Proc. Indian Acad. Sci. (Math. Sci.) Vol. 122, No. 2, May 2012, pp. 175–179.c© Indian Academy of Sciences

Permutation groups with bounded movementhaving maximum orbits

MEHDI ALAEIYAN and BEHNAM RAZZAGHMANESHI

Department of Mathematics, Islamic Azad University-Karaj Branch, Karaj, IranE-mail: alaeiyan@iust.ac.ir

MS received 14 April 2010; revised 12 June 2011

Abstract. Let G be a permutation group on a set � with no fixed points in � andlet m be a positive integer. If no element of G moves any subset of � by more than mpoints (that is, |�g\�| ≤ m for every � ⊆ � and g ∈ G), and also if each G-orbit hassize greater than 2, then the number t of G-orbits in � is at most 1

2 (3m − 1). Moreover,the equality holds if and only if G is an elementary abelian 3-group.

Keywords. Permutation group; bounded movement; orbit.

1. Introduction

Let G be a permutation group on a set � with no fixed points in � and let m be a positiveinteger. If for a subset � of � the size |�g\�| is bounded, for g ∈ G, we define themovement of � as move(�) = maxg∈G |�g\�|. If move(�) ≤ m for all � ⊆ �, then Gis said to have bounded movement and the movement of G is defined as the maximum ofmove(�) over all subsets �, that is,

m := move(G) := sup{|�g\�||� ⊆ �, g ∈ G}.This notion was introduced in [1,2,5]. By Theorem 1 of [5] if G has bounded movementequal to m, then � is finite. Moreover both the number of G-orbits in � and the lengthof each G-orbit are bounded above by linear functions of m. In particular, it was shownin Lemma 3.5 of [3] that the number of G-orbits is at most 2m − 1. In this paper we willimprove this to 1

2 (3m − 1), when each G-orbit has size ≥3. If m = 1, then t = 1, |�| = 3and G is Z3 or S3. So in the rest of this paper we suppose that m is greater than 1. Wepresent here a classification of all groups for which the bound 1

2 (3m − 1) is attained. Weshall say that an orbit of permutation group is nontrivial if its length is greater than 1. Themain result is the following theorem.

Theorem 1.1. Let m be a positive integer and suppose that G is a permutation groupon a set � such that G has no fixed points in �, and G has bounded movement equalto m. If each G-orbit has size greater than 2, then the number t of G-orbits in � is atmost 1

2 (3m − 1). Moreover, t = 12 (3m − 1), if and only if m is a power of 3, and G is

elementary abelian of order 3m, all G-orbits have length 3, and the pointwise stabilizersof the G-orbits are precisely 1

2 (3m − 1) distinct subgroups of G of index 3.

175

176 Mehdi Alaeiyan and Behnam Razzaghmaneshi

Note that an orbit of a permutation group is non trivial if its length is greater than 1. Thegroups described in the next section are examples of permutation groups with boundedmovement equal to m which have exactly 1

2 (3m − 1) nontrivial orbits.

2. Examples and preliminaries

Let 1 �= g ∈ G and suppose that g in its disjoint cycle representations has t nontrivialcycles of lengths l1, . . . , lt , say. We might represent g as

g = (a1a2 . . . al1)(b1b2 . . . bl2) . . . (z1z2 . . . zlt ).

Let �(g) denote a subset of � consisting of �li/2 points from the i-th cycle, for each i ,chosen in such a way that �(g)g ⋂

�(g) = ∅. For example, we could choose

�(g) = {a2, a4, . . . , ak1 , b2, b4, . . . , bk2 , . . . , z2, z4, . . . , zkt },

where ki = li − 1 if li is odd and ki = li if li is even. Note that �(g) is not uniquelydetermined as it depends on the way each cycle is written. For any set �(g) consists ofevery second point of every cycle of g. From the definition of �(g) we see that

|�(g)g\�(g)| = |�(g)| =t∑

i=1

�li/2.

The next lemma shows that this quantity is an upper bound for |�g\�| for an arbitrarysubset � of �.

Lemma 2.1 (Lemma 2.1 of [4]). Let G be a permutation group on a set � and supposethat � ⊆ �. Then for each g ∈ G, |�g\�| ≤ ∑t

i=1�li/2, where li is the length of thei-th cycle of g and t is the number of nontrivial cycles of g in its disjoint cyclerepresentation. This upper bound is attained for the � = �(g) defined above.

Now we will show that there certainly is an infinite family of 3-groups for which themaximum bound obtained in Theorem 1.1 holds.

Example 2.2. Suppose that d is a positive integer, G := Zd3 , t := 1

2 (3d − 1), and letH1, . . . , Ht be an enumeration of the subgroups of index 3 in G. Define �i to be the cosetspace of Hi in G and � := �1 ∪ . . . ∪ �t . If g ∈ G, then g lies in 1

2 (3d−1 − 1) of thegroups Hi and therefore acts on � as a permutation with 1

2 (3d − 3) fixed points and 3d−1

disjoint 3-cycles. Taking every second point from each of these 3-cycles to form a set �,by Lemma 2.1 we see that move(G) = 3d−1. It follows that G has a bounded movementequal to m, and G has 1

2 (3move(G) − 1) nontrivial orbits in �.

Since m > 1, the classification in Theorem 1.1 follows immediately from the followingtheorem about subsets with movement m.

Permutation groups with bounded movement 177

DEFINITION

Let G be a permutation group on a set � with orbits �i , for i ∈ I . We shall say that asubset � ⊆ � cuts across each G-orbit if �i := � ∩ �i /∈ {∅,�i }, for all i ∈ I .

Now we have the following theorem which is basic.

Theorem 2.3. Let G ≤ Sym(�) be a permutation group with t orbits for positive integert , such that each orbit has length greater than 2. Moreover suppose that � ⊆ � such thatmove(�) = m > 1, and � cuts across each G-orbit. Then t ≤ 1

2 (3m − 1). Moreover,if t = 1

2 (3m − 1) then:

(1) G is an elementary abelian 3-group and every G-orbit has size 3;(2) If the rank of the group G is d, then d ≥ 3, t = 1

2 (3d − 1) and m = 3d−1;(3) The t different G-orbits are (G-equivalent to) the coset spaces of the 1

2 (3d−1) differentsubgroups of index 3 in G.

3. Proof of Theorem 2.3

Let �1, . . . , �t be t orbits of G of lengths n1, . . . , nt . Choose αi ∈ �i and let Ti := Gαi

so that |G : Ti | = ni . For g ∈ G, set �(g) := {αi |αgi �= αi } and let γ (g) := |�(g)|. Since

�(g) ∩ �(g)g = ∅ it follows that γ (g) ≤ m for all g ∈ G. Let � := �1 ∪ · · · ∪ �t , andlet G and T1, . . . , Tt denote the finite permutation groups on � induced by G and T1, . . . ,

Tt respectively. Then ni = |G : Ti |.For g ∈ G, let g ∈ G denote the permutation of � induced by g. Then as γ (1G) = 0,

we have∑

g∈G γ (g) < m|G|.Now, counting the pairs (g, i) such that g ∈ G and α

gi �= αi gives

∑g∈G γ (g) = ∑

i |{g ∈ G|αgi �= αi }| = ∑

i |{g ∈ G|g /∈ Ti }|= ∑

i (|G| − |Ti |) = |G| ∑i (1 − 1ni

).

It follows that∑

i (1 − 1ni

) < m. Since ni ≥ 3 for each i , it follows that∑

i (1 − 1ni

) ≥ 2t3

and hence 2t3 < m, that is, t ≤ 1

2 (3m − 1).Consequently G has at most 1

2 (3m − 1) orbits in �. Suppose that G has exactly torbits say, �1,�2, . . . , �t , where t = 1

2 (3m − 1). Suppose further that � ⊆ � has move(�) = m that cuts across each of the G-orbits �i . For each i , set �i = � ∩ �i . Note that0 < |�i | < ni . Then we have the following claim.

Claim 3.1. If Theorem 2.3 holds for the special case in which |�i | = 1 for eachi = 1, . . . (3m − 1)/2, then it holds in general.

Proof. Let Theorem 2.3 hold for the special case in which |�i | = 1. For each i =1, . . . , 1

2 (3m −1), define∑

i := {�gi |g ∈ G}, and note that |∑i | ≥ 2 since � cuts across

�i . Set � := ∪i≥1∑

i . Then G induces a natural action on � for which the G-orbits are�1, . . . , �t . Let G� denote the permutation group induced by G on �, and let K denotethe kernel of this action.

178 Mehdi Alaeiyan and Behnam Razzaghmaneshi

We claim that the t-element subset �� = {�1, . . . , �t } ⊆ � has movement equal to mrelative to G� , and that �� cuts across each G�-orbit �i . For each g ∈ G, |�g −�| ≤ mand hence |�g

� − �� | ≤ m. Thus move (��) ≤ m. Also, since |�i | ≥ 2 and �� ∩ �i

consists of the single element �i of �i , the set �� cuts across each of the 12 (3m −1) orbits

�i . However, it follows that the number of G�-orbits is at most 12 (3 move(��) − 1), and

hence move (��) = m.Thus the hypotheses of Theorem 2.3 hold for the subset �� ⊆ � relative to G� , and

�� meets each G�-orbit in exactly one point. By our assumption it follows that t =12 (3d − 1) = 1

2 (3m − 1) for some d > 1, and that G� = Zd3 and each |�i | = 3. Further,

the subgroups Hi of G fixing �i setwise range over the 12 (3d − 1) distinct subgroups

which have index 3 in G and which contain K. In particular, for each i , Hi is normal inG and hence the Hi -orbits in �i are blocks of imprimitivity for G, and their number isat most |G : H | = 3. Since Hi fixes �i setwise it follows that �i is an Hi -orbit andni = 3|�i |.

Let g ∈ G\K . Then as in Example 2.2, in its action on �, g moves exactly m of the�i . Since the �i are blocks of imprimitivity for G, each �

gi is either �i or �i ∩ �i

g = φ.It follows that |�g\�| is equal to the sum of the sizes of the m subsets �i moved by g.However, since move (�) = m, each of these m subsets �i must have size 1. Since foreach i we may choose an element g which moves �i , we deduce that each of the �i hassize 1, and that K is the identity subgroup. It follows that Theorem 2.3 holds for G. Thusthe claim is proved.

From now on we may and shall assume that each |�i | = 1. Let �i = {ωi }. Furtherwe may assume that n1 ≤ n2 ≤ · · · ≤ nt . For g ∈ G, let c(g) denote the number ofintegers i such that ω

gi = ωi . Note that since move (�) = m, we have

c(g) > t−m = 1

2(3m−1)−m = (m−1)/2 and also c(1G) = t > (m−1). (∗)

Lemma 3.2. With the above notion we have n1 = 3.

Proof. Let X denote the number of pairs (g, i) such that g ∈ G, 1 ≤ i ≤ t and ωgi = ωi .

Then X = ∑g∈G c(g), and by our observations, X > |G|. 1

2 (m − 1), because by therelation of (*) we have

X = ∑g∈G c(g)=∑

1 �=g∈G c(g) +c(1G)>(|G|−1) 12 (m − 1) + (m − 1)

= |G| 12 (m − 1) + 1

2 (m − 1) > |G| 12 (m − 1).

On the other hand, for each i , the number of elements of G which fix ωi is |Gωi | = |G|ni

,

and hence X = |G|∑ti=1 n−1

i . If all ni ≥ 4, then X ≤ |G| t4 ≤ |G|(m − 1)/2 (since

m ≥ 2 ) which is a contradiction. Hence ni = 3.A similar argument to this enables us to show that all ni = 3, for each (2 ≤ i ≤ t) and

hence G is an elementary abelian 3-group.

Lemma 3.3. Let G = Zr3 for some r ≥ 2. Moreover for each ni = 3, the stabilizers Gωi

(2 ≤ i ≤ t) are pairwise distinct subgroups of index 3 in G, and for each g �= 1,c(g) =(m − 1)/2, such that g ∈ G, 2 ≤ i ≤ t , and ω

gi = ωi .

Permutation groups with bounded movement 179

Proof. By Lemma 3.2, n1 = 3. Thus H := Gω1 is a subgroup of index 3. This time wecompute the number Y of pairs (g, i) such that g ∈ G\H, 2 ≤ i ≤ t , and ω

gi = ωi . For

each such g, ωg1 �= ω1 and hence there are c(g) of these pairs with first entry g. Thus

Y = ∑g∈G\H c(g) ≥ |G\H |(m −1)/2 = (|G|−|H |)(m −1)/2 = (2|H |/3)(m −1)/2 =

|G| · (m − 1)/3.On the other hand, for each i ≥ 2, the number of elements of G\H , which fix ωi is

|Gωi \H |. If Gωi = H , |Gωi \H | = 0 while if Gωi �= H , then |Gωi \H | = |Gωi \(H ∩Gωi )| = |Gωi | − |Gωi ∩ H | = |G|(1 − 1/3) = 2/3|G|. Hence

Y = ∑ti=2 |G\H | = 2

3 |G| ∑ti=2

1ni

≤ 29 |G|(t − 1)

= 29 |G|( 1

2 (3m − 1) − 1) = |G|3 (m − 1).

So we have

|G|m − 1

3≤ Y ≤ |G|m − 1

3.

It follows that equality holds in both of the displayed approximations for Y. This means,in particular, that each ni = 3. Further, for each i ≥ 2, Gωi �= H and so r ≥ 2. Arguingin the same way with H replaced by Gωi , for some i ≥ 2, we see that Gωi �= Gω j ifj �= i , and also if g ∈ Gωi then c(g) = (m − 1)/2. Thus the stabilizers Gωi (1 ≤ i ≤ t)are pairwise distinct, and if g �= 1, then c(g) = (m − 1)/2.

Finally we determine m.

Lemma 3.4. m = 3r−1.

Proof. We use the information in Lemma 3.3 to determine precisely the quantity X =∑g∈G c(g) = t + (|G| − 1).(m − 1)/2 = 1

2 (3m − 1) + (3r − 1)(m − 1)/2.On the other hand, from the proof of Lemma 3.2,

X = |G|t∑

i=1

n−1i = |G|

3t = 3r−1 .

1

2(3m − 1).

Thus (3r−1 − 1)( 12 (3m − 1)) = (3r − 1)(m − 1)/2, and this implies that m = 3r−1.

The proof of Theorem 2.3 now follows from Lemmas 3.2–3.4.

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