Representation of permutations as products of two cycles

Herzog, Marcel; Kaplan, Gil; Lev, Arieh

doi:10.1016/j.disc.2004.01.019

Cited by 8 publications

(7 citation statements)

References 6 publications

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“…The next lemma follows immediately from the following result in [6]. Then for each m-cycle σ ∈ S n there exist in S n cycles C i , 1 i t, of sizes l i , respectively, such that σ = C 1 C 2 · · · C t .…”

Section: Lemma 21 Let M N Be Even Integers Satisfying 2 M N − 2 Anmentioning

confidence: 85%

“…Given integers k, l 2, we ask for the largest integer n = n(k, l) such that every permutation in A n is a product of k cycles of length l. By the definition of A n , n(k, l) exists only if either l is odd or k is even. E. Bertram solved the problem for k = 2 in 1972 (see also [6] for another proof of this result). He proved the following theorem: It follows from Theorem 1.1 that if k = 2 and l > 2, then n(2, l) equals the largest integer n satisfying 3n 4 = l. Suppose that l = 3d + e, where e ∈ {0, 1, 2} and let n = 4l 3 + 1 = 4d + e + 1.…”

Section: Introductionmentioning

confidence: 89%

“…The covering numbers for the groups A n and S n were extensively studied by Brenner et al (see [4] for a survey), Dvir [5], Vishne [11] and many others. In particular, since the set of all cycles of a given odd length l (2 l < n − 1) constitutes a conjugacy class of A n (see [10, 11.1.5]), our results (as well as the results in [2,3,6]) deal with the covering numbers of these classes of l-cycles in A n . The covering numbers for various groups other then A n (or S n ) were also extensively studied.…”

Section: Introductionmentioning

confidence: 98%

See 2 more Smart Citations

Covering the alternating groups by products of cycle classes

Herzog¹,

Kaplan²,

Lev³

2008

Journal of Combinatorial Theory, Series A

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Given integers k, l 2, where either l is odd or k is even, we denote by n = n(k, l) the largest integer such that each element of A n is a product of k cycles of length l. For an odd l, k is the diameter of the undirected Cayley graph Cay(A n , C l ), where C l is the set of all l-cycles in A n . We prove that if k 2 and l 9 is odd and divisible by 3, then 2 3 kl n(k, l) 2 3 kl + 1. This extends earlier results by Bertram [E. Bertram, Even permutations as a product of two conjugate cycles, J. Combin. Theory 12 (1972) 368-380] and Bertram and Herzog [E. Bertram, M. Herzog, Powers of cycle-classes in symmetric groups, J. Combin. Theory Ser. A 94 (2001) 87-99].

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Section: Lemma 21 Let M N Be Even Integers Satisfying 2 M N − 2 Anmentioning

confidence: 85%

Section: Introductionmentioning

confidence: 89%

Section: Introductionmentioning

confidence: 98%

See 1 more Smart Citation

Covering the alternating groups by products of cycle classes

Herzog¹,

Kaplan²,

Lev³

2008

Journal of Combinatorial Theory, Series A

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“…It is easy to check, see for example, [, Lemma 2], that

π \in {sans-serifA}_{n}

if and only if

2 | (m_{π} - n_{π})

. Furthermore, given any two integers

l_{1}, l_{2} ⩾ 2

, by [, Theorem 7],

π

can be written as a product

u v

of an

l_{1}

‐cycle

u \in {sans-serifS}_{n}

and an

l_{2}

‐cycle

v \in {sans-serifS}_{n}

if and only if at least one of the following statements holds: (a)

n_{π} = 2

and

π

is a disjoint product of an

l_{1}

‐cycle and an

l_{2}

‐cycle; (b)

l_{1} + l_{2} = m_{π} + n_{π} + 2 s

for some

s \in {double-struckZ}_{⩾ 0}

and

false| l_{1} - l_{2} false| ⩽ m_{π} - n_{π}

. (i)First we consider the case

p = n

. If

σ

is an odd permutation, then by [, Corollary 3.1],

σ = z y^{- 1}

, where

z

is an

(n - 1)

‐cycle and

y \in {sans-serifA}_{n}

is an

…”

Section: Almost Simple Groups Imentioning

confidence: 99%

“…Clearly, m π 2n π . It is easy to check, see for example, [14,Lemma 2], that π ∈ A n if and only if 2|(m π − n π ). Furthermore, given any two integers l 1 , l 2 2, by [14,Theorem 7], π can be written as a product uv of an l 1 -cycle u ∈ S n and an l 2 -cycle v ∈ S n if and only if at least one of the following statements holds:…”

Section: Almost Simple Groups Imentioning

confidence: 99%