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A New Characterization of the Janko Group

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A New Characterization of the Janko Group Australian Journal of Basic and Applied Sciences, 6(2): 130-132, 2012 ISSN 1991-8178 A New Characterization of The Janko Group Seyed Sadegh Salehi Amiri, Alireza Khalili Asboei and Abolfazl Tehranian Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran. Abstract. Let be a finite group and be the set of elements of order . Let and be the number of elements of order in . Set nse():=| . It is proved that Mathieu groups are uniquely determined by its nse and order. In this paper as the main result, it is proved that if is a group such that nse()=nse, then . Key words: Janko group, element order, simple group, Sylow subgroup. INTRODUCTION If is an integer, then we denote by the set of all prime divisors of . Let be a finite group. Denote by the set of primes such that contains an element of order . Also the set of element orders of is denoted by . A finite group is called a simple group, if is a simple group with || . Set =| | the order of is and nse: | . In fact, is the number of elements of order in and nse is the set of sizes of elements with the same order. In (Shao, C.G., et al., 2008) and (Shao, C.G., Q.H. Jiang. 2010), it is proved that all simple groups and Mathieu groups can be uniquely determined by nse() and order . In (Khatami, M. et al., 2009) and (Shen, R., et al., 2010), it is proved that if is one of the groups , , and 2, , for 7,8,11,13, then it can be uniquely determined by only nse(). In this paper, by new method we show that the Janko group is characterizable by only nse(). In fact, the main theorem of our paper is as follow: Main Theorem: Let be a group such that nse() = nse(), then . We note that there are finite groups which are not characterizable by nse(G) and ||. In 1987, Thompson gave an example as follows: Let C C C CAand L4 C be the maximal subgroups of . Then nse() = nse() and ||=||, but . Throughout this paper, we denote by the Euler totient function. If is a finite group, then we denote by a Sylow subgroup of and : is the number of Sylow subgroup of , that is., =||. All further unexplained notations are standard and refer to (Conway, J. H., et al., 1985), for example. 2. Preliminary Results: In this section we bring some preliminary lemmas to be used in the proof of main theorem theorem. Lemma 2.1. (Frobenius, G., 1895) Let be a finite group and be a positive integer dividing ||. If | 1, then |||. Lemma 2.2. (Khosravi, B. and B. Khosravi. 2005) Let be a sporadic simple group and be the greatest element of . Then is uniquely determined by || and nS). Lemma 2.3. (Shen, R., et al., 2010) Let be a group containing more than two elements. Let and be the number of elements of order in . If = | is finite, then G is finite and || 1. Let be a group such that nse(G) = nse(). By Lemma 2.3, we can assume that is finite. Let be the number of elements of order . We note that = . , where is the number of cyclic subgroups of order in . Also we note that if 2, then φ is even. If , then by Lemma 2.1 and the above notations we have: | | ∑| In the proof of the main theorem, we almost apply and the above comments. Corresponding Author: Seyed Sadegh Salehi Amiri, Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran E-mail: [email protected] 130 Aust. J. Basic & Appl. Sci., 6(2): 130-132, 2012 3. Proof of the Main Theorem: Let be a group, such that nse () = nse() = {1,1463,5852,11704,15960,23408,25080,27720,29260, 35112}. At first we prove that 2,3,5,7,11,19. Since 1463 nse, it follows that by , 2 and 1463 . Let 2 , by , |1 and 1| , which implies that 3,5,7,11,13,19. If 13 , then by , 35112. On the other hand, by , we conclude that if 26 , then 15960, 25080, 27720, 35112. Since 26|1 , then 26|52536, 26|61656, 26|64296 or 26|71688, which is a contradiction. That is 26 . Thus acts fixed point freely on the set of elements of order 2, and |||, which is a contradiction. Hence 13 . Therefore 2,3,5,7,11,19. If 3, 5, 7, 11 and 19 , then 5852,23408, 11704, 25080, 15960 and 27720, by . Also we can see easily that does not contain any elements of order 2 , 3 , 5 , 7 , 11 and 19 . Similarly, we can see that if 2 , where 24, then 5852, 11704, 15960, 23408, 25080, 27720, 29260, 35112 , if 32 , then 23408 , if 9, then 25080,27720, if 25 , then 27720 and if 121 , then 27720. In follow, we show that could not be 2, 3 and must be equal to 2,3,5,7,11,19. Let 2. Since 2 , then 1,2, 2 ,…,2 . On the other hand, nse() has ten elements, which is a contradiction. We know that 2. We claim that 3. Suppose that 3. If 5, 7, 11 and 19 , then 2, which is a contradiction. Hence 5, 7, 11 or 19 . If 11 , since 11 , then 11 or 121. Suppose that 11. By Lemma 2.1, by considering ||, we have |||1 15961. Hence ||11 and so /11 15960/10 1596. We know that |||, since 3 we get a contradiction. Now suppose that 121. We have |||1 , and so |||1 15960 27720. Hence || 121 and /121 27720/110 252, since 3, we get a contradiction. Therefore 11 . If 5, since 125 , then 5 or 25 . If 5, then |||1 11705, by Lemma 2.1. Hence ||5 and /5 11704/4 2926. So 11 , which is a contradiction. Similarly if 25, then 11 . Thus 5. If 7, since 49 , then 7. We can see easily ||7 and 4180. Thus 11 , which is a contradiction. Also, if 19 , then 11 , which is a contradiction. Therefore 3. Let 2, 3. Since 3 , then 3 or 9. At first, suppose that 3. We have |||1 , where 5852, 23408. If 5852, then ||3 and /3 2926. Then 7|||, which is a contradiction. If 23408, then |||1 23409. So |||3 . We consider all possibility cases. Case1. If ||3, then /3 11704. Since 7, we get a contradiction. Case2. If ||9 , then || 2 9 175560 5852 11704 15960 23408 25080 27720 29260 35112, where , , , , , , , , and are non-negative 2 5 2 5 integers. Since 2 and 3, then 1,2,3, 2 ,...,2 2 3, 2 3,...,2 3 . Hence || 12 and 0 1 2 . By above equation we have 175560 2 9 175560 2 35512. So 19506 2 27309. Since is a non-negative integer, then there is no possibility for . Then we get a contradiction. Case3. If ||27 , then |G| 2 27 175560 5852 11704 15960 23408 25080 27720 29260 35112, where , , , , , , , , and are non-negative integers. As above we have || 12 and 0 1 2. Hence 175560 2 27 175560 2 35512. So 6502 2 9103. The only possibility for m is 13. If m13, 13 then |G| 2 27 175560 5852 11704 15960 23408 25080 27720 29260 35112, where 0 1 2 . Hence we have 5852 11704 15960 23408 25080 27720 29260 35112 45624 , where 0 1 2. It is easy to check that this equation has no solution, which is a contradiction. Case4. If |P|81 , then |G| 2 81 175560 5852 11704 15960 23408 25080 27720 29260 35112, where , , , , , , , , and are non-negative integers. Again as above || 12 and 0 1 2. Hence 175560 2 81 175560 2 35512. So 2167 2 3034. Since is a non-negative integer, then there is no possibility for m. Then we get a contradiction. Now Suppose that 9. We have |P||1 m3 9, where 3 5852, 23408 and 9 25080, 27720. It is easy to see |P3||9. Since 9. then |3|9. On the other hand, we have n3 9/φ9. So 3 4180 or 4620. Since 11|3 and 3||G|, we get a contradiction. Therefore can not be equal to 2,3. 131 Aust. J. Basic & Appl. Sci., 6(2): 130-132, 2012 Until now, we prove that 2,3 2,3,5,7,11,19. So at least on of the numbers 5,7,11 or 19 belong to . If 7, then n 4180. Hence 5 . Similarly if 19 , then 5. If 11 , then n 252 or 1596. Hence 7 and so 5. Therfore in any cases we can assume that 5 . If 5 , then n 1386 or 2926. Since 5||G| , then 2, 3, 5, 7, 11 or 2, 3, 5, 7, 11, 19. If 2, 3, 5, 7, 11, then 4180. So 19||G|, which is a contradiction. Hence we have 2 2 2, 3, 5, 7, 11, 19. Since 7 and 19 , we have |7|7 and |P|19. We prove that 21, 95 and 209 . Suppose that 95 , we know that if and are Sylow 19 subgroups of , then and are conjugate, which implies that and are conjugate in . Therefore φ95. n.k, where k is the number of cyclic subgroups of order 5 in P. Since 19/φ191540, we have 72 1540|95, which is a contradiction. Hence 95 . Similarly we can prove that 21 and 209 . Since 95 , then the P5 acts fixed point freely on the set of elements of order 19, and so |P||m 27720 , which implies that |P|5.
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