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Commutator Subgroups of Generalized Hecke

and Extended Generalized Hecke Groups

S¸ule Kaymak (Sarıca), Bilal Demir, ¨Ozden Koruo˘glu and Recep S¸ahin

Abstract

Let p and q be integers such that 2 ≤ p ≤ q, p + q > 4 and let Hp,q be the generalized Hecke group associated to p and q. The generalized Hecke group Hp,qis generated by X(z) = −(z −λp)−1and Y (z) = −(z + λq)−1 where λp = 2 cosπp and λq = 2 cosπq. The extended generalized Hecke group Hp,qis obtained by adding the reflection R(z) = 1/z to the generators of generalized Hecke group Hp,q. In this paper, we study the commutator subgroups of generalized Hecke groups Hp,q and extended generalized Hecke groups Hp,q.

1

Introduction

In [5], Hecke introduced the groups H(λ) generated by two linear fractional transformations

T (z) = −1

z and U (z) = z + λ, where λ is a fixed positive real number. Let S = T U , i.e.,

S(z) = − 1 z + λ.

Key Words: Generalized Hecke groups, Extended generalized Hecke groups, commutator subgroups.

2010 Mathematics Subject Classification: 20H10, 11F06, 30F35 Received: 06.02.2016

Revised: 15.06.2017 Accepted: 20.06.2017

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Hecke showed that H(λ) is discrete if and only if λ = λq = 2 cos(πq), q ≥ 3

integer, or λ ≥ 2. We consider the former case q ≥ 3 integer and we denote it by Hq = H(λq). The Hecke group Hq is isomorphic to the free product of

two finite cyclic groups of orders 2 and q,

Hq=< T, S : T2= Sq = I >' C2∗ Cq.

The first few Hecke groups Hq are H3 = Γ = P SL(2, Z) (the modular

group), H4= H( √ 2), H5 = H(1+ √ 5 2 ), and H6= H( √

3). It is clear from the above that Hq ⊂ P SL(2, Z [λq]) unlike in the modular group case (the case

q = 3), the inclusion is strict and the index |P SL(2, Z [λq]) : Hq| is infinite as

Hq is discrete whereas P SL(2, Z [λq]) is not for q ≥ 4.

Lehner studied in [11] a more general class Hp,q of Hecke groups Hq, by

taking

X = −1 z − λp

and V = z + λp+ λq,

where 2 ≤ p ≤ q, p + q > 4. Here if we take Y = XV = −z+λ1

q, then we have

the presentation,

Hp,q=< X, Y : Xp= Yq = I >' Cp∗ Cq.

In particular Hp,q has the signature (0; p, q, ∞). We call these groups

as generalized Hecke groups Hp,q. We know from [11] that H2,q = Hq,

|Hq : Hq,q| = 2, and there is no group H2,2. Furthermore all Hecke groups

Hq are included in generalized Hecke groups Hp,q. Generalized Hecke groups

Hp,q have been also studied by Calta and Schmidt in [2] and [3].

Extended generalized Hecke groups Hp,qcan be defined similar to extended

Hecke groups Hq, by adding the reflection R(z) = 1/z to the generators of

generalized Hecke group Hp,q. Hence extended generalized Hecke groups Hp,q

have a presentation

Hp,q =< X, Y, R : Xp= Yq = R2= I, RX = X−1R, RY = Y−1R >,

that is

Hp,q=< X, Y, R : Xp = Yq = R2= (XR)2= (Y R)2= I >∼= Dp∗Z2Dq.

The groups Hp,q is a subgroup of index 2 in Hp,q.

Now we focus on the commutator subgroups of Hp,q and Hp,q. The

com-mutator subgroups of Hq, Hq, Hqm (m−th power subgroup of Hq) and Hp,q0

(p and q are relatively prime) have been studied by many authors see [1], [16], [18], [19], [20] and [22].

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In this paper, we study the commutator subgroups of Hp,q and Hp,q. We

give the generators, the group structures and the signatures of the commutator subgroups of Hp,q and Hp,q. Here we use the Reidemeister-Schreier method,

the permutation method (see, [21]) and the extended Riemann-Hurwitz con-dition (see, [10]) to get our results.

Remark 1. i) The Hecke groups Hq, the extended Hecke groups Hq and

their normal subgroups have been studied extensively for many aspects in the literature. For examples, see [4], [8], [9], [12] and [14]. Also, there are many relationships between the groups Hq (or Hq) and the automorphism groups of

Riemann (or Klein) surfaces or of regular maps {p, q}. Naturally, many results related with Hq and Hq can be generalized to the groups Hp,q and Hp,q.

ii) Generalized Hecke groups H(p1, p2, ..., pn) and extended generalized Hecke

groups H∗(p1, p2, ..., pn) have been introduced first by Huang in [6]. Our

stud-ied groups are the special cases (n = 2) of these groups given in [6] and [7].

2

Commutator Subgroups of Generalized Hecke Groups

H

p,q

In this section, we study the commutator subgroup Hp,q0 of generalized Hecke groups Hp,q. We use standard notation, in particular, G(n) denotes the nth

derived group of a group G.

Theorem 1. Let p and q be integers such that 2 ≤ p ≤ q, p + q > 4. i) Hp,q: Hp,q0

= pq.

ii) The commutator subgroup Hp,q0 of Hp,qis a free group of rank (p−1)(q −

1) with basis [X, Y ], [X, Y2], ..., [X, Yq−1], [X2, Y ], [X2, Y2], ..., [X2, Yq−1],

..., [Xp−1, Y ], [Xp−1, Y2], ..., [Xp−1, Yq−1]. And the signature of H0 p,q is

(pq−p−q−(p,q)+22 ; ∞(p,q)) where (p, q) is the greatest common divisor of p and

q. iii)For n ≥ 2, Hp,q: H (n) p,q = ∞.

Proof. i) Firstly, we set up the quotient group Hp,q/Hp,q0 . The quotient group

Hp,q/Hp,q0 is the group obtained by adding the relation XY = Y X to the

relations of Hp,q. Thus we have

Hp,q/Hp,q0 =< X, Y : X

p= Yq= I, XY = Y X >' C p× Cq.

ii) Now we can determine the generators of Hp,q0 by the Reidemeister-Schreier method. To do this, we choose the set Σ = {I, X, X2, ..., Xp−1, Y, Y2, ...,

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Yq−1, XY, XY2, ..., XYq−1, X2Y, X2Y2, ..., X2Yq−1, ..., Xp−1Y, Xp−1Y2, ..., Xp−1Yq−1} as a Schreier transversal. All possible products are

I.X.(X)−1= I, X.X.(X2)−1= I, .. . Xp−1.X.(I)−1= I, Y.X.(XY )−1= Y XY−1X−1= [Y, X], Y2.X.(XY2)−1= Y2X(XY2)−1= [Y2, X], .. . Yq−1.X.(XYq−1)−1= Yq−1XY−(q−1)X−1= [Yq−1, X], XY.X.(X2Y )−1= XY XY−1X−2= [X, Y ].[Y, X2], XY2.X.(X2Y2)−1= XY2XY−2X−2 = [X, Y2].[Y2, X2], .. . XYq−1.X.(X2Yq−1)−1 = XYq−1XY−(q−1)X−2 = [X, Yq−1].[Yq−1, X2], X2Y.X.(X3Y )−1= X2Y XY−1X−3= [X2, Y ].[Y, X3], X2Y2.X.(X3Y2)−1= X2Y2XY−2X−3= [X2, Y2].[Y2, X3], .. . X2Yq−1.X.(X3Yq−1)−1= X2Yq−1XY−(q−1)X−3= [X2, Yq−1].[Yq−1, X3], .. . Xp−1Y.X.(Y )−1= Xp−1Y XY−1 = [X−1, Y ], Xp−1Y2.X.(Y2)−1= Xp−1Y2XY−2= [X−1, Y2], .. . Xp−1Yq−1.X.(Yq−1)−1 = Xp−1Yq−1XY−(q−1)= [X−1, Yq−1].

The other products are equal to the identity. Thus, we find the generators of H0

p,q as [X, Y ], [X, Y2], ..., [X, Yq−1], [X2, Y ], [X2, Y2], ..., [X2, Yq−1], ...,

[Xp−1, Y ], [Xp−1, Y2], ..., [Xp−1, Yq−1]. The signature of H0

p,qcan be obtained

by permutation method and Riemann-Hurtwitz formula.

iii) If we take relations and abelianizing, we find that the resulting quotient is infinite. Thus, it follows that Hp,q00 has infinite index in Hp,q0 . Further since this has infinite index it follows that in each group in the derived series from this point on have infinite index.

Example 1. If p = 3 and q = 4, then H3,4 : H3,40

= 12. We choose Σ = {I, X, X2, Y, Y2, Y3, XY, XY2, XY3, X2Y, X2Y2, X2Y3} as a Schreier

transversal for H3,40 . Using the Reidemeister-Schreier method, we get the gen-erators of H3,40 as [X, Y ], [X, Y2], [X, Y3], [X2, Y ], [X2, Y2], [X2, Y3]. Also

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Corollary 1. If p = 2, then the generators of H2,q0 are [X, Y ], [X, Y2], ..., [X, Yq−1]. Also, the signatures of H0

2,q is either ( q−1

2 ; ∞) if q is odd, or

(q−22 ; ∞(2)) if q is even. These results coincide with the ones given in [1] and

[20], for Hecke groups Hq.

3

Commutator Subgroups of Extended Generalized Hecke

Groups H

p,q

In this section, we study the first commutator subgroups H0p,q of extended generalized Hecke groups Hp,q.

Theorem 2. Let p and q be integers such that 2 ≤ p ≤ q, p + q > 4.

i) If p and q are both odd numbers, then H0p,q=< X, Y : Xp= Yq = I >'

Cp∗ Cq.

ii) If p is an odd number and q is an even number, then H0p,q =< X, Y XY−1, Y2: Xp = (Y XY−1)p = (Y2)

q

2 = I >' Cp∗ Cp∗ Cq/2.

iii) If p is an even number and q is an odd number, then H0p,q =< X2, Y, XY X−1: (X2)p2 = Yq = (XY X−1)q = I >' C

p/2∗ Cq∗ Cq.

iv) If p and q are both even numbers, then H0p,q =< X2, Y X2Y−1, Y2,

XY2X−1, XY XY−1: (X2)p/2= (Y X2Y−1)p/2= (Y2)q/2= (XY2X−1)q/2=

I >' Cp/2∗ Cp/2∗ Cq/2∗ Cq/2∗ Z.

Proof. The quotient group Hp,q/H 0 p,q is Hp,q/H 0 p,q =< X, Y, R : Xp= Yq = R2= I, XR = RX−1, Y R = RY−1, XY = Y X, XR = RX, Y R = RY > . (3.1) i) If p and q are both odd numbers, then from (3.1), we get

Hp,q/H 0

p,q =< R : R

2= I >' C 2.

Now we can determine the Schreier transversal as Σ = {I, R}. According to the Reidemeister-Schreier method, we can form all possible products :

I.X.(I)−1= X, I.Y.(I)−1= Y, I.R.(R)−1= I, R.X.(R)−1= RXR, R.Y.(R)−1= RY R, R.R.(I)−1= I.

Since RXR = X−1 and RY R = Y−1, the generators of H0p,q are X and Y. Thus we have H0p,q =< X, Y : Xp= Yq = I >' H

p,q. The signature of H 0 p,q

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ii) If p is an odd number and q is an even number, then from (3.1), we obtain Hp,q/H 0 p,q =< Y, R : Y 2= R2= I, Y R = RY >' C 2× C2.

Now we can determine the Schreier transversal as Σ = {I, Y, R, Y R}. Accord-ing to the Reidemeister-Schreier method, we can form all possible products :

I.X.(I)−1= X, I.Y.(Y )−1 = I, I.R.(R)−1= I, Y.X.(Y )−1= Y XY−1, Y.Y.(I)−1= Y2, Y.R.(Y R)−1= I, R.X.(R)−1= RXR, R.Y.(Y R)−1= RY RY−1, R.R.(I)−1= I,

Y R.X.(Y R)−1= Y RXRY−1, Y R.Y.(R)−1= Y RY R, Y R.R.(Y )−1 = I.

After required calculations, we get the generators of H0p,q as X, Y XY−1 and Y2. Thus we obtain H0p,q =< X, Y XY−1, Y2 : Xp = (Y XY−1)p =

(Y2)q2 = I >' C

p∗ Cp∗ Cq/2. The signature of H 0

p,q is (0; p, p, (q/2), ∞).

iii) The proof is similar to ii). Also, the signature of H0p,qis (0; (p/2), q, q, ∞).

iv) If p and q are both even numbers, then from (3.1), we find

Hp,q/H 0

p,q =< X, Y, R : X

2= Y2= R2= (XY )2= (XR)2= (Y R)2= I >

' C2× C2× C2.

Now we can determine the Schreier transversal as Σ = {I, X, Y, R, XY, XR, Y R, XY R}. According to the Reidemeister-Schreier method, we can form all possible products : I.X.(X)−1= I, I.Y.(Y )−1= I, X.X.(I)−1= X2, X.Y.(XY )−1= I, Y.X.(XY )−1= Y XY−1X−1, Y.Y.(I)−1= Y2, R.X.(XR)−1= RXRX−1, R.Y.(Y R)−1= RY RY−1, XY.X.(Y )−1= XY XY−1, XY.Y.(X)−1 = XY2X−1, XR.X.(R)−1= XRXR, XR.Y.(XY R)−1= XRY RY−1X−1, Y R.X.(XY R)−1= Y RXRY−1X−1, Y R.Y.(R)−1= Y RY R,

XY R.X.(Y R)−1= XY RXRY−1, XY R.Y.(XR)−1= XY RY RX−1, The other products are equal to the identity. Since RXRX−1 = X−2, XRXR = I, Y RXRY−1X−1 = Y XY−1X−1, XY RXRY−1 = XY XY−1, RY RY−1 = Y−2, XRY RY−1X−1 = I, Y RY R = I and XY RY RX−1 = XY2X−1, we have the generators of H0p,q as X2, Y X2Y−1, Y2, XY2X−1,

XY XY−1. Then we get H0p,q =< X2, Y X2Y−1, Y2, XY2X−1, XY XY−1 :

(X2)p/2 = (Y X2Y−1)p/2 = (Y2)q/2 = (XY2X−1)q/2 = I > ' Cp/2∗ Cp/2∗

Cq/2∗ Cq/2∗ Z. Also, the signature of H 0

p,q is (0; (p/2)

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Corollary 2. 1) The first commutator subgroup of Hp,q does not contain any

reflection and so H0p,q is a subgroup of Hp,q.

2) If p = 2, then the first commutator subgroups H02,q coincide with the ones given in [16] for extended Hecke groups Hq.

3) Hp,q0 ≤ H0p,q≤ Hp,q.

Now we study the second commutator subgroup H00p,q of Hp,q, except the

case p and q are both even numbers, since in this case H00p,q has infinite index in Hp,q.

Theorem 3. Let p and q be integers such that 2 ≤ p ≤ q, p + q > 4. i) If p and q are both odd numbers, then H00p,q= Hp,q0 .

ii) If p is an odd number and q is an even number, then H00p,q is a free

group of rank (q − 1)p2− pq + 1.

iii) If p is an even number and q is an odd number, then H00p,q is a free group of rank (p − 1)q2− pq + 1.

Proof. The case i) can be seen from the Theorem 2.1. Since the cases ii) and iii) are similar, we prove only the case iii). If we take a = X2, b = Y, c = XY X−1, then the quotient group H0p,q/H00p,q is the group obtained by adding the relations ab = ba, ac = ca and bc = cb to the relations of H0p,q.Then

H0p,q/H00p,q ∼= Cp/2× Cq× Cq. Therefore, we obtain H 0 p,q: H 00 p,q = pq 2/2. Now we choose Σ = {I, a, a2, · · · , a(p/2)−1, b, b2, · · · , bq−1, c, c2, · · · , cq−1, ab, ab2, · · · , abq−1, a2b, a2b2, · · · , a2bq−1, · · · , a(p/2)−1b, a(p/2)−1b2, · · · , a(p/2)−1bq−1, ac, ac2, · · · , acq−1, a2c, a2c2, · · · , a2cq−1, · · · , a(p/2)−1c, a(p/2)−1c2, · · · , a(p/2)−1cq−1, bc, bc2, · · · , bcq−1, b2c, b2c2, · · · , b2cq−1, · · · , bq−1c, bq−1c2, · · · , bq−1cq−1, abc, abc2, · · · , abcq−1, ab2c, ab2c2, · · · , ab2cq−1, · · · , abq−1c, abq−1c2, · · · , abq−1cq−1, · · · , a(p/2)−1bc, a(p/2)−1bc2, · · · , a(p/2)−1

bcq−1, a(p/2)−1b2c, a(p/2)−1b2c2, · · · , a(p/2)−1b2cq−1, · · · , a(p/2)−1bq−1c, a(p/2)−1bq−1c2, · · · , a(p/2)−1bq−1cq−1} as a Schreier transversal for H00p,q.

According to the Reidemeister-Schreier method, we get the generators of H00p,q as follows. There are total (p − 1)q2− pq + 1 generators obtained by using

the theorem of Nielsen in [13]. These generators are[a, b], [a, b2], · · · , [a, bq−1], [a2, b], [a2, b2], · · · , [a2, bq−1], · · · , [a(p/2)−1, b], [a(p/2)−1, b2], · · · , [a(p/2)−1, bq−1], [a, c], [a, c2], · · · , [a, cq−1], [a2, c], [a2, c2], · · · , [a2, cq−1], · · · , [a(p/2)−1, c], [a(p/2)−1, c2], · · · , [a(p/2)−1, cq−1], [b, c], [b, c2], · · · , [b, cq−1], [b2, c], [b2, c2], · · · , [b2, cq−1], · · · , [bq−1, c], [bq−1, c2], · · · , [bq−1, cq−1], [a, bc], [a, b2c], · · · , [a, bq−1c], [a, bc2], [a, b2c2], · · · , [a, bq−1

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[a2, bq−1c], [a2, bc2], [a2, b2c2], · · · , [a2, bq−1c2], · · · , [a2, bcq−1], [a2, b2cq−1], · · · , [a2, bq−1cq−1], · · · , [a(p/2)−1, bc], [a(p/2)−1, b2c], · · · , [a(p/2)−1, bq−1c], [a(p/2)−1, bc2], [a(p/2)−1, b2c2], · · · , [a(p/2)−1, bq−1c2], · · · , [a(p/2)−1, bcq−1], [a(p/2)−1, b2cq−1], · · · , [a(p/2)−1, bq−1cq−1], [ab, bc], [ab, b2c], · · · , [ab, bq−1c], [ab, bc2], [ab, b2c2], · · · , [ab, bq−1c2], · · · , [ab, bcq−1

], [ab, b2cq−1], · · · , [ab, bq−1cq−1], [a2b, bc], [a2b, b2c], · · · , [a2b, bq−1c], [a2b, bc2], [a2b, b2c2],· · · ,[a2b, bq−1c2], · · · , [a2b, bcq−1], [a2b, b2cq−1],· · · , [a2b, bq−1cq−1], · · · ,[a(p/2)−1b, bc],[a(p/2)−1b, b2c],· · · ,[a(p/2)−1b, bq−1c],[a(p/2)−1b, bc2],[a(p/2)−1b, b2c2], · · · ,[a(p/2)−1b, bq−1c2],· · · ,[a(p/2)−1b, bcq−1],[a(p/2)−1b, b2cq−1],· · · ,[a(p/2)−1b, bq−1cq−1].

Also, the signature of H00p,q is (q2(p−1)−q(p+1)+22 ; ∞(q)).

Example 2. Let p = 4 and q = 5. Then we have H 0 4,5: H 00 4,5 = 50. We choose Σ = {I, a, b, b2, b3, b4, c, c2, c3, c4, ab, ab2, ab3, ab4, ac, ac2,

ac3, ac4, bc, bc2, bc3, bc4, b2c, b2c2, b2c3, b2c4, b3c, b3c2, b3c3, b3c4, b4c,

b4c2, b4c3, b4c4, abc, abc2, abc3, abc4, ab2c, ab2c2, ab2c3, ab2c4, ab3c, ab3c2, ab3c3, ab3c4, ab4c, ab4c2, ab4c3, ab4c4 } as a Schreier transversal for H00

4,5.

According to the Reidemeister-Schreier method, we get total 56 generators of H004,5 as [a, b], [a, b2], [a, b3], [a, b4], [a, c], [a, c2], [a, c3], [a, c4], [b, c], [b, c2],

[b, b3], [b, c4], [b2, c], [b2, c2], [b2, c3], [b2, c4], [b3, c], [b3, c2], [b3, c3], [b3, c4], [b4, c],

[b4, c2], [b4, c3], [b4, c4], [a, bc], [a, b2c], [a, b3c], [a, b4c], [a, bc2], [a, b2c2], [a, b3c2],

[a, b4c2], [a, bc3], [a, b2c3], [a, b3c3], [a, b4c3], [a, bc4], [a, b2c4], [a, b3c4], [a, b4c4],

[ab, bc],[ab, b2c],[ab, b3c],[ab, b4c],[ab, bc2], [ab, b2c2], [ab, b3c2], [ab, b4c2], [ab, bc3],

[ab, b2c3], [ab, b3c3], [ab, b4c3], [ab, bc4], [ab, b2c4], [ab, b3c4], [ab, b4c4]. Also, the signature of H004,5 is (26; ∞(5)).

Corollary 3. Let p and q be integers such that 2 ≤ p ≤ q, p + q > 4. In case p and q are both even numbers,

Hp,q : H

(n) p,q

= ∞ for n ≥ 2 and, in other cases of p and q, Hp,q: H (n) p,q = ∞ for n ≥ 3.

Corollary 4. If p = q odd number, then the commutator subgroups H00q,q coincides with the ones given in [15] for extended Hecke groups Hq.

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[15] R. Sahin and O. Bizim, Some subgroups of the extended Hecke groups H( λq), Acta Math. Sci., Ser. B, Engl. Ed. 23, No.4 (2003), 497-502.

[16] R. Sahin, O. Bizim and I. N. Cang¨ul, Commutator subgroups of the ex-tended Hecke groups, Czech. Math. J. 54, No.1 (2004), 253-259.

[17] R. Sahin, S. Ikikardes and ¨O. Koruo˘glu, Some normal subgroups of the extended Hecke groups H( λp), Rocky Mountain J. Math. 36 (2006), no.

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[18] R. Sahin and ¨O. Koruo˘glu, Commutator subgroups of the power subgroups of some Hecke groups, Ramanujan J. 24 (2011), no. 2, 151–159.

[19] R. Sahin and ¨O. Koruo˘glu, Commutator subgroups of the power subgroups of Hecke groups H(λq) II, C. R. Math. Acad. Sci. Paris 349 (2011), no.

3-4, 127–130.

[20] T. A. Schmidt and M. Sheingorn, Covering the Hecke triangle surfaces, Ramanujan J. 1 (1997), no. 2, 155–163.

[21] D. Singerman, Subgroups of Fuchsian groups and finite permutation groups, Bull. Lond. Math. Soc. 2, 319-323 (1970).

[22] V. V. Tsanov, Valdemar, Triangle groups, automorphic forms, and torus knots, Enseign. Math. (2) 59 (2013), no. 1-2, 73–113.

S

¸ule KAYMAK (SARICA),

Institute of Science, Department of Mathematics, Balıkesir University,

C¸ a˘gı¸s Campus, 10145, Balıkesir, Turkey. Email: sulekaymak0@gmail.com Bilal DEM˙IR,

Necatibey Faculty of Education, Department of Mathematics, Balıkesir University,

Soma Street, 10100, Balıkesir, Turkey. Email: bdemir@balikesir.edu.tr ¨

Ozden Koruo˘glu,

Necatibey Faculty of Education, Department of Mathematics, Balıkesir University,

Soma Street, 10100, Balıkesir, Turkey. Email: ozdenk@balikesir.edu.tr Recep S¸ahin,

Faculty of Science and Arts, Department of Mathematics, Balıkesir University,

10145 C¸ a˘gı¸s Campus, Balıkesir, Turkey Email: rsahin@balikesir.edu.tr

Referanslar

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