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By Burkhard Külshammer

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Additional resources for Algebraische Kombinatorik [Lecture notes]

Example text

Dann operiert G in offensichtlicher Weise auf der Menge Ω der 81 Färbungen der Ecken mit 3 Farben (blau, rot, weiß). Die wesentlich verschiedenen Färbungen entsprechen genau den Bahnen von G auf Ω. 1 haben wir berechnet: |G\Ω| = 24. h. |G| = 4! = 24. Dann operiert G in offensichtlicher Weise auf der Menge Ω der 64 Graphen mit Eckenmenge V = {1, 2, 3, 4}. Die wesentlich verschiedenen Graphen entsprechen dabei genau den Bahnen von G auf Ω. Wir haben berechnet: |G\Ω| = 11. (iii) Sei K ein (endlicher) Körper und n ∈ N.

Für H ∈ L(G) hat die Bahn von H unter G die Form clG (H) := {gHg −1 : g ∈ G} (Konjugationsklasse von H in G) . Der Stabilisator von H in G hat die Form NG (H) := {g ∈ G : gHg −1 = H} (Normalisator von H in G) . Dann ist NG (H) ≤ G eine Untergruppe und |G : NG (H)| = |clG (H)|. Offenbar ist H ⊆ NG (H). Nach Definition ist H ein Normalteiler von NG (H). 9 Satz Sei p ∈ P und G eine endliche p-Gruppe19 . Aus G = {1} folgt Z(G) = {1}. Beweis. Sei R ein Repräsentantensystem für die Konjugationsklassen von G.

2m2 n! Elemente g ∈ Sym(n) vom Typ λ. m2 ! 3m3 m3 ! . Beispiel Für λ = (3, 2, 2) 7 sind m1 = 0, m2 = 2 und m3 = 1. Die Anzahl der g ∈ Sym(7) vom Typ λ beträgt 7! = 210 . 10 0! 22 2! 31 1!

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