Conjugacy-closed abelian Sylow implies retract

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This article gives the statement, and possibly proof, of a normal p-complement theorem: necessary and/or sufficient conditions for the existence of a Normal p-complement (?). In other words, it gives necessary and/or sufficient conditions for a given finite group to be a P-nilpotent group (?) for some prime number p.
View other normal p-complement theorems


Suppose G is a finite group, and p is a prime dividing the order of G. Further, suppose P is a p-Sylow subgroup, and P is Abelian as a group, and is also conjugacy-closed: any two elements of P that are conjugate in G are conjugate in P. (Note that this effectively means that no two distinct elements of P are conjugate in G).

Then, there exists a normal p-complement: a normal subgroup N such that NP = G, and N \cap P is trivial. In other words, P is a retract of G.

Related facts


  • Burnside's normal p-complement theorem: Burnside's normal p-complement theorem states that if a Sylow subgroup is central in its normalizer, it is a retract. The proof of Burnside's normal p-complement theorem relies on this fact, along with a local conjugacy-determination property in the normalizer.

Stronger facts

Opposite facts

Facts used

  1. Focal subgroup theorem
  2. Any Abelian group is a direct product of its Sylow subgroups. In particular, a Sylow subgroup of an Abelian group is a direct factor, and has a normal complement.


Given: A finite group G, a p-Sylow subgroup P that is conjugacy-closed in G and abelian.

To prove: There exists a normal p-complement N.


  1. P \cap [G,G] = [P,P]: By fact (1), P \cap [G,G] is generated by elements of the form x^{-1}y where x,y \in P are conjugate in G. Since P is conjugacy-closed in G, this is the same as saying that x,y are conjugate in P, so P \cap [G,G] = [P,P].
  2. P \cap [G,G] is trivial: This follows from the previous step, and the observation that P is abelian.
  3. Consider the quotient by [G,G], and denote the image of P as \overline{P}. Since G/[G,G], the abelianization of G, is an Abelian group, and \overline{P} 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, NP = G, since it contains [G,G] and its image is the whole of G/[G,G]. Further, N \cap P is trivial, because P does not intersect [G,G], and the images \overline{P} and \overline{N} intersect trivially. Thus, N is the required normal complement.