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Induction of Marks.

 

Let M be a subgroup of G. Then the marks of M can be induced to marks of G by means of the following trivial observation. A subgroup V of M is also a subgroup of G, and V has the same subgroups regardless of whether it is viewed as a subgroup of M or as a subgroup of G. But in general not every subgroup of V which in G is conjugate to a given subgroup U of V is conjugate to U in M. More precisely, a G-conjugacy class of subgroups of V is a disjoint union of M-conjugacy classes of subgroups of V. In terms of numbers of subgroups this means

  La339

This leads to the following induction formula for tables of marks.

  Thm343

proof356

Rem379

In order to compute the table of marks of G by induction of marks it is sufficient to know the tables of marks of representatives of the conjugacy classes of maximal subgroups of G since every proper subgroup of G is contained in a maximal subgroup M of G. The remaining problem is to determine the fusion from the maximals to G in order to know which representatives of subgroups U' of M are conjugate to a representative U of subgroups of G. This means to determine for each maximal subgroup a map from its conjugacy classes of subgroups to the conjugacy classes of subgroups of G. The next sections provide tools to determine the conjugacy classes of subgroups of G together with the fusion maps from its maximals.



Götz Pfeiffer Wed Oct 30 09:52:08 GMT 1996