Lattice-complemented not implies permutably complemented

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., lattice-complemented subgroup) need not satisfy the second subgroup property (i.e., permutably complemented subgroup)
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Statement

A lattice-complemented subgroup of a group need not be permutably complemented.

Definitions used

Lattice-complemented subgroup

Further information: Lattice-complemented subgroup

A subgroup H of a group G is termed lattice-complemented in G if there exists a subgroup K of G such that H \cap K is trivial and \langle H, K \rangle = G.

Permutably complemented subgroup

Further information: Permutably complemented subgroup

A subgroup H of a group G is termed permutably complemented in G if there exists a subgroup K of G such that H \cap K is trivial and HK = G.

Proof

Example of the alternating group

Further information: alternating group:A4

Let G be the alternating group on the set \{ 1,2,3,4\}. Consider the two-element subgroup:

H := \langle (1,3)(2,4) \rangle.

  • H is lattice-complemented in G: The cyclic subgroup generated by (1,2,3) is a lattice complement to H in G.
  • H is not permutably complemented in G: This can be seen by inspection. In fact, there is no subgroup of order six in G, and any permutable complement to H must have order six.

Example of the dihedral group

Further information: dihedral group:D16

Let G be the dihedral group of order 16:

G = \langle a,x \mid a^8 = x^2 = e, xax^{-1} = a^{-1} \rangle.

Let H be the subgroup of G generated by a^4 and x:

H = \langle a^4, x \rangle = \{ e, a^4, a^4x , x \}.

  • H is a lattice-complemented subgroup of G: Indeed, the subgroup \langle ax \rangle is a lattice complement to H in G.
  • H is not a permutably complemented subgroup of G: Any permutable complement K to H must have order four. Hence, the intersection of K with \langle a \rangle must have order either two or four. In either case, that intersection must contain the cyclic subgroup \langle a^4 \rangle. Hence, K does not intersect H trivially.

A generic example: simple non-Abelian group

Every simple non-Abelian group is a K-group: every subgroup is lattice-complemented. On the other hand, by Hall's theorem on solvability, there does not exist a p-Sylow complement for every p, so not every p-Sylow subgroup is permutably complemented. Thus, we can always find a lattice-complemented subgroup that is not permutably complemented.