Chen G.'s A Characterization of Alternating Groups by the Set of PDF By Chen G.

Read or Download A Characterization of Alternating Groups by the Set of Orders of Maximal Abelian Subgroups PDF

Best symmetry and group books

A dialogue of the origins of decorative paintings - illustrated via the oldest examples, relationship typically from the paleolithic and neolithic a while, and regarded from the theory-of-symmetry perspective. due to its multidisciplinary nature, it's going to curiosity a variety of readers, together with mathematicians, artists, paintings historians, architects, psychologists and anthropologists.

Extra resources for A Characterization of Alternating Groups by the Set of Orders of Maximal Abelian Subgroups

Example text

6 ([Laf84a]). In a group G, the relation of being G-connected is an equivalence relation on the set of all chief factors of G. 2 A generalisation of the Jordan-H¨ older theorem 43 Proof. The only non-obvious property to prove is transitivity. Let F1 , F2 , F3 be chief factors of G such that F1 is G-connected to F2 and F2 is G-connected to F3 . We may suppose that no two are G-isomorphic. Therefore 1. there exists a normal subgroup N of G such that G/N is group of type 3 whose minimal normal subgroups are A/N B/N ∼ =G F2 , and 2.

2a. Clearly L/K is a supplement of M/K in N/K. By 1c, the subgroup H is determined, up to conjugacy in K, by L. Suppose that G = RM and R ∩ N is conjugate to a subgroup of L in N . Since N = RM ∩ N = (R ∩ N )M , there is an element m ∈ M such that (R ∩ N )m ≤ L. Write H0 = Rm . Then G = H0 M and H0 ∩ N ≤ L. It follows, from 1b, that H0 is conjugate to a subgroup of H. Hence R is conjugate to a subgroup of H. Conversely, if R is conjugate to a subgroup of H, then, since G = RM , we have that Rm ≤ H, for some m ∈ M .

Then G = H0 M and H0 ∩ N ≤ L. It follows, from 1b, that H0 is conjugate to a subgroup of H. Hence R is conjugate to a subgroup of H. Conversely, if R is conjugate to a subgroup of H, then, since G = RM , we have that Rm ≤ H, for some m ∈ M . Then (R ∩ N )m = Rm ∩ N ≤ H ∩ N ≤ L. 2b. If G = RM and L is conjugate to (R ∩ N )K in N , there is an element m = (R ∩ N )m K = (Rm ∩ N )K. If m ∈ M such that L = (R ∩ N )K R ∩ M = (R ∩ S1 ) × · · · × (R ∩ Sn ), by 1b, we deduce that H0 = Rm is conjugate to H.