Conjugacy-closed abelian Hall implies retract

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Suppose G is a finite group and H is an abelian conjugacy-closed Hall subgroup of G. Then, H is a retract of G. In other words, H possesses a normal complement N in G: a normal subgroup N of G such that N \cap H is trivial and NH = G.

Related facts

Facts used

  1. Conjugacy-closed and Hall implies commutator subgroup equals intersection with whole commutator subgroup (this is in turn an immediate corollary of the analogue of focal subgroup theorem for Hall subgroups.


Given: A finite group G and an abelian conjugacy-closed Hall subgroup H of G.

To prove: H has a normal complement in G.


  1. H \cap [G,G] is trivial: This follows from fact (1), and the given fact that H is abelian.
  2. H has a normal complement: Consider the quotient by [G,G], and denote the image of H as \overline{H}. Since G/[G,G], the abelianization of G, is an Abelian group, and \overline{H} is a Sylow subgroup, there exists a normal complement \overline{N}. Let N be the inverse image of \overline{N} under the quotient map. Clearly, NH = G, since it contains [G,G] and its image is the whole of G/[G,G]. Further, N \cap H is trivial, because P does not intersect [G,G], and the images \overline{H} and \overline{N} intersect trivially. Thus, N is the required normal complement.