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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1901 Excerpt: ...103--104, G has as maximal invariant subgroup the group generated by the substitution where ft is a primitive root of ftd=l, d being the greatest common divisor of 4 and pn--l. The quotient-group LF(4, pn) is a simple group of order j O4 -i)i3 O3 -i)i8 08n-I)jp To Mft there corresponds in G't the substitution which multiplies every index by ft2 and therefore the identity if p = 2 or p =4Z + 3; while, for p =Al-f 1, it is the substitution T multiplying each of the six indices by--1. We may state the theorem: i/, and only if, v be a square in the field. To the substitution (ay) of C?4 corresponds in Crs the substitution a2: If 8 be the particular substitution 140), the partial substitution 141) becomes ( 0 0 of determinant vl. Hence if 140) belong to Gi v must be a square in the field. Inyersely, if v be a square, 140) is the second compound of the following substitution of determinant unity: fi/V. 0 0 0 0 v'A 0 0 0 0 -0 0 0 0 v-V., Note.--The second compound contains the substitution-MS=I;-M2 Ml lS-H=/M4-M: =-2S U=V YU) lSi=V YSi In fact, the latter is the second compound of the substitution = 1 (mod2), formed by multiplying each coefficient of the partial substitution 141) by that coefficient of the matrix as tvhich lies symmetrical to it. Gt does not contain the substitution M1 = (Ynsi) The left member of our relation is seen to be the expansion of the expression and is therefore =1 (mod 2), since a, y =1. The substitution Mt does not-J-: -c 7 the relation and so does not belong to the group Glt2. 167. Theorem.--Upon applying to f a linear m-ary transformation of determinant I), tlie determinant A of f is multiplied by Ds. In view of 100, it suffices to prove the ...

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