Download Algebra I. Basic notions of algebra by A. I. Kostrikin, I. R. Shafarevich PDF

By A. I. Kostrikin, I. R. Shafarevich

ISBN-10: 0387170065

ISBN-13: 9780387170060

From the experiences: "... this can be one of many few mathematical books, the reviewer has learn from disguise to hide ...The major benefit is that almost on each web page you will discover a few unforeseen insights... " Zentralblatt für Mathematik "... There are few proofs in complete, yet there's a thrilling blend of sureness of foot and lightness of contact within the exposition... which transports the reader without problems around the entire spectrum of algebra...Shafarevich's ebook - which reads as very easily as a longer essay - breathes lifestyles into the skeleton and should be of curiosity to many sessions of readers; definitely starting postgraduate scholars could achieve a most respected point of view from it but... either the adventurous undergraduate and the proven specialist mathematician will discover a lot to enjoy..."

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Ii) Daher folgt aus vollinvariant die Eigenschaft charakteristisch und daraus die Eigenschaft normal. (iii) Für jede charakteristische Untergruppe U f(f−1 (U)) ⊆ f(U), d. h. f(U) = U. 3 (i) Für jede Gruppe G ist das Zentrum von G charakteristisch in G. Denn für z ∈ Z(G), g ∈ G und f ∈ Aut G gilt: f(z)f(g) = f(zg) = f(gz) = f(g)f(z), d. h. f(z) ∈ Z(G) wegen f(G) = G. Im Allgemeinen ist Z(G) nicht vollinvariant in G (siehe Übung). (ii) Für jede Gruppe G ist U = g2 : g ∈ G vollinvariant in G. Denn für g ∈ G und f ∈ End G ist f(g2 ) = f(g)2 ∈ U.

H. x − y ∈ A. 1 (Torsionsgruppe, torsionsfrei) Man bezeichnet T (A) als Torsionsgruppe von A. Falls T (A) = A heißt A selbst Torsionsgruppe und falls T (A) = 0 heißt A torsionsfrei. 2 Sei A eine abelsche Gruppe. Dann ist T (A) die Torsionsgruppe und A/T (A) torsionsfrei. B EWEIS : Die erste Aussage ist trivial. Zum Beweis der zweiten Aussage habe a + T (A) ∈ A/T (A) die Ordnung n < ∞. Dann ist 0 = n(a + T (A)) = na + T (A), d. h. na ∈ T (A). Folglich ist m := | na | < ∞. Daher mna = 0, d. h. a ∈ T (A) und a + T (A) = 0.

Bn ∈ B mit f(b1 ) = a1 , . . , f(bn ) = an sowie b ∈ B. Dann existieren z1 , . . , zn ∈ Z mit f(b) = z1 a1 + · · · + zn an = z1 f(b1 ) + · · · + zn f(bn ) = f(z1 b1 + · · · + zn bn ) = : f(n). Also ist u ∈ U := b1 , . . , bn mit f(b − u) = 0. Folglich: b − u ∈ ker f und b = u + (b − u) ∈ U + ker f. Damit ist gezeigt: B = U + ker f. Sei andererseits y ∈ U ∩ ker f und y = y1 b1 + · · · + yn bn . Dann ist 0 = f(y) = y1 f(b1 ) + · · · + yn f(bn ) = y1 a1 + · · · + yn an . Da a1 , . . , an linear unabhängig sind, folgt, y1 = · · · = yn = 0 und y = 0.

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Algebra I. Basic notions of algebra by A. I. Kostrikin, I. R. Shafarevich

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