The ¯rst Janko group, J1, was discovered by Zvonimir Janko in 1965. It was ¯rst introduced (without proof) in [1] and a complete description with proofs can be found in [2]. 3 Order: jJ1j = 175; 560 = 2 ¢ 3 ¢ 5 ¢ 7 ¢ 11 ¢ 19 De¯nition: We de¯ne J1 to be the subgroup of GL7(11) generated by the matrices 0 1 0 1 0 1 0 0 0 0 0 ¡3 2 ¡1 ¡1 ¡3 ¡1 ¡3 B C B C B C B C B0 0 1 0 0 0 0C B¡2 1 1 3 1 3 3 C B C B C B0 0 0 1 0 0 0C B¡1 ¡1 ¡3 ¡1 ¡3 ¡3 2 C B C B C B C B C Y = B0 0 0 0 1 0 0C Z = B¡1 ¡3 ¡1 ¡3 ¡3 2 ¡1C B C B C B0 0 0 0 0 1 0C B¡3 ¡1 ¡3 ¡3 2 ¡1 1 C B C B C B C B C @0 0 0 0 0 0 1A @ 1 3 3 ¡2 1 1 3 A 1 0 0 0 0 0 0 3 3 ¡2 1 1 3 1 This group has the following properties: » 1. J1 contains an involution t such that CG(t) = hti £ A5 2. The Sylow 2-subgroups of J1 are elementary abelian of order 8. Simplicity: Janko proved that G is a group such that the Sylow 2-subgroups are abelian, G has no subgroup of index 2, and G contains an involution t with the centralizer hti£A5 then G is a simple group. Proving that such a group exists (and is isomorphic to hY; Zi) was done in [2]. Here is the proof of simplicity of such a group G (hence, a proof of the simplicity of J1). We begin by mentioning a few necessary results to the proof: Theorem 0.1. If a Sylow p¡subgroup of G is abelian and N = NG(P ) then the maximal p¡factor group of G is isomorphic to P \ Z(N). Theorem 0.2. The involutions of a group J1 are all conjugate. Proof. Let S be a Sylow 2-subgroup of J1,N = NG(S) and C = CG(S). (Janko 3 shows that the Sylow 2-normalizer is the holomorph of (Z2) by the non-cyclic group of order 21. So we assume that N 6= C). If S is elementary abelian, then S contains seven conjugacy classes of involutions. Since jN=Cj = 21 we have some element of order 7 which cyclically permutes these involutions. Hence the involutions of S are conjugate in N, therefore, they are conjugate in G. Since the Sylow subgroups are conjugate, we are done. ¤ 1 2 We now prove that J1 is simple. Proof. We de¯ne O(G)(= O20 (G)) to be the largest normal subgroup of G of odd order. First suppose that O(G) 6= 1. Let C = CG(t). Since C has no odd order normal subgroups (because any odd order normal subgroup must also be an odd order normal subgroup of A5 which is simple), we must have C\O(G) = 1. Since the involutions of J1 are conjugate. We want to show that some subgroup isomorphic to Z2 £ Z2 acts ¯xed-point-free on O(G). Since O(G) \ C = 1 this means that no element of O(G) commutes with x for any involution x. Hence some 4-group acts ¯xed point free. However, by theorem 10.1.4 of Gorenstein, a ¯xed-point-free ¡1 automorphism of order 2 must send y to y . However, if x1 and x2 are generators of the four-group, this would imply that yx1x2 = y, a contradiction. This says that O(G) = 1. Now suppose that G1 C G and jG=G1j has odd order. If C is not contain in G1 then G1 \ C is an odd index normal subgroup of C, which is impossible, hence 2 C · G1. Suppose ¯rst that G1 has a normal subgroup of index 2. Then O (G) 2 is non-trivial and characteristic. Now O (G1) C G. Since all the involutions of G 2 2 are conjugate, O (G1) must have odd order, therefore O (G1) = 1, which says that G1 is a normal Sylow 2-subgroup. This contradicts Janko's description of a Sylow 2-normalizer. Hence we have that G1 has no subgroup of index 2, hence using the same argument as above, all involutions of G1 are conjugate. Using a variation on the typical Frattini argument (with the same proof), we must have G = CG1, which is impossible since C · G1. So G has no non-trivial odd order factor groups. This says that for any normal subgroup N, both jNj and jG=Nj are abelian. This implies that N must be a 2-subgroup of G. However, all involutions of G are conjugate, so this leads to the ¯nal contradiction. ¤ References: (1) Janko, Zvonimir A new ¯nite simple group with abelian 2-Sylow subgroups. Proc. Nat. Acad. Sci. U.S.A. 53 1965 657{658. (2) Janko, Zvonimir A new ¯nite simple group with abelian Sylow 2-subgroups and its characterization. J. Algebra 3 1966 147{186..