Permutably complemented is not finite-intersection-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., permutably complemented subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).
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Statement

It is possible to have a group G and two permutably complemented subgroups H, K \le G such that H \cap K is not permutably complemented in G.

Proof

Example of the dihedral group

Further information: dihedral group:D8

Suppose G is the dihedral group of order eight:

G = \langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle.

Consider the two subgroups:

H = \langle a^2, x \rangle = \{ e, a^2, x, a^2x \}, \qquad K = \langle a^2, ax \rangle = \{ e, a^2, ax, a^3x \}.

Then:

H \cap K = \{ e, a^2 \}.

We have:

  • H is a permutably complemented subgroup of G: The subgroup \{ e, ax \} is a permutable complement to H in G.
  • K is a permutably complemented subgroup of G: The subgroup \{ e, x\} is a permutable complement to K in G.
  • H \cap K is not permutably complemented in G: This can be seen by direct inspection, but also follows from the more general fact that in a nilpotent group any nontrivial normal subgroup intersects the center nontrivially. Here, H \cap K is the center, and if it has a permutable complement, that subgroup must be a nontrivial normal subgroup, leading to a contradiction.