By Garrett P.
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G = Since (d 2, f2) = \ffv, 16, a contra- 41 + 41. 45) Arfm(mb12)- 8 and Therefore 1 V Supp(mb12). Argi fbl2,kl2l7 Supp(lb12) 16. that which 12 24 Together with implies Igj1, 111 1) B of degree suitable for E a obtain we W4 19,W41 gJ1,11 gJ1,11, = 1 + g EE 4. = V4 E B and 24 =. and d 2 Now we may rewrite gg 4V4) 24. = = = = Clearly, 9 W41. Now we = 99 Since W474 either = can write A W47-12 4 = W474 --(31+21). b12 appears in g-g, either b12 EE W4T4_ or b12 E lW4U4- In the first case, 0. Thus, in any case, 41 + b12- In the second one, b121 nW4T4 W4U4 41 + b12 or W4T4 41 + k12= = = In the first case, Cg = (41 b12)(31 + + 21) 5b12 + 4k12 + 81 121 + = and, therefore, ,\q-,,,(Tnb12) In the = 4b12 + 2k12- second case, Cg = 4b12 121 + + 5k12 + 81 and ,\rfm(mb12) It follows sible.
Algebra distinct with (c) U a (c) Icl = 1, Then c. 1, 1XI + pairwise of subset say table are b such that 6_5 are 1, 11, N := 117 1, b121 M b2 c, d E B is the order following order, that 1, (c), a(c), G b(c), X E x E x U b (c). , ac 77 c":: 7 a&: 7 I b (c) n. that i = I - are = - following (m - 1) 1 + (m - 2) a holds: table b& (m (m b&+f' 1)c0'+0 - 1) - (m + -)-a-&--+ 7 2 - C is (m - I) b&+# m&+f'+' bc+# shows that table of B. Since subset table a + mac"+O+ b is faithful, C B follows. and the claim Consider (m 1) b, a2 ab ac'+O [FC-T.
43) E x assume It sides = 2f7 d + EM4 both follows. 44) -jM4 = = and 28. Ther efore comparing Tn47ff4- modulo 2, = we 41 + b 12 obtain bV4(mod 2). in dM4 are ones. (dU, M4774) 16, all non-zero coefficient bV4 and (dV4, bTn4) (UM4) EV4) 16. Since (bTn4, bTn4) 28, 16 bm4 X4 + 2X6 -===>4AdV4X4 + 2AdX4x66 The latter equation has solution 1 which contradicts a unique the fact that 11 XdX4X6 AdV4X4 (UM4, UM4) Since Hence UM4 = = = = = = = = . = 6AdX4X6 7- Starting = 0(mod 4) from now we El shall assume that Ill = 3.