Theory of Groups of Finite Order - Tapa blanda

Sinopsis

This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1897 edition. Excerpt: ... Corollary. If H is Abelian, no two operations of H which are not conjugate in / can be conjugate in G. Hence the number of distinct sets of conjugate operations in G, which have powers of p for their orders, is the same as the number of such sets in /. 82. Suppose next that Q is not self-conjugate in H. Then every operation that transforms Q into P must transform H into a sub-group of order pa in which P is not self-conjugate. Of the sub-groups of order pa, to which P belongs and in which P is not self-conjugate, choose H' so that, in H', P forms one of as small a number of conjugate operations or sub-groups as possible. Let g be the greatest sub-group of H' that contains P self-conjugately. Among the sub-groups of order pa that contain P self-conjugately, there must be one or more to which g belongs. Let H be one of these; and suppose that h and h! are the greatest sub-groups of H and H' respectively that contain g self-conjugately. The orders of both h and h' must (Theorem II, § 55) be greater than the order of g; and in consequence of the assumption made with respect to H every sub-group, having a power of p for its order and containing h, must contain P self-conjugately. Now consider the sub-group h, h'. Since it does not contain P self-conjugately, its order cannot be a power of p. Also if pP is the highest power of p that divides its order, it must contain more than one sub-group of order pP. For any sub-group of order pP, to which h belongs, contains P selfconjugately; and any sub-group of order pP, to which h' belongs, does not. Suppose now that S is an operation of h, h', having its order prime to p and transforming a sub-group of h, h' of order pP, to which h belongs, into one to which h' belongs. Then S cannot be...

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