Let $I_n(G)$ denote the number of elements of order $n$ in a finite group $G$. Malinowska recently asked “what is the smallest positive integer $k$ such that whenever there exist two nonabelian finite simple groups $S$ and $G$ with prime divisors $p_1,\,\cdots ,\,p_k$ of $|G|$ and $|S|$ satisfying $2=p_1<\,\cdots \

Chimere Stanley Anabanti

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Anabanti, Chimere Stanley

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Comptes RendusMathématiqueChimere Stanley AnabantiA question of Malinowska on sizes of finite nonabelian simple groups inrelation to involution sizesVolume 358, issue 11-12 (2020), p. 1135-1138Published online: 25 January 2021https://doi.org/10.5802/crmath.130This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/Les Comptes Rendus. Mathématique sont membres duCentre Mersenne pour l’édition scientifique ouvertewww.centre-mersenne.orge-ISSN : 1778-3569Comptes RendusMathématique2020, 358, n 11-12, p. 1135-1138https://doi.org/10.5802/crmath.130Group Theory / Théorie des groupesA question of Malinowska on sizes of finitenonabelian simple groups in relation toinvolution sizesChimere Stanley Anabantiaa Institut für Analysis und Zahlentheorie, Technische Universität Graz, Austria.E-mail: anabanti@math.tugraz.at, chimere.anabanti@unn.edu.ngAbstract. Let In (G) denote the number of elements of order n in a finite group G . Malinowska recently asked“what is the smallest positive integer k such that whenever there exist two nonabelian finite simple groupsS and G with prime divisors p1, · · · , pk of |G| and |S| satisfying 2 = p1 < ·· · < pk and Ipi (G) = Ipi (S) for alli ∈ {1, · · · , k}, we have that |G| = |S|?”. This paper resolves Malinowska’s question.2020 Mathematics Subject Classification. 20D60,20D06.Funding. The author is supported by both TU Graz and partial funding from the Austrian Science Fund(FWF): P30934-N35, F05503, F05510. He is also at the University of Nigeria, Nsukka.Manuscript received 25th May 2020, revised 6th October 2020, accepted 7th October 2020.1. IntroductionIn 1979, Herzog [6] conjectured that two finite simple groups containing the same numberof involutions have the same order. Zarrin [9], in 2018, disproved Herzog’s conjecture with acounterexample. Then he conjectured that “if S is a non-abelian simple group and G a groupsuch that I2(G) = I2(S) and Ip (G) = Ip (S) for some odd prime divisor p, then |G| = |S|”. Zarrin’sconjecture was disproved by the recent author in [1]. In an attempt to reformulate the mentionedconjecture of Zarrin (considering the works in [1,6,9]), Malinowska [8] asked: “what is the smallestpositive integer k such that whenever there exist two nonabelian finite simple groups S and Gwith prime divisors p1, · · · , pk of |G| and |S| satisfying 2 = p1 < ·· · < pk and Ipi (G) = Ipi (S) for alli ∈ {1, · · · ,k}, we have that |G| = |S|?”.A simple computational check in GAP [5] or Magma [3] (see [1] for example) tells us thatI2(A8) = 315 = I2(PSL(3,4)), I7(A8) = 5760 = I7(PSL(3,4)) and |A8| = 20160 = |PSL(3,4)|. Also,the only prime divisors of 20160 are 2,3,5 and 7, and that I3(A8) = 1232 < 2240 = I3(PSL(3,4))and I5(A8) = 1344 < 8064 = I5(PSL(3,4)). One might think that the smallest such integer k inMalinowska’s question is 2 (since k > 1 by hypothesis). This is not true since the results in [1]implies that k > 2. In particular, PSL(4,3) and PSL(3,9) are nonabelian simple groups. AssumeISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/1136 Chimere Stanley Anabantifor contradiction that k = 2. Then there exist prime divisors p1 and p2 of the orders of the groupsPSL(4,3) and PSL(3,9) such that 2 = p1 < p2 = 13 and I2(PSL(4,3)) = 7371 = I2(PSL(3,9)),I13(PSL(4,3)) = 1866240 = I13(PSL(3,9)) but |PSL(4,3)| = 6065280 < 42456960 = |PSL(3,9)|.Thus, the smallest value of k in Malinowska’s question cannot be 2. The goal of this paper is toresolve Malinowska’s question.For clarity, we reinstate Malinowska’s question as follows: “What is the smallest positive integerk such that whenever there exist two nonabelian finite simple groups S and G that satisfyHypothesis(k), we have that |G| = |S|?”. We note that Hypothesis(k) says that there exist primedivisors p1, · · · , pk of |G| and |S| satisfying 2 = p1 < ·· · < pk such that G and S have the samenumbers of elements of order pi for each i ∈ [1,k].For the rest of this section, we state the main result of this paper.Theorem 1. If G and S are two non-isomorphic finite simple nonabelian groups such that|G| = |S|, then Hypothesis(k) does not hold for any k > 2.2. Proof of Theorem 1Before we give a proof of Theorem 1, we first recall some important results.Theorem 2 ([7, Theorem 5.1]). If G and S are non-isomorphic finite nonabelian simple groupssuch that |S| = |G|, then either G = PSL3(4) and S = PSL4(2) or G =Ω2n+1(q) and S = PSp2n(q) forsome odd prime power q, and some n ≥ 3.As at 1955, Artin [2] gave a prototype of Theorem 2 for all the finite nonabelian simple groupsknown then. The groups (PSL3(4) and PSL4(2) or Ω2n+1(q) and PSp2n(q) for some odd primepower q , and some n ≥ 3) mentioned in Theorem 2 above were also given in Artin’s paper sincethey were known at that time. As more finite simple groups were discovered, Tits et al. in manypapers reaffirmed that no other pair of finite simple groups other than the ones found by Artinsatisfies the hypothesis of Theorem 2.Remark 3.(i) The groups Ω2n+1(q) and PSp2n(q) are simple for all odd prime power q and n > 2.Moreover, ∣∣Ω2n+1(q)∣∣= qn2 ∏ni=1(q2i −1)2= ∣∣PSp2n(q)∣∣ .(ii) The involutions in Ω2n+1(q) and PSp2n(q) arise from subspace configurations of theFq -space V that O(V ) or Sp(V ) naturally acts on. The group Ω2n+1(q) has n conjugacyclasses of involutions and the group PSp2n(q) has bn2 c+ 1 conjugacy classes of involu-tions. One can then use [4, Table 4.5.1] or otherwise to obtain the values of I2(Ω2n+1(q))and I2(PSp2n(q)) according as q ≡ 1 mod 4 or q ≡ 3 mod 4. We give a summary of suchresults in Proposition 4 below.Proposition 4. Let q be any odd prime power and n ≥ 3. Then I2(Ω2n+1(q)) 6= I2(PSp2n(q)). Inparticular, the following holds:(a) if q ≡ 3 mod 4, then I2(Ω2n+1(q)) > I2(PSp2n(q));(b) if q ≡ 1 mod 4, then I2(Ω2n+1(q)) < I2(PSp2n(q)).Proof. Follows from computations using [4, Table 4.5.1]. We now give a proof of Theorem 1.C. R. Mathématique, 2020, 358, n 11-12, 1135-1138Chimere Stanley Anabanti 1137Proof of Theorem 1. Let G and S be two non-isomorphic finite simple nonabelian groups suchthat |G| = |S|. Suppose there exists an integer k > 2 satisfying Hypothesis(k). In the light ofTheorem 2, (G ,S) ∈ {(PSL3(4),PSL4(2)), (Ω2n+1(q),PSp2n(q))} for some odd prime power q , andn ≥ 3. The exposition above tells us that G and S cannot be PSL3(4) and PSL4(2). So the onlypossibilty is that G and S areΩ2n+1(q) and PSp2n(q) for some odd prime power q , and n ≥ 3. ByProposition 4, I2(Ω2n+1(q)) 6= I2(PSp2n(q)) for all such q and n; a contradiction to Hypothesis(k).Therefore k 6> 2. An immediate consequence of Theorems 1 and 2 is the following:Corollary 5. k = 3 in Malinowska’s question.3. An observationWe close this paper with Observation 6 below which was motivated by the following question:Are there non-isomorphic simple groups S and G with In(G) = In(S) for all orders of elements n?The answer to this question is clearly ‘yes’ if we remove the word “simple” from the question. Forinstance, the Heisenberg group mod 3 (given by the presentation: 〈x, y, z | x3 = y3 = z3 = [x, z] =[y, z] = 1,[x, y] = z〉) and the elementary abelian 3-group of rank 3 are the only two groups ofsize 27 whose exponent is 3; so they give rise to a positive answer to the question when the word“simple” is removed from the question. However, if we leave the question ‘as it is’, then an answeris given in the following result:Observation 6. There are no non-isomorphic simple groups S and G with In(G) = In(S) for allorders of elements n.Proof. Suppose there are two non-isomorphic simple groups S and G with In(G) = In(S) forall orders of elements n. Clearly, |S| = |G|. In the light of Theorem 2, either G = PSL(3,4) andS = PSL(4,2) or G = Ω2n+1(q) and S = PSp2n(q) for some odd prime power q, and some n > 2.ButI3(PSL(4,2)) = 1232 < 2240 = I3(PSL(3,4)).So G and S cannot be PSL(3,4) and PSL(4,2); whence we take G =Ω2n+1(q) and S = PSp2n(q) forsome odd prime power q , and some n > 2. This choice is not possible by Proposition 4. Thus, nosuch non-isomorphic simple groups exist. AcknowledgementsThe author is grateful to Professors Sarah Hart, Richard Lyons and Geoff Robinson for usefuldiscussion during the writing of this paper.References[1] C. S. Anabanti, “A counterexample to Zarrin’s Conjecture on sizes of finite nonabelian simple groups in relation toinvolution sizes”, Arch. Math. 112 (2019), no. 3, p. 225-226.[2] E. Artin, “The Orders of the Classical Simple Groups”, Commun. Pure Appl. Math. 8 (1955), p. 455-472.[3] W. Bosma, J. Cannon, C. Playoust, “The Magma algebra system”, 2019, Version 2.24–5, http://magma.maths.usyd.edu.au/calc/.[4] D. Gorenstein, R. Lyons, R. Solomon, The Classification of the Finite Simple Groups. Part I, Chapter A: Almost simpleK -groups, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, 1998.[5] T. G. Group, “GAP – Groups, Algorithms, and Programming”, 2020, Version 4.11.0, https://www.gap-system.org.[6] M. Herzog, “On the classification of finite simple groups by the number of involutions”, Proc. Am. Math. Soc. 77 (1979),no. 3, p. 313-314.C. R. Mathématique, 2020, 358, n 11-12, 1135-11381138 Chimere Stanley Anabanti[7] W. Kimmerle, R. Lyons, S. Robert, D. N. Teague, “Composition factors from the group ring and Artin’s theorem onorders of simple groups”, vol. 60, 3, no. 1, London Mathematical Society, 1990, p. 89-122.[8] I. A. Malinowska, “Finite groups with few normalizers or involutions”, Arch. Math. 112 (2019), no. 5, p. 459-465.[9] M. Zarrin, “A counterexample to Herzog’s Conjecture on the number of involutions”, Arch. Math. 111 (2018), no. 4,p. 349-351.C. R. Mathématique, 2020, 358, n 11-12, 1135-1138

A question of Malinowska on sizes of finite nonabelian simple groups in relation to involution sizes

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Comptes RendusMathématiqueChimere Stanley AnabantiA question of Malinowska on sizes of finite nonabelian simple groups inrelation to involution sizesVolume 358, issue 11-12 (2020), p. 1135-1138.<https://doi.org/10.5802/crmath.130>© Académie des sciences, Paris and the authors, 2020.Some rights reserved.This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/Les Comptes Rendus. Mathématique sont membres duCentre Mersenne pour l’édition scientifique ouvertewww.centre-mersenne.orgComptes RendusMathématique2020, 358, n 11-12, p. 1135-1138https://doi.org/10.5802/crmath.130Group Theory / Théorie des groupesA question of Malinowska on sizes of finitenonabelian simple groups in relation toinvolution sizesChimere Stanley Anabantiaa Institut für Analysis und Zahlentheorie, Technische Universität Graz, Austria.E-mail: anabanti@math.tugraz.at, chimere.anabanti@unn.edu.ngAbstract. Let In (G) denote the number of elements of order n in a finite group G . Malinowska recently asked“what is the smallest positive integer k such that whenever there exist two nonabelian finite simple groupsS and G with prime divisors p1, · · · , pk of |G| and |S| satisfying 2 = p1 < ·· · < pk and Ipi (G) = Ipi (S) for alli ∈ {1, · · · , k}, we have that |G| = |S|?”. This paper resolves Malinowska’s question.2020 Mathematics Subject Classification. 20D60,20D06.Funding. The author is supported by both TU Graz and partial funding from the Austrian Science Fund(FWF): P30934-N35, F05503, F05510. He is also at the University of Nigeria, Nsukka.Manuscript received 25th May 2020, revised 6th October 2020, accepted 7th October 2020.1. IntroductionIn 1979, Herzog [6] conjectured that two finite simple groups containing the same numberof involutions have the same order. Zarrin [9], in 2018, disproved Herzog’s conjecture with acounterexample. Then he conjectured that “if S is a non-abelian simple group and G a groupsuch that I2(G) = I2(S) and Ip (G) = Ip (S) for some odd prime divisor p, then |G| = |S|”. Zarrin’sconjecture was disproved by the recent author in [1]. In an attempt to reformulate the mentionedconjecture of Zarrin (considering the works in [1,6,9]), Malinowska [8] asked: “what is the smallestpositive integer k such that whenever there exist two nonabelian finite simple groups S and Gwith prime divisors p1, · · · , pk of |G| and |S| satisfying 2 = p1 < ·· · < pk and Ipi (G) = Ipi (S) for alli ∈ {1, · · · ,k}, we have that |G| = |S|?”.A simple computational check in GAP [5] or Magma [3] (see [1] for example) tells us thatI2(A8) = 315 = I2(PSL(3,4)), I7(A8) = 5760 = I7(PSL(3,4)) and |A8| = 20160 = |PSL(3,4)|. Also,the only prime divisors of 20160 are 2,3,5 and 7, and that I3(A8) = 1232 < 2240 = I3(PSL(3,4))and I5(A8) = 1344 < 8064 = I5(PSL(3,4)). One might think that the smallest such integer k inMalinowska’s question is 2 (since k > 1 by hypothesis). This is not true since the results in [1]implies that k > 2. In particular, PSL(4,3) and PSL(3,9) are nonabelian simple groups. AssumeISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/1136 Chimere Stanley Anabantifor contradiction that k = 2. Then there exist prime divisors p1 and p2 of the orders of the groupsPSL(4,3) and PSL(3,9) such that 2 = p1 < p2 = 13 and I2(PSL(4,3)) = 7371 = I2(PSL(3,9)),I13(PSL(4,3)) = 1866240 = I13(PSL(3,9)) but |PSL(4,3)| = 6065280 < 42456960 = |PSL(3,9)|.Thus, the smallest value of k in Malinowska’s question cannot be 2. The goal of this paper is toresolve Malinowska’s question.For clarity, we reinstate Malinowska’s question as follows: “What is the smallest positive integerk such that whenever there exist two nonabelian finite simple groups S and G that satisfyHypothesis(k), we have that |G| = |S|?”. We note that Hypothesis(k) says that there exist primedivisors p1, · · · , pk of |G| and |S| satisfying 2 = p1 < ·· · < pk such that G and S have the samenumbers of elements of order pi for each i ∈ [1,k].For the rest of this section, we state the main result of this paper.Theorem 1. If G and S are two non-isomorphic finite simple nonabelian groups such that|G| = |S|, then Hypothesis(k) does not hold for any k > 2.2. Proof of Theorem 1Before we give a proof of Theorem 1, we first recall some important results.Theorem 2 ([7, Theorem 5.1]). If G and S are non-isomorphic finite nonabelian simple groupssuch that |S| = |G|, then either G = PSL3(4) and S = PSL4(2) or G =Ω2n+1(q) and S = PSp2n(q) forsome odd prime power q, and some n ≥ 3.As at 1955, Artin [2] gave a prototype of Theorem 2 for all the finite nonabelian simple groupsknown then. The groups (PSL3(4) and PSL4(2) or Ω2n+1(q) and PSp2n(q) for some odd primepower q , and some n ≥ 3) mentioned in Theorem 2 above were also given in Artin’s paper sincethey were known at that time. As more finite simple groups were discovered, Tits et al. in manypapers reaffirmed that no other pair of finite simple groups other than the ones found by Artinsatisfies the hypothesis of Theorem 2.Remark 3.(i) The groups Ω2n+1(q) and PSp2n(q) are simple for all odd prime power q and n > 2.Moreover, ∣∣Ω2n+1(q)∣∣= qn2 ∏ni=1(q2i −1)2= ∣∣PSp2n(q)∣∣ .(ii) The involutions in Ω2n+1(q) and PSp2n(q) arise from subspace configurations of theFq -space V that O(V ) or Sp(V ) naturally acts on. The group Ω2n+1(q) has n conjugacyclasses of involutions and the group PSp2n(q) has bn2 c+ 1 conjugacy classes of involu-tions. One can then use [4, Table 4.5.1] or otherwise to obtain the values of I2(Ω2n+1(q))and I2(PSp2n(q)) according as q ≡ 1 mod 4 or q ≡ 3 mod 4. We give a summary of suchresults in Proposition 4 below.Proposition 4. Let q be any odd prime power and n ≥ 3. Then I2(Ω2n+1(q)) 6= I2(PSp2n(q)). Inparticular, the following holds:(a) if q ≡ 3 mod 4, then I2(Ω2n+1(q)) > I2(PSp2n(q));(b) if q ≡ 1 mod 4, then I2(Ω2n+1(q)) < I2(PSp2n(q)).Proof. Follows from computations using [4, Table 4.5.1]. We now give a proof of Theorem 1.C. R. Mathématique, 2020, 358, n 11-12, 1135-1138Chimere Stanley Anabanti 1137Proof of Theorem 1. Let G and S be two non-isomorphic finite simple nonabelian groups suchthat |G| = |S|. Suppose there exists an integer k > 2 satisfying Hypothesis(k). In the light ofTheorem 2, (G ,S) ∈ {(PSL3(4),PSL4(2)), (Ω2n+1(q),PSp2n(q))} for some odd prime power q , andn ≥ 3. The exposition above tells us that G and S cannot be PSL3(4) and PSL4(2). So the onlypossibilty is that G and S areΩ2n+1(q) and PSp2n(q) for some odd prime power q , and n ≥ 3. ByProposition 4, I2(Ω2n+1(q)) 6= I2(PSp2n(q)) for all such q and n; a contradiction to Hypothesis(k).Therefore k 6> 2. An immediate consequence of Theorems 1 and 2 is the following:Corollary 5. k = 3 in Malinowska’s question.3. An observationWe close this paper with Observation 6 below which was motivated by the following question:Are there non-isomorphic simple groups S and G with In(G) = In(S) for all orders of elements n?The answer to this question is clearly ‘yes’ if we remove the word “simple” from the question. Forinstance, the Heisenberg group mod 3 (given by the presentation: 〈x, y, z | x3 = y3 = z3 = [x, z] =[y, z] = 1,[x, y] = z〉) and the elementary abelian 3-group of rank 3 are the only two groups ofsize 27 whose exponent is 3; so they give rise to a positive answer to the question when the word“simple” is removed from the question. However, if we leave the question ‘as it is’, then an answeris given in the following result:Observation 6. There are no non-isomorphic simple groups S and G with In(G) = In(S) for allorders of elements n.Proof. Suppose there are two non-isomorphic simple groups S and G with In(G) = In(S) forall orders of elements n. Clearly, |S| = |G|. In the light of Theorem 2, either G = PSL(3,4) andS = PSL(4,2) or G = Ω2n+1(q) and S = PSp2n(q) for some odd prime power q, and some n > 2.ButI3(PSL(4,2)) = 1232 < 2240 = I3(PSL(3,4)).So G and S cannot be PSL(3,4) and PSL(4,2); whence we take G =Ω2n+1(q) and S = PSp2n(q) forsome odd prime power q , and some n > 2. This choice is not possible by Proposition 4. Thus, nosuch non-isomorphic simple groups exist. AcknowledgementsThe author is grateful to Professors Sarah Hart, Richard Lyons and Geoff Robinson for usefuldiscussion during the writing of this paper.References[1] C. S. Anabanti, “A counterexample to Zarrin’s Conjecture on sizes of finite nonabelian simple groups in relation toinvolution sizes”, Arch. Math. 112 (2019), no. 3, p. 225-226.[2] E. Artin, “The Orders of the Classical Simple Groups”, Commun. Pure Appl. Math. 8 (1955), p. 455-472.[3] W. Bosma, J. Cannon, C. Playoust, “The Magma algebra system”, 2019, Version 2.24–5, http://magma.maths.usyd.edu.au/calc/.[4] D. Gorenstein, R. Lyons, R. Solomon, The Classification of the Finite Simple Groups. Part I, Chapter A: Almost simpleK -groups, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, 1998.[5] T. G. Group, “GAP – Groups, Algorithms, and Programming”, 2020, Version 4.11.0, https://www.gap-system.org.[6] M. Herzog, “On the classification of finite simple groups by the number of involutions”, Proc. Am. Math. Soc. 77 (1979),no. 3, p. 313-314.C. R. Mathématique, 2020, 358, n 11-12, 1135-11381138 Chimere Stanley Anabanti[7] W. Kimmerle, R. Lyons, S. Robert, D. N. Teague, “Composition factors from the group ring and Artin’s theorem onorders of simple groups”, vol. 60, 3, no. 1, London Mathematical Society, 1990, p. 89-122.[8] I. A. Malinowska, “Finite groups with few normalizers or involutions”, Arch. Math. 112 (2019), no. 5, p. 459-465.[9] M. Zarrin, “A counterexample to Herzog’s Conjecture on the number of involutions”, Arch. Math. 111 (2018), no. 4,p. 349-351.C. R. Mathématique, 2020, 358, n 11-12, 1135-1138

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A question of Malinowska on sizes of finite nonabelian simple groups in relation to involution sizes

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