By Istituto Nazionale di Alta Matematicia (Auth.)
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Example text
5*, where —— is an integer prime to 5. Here we have used %i (i + yt) also the fact that yi and y2 = — (1 -\- yx) are prime to 5 since Yl and Y2 do not vanish or zr. Let 2s be the maximal power of 2 dividing yi (1 + Vi)- Then, clearly, « must be even. This fact will be used to rule out many cases. , where i = 1 or 2 and not both Yi and Y2 could appear with the non zero multiplicity in ψ*. We note (using also ψ*(1)==0) that every irreducible character of G appears with the multiplicity 0 or ± 1 in V-'f, i = l , 2, 3, 4.
Then for any element o in the set of special classes of N (B) we have : X(a) = ni — n2 ξ4 (σ) + % f9 (σ) + n4 fu (σ) + % f6 (σ) + η 6 f5 (σ). In particular we have : X {zr) = n, - i - ( - 1 - f;5) n 2 + i - (1 - |fô) n 3 - *i6 . 44 zvoNiMiit JANKO - Some new simple groups 0 etc. -a 5 8 lift + + ~-H | Clearly P 4 is also < τ > admissible and P 4 < τ > is a non abelian group of order 6. Suppose that P 3 φ ζ>. Then the cyclic group <τ> X P 3 of order 6 acts faithfully on Ζ(Τ±) which is not possible. Hence we have Q=CP{T). The dihedral group I) = < i, z, % > of order 8 centralizes Ç and so looking at C (Z {D)\ we see that τ must be conjugate in G to 2. 2. 1 we have that C(Q)/Q is isomorphic to L2(7) or L2(9). An 82subgroup of N (Q) is dihedral of order 16 and so N(Q)/Q is isomorphic to PGL(2,7) or PGP (2, 9) and P c C (Q).