Revisiting the Leinster groups

C. R. Acad. Sci. Paris, Ser. I 352 (2014) 1–6 Contents lists available at ScienceDirect C. R. Acad. Sci. Paris, Ser. I www.sciencedirect.com Number theory/Group theory Revisiting the Leinster groups Quelques résultats sur les groupes de Leinster Sekhar Jyoti Baishya Department of Mathematics, North-Eastern Hill University, Permanent Campus, Shillong-793022, Meghalaya, India a r t i c l e i n f o Article history: Received 3 July 2013 Accepted after revision 12 November 2013 Available online 22 December 2013 Presented by the Editorial Board a b s t r a c t A finite group is said to be a Leinster group if the sum of the orders of its normal subgroups equals twice the order of the group itself. In this paper we give some new results concerning Leinster groups.  2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved. r é s u m é Un groupe de Leinster est un groupe fini tel que la somme des cardinaux de ses sousgroupes distingués soit égale au double du cardinal de G. Dans cette note, nous donnons quelques résultats nouveaux sur les groupes de Leinster.  2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved. 1. Introduction A number is perfect if the sum of its divisors equals twice the number itself. In 2001, T. Leinster [6], developed and studied a group-theoretic analogue of perfect numbers. A finite group is said to be a perfect group (not to be confused with the one which is equal to its commutator subgroup) or an immaculate group or a Leinster group if the sum of the orders of its normal subgroups equals twice the order of the group itself. Clearly, a finite cyclic group C n is Leinster if and only if its order n is a perfect number. In fact, the abelian Leinster groups are precisely the finite cyclic groups whose orders are perfect numbers. It may be mentioned here that up to now, only one Leinster group of odd order is known, namely (C 127 ⋊ C 7 ) × C 34 .112 .192 .113 . It was discovered by F. Brunault [8]. More information on this and the related concepts can be found in the works of S.J. Baishya and A.K. Das [3], A.K. Das [4], M. Tărnăuceanu [10,12], T.D. Medts and A. Maróti [9], etc. Given a finite group G, let τ (G ) denote the number of normal subgroups of G and σ (G ) denote the sum of the orders of the normal subgroups of G. In this paper, among other results, we classify Leinster groups G with τ (G )  7. 2. Some basic results The Leinster groups among the dihedral groups are in one-to-one correspondence with the odd perfect numbers [6, Example 2.4] and so it is an open question as to whether there are any. It would be interesting to find the Leinster groups among the well-known families of groups. In the following result, we classify the Leinster groups among the generalized quaternion group Q 4m of order 4m, m  2 given by a, b | a2m = 1, b2 = am , bab−1 = a−1 . Proposition 2.1. The generalized quaternion group Q 4m , m  2 is Leinster if and only if m = 3. E-mail address: [email protected]. 1631-073X/$ – see front matter  2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved. http://dx.doi.org/10.1016/j.crma.2013.11.009 2 S.J. Baishya / C. R. Acad. Sci. Paris, Ser. I 352 (2014) 1–6 Proof. It is well known that Q 4m Z ( Q 4m ) σ (Q ) σ (D ) ∼ = D 2m , for any integer m  2. By [9, Observation 3.1], we have | Q 4m4m|  | D 2m2m| . Now, σ (D ) if m is even, then by [6, Example 2.4], we have | D 2m| > 2 and so Q 4m is not Leinster. Next, suppose m is odd. In this 2m situation, it can be easily proved that the proper normal subgroups of Q 4m are precisely the subgroups of the cyclic group generated by a. Consequently, σ ( Q 4m ) = 4m + σ (2m), where σ (2m) is the sum of the positive divisors of 2m. We know that 6 is the only perfect number of the form 2m, where m is odd. Therefore Q 4m , m  2 is Leinster if and only if m = 3. ✷ Let I (G ) denote the set of all solutions of the equation x2 = 1 in G. If n is an odd perfect number, then D 2n is a Leinster group [6, Example 2.4] with | I ( D 2n )| > n. The following theorem shows that if G is a Leinster group such that | I (G )| > |G2 | , then G ∼ = D 2n , where n is an odd perfect number. In the following theorem |Cent(G )| denotes the number of distinct centralizers of G. Recall that a finite group G is said to be a CA-group if the centralizer C (x) is abelian for every x ∈ G \ Z (G ) (see [5]). Theorem 2.2. If G is a Leinster group with | I (G )| > |G | 2 , then G ∼ = D 2n , where n is an odd perfect number. |G | Proof. Let G be a Leinster group with | I (G )| > 2 . Clearly, |G | is even, otherwise | I (G )| = 1 and hence G is trivial, which is not possible. Now, suppose G is abelian. Then G is cyclic by [6, Corollary 4.2] and hence | I (G )| = 2. It follows that |G | = 2, which is not possible. Therefore, G is non-abelian and without any loss we can assume that |G | = 2m n, where m  1 is an integer and n > 1 is an odd integer, noting that a 2-group is not Leinster [6, Example 2.3]. Let G ′ be the commutator subgroup of G. By [14, p. 251], we have GG′ is an elementary abelian 2-group and therefore by [6, Theorem 4.1], we have | GG′ | = 2. Again, by [14, Theorem 5], we have G ′ is abelian and hence by [2, Theorem 2.3], we get:     Cent(G ) = G ′  + 2 = |G | + 2. (1) 2 |G | It is easy to see that Z (G )  G ′ . Now, it follows from (1) that the elements of G \ G ′ will produce exactly 2 distinct centralizers, noting that elements of G ′ produce exactly two distinct centralizers, namely G and G ′ = C ( g ), where g ∈ |G | G ′ \ Z (G ). Let a, b ∈ G \ G ′ such that a = b. Since |G \ G ′ | = 2 , it follows that: C (a) = C (b). (2) Again, since G ′ is an abelian normal subgroup of G of index 2, therefore by [5, Theorem A], we have G is a CA-group and hence by [5, Proposition 3.2], we get: ab = ba. (3) Now, suppose | Z (G )| =  1. Let x ∈ G \ G ′ . Since Z (G )  C (x), therefore by (3), there exists y ∈ G ′ \ Z (G ) such that y ∈ C (x). It follows that x ∈ C ( y ) = G ′ , which is a contradiction. Hence | Z (G )| = 1. Again, by [14, Lemma 9], we have I (G ′ ) ⊆ Z (G ) and hence |G ′ | = n, noting that | GG′ | = 2. Let z ∈ G \ G ′ . If |C ( z)| =  2, then there exists w ∈ C (z) such that w = z and w = 1. It follows from (3) that w ∈ G ′ \ Z (G ). But then z ∈ C ( w ) = G ′ , which is a contradiction. Therefore |C ( z)| = 2 and hence:   Cl( z) = |G | . (4) 2 Now, suppose N ✂ G, N = G. If | N | is even, then there exists u ∈ N such that o(u ) = 2. Clearly, u ∈ G \ G ′ and hence |G | by (4), we have |Cl(u )| = 2 . But Cl(u )  N, which is not possible. Therefore | N | is odd and hence N ✂ G ′ . Again, note that v ∼ v −1 for any v ∈ G ′ . For if, v ≁ v −1 for some v ∈ G ′ then the map: φ v : I (G ) −→ G \ I (G ) given by h −→ hv |G | is one–one, which is not possible since | I (G )| > 2 . Now, suppose N ✂ G ′ . Since G ′ is abelian of index 2, therefore Cl(r ) = {r , r −1 } for any r ∈ G ′ \ Z (G ). Hence N ✂ G. Therefore, the proper normal subgroups of G are precisely the normal subgroups of G ′ . In other words, σ (G ) = |G | + σ (G ′ ). Now, since G is a Leinster group, it follows that |G | = 2|G ′ | = σ (G ′ ). Therefore by [6, Corollary 4.2], G ′ is cyclic and |G ′ | = n is an odd perfect number. Hence G ∼ = D 2n . ✷ A group is said to be semi-simple if it is a direct product of non-abelian simple groups. In this connection, we have the following result. Proposition 2.3. No finite semi-simple group is Leinster. S.J. Baishya / C. R. Acad. Sci. Paris, Ser. I 352 (2014) 1–6 3 Proof. Let G be a finite semi-simple group. Then G = H 1 × H 2 ×· · · × H n , where each H i , 1  i  n is a finite non-abelian simple group. By [4, Corollary 4.6], we have σ (G ) = i σ ( H i ) = i (1 + | H i |). Since finite non-abelian simple groups are of even order, σ (G ) is odd. Therefore, G is not Leinster. ✷ Continuing with the finite groups that are not Leinster, we also have the following result. Proposition 2.4. There is no Leinster group of order p 2 q2 , where p, q are primes. Proof. Let G be a Leinster group of order p 2 q2 , where p, q are primes. Since no p-group is Leinster [6, Example 2.3], therefore without any loss we may assume that p < q. Now, if G is abelian, then G is cyclic [6, Corollary 4.2] and hence |G | is a perfect number, which is not possible. Next, suppose that G is non-abelian. Now, if q > 3, then any q-Sylow subgroup of G is normal. Suppose G has no normal subgroup of order p. Then p 2 q2 = 1 + qn for some positive integer n, which is not possible. Again, note that G can have at the most one normal subgroup of order p; otherwise the p-Sylow subgroup of G will be normal and G will be an abelian group, which is not possible. It follows that p 2 q2 = 1 + p + qn for some positive integer n, which is again not possible, since q > 3. Therefore q = 3 and so |G | = 36, which is a contradiction by GAP [13]. ✷ We need the following remark for proving the next proposition. m Remark 2.5. Consider the group G m = a, b | a3 = b2 = 1, bab−1 = a−1 . Using [9, Proposition 3.8], one can easily see that, if G m is a Leinster group, then m = 2 and hence G m ∼ = Q 12 . Proposition 2.6. Let G be a Leinster group such that Z (GG ) ∼ = S 3 . Then G ∼ = S 3 × C 5 or G ∼ = Q 12 . Proof. Let G be a Leinster group such that Z (GG ) ∼ = S 3 . By [7, Corollary 2.2], we have G = G m × A, where m  1 and A is an abelian group. Now, if | A | = 1, then G ∼ = G m and hence by Remark 2.5, G ∼ = Q 12 . Next, suppose | A | > 1. We have ×A | = 2m . Suppose m  3. Then by [9, Corollary 3.3], there exist N 1 ✂ G, N 2 ✂ G and N 3 ✂ G a × A ✂ G m × A and | Gam× A |G | |G | |G | such that | N 1 | = 2 , | N 2 | = 4 and | N 3 | = 8 . It follows that σ (G ) > 2|G |, which is impossible. Next, suppose m = 2. Then G = Q 12 × A. Note that Q 12 have normal subgroups of order 2, 3 and 6, say N 1 , N 2 and N 3 , respectively. Therefore {1} × A , N 1 × A, N 2 × A and N 3 × A are normal subgroups of Q 12 × A of indices 12, 6, 4 and 2 respectively, which is again contradictory to the definition of Leinster groups. Therefore, G = S 3 × A. In the present situation, one can verify that gcd(| S 3 |, | A |) = 1 and by [6, Corollary 3.2], σ ( S 3 × A ) = σ ( S 3 ) × σ ( A ). Now, since G = S 3 × A is Leinster, it follows that 6| A | = 5σ ( A ) and hence A = C 5 . ✷ As an immediate corollary, we have the following result. Corollary 2.7. Consider the group G = a, b | a2n = b3 = 1, a−1 ba = b−1 , n  1. If G is Leinster, then n = 2 or 5 and hence G ∼ = Q 12 or G ∼ = S3 × C5. Proof. By [1, Lemma 2.7], we have |G | = 6n and Z (G ) = a2 . Therefore or G ∼ = S3 × C5. ✷ G Z (G ) ∼ = S 3 and by Proposition 2.6, we have G ∼ = Q 12 3. Leinster groups with τ ( G )  7 In this section, we classify Leinster groups with at the most seven normal subgroups. We begin with the following result which gives a necessary condition for a finite group to be Leinster. Proposition 3.1. Let G be a Leinster group. Then τ (G )  4, where the equality holds if and only if G ∼ = C6. Proof. If τ (G ) < 4, then one can easily see that σ (G ) < 2|G |. For the second part, suppose G is a Leinster group with τ (G ) = 4. Let M and N be the proper non-trivial normal subgroups of G. Let | M | = m and | N | = n. Then we have: | G | = 1 + m + n. (5) From (5), it follows that G = M N and M ∩ N = {1}. Without any loss, we may assume that m < n. Then, by (5), we have m < n  m + 1, forcing n = m + 1. Now, again using (5), we get |G | = 6 and hence G ∼ = C 6 . The converse is trivial. ✷ As a consequence, we have the following result. 4 S.J. Baishya / C. R. Acad. Sci. Paris, Ser. I 352 (2014) 1–6 Corollary 3.2. Let G be a finite group satisfying one of the following conditions: (a) G ∼ = H × K , where H and K are simple groups. (b) |G | = pq, where p and q are primes. Then G is a Leinster group if and only if G ∼ = C6. Proof. Let G be a Leinster group satisfying one of the given conditions. Since p-groups are not Leinster [6, Example 2.3], the simple groups H and K (if (a) holds) are almost coprime in the sense of [4], and the primes p and q (if (b) holds) are distinct. Thus, if (a) is satisfied, then τ (G ) = τ ( H )τ ( K ) = 4 (by [4, Corollary 4.6]), and so, G ∼ = C 6 (by Proposition 3.1). On the other hand, if (b) is satisfied, then G ∼ = C p × C q (otherwise, τ (G ) = 3), that is, G satisfies (a), and so we have G ∼ = C6 once again. This completes the proof as C 6 is already seen to be Leinster. ✷ As an immediate consequence, we also have the following result. Corollary 3.3. Let G be a finite group such that the proper non-trivial normal subgroups have the same order. Then G is not Leinster. Proof. In view of Proposition 3.1, we have τ (G )  4. Let M and N be two different normal subgroups of G having the same order. Then we have M ∩ N = {1} and G = M N. It follows that G = M × N and both M and N are simple groups. Now, the result follows using Corollary 3.2. ✷ The following theorem classifies all Leinster groups with exactly five normal subgroups. Theorem 3.4. If G is a Leinster group with τ (G ) = 5, then G ∼ = Q 12 . Proof. Let G be a Leinster group with τ (G ) = 5. Let N 1 , N 2 , N 3 be the proper non-trivial normal subgroups of G and n1 , n2 , n3 denote their order respectively. Without any loss, we may assume that 1 < n1  n2  n3 . From the definition of Leinster groups, we have: | G | = 1 + n1 + n2 + n3 . It follows using (6) that n3 ∈ { (6) |G | |G | 2 , 3 }. Now, suppose n3 = |G | 3 . Then by (6), we have: 2|G | = 3 + 3n1 + 3n2 . (7) Clearly, | N 1 | = | N 2 |, otherwise n1 = n2 = 3, which is impossible. Now, if N 1  N 2 , then by (7), we have n1 = 3 and hence n2 = 6 or 12 and consequently |G | = 15 or 24, which is a contradiction to the choice of G. Next, suppose N 1  N 2 . Then N 1 ∩ N 2 = {1} and hence N 1 N 2 = N 3 , otherwise G = N 1 × N 2 , which implies τ (G ) = 5. Thus we have n1 n2 = n3 = n3 = |G | 2 |G | 3 . It follows from (7) that 2n1 n2 = 1 + n1 + n2 , which is again impossible. Therefore, we have and by (6), we have: |G | = 2 + 2n1 + 2n2 . (8) In this situation also clearly | N 1 | =  | N 2 |, otherwise n1 = n2 = 2, which is impossible. Now, if N 1  N 2 , then by (8), we have n1 = 2 and n2 = 6, which is not possible to the choice of G. Therefore N 1  N 2 . It follows that N 1 ∩ N 2 = {1} and |G | hence N 1 N 2 = N 3 , otherwise G = N 1 × N 2 , which implies τ (G ) = 5. Thus we have n1 n2 = n3 = 2 . Now, using (8) we get n1 n2 = 1 + n1 + n2 . It follows that n1 < n2  n1 + 1 and hence n2 = n1 + 1. Thus we get from (8) that |G | = 4 + 4n1 . Consequently, n1 = 2 or n1 = 4. Therefore, in view of the choice of G, it follows that n1 = 2 and so |G | = 12. Now, it is a routine matter to see that G ∼ = Q 12 . ✷ We now classify Leinster groups with exactly six normal subgroups. We have used GAP [13] to verify some of the steps. Theorem 3.5. If G is a Leinster group with τ (G ) = 6, then G ∼ = C 28 or S 3 × C 5 . Proof. Let G be a Leinster group with τ (G ) = 6. Let N 1 , N 2 , N 3 , N 4 be the proper non-trivial normal subgroups of G and n1 , n2 , n3 , n4 denote their order respectively. Without any loss, we may assume that 1 < n1  n2  n3  n4 . From the definition of Leinster groups, we have: | G | = 1 + n1 + n2 + n3 + n4 . It follows from (9), that n4 ∈ { (9) |G | |G | 2 , 3 }. Now, suppose n4 = contradiction to the choice of G. Therefore n4 = |G | 2 |G | 3 . Then again using (9), we have n3 ∈ { and again using (9), we have n3 ∈ { |G | |G | , 3 }, which is a }, noting that G 7 5 |G | |G | |G | |G | 4 , 5 , 6 , S.J. Baishya / C. R. Acad. Sci. Paris, Ser. I 352 (2014) 1–6 cannot have a normal subgroup of index 3. Now, if n3 = |G | Similarly, if n3 = we have n2 = |G | 6 |G | 7 5 |G | , then by (9), we have n2 > again using (9), we have n2 = |G | 7 . It follows that |G | = 28 and hence G ∼ = C 28 . or |G | 7 , which is a contradiction. 5 4 . Now, if n3 = |G | , then by (9), 5 |G | and so |G | = 30. Now, it is a routine mater to check that G ∼ = S 3 × C 5 . Next, suppose that n3 = 4 . Then 6 , then also by (9), we end with a contradiction. Therefore n3 = |G | ✷ As a consequence we get the following result on Leinster groups of order pqr, where p < q < r are primes. In the following result, G n denotes the unique normal subgroup of order n, where n is a positive integer. Theorem 3.6. If G is a Leinster group of order pqr, where p < q < r are primes, then G ∼ = S3 × C5. Proof. One can easily verify that every normal subgroup of G is uniquely determined by its order. It follows using Proposition 3.1 and Theorem 3.4 that 6  τ (G )  8. If τ (G ) = 8, then G ∼ = C pqr . Therefore pqr is a perfect number, which is impossible (see [11]). Next suppose that τ (G ) = 7. Note that by Sylow theorem, we already have four normal subgroups, namely G 1 , G pqr , G r and G qr . Now, if G p ✂ G and G q ✂ G, then G ∼ = C pqr and τ (G ) = 8, which is a contradiction. Therefore G pq and G pr must be normal in G. But then G pq ∩ G pr = G p ✂ G and G pq ∩ G qr = G q ✂ G, which is again a contradiction. Therefore we have τ (G ) = 6. Now, the result follows using Theorem 3.5. ✷ We conclude this section with the following result on Leinster groups with exactly seven normal subgroups. Theorem 3.7. If G is a Leinster group with τ (G ) = 7, then G ∼ = C7 ⋊ C8. Proof. Let G be a Leinster group with τ (G ) = 7. Let N 1 , N 2 , N 3 , N 4 , N 5 be the proper non-trivial normal subgroups of G and n1 , n2 , n3 , n4 , n5 denote their order respectively. Without any loss, we may assume that 1 < n1  n2  n3  n4  n5 . From the definition of Leinster groups, we have: | G | = 1 + n1 + n2 + n3 + n4 + n5 . (10) /{ In view of (10), using GAP [13] and certain standard results from the theory of finite groups, we can see that ni ∈ for any i ∈ {1, 2, 3, 4, 5} and ni = Now, (10) becomes: |G | 4 for some i ∈ {1, 2, 3, 4, 5}. Therefore, by correspondence theorem, n5 = |G | 2 Since n3 ∈ /{ 3 , 5 }, it follows from (11) that n3 ∈ { 14 , 13 , 12 , 11 , 10 , 8 , 7 , 6 3 , 5 5 } 4 . |G | }. In view of (11), again using GAP [13] |G | 8 . Therefore (11) becomes: |G | = 8(1 + n1 + n2 ). |G | |G | , (11) |G | |G | |G | |G | |G | |G | |G | |G | and certain standard results from the theory of finite groups, we can see that n3 = Since n2 ∈ /{ 3 and n4 = |G | = 4(1 + n1 + n2 + n3 ). |G | |G | |G | |G | (12) G | |G | |G | |G | |G | |G | |G | }, it follows from (12) that n2 ∈ { |23 , 22 , 20 , 19 , 18 , 17 , 16 , |G | |G | |G | |G | 14 , 13 , 11 , 10 GAP [13] and certain standard results from the theory of finite groups, we can see that n2 = 3|G | = 56(1 + n1 ). }. In view of (12), again using |G | 14 . Therefore (12) becomes (13) 3 From (13) it follows that n1 | 56. Now, if n1 = 56, then |G | = 2 .7.19. But then the Sylow 19-subgroup of G is normal in G, which is not possible. Next, if 2 < n1 < 56, then in view of (13), using GAP [13], we get a contradiction. Hence n1 = 2 and by (13) we get |G | = 56. Therefore using GAP [13], we have G ∼ = C7 ⋊ C8. ✷ Acknowledgement I would like to express my gratitude to my supervisor Prof. Ashish Kumar Das for his valuable assistance during the preparation of this paper. I am also grateful to the referee for a number of useful suggestions and comments. References [1] A.R. Ashrafi, Counting the centralizers of some finite groups, Korean J. Comput. Appl. Math. 7 (1) (2000) 115–124. [2] S.J. Baishya, On finite groups with specific number of centralizers, Int. Electron. J. Algebra 13 (2013) 53–62. [3] S.J. Baishya, A.K. Das, Harmonic numbers and finite groups, Rend. Semin. Mat. Univ. Padova (2013), in press, http://rendiconti.math.unipd.it/ forthcoming.php?lan=english#BaishyaDas. [4] A.K. Das, On arithmetic functions of finite groups, Bull. Aust. Math. Soc. 75 (2007) 45–58. [5] S. Dolfi, M. Herzog, E. Jabara, Finite groups whose non-central commuting elements have centralizers of equal size, Bull. Aust. Math. Soc. 82 (2010) 293–304. [6] T. 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