Permutability is not upper join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., permutable subgroup) not satisfying a subgroup metaproperty (i.e., upper join-closed subgroup property).
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Statement

It is possible to have a group G, a subgroup H, and intermediate subgroups K_1, K_2 of G containing H such that H is a permutable subgroup in both K_1 and K_2, but H is not a permutable subgroup in \langle K_1, K_2 \rangle.

Related facts

Related facts that don't hold for permutable subgroups

Related facts that do hold for permutable subgroups

Proof

Construction of the counterexample

Setup: Let p be an odd prime.

  • A is a group generated by two elements a,b subject to the relations a^{p^2} = 1, b^p = 1 and ab = ba^{p+1}. Alternatively A is the semidirect product of the additive group modulo p^2 by the multiplicative group of order p in the multiplicative group of automorphisms. Note that A is a non-Abelian group of order p^3.
  • C is a cyclic group of order p^2, generated by an element c.
  • G = A \times C.
  • B = \{ b \}.
  • K_1 = A \times \{ e \}, and K_2 = B \times C = \{b , c\}.
  • H = B \times \{ e \}.

We claim that H is permutable in both K_1 and K_2 but not in G.

  • H is permutable in K_1: This follows from the fact that B is permutable in A. This can be verified using the fact that it is in the Baer norm, or equivalently, that it commutes with all cyclic subgroups. The proof details are given in the example for permutable not implies normal.
  • H is permutable in K_2: This follows from the fact that H is a direct factor of K_2, hence normal in K_2, hence permutable in K_2.
  • H is not permutable in G: Consider the cyclic subgroup D generated by (a,c). The claim is that HD \ne DH. To prove this notice that DH \ni (a,c)(b,e) = (ab,c) = (ba^{p+1},c). This is clearly not in HD.