By Biasi C., de Mattos D.

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2 A generalisation of the Jordan-H¨ older theorem 41 in fact, proving the generalised Jordan-H¨ older Theorem is reduced to proving that, in the above situation, N1 N2 /N1 and N1 N2 /N2 are simultaneously Frattini chief factors of G. For this reason J. Lafuente, in [Laf89], wonders about the precise condition on a set X of maximal subgroups of a group G which allows a proof that, in the above situation, if N1 and N2 have supplements in X, then N1 N2 /N1 and N1 N2 /N1 possess simultaneously supplements in X, or, in other words, which is the precise condition on X to prove a Jordan-H¨ older-type Theorem.

Then L0 /K is a supplement of M/K in N/K. Clearly L = (U ∩ N )K ≤ L0 . By maximality of L, we have that L = L0 . But then H ∩ N ≤ L and, by 1b, H k ≤ U , for some k ∈ K. Clearly, this implies that U = H. Hence U is maximal in G. Conversely, let U be a maximal subgroup of G which supplements M in G such that U ∩ M = (U ∩ S1 ) × · · · × (U ∩ Sn ). Write L = (U ∩ N )K. Suppose that L ≤ L0 < N . Consider a supplement R of M in G determined by L0 under the bijection. Then L0 = (R ∩ N )K. Since U ∩ N ≤ L0 , then U k ≤ R, for some k ∈ K.

Hence U ∩ Soc(G) ≤ R1 × · · · × Rn = R1 × R1t2 × · · · × R1tn . 18. 18 (4), if y ∈ U ∩ Soc(G) and g ∈ V , then (y g )π1 = π1 g (y ) . This is to say that R1 is a V -invariant subgroup of S1 . Therefore R1 × · · · × Rn = R1 × R1t2 × · · · × R1tn is a V -invariant subgroup of Soc(G). 19. e. the projections of U ∩ Soc(G) on each Si are surjective. 20. Let us deal ﬁrst with the Case 19a: suppose that R1 is a proper subgroup of S1 . Suppose that R1 ≤ T1 < S1 and T1 is a V -invariant subgroup of S1 .

### A Borsuk-Ulam Theorem for compact Lie group actions by Biasi C., de Mattos D.

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