A Simple Classification of Finite Groups of Order p 2 q 2

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1 Mathematics Interdisciplinary Research (017, xx yy A Simple Classification of Finite Groups of Order p Aziz Seyyed Hadi, Modjtaba Ghorbani and Farzaneh Nowroozi Larki Abstract Suppose G is a group of order p where p > are prime numbers and suppose P and Q are Sylow p-subgroup and Sylow -subgroup of G, respectively. In this paper, we show that up to isomorphism, there are four groups of order p when Q and P are cyclic, three groups when Q is a cyclic and P is an elementary abelian group, p + 3p/ + 7 groups when Q is an elementary abelian group and P is a cyclic group and finally, p+5 groups when both Q and P are elementary abelian groups. Keywords: Semi-direct product, p-group, Sylow subgroup. 010 Mathematics Subject Classification: 0D0. 1. Introduction All finite groups can be broken down into a series of finite simple groups which have been called the building blocks of finite groups. The history of finite simple groups originated in the 1830 s with Evariste Galois and the solution of fifth degree polynomial euations. In the twentieth century, the recognition of the importance of finite simple groups inspired a huge effort to find all finite simple groups, see [9] for more details. But the classification of finite groups is still an open problem. In this paper, by using the notation of [1], we determine all groups of order p by a simple method.. Preliminaries and Notations In this section, we first recall some concepts, notations and results in group theory, which are used in the next section. We use standard basic group theory terminol- Corresponding author ( mghorbani@srttu.edu Academic Editor: Ali Reza Ashrafi Received 10 November 015, Accepted 31 May 016 DOI: 10.05/mir c 017 University of Kashan

2 74 A. Seyyed Hadi, M. Ghorbani and F. Nowroozi Larki ogy and notations, see [, 4, 5, 7, 10, 11] as well as [6, 8]. The set of all Sylow p-subgroups of G are denoted by Syl p (G. A semi-direct product is a generalization of a direct product. Let N be a normal subgroup of G, each element g G defines an automorphism of N, n gng 1, and this defines a homomorphism θ : G Aut(N, g i g N. If there exists a subgroup Q of G such that G G/N maps Q isomorphically onto G/N, then we can reconstruct G from N, Q and the restriction of θ to Q. Indeed, an element g G can be written uniuely in the form g = n, n N, Q. must be the uniue element of Q mapping to gn G/N and n must be g 1. Thus, we have a one-to-one correspondence of sets If g = n and g = n, then G N Q. gg = (n(n = n(n 1 = nθ((n.. Euivalently, G is a semi-direct product of subgroups N and Q if N G, NQ = G, N Q = {1}. Note that Q need not be a normal subgroup of G. When G is the semi-direct product of subgroups N and Q, we write G = N Q or N ϕ Q. Theorem.1. [8] Let P be a p-group, Q be a -group and ψ, ϕ : Q Aut(P be two homomorphisms. Then Q ϕ P = Q ψ P if and only if ψ γ and ϕ are conjugate in Aut(P for some γ Aut(Q. Lemma.. [8] Up to isomorphism, there is a 1-1 correspondence between groups G = Q ϕ P and the number of orbits of action Aut(P Aut(Q on the set Hom(Q, Aut(P where for all α Aut(P, β Aut(Q and y Q, we have ϕ (α,β (y = α [ϕ β 1 (y] α 1. Suppose Q = y is a cyclic -group. For every non-trivial homomorphism ϕ : Q Aut(P, Q y = ϕ(y is a subgroup of Aut(P of order Q. On the other hand, all automorphisms of Q map y to y j for some j and hence Q y j = (Q y j = Q y. Thus, we can deduce the following result. Lemma.3. Let Q be a cyclic group, then up to isomorphism, all groups of form G = Q P are corresponding to conjugacy classes of subgroups of Aut(P of order dividing Q.

3 A Simple Classification of Finite Groups of Order p 75 The general linear group of degree n is the set of n n non-singular matrices, together with ordinary multiplication of matrices as its binary operation. Because the product of two non-singular matrices is again non-singular, and the inverse of a non-singular matrix is non-singular, then it forms a group. Generally, the general linear group of degree n over any field F is the set of n n non-singular matrices with entries from F denoted by GL n (F or GL(n, F. A field F that contains a finite number of elements is called Galois field. A finite field of order exists if and only if the is a prime power p k where p is a prime number and k is a positive integer. All fields of a given order are isomorphic. By this notation, if F is a finite field with = p n elements, then the general linear group of degree n over the filed F is denoted by GL(n,. Theorem.4. [8] The conjugacy classes of GL(, p are as reported in Table 1. In this table ρ, σ are primitive elements of GF (p and GF (p, respectively. Table 1: Conjugacy classes of group GL(, p. Conjugacy ( classes Number of such classes No. Elements ρ a 0 0 ρ a (p 1 1 ( ρ a 1 0 ρ a (p 1 (p 1(p + 1 ( ρ a 0 1 (p 1(p p(p + 1 ( 0 ρ b σ a 0 0 σ ap 1 p(p 1 p(p 1 3. Main Results In this section, the conseuences of the Sylow Theorems are cases where the size of G forces it to have a non-trivial normal subgroup. Theorem 3.1. Let G be a group of order p. If p = 6, then G is isomorphic with one of the following groups: 1. C 36,. C 18 C, 3. C 6 C 6, 4. C 1 C 3, 5. D 6 C 6 = S3 C 6 = D1 C 3,

4 76 A. Seyyed Hadi, M. Ghorbani and F. Nowroozi Larki 6. D 6 D 6 = S3 S 3, 7. D 18 C, 8. A 4 C 3, 9. D 36, 10. H C 3, 11. K C, 1. a, b, c a 3 = b 3 = c 4 = [a, b] = 1, c 1 ac = b, c 1 bc = a 1, 13. a, b, c a = b = c 9 = [a, b] = 1, c 1 ac = b, c 1 bc = ab, 14. a, b, c a 3 = b 3 = c 4 = [a, b] = 1, c 1 ac = a 1, c 1 bc = b 1, where and H = x, y x 4 = y 3 = 1, x 1 yx = y 1 K = a, b, c a = b = c = (abc = (ab 3 = (ac 3 = 1. If p 6, then G has one of the following structures: C C p, C (C p C p, (C C C p, (C C (C p C p. Proof. If p = 6, then the proof is clear. Suppose p 6, since p > according to Sylow theorem, it is clear that Sylow p-subgroup of G is normal. We assume that P Syl p (G and Q Syl (G. This implies that Q P = 1, P G, G = P Q. Hence, G is isomorphic with semi-direct product Q ϕ P, where ϕ : Q Aut(P is a homomorphism. Since, every group of order p is abelian, then Syl p (G = {C p, C p C p } and Syl (G = {C, C C }. By using Lemma. and Theorem.1, the group G is isomorphic with one of the following presentation: C ϕ C p, (C C ϕ C p, C ϕ (C p C p, (C C ϕ (C p C p. In continuing, we determine the presentation of all these groups. To do this, we can consider the following cases: 1. G = C ϕ C p, then Aut(C p = C p(p 1 and consider the homomorphism ϕ : C C p(p 1. If Imϕ = 1, then G = C C p. If Imϕ = C and (p 1, since Aut(P is cyclic, according to Lemma.3 there exists only one group with the following presentation: G = a, b a = b p = 1, a 1 ba = b r, r 1(mod p = a, b a = b p = 1, a 1 ba = b r (p 1p, r = r0,

5 A Simple Classification of Finite Groups of Order p 77 where r 0 is p -th root of unity. If Imϕ = C and (p 1, since Aut(P is cyclic, by Lemma.3, there exists only one group with the following presentation: G = a, b a = b p = 1, a 1 ba = b r, r 1(mod p (p 1p = a, b a = b p = 1, a 1 ba = b r, r = r 0, where r 0 is p -th root of unity.. G = (C C ϕ C p and assume that the image of ϕ is not trivial where ϕ : C C Aut(C p. If Imϕ = 1, then G = C C C p. Let Imϕ = C and (p 1. Since Aut(P is cyclic, by using Lemma 1.3, there exists only one group with the following presentation: G = a, b, c a = b = c p = 1, [a, b] = [b, c] = 1, a 1 ca = c r, where r 0 is p -th root of unity and r 1(mod p. 3. G = C ϕ (C p C p Aut(C p C p = GL(, p. If Imϕ = 1, then G = C (C p C p G is abelian. If Imϕ = C, then for given homomorphism ϕ : C GL(, p, Imϕ = 1 or C or C. Suppose the subgroups of order are in the first class in Table 1. Hence, p 1 and α is a primitive root of unity of F p This implies that Imϕ = C = ( β 0 0 β β = α p 1. G = a, b, c a = b p = c p = 1, a 1 ba = b β, a 1 ca = c β, bc = cb. Let the subgroups of order be in the second class of Table 1. Clearly, in this case, one can not find a new presentation for G. Let subgroups of order be in the third class. If =, then Imϕ = C = ( G = a, b, c a = b p = c p = 1, ab = ba, a 1 ca = c 1, bc = cb. Let (p 1 and, then α is a p-th root of unity and β = α p 1. All solutions of euation x 1( mod p are x 1 = 1, x = β, x 3 = β,..., x = β 1.

6 78 A. Seyyed Hadi, M. Ghorbani and F. Nowroozi Larki This means that there are +1 non-conjugate cyclic groups of order as follows: ( β ( β 0, 0 β i where i =, 3,..., 1 and 1. Hence in this case, one can verify that G = a, b, c a = b p = c p = 1, ab = ba, a 1 ca = c β, bc = cb, G = a, b, c a = b p = c p = 1, a 1 ba = b β, a 1 ca = c βi, bc = cb. Finally, let the subgroups of order be in the fourth row in Table 1. For = there is no group of order two. If and p 1, then there is no a subgroup of order, but if divides p + 1, then we can construct a subgroup of order as follows: p 1 Suppose σ be a primitive root of GF (p and α 0 = σ, then p 1 a 0 = σ = α + β D : α, β, D GF (p, β 0, D is not suare ( α0 0 0 α p 0 ( α βd [ β α ] where for an element g G, [g] means the conjugacy class of g in G. This means that G = a, b, c a = b p = c p = 1, a 1 ba = b α c βd, a 1 ca = b β c α, bc = cb. Suppose Imϕ = C, then all non-conjugate cyclic subgroups of order in GL(, p are as follows: (a The cyclic subgroups of order are in the first row in Table 1. Further, divides p 1 and α is a p-primitive root of unity in F p, thus Hence Imϕ = C = ( β 0 0 β, β = α p 1. G = a, b, c a = b p = c p = 1, a 1 ba = b β, a 1 ca = c β, bc = cb. (b The cyclic subgroups of order are in the second row. In this case, there is no a cyclic subgroup of order.

7 A Simple Classification of Finite Groups of Order p 79 (c The cyclic subgroups of order are in the third row. If divides p 1, then the euation x 1 (mod p has exactly roots. Among them, ( 1 subgroups are of order there are + non-conjugate subgroups of order as follows: Let α be a p-primitive root of unity, then ( β = α p 1 β 0 and C i = 0 β i, where i ( 1/ or i = k (k ( + 1/ or i = 1. Thus, there are ( + / groups with the following presentation: G i = a, b, c a = b p = c p = 1, a 1 ba = b β, a 1 ca = c βi, bc = cb. (d The cyclic subgroups of order are in the fourth row. If =, then two following cases hold: 1 p = 4k + 1 (p 1/ = k(k + 1. Hence, there is not a cyclic group of order. p = 4k + 3 p 1 = 8(k + 1(k + 1. This leads us to verify that (p 1/ = (k + 1(k + 1. In this case, we can construct a cyclic subgroup as follows: Let σ be a generator of multiplicative group GF (p, then p 1 4 σ = α + β D : α, β, D GF (p, β 0, D is not suare G = a, b, c a 4 = b p = c p = 1, a 1 ba = b α c βd, a 1 ca = b β c α, bc = cb. Let and p + 1 or p 1. If p 1, then p 1 we can not construct a new presentation. Suppose p + 1 and σ is a multiplicative subgroup of GF (p. Then G = a, b, c a = b p = c p = 1, a 1 ba = b α c βd, a 1 ca = b β c α, bc = cb. 4. G = (C C ϕ (C p C p, first notice that Aut(C p C p = GL(, p. We are interested about all automorphism of the form ϕ : (C C Aut(C p C p = GL(, p. If Imϕ = 1, then G = C C C p C p is an abelian group of order p. If Imϕ = C, then all non-conjugate subgroups of order in GL(, p are as follows: (a Subgroups belonging to the first row in Table 1. If (p 1 and α is a p-primitive root of unity, then ( Imϕ = C β 0 =, β = α p 1 0 β

8 80 A. Seyyed Hadi, M. Ghorbani and F. Nowroozi Larki and hence G = a, b, c, d a = b = c p = d p = 1, b 1 cb = c β, b 1 db = d β where [a, c] = [a, d] = [a, b] = [c, d] = 1. (b Subgroups belonging to the second row in Table 1. In this case there is no element of order. (c Subgroups belonging to the third row in Table 1. If =, then ( Imϕ = C 1 0 = 0 1 G = a, b, c, d a = b = c p = d p = 1, b 1 cb = c 1, where [a, b] = [a, c] = [a, d] = [b, d] = [c, d] = 1. If, (p 1, then α is a root of unity and β = α p 1. Thus, all roots of the euation x 1(mod p are x 1 = 1, x = β, x 3 = β,..., x = β 1. Hence, there are ( + 1/ non-conjugate cyclic subgroups ( ( 1 0 β 0, 0 β 0 β i, where i =, 3, 4,..., 1 or i = 1. Similar to our last discussion, we get two presentations for G which are not new. (d Subgroup of order belonging to the fourth row in Table 1. It is not difficult to see that in this case a group of order exists if, p 1 k and p + 1. Let σ be a root of unity in GF (p and α 0 = σ p 1 = α + β D : α, β, D GF (p, β 0, D is not suare. Then ( α0 0 0 α p 0 and hence ( α βd [ β α G = a, b, c, d a = b = c p = d p = 1, b 1 cb = c α d βd, b 1 db = c β d α, where [a, b] = [a, c] = [a, d] = [c, d] = 1. Finally, suppose Imϕ = C C and α is p-th primitive root of unity, β = α p 1, and p 1. Then ( ( C C β =, β ]

9 A Simple Classification of Finite Groups of Order p 81 G = a, b, c, d a = b = c p = d p = 1, a 1 ca = c β, b 1 db = d β, where [a, b] = [b, c] = [c, d] = [a, d] = 1. This completes the proof. References [1] W. Burnside, Theory of Groups of Finite Order, d ed. Dover Publications, Inc., New York, [] D. S. Dummit, R. M. Foote, Abstract Algebra, Third Edition, John Wiley & Sons, Inc., Hoboken, NJ, 004. [3] M. Hall, On the number of Sylow subgroups in a finite group, J. Algebra 7 ( [4] B. Huppert, Endliche Gruppen I., Springer-Verlag, Berlin, [5] C. Leedham-Green, S. McKay, The Structure of Groups of Prime Power Order, London Mathematical Society Monographs. New Series, 7. Oxford Science Publications. Oxford University Press, Oxford, 00. [6] L. Pyber, Enumerating finite groups of given order, Ann. of Math. 137 ( [7] D. J. S. Robinson, A Course in the Theory of Groups, Springer-Verlag, New York, 198. [8] J. J. Rotman, An Introduction to the Theory of Groups, Fourth edition, Graduate Texts in Mathematics, 148, Springer-Verlag, New York, [9] R. Solomon, A brief history of the classification of the finite simple groups, Bull. Amer. Math. Soc. (N.S. 38 ( [10] M. Suzuki, Group Theory I., Springer-Verlag, Berlin-New York, 198. [11] M. Suzuki, Group Theory II., Springer-Verlag, Berlin-New York, Aziz Seyyed Hadi, Modjtaba Ghorbani and Farzaneh Nowroozi Larki Department of Mathematics, Faculty of Science, Shahid Rajaee Teacher Training University, Tehran, , I. R. Iran aziz.saidhadi@gmail.com mghorbani@srttu.edu fnowroozi@gmail.com