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By Harm Pralle

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Dazu gen¨ ugt e ∈ Hf H zu zeigen. Denn wegen r ∈ HeH existieren h, h′ ∈ H mit r ∈ heh′ und somit e ∈ h∗ r(h′ )∗ . Ferner existieren l, l′ ∈ H mit r ∈ lf l′ . Zusammen folgt e ∈ h∗ r(h′ )∗ ⊆ h∗ lf l′ (h′ )∗ ⊆ Hf H. Die beiden anderen Implikationen sind offenbar. ✷ F¨ ur H ∈ C und S ⊆ R setzen wir S//H := {sH | s ∈ S} und definieren damit das Quotientenschema von (X, R) ¨ uber H als (X, R)H := (X/H, R//H) . 8 Es seien (X, R) ein endliches Assoziationsschema und H ∈ C. Dann gelten: (i) 1X/H = 1H (ii) (r H )∗ = (r ∗ )H (iii) F¨ ur d, e, f ∈ R gilt adH eH f H = 1 nH b∈HdH c∈HeH abcf , wobei nH := h∈H nh .

Dann ist Gx transitiv auf xr f¨ ur alle r ∈ L , und G ist transitiv auf x L . Wir nehmen, es gebe eine Menge R ⊆ L , so dass Gx nicht transitiv auf xr ist f¨ ur alle r ∈ R. Es sei r ∈ R minimaler L¨ange l(r). 14). Nun seien y, z ∈ xr ⊆ xql. Also existieren v, w ∈ xq mit y ∈ vl und z ∈ wl. Weil r minimale L¨ ange in R hat, gilt q ∈ / R. Also operiert Gx transitiv auf xq und es existiert e ∈ Gx mit ve = w. Daher gilt ye ∈ vle ⊆ vel = wl. Nach Schritt 3 existiert wegen z ∈ wl ein f ∈ Gxw , so dass z = yef .

Dann hat G die Orbitale 1 := {1, 1}G, a := {{1, 2}, {3, 4}} und b := {{1, 3}, {1, 4}, {2, 3}, {2, 4}}. Die Menge der Relationen R := {1, a, b} bildet ein Assoziationsschema (R, X) mit X := {1, . . , 4}. Die Adjazenzalgebra KR hat die Standardbasis   0 1 0 0  1 0 0 0  σ1 := I4 , σa :=  , σb := 0 0 0 1  0 0 0 J2 J2 0 . 5 Es seien (X, R) ein endliches Assoziationsschema und K ein K¨ orper. (i) F¨ ur die Menge J := {r ∈ R | char(K) | |r ∗ |} gilt J(KR) ⊆ KJ. (ii) Falls char(K) | |r| f¨ ur alle r ∈ R, dann ist KR halbeinfach.

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Algebraische Kombinatorik by Harm Pralle


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