Permutability is not finite-intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., permutable subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about permutable subgroup|Get more facts about finite-intersection-closed subgroup property|

Statement

Verbal statement

The intersection of two permutable subgroups of a group need not be permutable.

Symbolic statement

It is possible to find a group $G$ and subgroups $H$ and $K$ of $G$ such that $H$ and $K$ are both permutable subgroups (viz quasinormal subgroups) but $H \cap K$ is not.

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 $a b = b a^{p + 1}$ . 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 $c$ .
$G = A \times C$ .
$B = {b}$ .

We claim that: $H = A \times {e}$ , and $K = B \times C = {b, c}$ , give the required counterexample.

Proof of the claim

$H = A \times {e}$ is permutable: $H$ is a direct factor of $G$ so it is clearly a normal subgroup and hence a permutable subgroup.

$K = B \times C = {b, c}$ is permutable: Since permutability satisfies the inverse image condition, we see that if $B$ is permutable in $A$ , then $B \times C = {b, c}$ is permutable in $G$ . Thus, it suffices to show that $B$ is permutable as a subgroup of $A$ . This can easily be checked by verifying that $B$ commutes with all the cyclic subgroups of $A$ . (a proof of this is provided in an example for permutable not implies normal).

$B$ is not permutable: Consider the cyclic subgroup $D$ generated by $(a, c)$ . The claim is that $B D \neq D B$ . To prove this notice that $(a, c) (b, 1) = (a b, c) = (b a^{p + 1}, c)$ . This is clearly not in $B D$ .

Further fact shown by the example

This example shows some further facts:

The intersection of a permutable subgroup with a direct factor need not be a permutable subgroup. In this example, for instance, $A$ is a direct factor, but its intersection with $C$ is still not a permutable subgroup.
A permutable subgroup of a direct factor need not be a permutable subgroup. In this case $B = A \cap C$ is a permutable subgroup inside $A$ , which itself is a direct factor.
Permutability is not a direct product-closed subgroup property