Intersection graph of subgroups of some non-abelian groups

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Intersection graph of subgroups of some non-abelian groups

R. Rajkumar

a,∗

and P. Devi

b

a,b_{Department of Mathematics, The Gandhigram Rural Institute – Deemed University, Gandhigram – 624 302, Tamil Nadu, India.}

Abstract

The intersection graph of subgroups of a group G is a graph whose vertex set is the set of all proper

subgroups ofGand two distinct vertices are adjacent if and only if their intersection is non-trivial. In this

paper, we obtain the clique number and degree of vertices of intersection graph of subgroups of dihedral group, quaternion group and quasi-dihedral group.

Keywords: Intersection graph, subgroups, non-abelian groups.

1

Introduction

There are several graphs associated with algebraic structures to investigate some specific properties of algebraic structures. Among them the intersection graphs have its own importance, which have been studied in the literature over the past fifty years. In 1964, Bosak [1] initiated the study of the intersection graphs

of semigroups. Later, Cs ák ány and Poll ák [5] defined the intersection graph of subgroups of finite group.

Followed by this, Zelinca investigated the intersection graph of subgroups of a finite abelian group [7]. In the recent years, several interesting properties of the intersection graphs of subgroups groups have been obtained in the literature, see for instance [2], [4], [5], [6] and the references therein.

LetGbe a group.The intersection graph of subgroups of G, denoted byI(G), is a graph with all the proper

subgroups ofGas its vertices and two distinct vertices inI(G)are adjacent if and only if the corresponding

subgroups have a non-trivial intersection inG.

LetGbe a simple graph. Thedegreeof a vertex vinG, denoted bydegG(v)is the number of vertices to

whichvis adjacent. AcliqueofGis a complete subgraph ofG. Theclique numberofGis the is the cardinality

of a largest clique inGand it is denoted byω(G).

For a positive integern,τ(n)denotes the number of positive divisor ofn;σ(n)denotes the sum of all the

positive divisors ofn.

The aim of this paper is to find the clique number and degree of vertices of the intersection graph of subgroups of dihedral group, quaternion group and quasi-dihedral group.

We will use the following result of Chakrabartyet al.in the subsequent section.

Theorem 1.1. ([3]) Let n= pα1

1 p α2

2 . . .p αk

k , where pi’s are distinct primes andαi ≥1. If H is a proper subgroup ofZn

with|H|=pβi1

i1 p βi₂

i2 . . .p β_ir

ir , thendegI(Zn)(H) =τ(n)−

∏

j/∈{i₁,i2,...ir}

(αj+1)−3.

2 Properties of

I

(

D

n

)

,

I

(

Q

n

)

,

I

(

QD

2α

)

First, we start with the dihedral group. The dihedral group of order 2n(n≥3)is defined by

Dn=ha,b|an =b2=1,ab=ba−1i.

∗_{Corresponding author.}

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The subgroups ofDnare listed below:

(i) cyclic groupsH₀r :=hanriof orderr, whereris a divisor ofn;

(ii) cyclic groupsH_i1:=haibiof order 2, wherei=1, 2, . . . ,n;

(iii) dihedral groupsH_ir :=hanr,aibiof order 2r, whereris a divisor ofn,r6=1,nandi=1, 2, . . . ,n

r.

The number of subgroups ofDnlisted in (i), (ii), (iii) areτ(n)−1,n,σ(n)−n−1 respectively and so the total

number of proper subgroups ofDnisτ(n) +σ(n)−2.

Theorem 2.2. Let n ≥ 3be an integer with n = pα1

1 p α2

2 . . .p αk

k , where pi’s are distinct primes and αi ≥ 1, and let

r=pβi1

i1 p βi₂

i2 . . .p β_ir

ir be a divisor of n. Then

(1) deg_I₍_D_n₎(Hr

0) =τ(n) +σ(n)−n−3−

∏

j/∈{i1,i2,...,ir}

(αj+1)−

∑

d|s d6=1

n

d, where s=

n pαi1

i₁ p

α_i

2

i₂ ...p

α

ik ik

;

(2) for each r6=1,n and i=1, 2, . . . ,n_r,deg_I₍_D_n₎(H_ir) =τ(n) +σ(n)−n−2−

∏

j∈{/ i1,i2,...ir}

(αj+1);

(3) for each i=1, 2, . . . ,n,deg_I₍_D_n₎(H_i1) =τ(n)−2.

Proof. (1): First we count the number of subgroups listed in (i) to which H₀r is adjacent in I(Dn). Here

hai ∼= Zn, so by Theorem 1.1, H₀r is adjacent withτ(n)−

∏

j∈{/ i1,i2,...ir}

(αj+1)−2 subgroups ofZn including

Zn. ClearlyH₀r is not adjacent with all thensubgroups ofDn listed in (ii). Finally, we count the number of

subgroups listed in (iii) to whichH₀ris adjacent. For every divisord6=1 ofs= n

pα_ii1

1 p

α_i

2

i₂ ...p

α_ik

ik

,H₀ris not adjacent

with H_id,i =1, 2, . . . ,_dn;H₀r is adjacent with each of the remaining proper subgroups ofDnlisted in (iii). The

total number of such subgroups isσ(n)−n−1−

∑

d|s d6=1

n

d. Summing up all these values gives the degree ofH

r

0.

(2): For each r 6= 1,n andi = 1, 2, . . . ,n_r, H₀r is the maximal cyclic subgroup of H_ir and so the number of

subgroups listed in (i) to which Hr

i is adjacent is the same as the number of subgroups listed in (i) to which

Hr₀ is adjacent including H₀r. The number of such subgroup is τ(n)−

∏

j∈{/ i1,i2,...ir}

(αj+1)−1. Among the

subgroups of Dn listed in (ii), H_ir has exactlyrsubgroups as its subgroups and soH_ir is adjacent with only

these subgroups in the list. Finally, we count the number of subgroups listed in (iii) to which H_ir is adjacent.

For every divisorlofr,H_iris intersect withH_il; for every divisordofs,(d,r) =1 and so by chinese remainder

theorem there exist an integer, let it betsuch thatH1

t is a subgroup of bothHirandHdi. SoHiradjacent with all

the subgroups ofDn listed in (iii). The total number of such subgroups isσ(n)−n−1. The degree of is just

the sum of these three values.

(3): For eachi=1, 2, . . . ,n, the order ofH_i1is 2. The number of subgroups ofDn containsH_i1isτ(n)−2 and

H1

i is not intersect with remaining proper subgroups ofDn, since order ofHi1is prime. This completes the

proof.

Theorem 2.3. For n≥3,ω(I(Dn)) =σ(n)−n−1+

k

∏

i=1 αi.

Proof. TakeA := C1∪ C2, whereC1 := {Hir |r | n,r 6= 1,n,i = 1, 2, . . . ,nr}andC2 := S{ha

n

ri | _r | _n_,_r 6₌ ₁

with r has every prime divisor of n as a factor}. Clearly A is a maximal clique and|A| = |C1|+|C2| =

(σ(n)−n−1) +

k

∏

i=1

αi. LetB be another clique different from A. ThenB should contains eitherha

n ri, for

somer| n,r 6= 1 andrdoes not contains all the prime divisors ofn orhaibi, for somei = 1, 2, . . .,n. IfB

contains the subgrouphanri_{, for some}_r|_n_,_r 6₌_{1 and}_r_{does not contains all the prime divisors of}_n_{, then let}

pjbe the prime divisor ofnwhich is not a divisor ofr. HereGhas at least two subgroups of order 2pjand so

we cannot take the subgroups of order 2pjinB. It follows that|B|<|A|. IfBcontains the subgrouphaibi, for

somei=1, 2, . . .,n, thenhaibiadjacent withτ(n)−2 and so we cannot takeσ(n)−τ(n) +1 subgroups inB.

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Next, we consider the quaternion group. For any integer n > 1, the quaternion group of order 4n, is defined by

Qn =ha,b|a2n=b4=1,b2=an,ab=ba−1i.

The subgroups ofQnare listed below:

(i) cyclic groupsH0,r:=ha

2n

r i, of orderr, whereris a divisor of 2n;

(ii) cyclic groupsHi,1:=haibiof order 4, wherei=1, . . . ,n;

(iii) quaternion groupsHi,r :=ha

n

r_,_ai_bi_{of order 4}_r_{, where}_r_{is a divisor of}_n_,_i₌_{1, . . . ,}n

r.

The number of subgroups ofQnlisted in (i), (ii), (iii) areτ(2n)−1,n,σ(n)−n−1 and so the total number of

proper subgroups ofQnisτ(2n) +σ(n)−2.

Theorem 2.4. Let n > 1be an integer with n = pα1

1 p α2

2 . . .p αk

k , where pi’s are distinct primes andαi ≥ 1, and let

r= pβi1

i1 p βi₂

i2 . . .p βir

ir be a divisor of n.

(1) If r is even, thendeg_I₍_Q_n₎(H0,r) =τ(2n) +σ(n)−3−

∏

j/∈{i1,i2,...ir}

(αj+1);

(2) If r is odd, thendeg_I₍_Q_n₎(H0,r) =τ(2n) +σ(n)−n−3−

∏

j/∈{i1,i2,...ir}

(αj+1)−

∑

d|s d6=1

n

d, where s=

n pαi1

i₁ p

α_i

2

i₂ ...p

α_ik

ik ;

(3) For each i=1, . . . ,n_r,deg_I₍_Q_n₎(Hi,r) =τ(2n) +σ(n)−3−

∏

j/∈{i1,i2,...ir}

(αj+1);

(4) For each i = 1, . . . ,n,deg_I₍_Q_n₎(Hi,1) = τ(2n) +σ(n)−3−

∏

j∈{/ i1,i2,...ir}

(αj+1), whereαj’s are powers of odd

prime factors of n.

Proof. (1)-(2): First we count the number of subgroups listed in (i) to whichH0,ris adjacent. Herehai ∼=Z2n,

by Theorem 1.1,H0,r adjacent withτ(2n)−

∏

j∈{/ i1,i2,...ir}

(αj+1)−2 subgroups ofZ2n includingZ2n. Now we

consider the following two cases:

Case a:ris even. HereQnhas an unique subgroup of order 2 and so every subgroup of even order inQnare

adjacent with each other, soH0,r is adjacent withσ(n)−1 subgroups ofQn excludingQn listed in (ii), (iii).

This completes the proof of part (1).

Case b:ris odd. ClearlyH0,r is not adjacent with all thensubgroups ofQnlisted in (ii), since order of H1,r

is 4. Finally we count the number of subgroups listed in (iii) to which H0,r is adjacent. For every divisor

d6=1 ofs= n pαi1

i₁ p

α_i

2

i₂ ...p

α_ik

ik

,H0,r is not adjacent withHi,d,i=1, 2, . . ., nd;H0,ris adjacent with remaining proper

subgroups ofQn listed in (iii). The total number of such subgroups isσ(n)−n−1−

∑

d|s d6=1

n

d. This completes

the proof of part (2).

(3): For eachi= 1, . . . ,n_r,H0,r is the maximal cyclic subgroup ofHi,rand so the number of subgroups listed

in (i) to whichHi,r is adjacent is the same as the number of subgroups listed in (i) to which H0,r is adjacent

includingH0,r. The number of such subgroups isτ(n)−

∏

j/∈{i1,i2,...ir}

(αj+1)−1. AlsoQnhas a unique subgroup

of order 2 and soHi,ris adjacent with all the subgroups listed in (ii), (iii), since order of subgroups ofQnlisted

in (ii), (iii) is even. The total number of such subgroups isσ(n)−1. This completes the proof of part (3).

(4): SinceQn has an unique subgroup of order 2, soHi,1is adjacent with all the subgroups listed in (ii), (iii).

AlsoHi,1is adjacent all the even order subgroups ofQnlisted in (i). ButHi,1is not adjacent with an odd order

subgroups ofQn listed in (i). The number of such subgroups isτ(2n) +σ(n)−

∏

j∈{/ i1,i2,...ir}

(αj+1)−2, where

αj’s are powers of odd prime factors ofn. This completes the proof of part (4).

Theorem 2.5. Let n > 1 be an integer and2n = 2α1_pα2

2 . . .p αk

k , where pi’s are distinct primes and αi ≥ 1. Then

ω(I(Qn)) =σ(n) +α1

k

∏

i=2

(4)

Proof. LetAbe the set of all even order subgroups of Qn. Then|A| = σ(n) +α1

k

∏

i=2

(αi+1)−1 andAis a

maximal clique inI(Qn). LetBbe another clique different fromA. ThenBshould containsha

2n

r i, for some

an odd divisorrofn,r6=1. ThenBcannot contain the subgroups of order 4. It follows that|B|<|A|. This

completes the proof.

Finally, we consider the quasi-dihedral group. For any positive integerα>3, the quasi-dihedral group of

order 2α_{, is defined by}

QD2α =ha,b|a2 α−1

=b2=1,bab−1=a2α−2−1i.

The proper subgroups ofQD2α are listed below:

(i) cyclic groupsH₀r =ha2α

−1

r i, whereris a divisor of 2α−1,r6=1;

(ii) the dihedral groupH₁2α−2 =ha2,bi ∼=D2α−2and the dihedral subgroupsHr_i ofH2

α−2

1 , whereris a divisor

of 2α−2_,_r₆₌₂α−2_,_i_{∈ {}_{1, 2, . . . ,}2α−2

r };

(iii) the quaternion groupH_2,2α−3 =ha2,abi ∼=Q₂α−3and the quaternion subgroupsHi,rofH2,2α−3, whereris

a divisor of 2α−3_,_r₆₌₂α−3_,_i_{∈ {}_{1, 2, . . . ,}2α−3

r }.

The number of subgroups of QD2α listed in (i), (ii), (iii) areτ(2α−1)−1, 2α−1−1, 2α−2−1 and so the total

number of proper subgroups ofQD2αisα+3(2α−2−1).

Theorem 2.6. Ifα≥4, then

(1) for each divisor r of2α−1_{, r}₆₌₁_,_deg

I(QD₂α)(H r

0) =α+2α−1−4;

(2) for each divisor r of2α−2_{, r}₆₌₁_{, i}₌₁_{, 2,}_{. . .}_,2α−2

r ,degI(QD₂α)(H r

i) =α+2α

−1₊_r₋₄_;

(3) for each divisor r of2α−3_{, r}₆₌₁_{, i}₌₁_{, 2,}_{. . .}_,2α−2

r ,degI(QD₂α)(Hi,r) =α+2

α−1₋₄_;

(4) for i=2,22,. . .,2α−2_,_deg

I(QD₂α)(H 1

i) =α−2;

(5) for i=1, 3,. . .,2α−3_,_deg

I(QD₂α)(Hi,1) =α+2

α−1₋₄_.

Proof. The only maximal subgroups ofQDα 2areH2

α−1

0 , the dihedral subgroupH2

α−2

1 and quaternion subgroup

H2,2α−3. Here H2₀ is a subgroup of all the subgroup of QD2α other than H_i1, i = 2, 22, . . ., 2α−2; also no

subgroups listed in (i), (iii) are adjacent with H1_i, i = 2, 22, . . ., 2α−2_{. It follows that deg}

I(QD2α)(H r

0) =

α+3(2α−2−1)−2α−2−1=α+2α−1−4. Proofs of parts (3) and (5) are similar to the above.

Next, we count the number of subgroups ofQD2α to whichHr_i is adjacent. By the above argumentH_iris

adjacent with all the subgroups listed in (i), (iii) and the dihedral subgroups ofH₁2α−2; alsoHr_i hasrsubgroups

of order 2 as its subgroups and soHr_i adjacent with these subgroups, so deg_I₍_QD

2α)(H r

i) =α+3(2α−2−1)− (2α−2₋_r₎₋₁₌_α₊₂α−1₊_r₋_4.

Finally, we count the number of subgroups ofQD2αto whichH1_i is adjacent. Note that deg_I₍_QD

2α)(H 1

i) =

deg_I₍_D_n₎(H_i1) +1, since order ofH_i1is 2, and it is not a subgroup of any subgroups ofH_i_,2α−3; H2

α−2

1 is also

a vertex ofI(QD2α). So by Theorem 2.2(3), we have deg_I₍_QD

2α)(H 1

i) = τ(2α−2)−1 = α−2. Hence the

proof.

Theorem 2.7. Forα≥3,ω(I(QD2α)) =α+2α−1−3.

Proof. LetAbe the set of all subgroups ofQD2αother thanhbaii,i =2, 22, . . ., 2α−2. ClearlyAis a maximal

clique inI(QD2α)and|A| = α+3(2α−2−1)−2α−2 = α+2α−1−3. LetBbe another clique inI(QD2α).

ThenBcontains exactly one subgroup of the formhbaii,i=2, 22, . . ., 2α−2_{. It follow that}_|B|_<_|A|_{, since for}

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References

[1] J. Bosak, The graphs of semigroups in: Theory of graphs and application, Academic Press, New York, 1964, pp. 119–125.

[2] B. Cs ák ány, G. Poll ák, The graph of subgroups of a finite group (Russian),Czechoslovak Math. J.19 (1969)

241–247.

[3] I. S. Chakrabarty, T. K. Ghosh, Mukherjee, M. K. Sen, Intersection graphs of ideals of rings,Discrete Math.

309 (2009) 5381–5392.

[4] R. Rajkumar and P. Devi, Toroidality and projective-planarity of intersection graphs of subgroups of finite groups, preprint available at arXiv:1505.08094v2 [math.GR] (2015).

[5] R. Rajkumar and P. Devi, Intersection graph of cyclic subgroups of groups,Electron. Notes Discrete Math.

(accepted).

[6] Rulin Shen, Intersection graphs of subgroups of finite groups,Czechoslovak Math J.60 (2010) 945–950.

[7] B. Zelinka, Intersection graphs of finite abelian groups,Czech. Math. J.25 (1975) 171–174.

Received: October 22, 2015;Accepted: March 24, 2016

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