , then p k r ′ −ν r ′ r ′ | f t also. Repeating this argument for each prime p r ∈ R, we have a = (α i ′ 1 , . . . , α i ′ s , α τ+e 1 Q p ν r r β, . . . , α τ+e t Q p ν r r β) ∼ (α j 1 , . . . , α j s , α τ+g 1 Q p ν r r g 1 β, . . . , α τ+g t−1 Q p ν r r β, α τ+g t Q p ν r r β), where p k r −ν r r | g t for every prime p r ∈ R. Now consider p r ∈ S 0 ∪ S 1 . Assume that g l ≡ g l+1 ≡ · · · ≡ g t (mod p r ). By (2.6) and Lemma 2 .1 ( iv), we have (α τ+g l Q p ν r r β, α τ+g l+1 Q p ν r r β, . . . , α τ+g t Q p ν r r β) ∼ (α τ+g ′ l Q p ν r r β, α τ+g l Q p ν r r β, . . . , α τ+g t Q p ν r r β) ∼ · · · ∼ (α τ+g ′ l Q p ν r r β, . . . , α τ+g ′ t−2 Q p ν r r β, α τ+g l Q p ν r r β, α τ+g t Q p ν r r β) ∼ (α τ+g ′ l Q p ν r r β, . . . , α τ+g ′ t−2 Q p ν r r β, α τ+h t−1 Q p ν r r β, α τ+h t Q p k r r β), for some h t such that if p r ′ is another prime divisor of N such that p k r ′ −ν r ′ r ′ | g t , then p k r ′ −ν r ′ r ′ | h t also. Repeating this argument for each prime p r ∈ S 0 ∪ S 1 , we obtain a = (α i ′ 1 , . . . , α i ′ s , α τ+e 1 Q p ν r r β, . . . , α τ+e t Q p ν r r β) ∼ (α j 1 , . . . , α j s , α τ+h 1 Q p ν r r β, . . . , α τ+h t−1 Q p ν r r , α τ β) = b. If h 1 , . . . , h t−1 are not all the same modulo p r for any prime divisor p r of N, then the induction hypothesis applies to b = (α j 1 , . . . , α j s , α τ+h 1 p ν r r β, . . . , α τ+h t−1 p ν r r , α τ β). So assume that the set I o f prime divisors p r of N such that h 1 ≡ · · · ≡ h t−1 ≡ 0 (mod p r ) is nonempty. Let J be the set of prime divisors of N that are not in I. By the Chinese Remainder Theorem, we can find an integer M satisfying the system of congruences M ≡ 0 (mod p k s s ) for each p s ∈ J, M  p∈I p=p r p ≡ 1 (mod p k r r ) for each p r ∈ I. Write b as (α j 1 , . . . , α j s , α τ+h ′ 1 Q r ∈I p ν r r β, . . . , α τ+h ′ t−1 Q r ∈I p ν r r , α τ β). Let x ∈ Z be such that x ≡ −h ′ t−1 (mod p r ) for each p r ∈ I and x ≡ 0 (mod p k s s ) for each p s ∈ J. Then, using Lemma 2.1(iii) repeatedly, we have (α τ+h ′ t−2 Q p r ∈I p ν r r β, α τ+h ′ t−1 Q p r ∈I p ν r r β, α τ β) ∼ (α τ+h ′ t−2 Q p r ∈I p ν r r β, α τ+(h ′ t−1 +M) Q p r ∈I p ν r r β, α τ+M Q p r ∈I p ν r r β) ∼ (α τ+(h ′ t−2 +x) Q p r ∈I p ν r r β, α τ+(h ′ t−1 +x+M) Q p r ∈I p ν r r β, α τ+M Q p r ∈I p ν r r β) ∼ (α τ+(h ′ t−2 +x) Q p r ∈I p ν r r β, α τ+(h ′ t−1 +x) Q p r ∈I p ν r r β, α τ β). (3.6) the electronic journal of combinatorics 16 (2009), #R95 10 [...]... T Ben-Itzhak and M Teicher, Graph theoretic method for determining Hurwitz equivalence in the symmetric group, Israel J Math 135 (2003) 83–91 [2] D Gorenstein, Finite Groups, 2nd ed., Chelsea Publishing Company, New York, 1980 [3] X Hou, Hurwitz equivalence in tuples of generalized quaternion groups and dihedral groups, Electron J Combin 15 (2008) #R80, 10pp [4] S P Humphries, Finite Hurwitz braid... such that a, b ∈ Cν,τ , and u(a) = u(b) if 2 | N 4 Bn -orbits in Tuples of Dicyclic and Semidihedral Groups The results in the previous section can also be applied to classify the Bn -orbits in dicyclic groups, which are closely related to dihedral groups The similarity between dihedral groups and dicyclic groups can be seen from the presentation of the dicyclic group Dic 4M of order 4M: Dic 4M = α,... representatives of the Bn -orbits in each of A, Bν,τ , and Cν,τ the electronic journal of combinatorics 16 (2009), #R95 14 For a = (αi1 β j1 , , in β jn ) ∈ Cν,τ , where 0 ik < 2m−1 and 0 jk 1, let u(a) = #{k : jk = 1 and ik ≡ τ (mod 2v+1 )} Again, it is easy to see that u(a) is an invariant of the Hurwitz action n The following theorem classifies the Bn -orbits in SD2m Theorem 4.1 Let m 4, and let the semidihedral. .. while different n -tuples in (4.2) have different combinations of invariants λ(a), π(a), and u(a) This allows us to establish the following criterion for two n n -tuples in SD2m to be equivalent Corollary 4.2 Let m 4, and let the semidihedral group SD2m be partitioned into sets A, Bν,τ , and Cν,τ as above (i) Two n -tuples a, b ∈ A are equivalent if and only if a is a permutation of b (ii) Two n -tuples a, b... -orbits in SD2m and M2m The proofs of our results are very similar to those in [3] and in Section 3, hence we omit them 4.1 n Bn -orbits in SD2m The semidihedral group SD2m of order 2m is defined for any m 3 When m = 3, SD8 is isomorphic to the abelian group Z2 × Z4 , so the problem of determining the Bn -orbits in SD8 is trivial In what follows, we concentrate on the case m 4 Like the dihedral group and. .. H98230-061-001), and the Massachusetts Institute of Technology Department of Mathematics The author would like to thank Joseph Gallian for his support and encouragement, as well as Ricky Liu for his assistance in proofreading this paper Finally, the author would like to thank the referee for pointing out an error in Theorem 4.3 in a previous version of this paper, and for several other useful comments and suggestions... αis+1 β, , in β), where 0 r s < n, {i1 , , ir } ⊂ 2Z, {ir+1 , , is } ⊂ 1 + 2Z, 0 ir < 2m−1 , 0 ir+1 ··· is < 2m−2 , 0 is+1 ··· in 1 m−1 in 1 in < 2 (4.3) i1 ··· 2m−2 , and As before, the invariants Φ(a), Ψ(a) and π(a) show that distinct n -tuples in (4.3) are n inequivalent This yields the following criterion for two n -tuples in M2m to be equivalent Corollary 4.4 Let m above 3, and let the group... partitioned into sets D and Dc as (i) Two n -tuples a, b ∈ D are equivalent if and only if a is a permutation of b (ii) Two n -tuples a, b ∈ Dc are equivalent if and only if Φ(a) = Φ(b), Ψ(a) = Ψ(b) and π(a) = π(b) the electronic journal of combinatorics 16 (2009), #R95 16 Acknowledgments This research was carried out at the University of Minnesota Duluth under the supervision of Joseph Gallian Financial... equivalent if and only if λ(a) = λ(b) and π(a) = π(b) (iii) Two n -tuples a, b ∈ Cν,τ are equivalent if and only if λ(a) = λ(b), u(a) = u(b), and π(a) = π(b) the electronic journal of combinatorics 16 (2009), #R95 15 n Bn -orbits in M2m 4.2 Let m 3 Recall that M2m has the following representation in terms of generators and relations: m−1 m−2 M2m = α, β | α2 = β 2 = 1, βαβ −1 = α2 +1 Like the dihedral group,... the dicyclic group, and the semidihedral group, every element of M2m can be uniquely written in the form αi β j , where 0 i < 2m−1 and 0 j 1 n For a = (αi1 β j1 , , in β jn ) ∈ M2m , let Φ(a) = the multiset{i′k : jk = 0}, where i′k = ik , if ik is even; m−2 ik mod 2 , if ik is odd; and let Ψ(a) = the multiset{i′′ : jk = 1}, where i′′ = ik mod 2m−2 k k n Then Φ(a) and Ψ(a) are invariants of the Hurwitz . Hurwitz Equivalence in Tuples of Dihedral Groups, Dicyclic Groups, and Semidihedral Groups Charmaine Sia Department of Mathematics Massachusetts Institute of Technology Cambridge,. each of which is invariant under the Hurwitz action. We then find a number of invariants of the Hurwitz action and show that these invariants completely determine the equivalence classes within each. C p ν,τ , and u(a) = u(b) if 2 | N. 4 B n -orbits in Tuples of Dicyclic and Semidihedral Groups The results in the previous section can also be applied to classify the B n -orbits in dicyclic groups,